Published by Rankguru Technology Solutions Private Limited
IL Achiever Series Chemistry for JEE Module 1
ISBN 978-81-985634-8-4 [SECOND
EDITION]
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Dr. B. S. Rao, the visionary behind Sri Chaitanya Educational Institutions, is widely recognised for his significant contributions to education. His focus on providing high-quality education, especially in preparing students for JEE and NEET entrance exams, has positively impacted numerous lives. The creation of the IL Achiever Series is inspired by Dr. Rao’s vision. It aims to assist aspirants in realising their ambitions.
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With gratitude and inspiration
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Preface
The Joint Entrance Examination (JEE) is a critical threshold that students aspiring to study at the prestigious IITs need to cross. Cracking this examination requires rigorous preparation, and good study materials are as indispensable in this journey as hard work and determination.
The quality of study materials can be determined by how well they cater to the needs of the aspirants and how comprehensive they are in their scope and reach. IL Achiever Series for JEE was first published in 2024 for long-term students and was quickly picked up by JEE aspirants. We actively reached out to the users and sought critical feedback. User feedback started pouring in and the relevant and valuable ones were incorporated into the books. Hence, the revised edition of IL Achiever Series was shaped and honed by the expert suggestions and recommendations of practising teachers and subject matter experts.
The second edition of IL Achiever Series includes substantial changes and improvements. New questions have been added where required, and all the JEE Main questions have been categorised into Levels 1, 2, and 3. While Level 1 questions are more aligned to NCERT, Levels 2 and 3 include more challenging, multi-concept questions of the same rigour as in JEE Main exam. Relevant topic-level changes have also been introduced to make the content clearer and more accessible.
The second edition of IL Achiever Series stands out because it is built to propel longterm JEE aspirants towards achieving their dream of acing the joint entrance examination. A variety of features in the books and an array of practice questions cater to their learning needs and prepare them thoroughly for the test.
The comprehensive coverage of the NCERT syllabus provides a strong foundation and helps in holistic preparation for JEE. The series includes diverse question types for JEE Main and JEE Advanced exams. Expert tips, theory-based questions, and a mix of difficulty levels aim to improve students’ conceptual understanding, cross-topic synthesis, and retention of key concepts.
The second edition of IL Achiever Series is more than just a set of books. It is a commitment to helping students learn, grow, and succeed. We have designed it to ignite your passion for engineering and to foster a lifelong love for learning in you. With every page you turn, you will move one step closer to materialising your dream of cracking JEE with a good rank.
Key Features of the Book
Chapter Outline
1.1 Importance of Chemistry
1.2 Nature of Matter
This outlines topics or learning outcomes students can gain from studying the chapter. It sets a framework for study and a roadmap for learning.
Solved Examples
Specific problems are presented along with their solutions, explaining the application of principles covered in the textbook.
Try yourself:
1. How many significant f igures are there in ‘ p ’?
Ans: This is a non-terminating and nonrecurring value. Hence, the number of significant figures in ‘p ’ is infinity.
Solved example
1. A jug contains 2 L of milk. Calculate the volume of the milk in m3 .
Sol. Since 1 L = 1000 cm3 and 1 m = 100 cm, 1m100cm =1= 100cm1 m
Try Yourself enables the student to practice the concept learned immediately.
This comprehensive set of questions enables students to assess their learning. It helps them to identify areas for improvement and consolidate their mastery of the topic through active recall and practical application.
TEST YOURSELF
1. Which of the following pairs can be cited as an example to illustrate the law of multiple proportions?
(1) Na2O, K2O (2) CO, CO2
(3) SO2, SO3 (4) Both 2 and 3
Organised as per the topics covered in the chapter and divided into three levels, this series of questions enables rigorous practice and application of learning.
These questions deepen the understanding of concepts and strengthen the interpretation of theoretical learning.
JEE MAIN LEVEL
LEVEL 1, 2, and 3
Single Option Correct MCQs
Numerical Value Questions
THEORY-BASED QUESTIONS
Single Option Correct MCQs
Statement Type Questions
Assertion and Reason Questions
JEE ADVANCED LEVEL
FLASHBACK
CHAPTER TEST
This comprehensive test is modelled after the JEE exam format to evaluate students’ proficiency across all topics covered, replicating the structure and rigour of the JEE examination. By taking this chapter test, students undergo a final evaluation, identifying their strengths and areas needing improvement.
Level 1 questions test the fundamentals and help fortify the basics of concepts. Level 2 questions are higher in complexity and require deeper understanding of concepts. Level 3 questions perk up the rigour further with more complex and multi-concept questions.
This section contains special question types that focus on in-depth knowledge of concepts, analytical reasoning, and problem-solving skills needed to succeed in JEE Advanced.
Handpicked previous JEE questions familiarise students with the various question types, styles, and recent trends in JEE examinations, enhancing students’ overall preparedness for JEE.
CHAPTER
SOME BASIC CONCEPTS OF CHEMISTRY
Chapter Outline
1.1 Importance of Chemistry
1.2 Nature of Matter
1.3 Properties of Matter and Their Measurement
1.4 Uncertainity in Measurement
1.5 Laws of Chemical Combinations
1.6 Daltons Atomic Theory
1.7 Atomic and Molecular Mass
1.8 Mole Concept and Molar Mass
1.9 Percentage Composition
1.10 Stoichiometry and Stoichiometric Calculations
1.1 IM PORTANCE OF CHEMISTRY
■ Chemistry is the study of materials, their preparations, properties, mutual interactions, and uses in different fields of life.
■ Chemistry plays a crucial role in daily life, influencing medicine, industry, and technology.
■ It contributes to environmental safety by designing safer alternatives to hazardous substances.
■ Example: Environmentally safe refrigerants (alternatives to CFCs that cause ozone depletion) have been successfully synthesized.
1.2 NATURE OF MATTER
Definition of Matter
■ Anything that has mass and occupies space is called matter.
■ Exists in three physical states: solid, liquid, and gas.
■ Properties of Different States of Matter
Interconversion of States of Matter
■ By changing temperature and pressure, matter can change its state:
■ Solid Liquid Gas
■ Heat required for melting/boiling.
2
■ Cooling required for condensation/freezing.
Solid Definite
CHAPTER 1: Some Basic Concepts of Chemistry
Definite Particles closely packed in an orderly fashion Least movement
Liquid Indefinite (takes shape of container) Definite Particles are close but can move More movement than solids
Gas Indefinite (fills entire container)
Classification of Matter
Indefinite Particles far apart Maximum movement
Matter is classified into mixtures and pure substance
I. Mixtures
■ Homogeneous Mixtures
Completely uniform composition throughout and cannot be visually distinguished.
Example: Salt Solutions in water, air.
■ Heterogeneous Mixtures
Non-uniform composition, different components can be observed.
Example: Oil in water, sand in salt.
Components can be separated using physical methods like hand-picking, filtration, crystallization, distillation, etc.
II. Pure Substances
■ Have a fixed composition and cannot be separated by simple physical methods.
■ Further classified into elements and compounds.
1. Elements
■ Consist of only one type of atoms.
■ Example: H₂, Cl₂, O₂, O₃, P₄, S₈, C n
■ Elements with similar atoms on fine division are called elemental forms.
2. Compounds
■ Contain atoms of different elements in a fixed ratio.Can be ionic or covalent.
■ Properties differ from those of constituent elements.
■ Ex: HCl, NH₃, NaOH, H₂SO₄, C₆H₁₂O₆
Molecules and Atomicity
■ The smallest particle of an element or compound capable of independent existence.
■ Number of atoms present in a molecule is called as its atomicity.
Monoatomic: He, Ne, Ar, Kr, Xe, Rn.
Diatomic: H₂, N₂, O₂, Cl₂.
Triatomic: O₃, H₂O
Tetraatomic: White phosphorus (P₄).
Octaatomic: Rhombic sulfur (S₈).
Atomic Mass and Unified Scale (amu/u)
■ Atomic mass is defined as the number of times an atom of an element is heavier than 1 amu.
■ Based on the C-12 scale:
Mass of hydrogen atom = 1.007825 amu
Mass of oxygen atom = 15.995 amu
Mass of a hydrogen atom= 1.66 × 10–²⁴ g
C-12 scale is also called the unified scale, here the symbol ‘u’ is used instead of amu.
1.3 PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substance in nature exhibits physical and chemical properties.
■ Examples of physical properties: Colour, od our, melting point, boiling point, density, etc.
■ Measurement systems:
■ English system
■ Metric system (more convenient, based on d ecimal system)
1.3.1 The International System of Units (SI)
■ Established by the 11th General Conference on Weights and Measures (CGPM).
■ The SI system has seven base units (listed in the table below). SI base units are well-defined and used globally.
■ The SI system uses prefixes to indicate multiples or submultiples of a unit.
1.3.2 Mass
■ Mass is the amount of matter present in an object, while weight is the force due to gravity. Mass remains constant, but weight varies with gravity.
■ SI unit of mass: kg, 1 kg = 1000 g
1.3.3 Volume
■ SI unit of volume: cubic metre (m³)
1 m³ = 1000 dm³ = 1000 cm³, 1 L = 1000 mL
1.3.4 Density
■ Density is the mass per unit volume of a substance.
■ SI unit of density: kg·m ⁻³
1.3.5 Temperature
■ Three common temperature scales: 1. Degree Celsius (°C) 2. Degree Fahrenheit (°F) 3. Kelvin (K) (SI unit)
■ Fahrenheit to Celsius:
°C = (5/9) × (°F - 32)
■ Celsius to Fahrenheit:
°F = (9/5) × °C + 32
■ Celsius to Kelvin:
Kelvin (K) = Celsius (°C) + 273.15
1.4 UNCERTAINTY IN MEASUREMENT
■ Chemistry deals with atoms and molecules, which have extremely low masses and are present in very large numbers.
Large and Small Numbers in Chemistry:
Ex: Mass of 602,200,000,000,000,000,000,000 the molecules of hydrogen gas is 2 g
Ex: Mass of a hydrogen at om is 0.00000000000000000000000166 g
Physical Quantities and Their SI Units
Thermodynamic Temperature
Amount
Luminous Intensity
■ Challenges in Handling Large Numbers:
Dealing with constants like Planck’s constant, speed of light, charges on particles requires handling numbers of similar magnitude.
■ Difficulty in Mathematical Operations:
Operations like addition, subtraction, multiplication, or division are challenging with such large or small numbers.
■ Solution: Scientific Notation:
Use scientific notation (exponential notation) to represent large or small numbers as N × 10ⁿ.
Scientific notation format: N is a number between 1 and 10, and n is an exponent (positive or negative).
■ Examples:
462.207 can be written as 4.62207 × 10².
0.00032 can be written as 3.2 × 10 –⁴.
■ Explanation of Scientific Notation:
Moving the decimal point results in adjusting the exponent accordingly.
In 462.207 → Move decimal 2 places left → Exponent is +2.
In 0.00032 → Move decimal 4 places right → Exponent is -4.
■ Key Point for Mathematical Operations:
Keep the rules of scientific notation in mind when performing addition, subtraction, multiplication, or division.
Significant Figures
■ Significant figures (SF) indicate the precision of a measurement. They are important in expressing uncertainty and ensuring accuracy in experimental results.
Key Points:
■ Precision vs. Accuracy:
Precision: Closeness of multiple measurements.
Accuracy: Closeness to the true value.
Example:
If true value = 2.00g:
Student A: 1.95g and 1.93g (Precise, but not accurate)
Student B: 1.94g and 2.05g (Neither precise nor accurate)
Student C: 2.01g and 1.99g (Both precise and accurate)
■ Uncertainty in Measurements:
The last digit in a measurement is always uncertain, e.g., 11.2 mL: 11 is certain, 2 is uncertain.
Rules for Determining Significant Figures:
■ Non-zero digits are always significant.
Example: 285 cm → 3 SF
Example: 0.25 mL → 2 SF
■ Leading zeros are not significant
Example: 0.03 → 1 SF
Example: 0.0052 → 2 SF
■ Zeros between non-zero digits are significant.
Example: 2.005 → 4 SF
■ Trailing zeros are significant if there is a decimal point.
Example: 0.200 g → 3 SF
Example: 100.0 → 4 SF
Without a decimal point, trailing zeros are not significant.
Example: 100 → 1 SF
■ Exact numbers (like counts) have infinite significant figures
Example: 2 balls → Infinite SF
■ Scientific notation:
Example: 4.01 × 10² → 3 SF
Example: 8.256 × 10–³ → 4 SF
Operations with Significant Figures:
Addition and Subtraction:
■ The result should have the same number of decimal places as the measurement with the least decimal places.
Example:
1. 2.6 kg (one decimal place) + 3.44 kg (two decimal places) = 6.04 kg
The final result is corrected to one decimal place as 6.0 kg.
2. 8.62 + 11.354 + 14.4052 = 34.3 792
The number should be reported up to two decimal places, and hence, is rounded off to 34.38
■ The result should have the same number of significant figures as the measurement with the least number of significant figures
2.5 × 1.25 = 3.125 It is 3.1
46.512 × 1.67 = 77.67504 It is 77.7
5.4221 × 15.6 = 84.58476 It is 84.6
312.6 ÷ 14.68 = 21.244277 It is 21.24
515.69 ÷ 2.6101 = 197.57480 It is 197.57
Rounding Off Numbers (Key Points)
■ When rounding off numbers to the required number of significant figures, follow these rules:
If the digit to be removed is greater than 5, increase the preceding digit by 1.
Example: 2.487 → 2.49 (Remove 7, increase 8 to 9)
CHAPTER 1: Some Basic Concepts of Chemistry
If the digit to be removed is less than 5, the preceding digit remains unchanged.
Example: 7.923 → 7.92 (Remove 3, keep 7.92)
■ If the digit to be removed is exactly 5:
If the preceding number is even, do not change it.
Example: 7.45 → 7.4
If the preceding number is odd, increase it by 1.
Example: 6.35 → 6.4
Dimensional Analysis
■ It is used to convert units from one system to another, ensuring consistency in units during calculations.
■ It is also known as the fa ctor-label method or unit factor method.
solved examples
1. A jug contains 2 L of milk. Calculate the volume of the milk in m 3 .
Sol. Since 1 L = 1000 cm3 and 1 m = 100 cm, 1m100cm =1= 100cm1 m
To get m3 from the above unit factors, the first unit factor is taken and it is cubed.
Now, 2L = 2 × 1000 cm3
The above is multiplied by the unit factor.
2. Density of pure gold is 19.3 g/cm3. When a gold ring is purchased from a jewellery shop, how will you check its purity?
Sol. Find the exact mass and volume of the ring and calculate its density. By comparing the density obtained with that of pure gold, one can get the idea of its purity.
Try yourself:
1. How many significant figures are there in ‘ p ’?
Ans: This is a non-terminating and non-recurring value. Hence, the number of significant figures in ‘ p ’ is infinity.
3. Give the answer for 2.5 × 1.25 in significant figures. (1) 3.1 (2) 3.12 (3) 3.13 (4) 3.125
4. Which one of the following has four significant figures? (1) 0.00050 (2) 6.300 (3) 0.052 (4) 6.02 × 1023
5. 0.00025 has how many significant figures? (1) 5 (2) 3 (3) –4 (4) 2
6. The number of significant figures in electronic charge 1.602 × 10 –19 C is (1) 1 (2) 2 (3) 3 (4) 4
7. The correctly reported answer of the addition of 4.523, 2.3, and 6.24 will have how many significant figures?
(1) Two (2) Three (3) Four (4) Five
8. After rounding off 1.235 and 1.225 to three significant figures, we will have their answers, respectively, as (1) 1.23, 1.22 (2) 1.24, 1.123 (3) 1.23, 1.23 (4) 1.24, 1.22
9. 8.281 has how many significant figures? (1) 1 (2) 2 (3) 3 (4) 4
10 Planck’s constant has the dimensions of (1) force (2) work (3) angular momentum (4) torque
11. Which of the following units represents the largest amount of energy? (1) calorie (2) erg (3) joule (4) electron-volt
Answer Key
1.5 LAWS OF CHEMICAL COMBINATIONS (5 LAWS)
1. Law of Conservation of Mass
Proposed by Antoine Lavoisier.
Matter can neither be created nor destroyed during a chemical change.
The total mass of reactants = Total mass of products.
Law is applicable only to balanced equations
AgNO3(aq) + KI(aq)→AgI(s)+ KNO3(aq)
Solved Examples
3. When 8.4 g of NaHCO3 is added to 3.65 g of HCl, then 5.85 g NaCl and 1.8 g water are formed. How many grams of CO2 gas escaped from the container?
Sol. Equation for the reaction:
NaHCO3 + HCl → NaCl + H2O + CO2
8.4 g 3.65 g 5.85 g 1.8 g x g
According to the law of conservation of mass, 8.4 + 3.65 = 5.85 + 1.8 + x grams of CO2
From this equation, x = 4.4 g.
Try yourself:
2. For a hypothetical equation, 2A(g) + 3B(g) → A2B3(g), 25 g of reactants are allowed to react. In the end, sum of the masses of the products formed and reactants unreacted is equal to 25 g. Why?
Ans: This is according to the law of conservation of mass.
2. Law of Definite Proportions
■ Proposed by Joseph Proust, also called the Law of Constant Proportions.
■ Statement: A chemical substance always contains the same elements in a fixed proportion by weight.
■ Example: Cupric carbonate samples (natural & synthetic) contain 51.35% Cu, 9.74% C, 38.91% O, proving the law.
■ Another Example: CO₂ composition remains constant, whether prepared chemically or by fermentation.
■ Carbon dioxide always contains carbon and oxygen in a fixed 3:8 ratio by weight, regardless of preparation method.
■ This law helps in determining chemical formulas of compounds.
■ Percent weight of a constituent element = Atomic100 weightofelementNumberofatoms Grammolecularweightofcompound × ×
Solved Examples
4. Two oxides of metal M have 27.6% and 30% oxygen by weight. If the formula of the first oxide is M3O4, what is the formula of the second oxide?
Sol. Oxide (A) : Oxygen 27.6%; 4’O’ atoms
Metal 72.4%; 3 ‘M’ atoms
Oxide (B) : Oxygen 30%; Metal 70%
Ratio of M and O atoms in oxide B = 2.9 : 4.35 = 2 : 3
Formula of the second oxide is M 2O3.
5. 60 g of compound (x) contains 25 g of element (y) and 35 g of element (z). What mass of compound (x) is formed when 5 g of (x) is combined with 9 g of (z), if the given data obeys the law of definite proportions?
Sol. Ratio of masses (y) and (z) in 60 g of compound (x) = 25 : 35 = 5 : 7 5 g of (y) reacts with 7 g of (z) to form 12 g of compound ( x) and 9 – 7 = 2 g of (z) is left unreacted.
Try yourself:
3. Why is the law of definite proportions not applicable to 12CO2 and CO18 2 , though both oxides are made up of the same elements C and O?
Combining ratio is different; thus, it is not applicable.
Combining ratio of masses of C and O in 18CO 2 is 12 : 36 = 3:9
Ans: Combining ratio of masses of C and O in 2CO12 is 12 : 32 = 3.8.
3. Law of Multiple Proportions
■ Proposed by John Dalton.
■ Statement: When two elements form multiple compounds, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios.
Examples:
Carbon and Oxygen: CO (C:O = 12:16) and CO₂ (C:O = 12:32) → O ratio 1:2.
Hydrogen & Oxygen: H₂O (O = 16 g) and H₂O₂ (O = 32 g) → O ratio 1:2.
Nitrogen & Oxygen: N₂O, NO, N₂O₃, NO₂, N₂O₅ → O ratio 1:2:3:4:5.
6. In two oxides of nitrogen, oxide-1 has 63.64% N and oxide-2 has 46.67% N. Prove that these figures illustrate the law of multiple proportions. Sol.
Number of parts by mass of nitrogen that combines with one part of oxygen:
Oxide-1: = 63.64 1.75 36.36
Oxide-2: 46.67 0.875 53.33 =
The ratio of masses of nitrogen that combine with the same mass (i.e., one part by mass) of oxygen = 1.75 : 0.875 = 2 : 1
This is a simple whole number ratio. Hence, the given figures illustrate the law of multiple proportions.
Try yourself:
4. Does the formation of hydrocarbons explain the law of multiple proportions?
Example: ,10H4C ,12H5C 14H6C
Ans: No, it does not explain the law of multiple proportions as the parts by mass of H that combine with 12 parts by mass of C is not in a simple whole number ratio.
4. Law of Combining Volumes
■ Proposed by Gay-Lussac.
■ Statement: When gases combine under similar conditions, their volumes follow a simple whole-number ratio, including the product if gaseous.
Examples: H₂ + Cl₂ → 2HCl (Volume ratio 1:1:2)
H₂ + ½O₂ → H₂O (100 mL H₂ + 50 mL O₂ → 100 mL H₂O)
■ Relation to Other Laws:
Gay-Lussac’s law applies to volume, while the law of definite proportions applies to mass.
Avogadro’s Law: Equal volumes of gases under the same conditions contain equal numbers of molecules.
Volume of gases is related to molecules, not atoms.
Example: N₂ + 3H₂ → 2NH₃
By experiment: 1:3:2 volume ratio
By Avogadro’s law: n + 3n → 2n molecules
Confirms that molecules contain atoms, aligning with Dalton’s atomic theory
IL ACHIEVER SERIES FOR JEE CHEMISTRY
Solved Examples
7. What is the volume ratio of the product gases in the decomposition of phosphorus pentachloride?
Sol. PCl5(g) → PCl3 (g) + Cl2( g)
Volume ratio of PCl3 and Cl2 is 1 : 1.
8. In what volume ratio does nitrogen gas and hydrogen gas combine to give ammonia under similar conditions of temperature and pressure?
Sol. N2 + 3H2 → 2NH3
Volume ratio of N2 and H2 = 1 : 3.
Try yourself:
5. Why does a balloon expand on blowing air?
Ans: When air is blown into a balloon, more air molecules enter into the balloon, thus expanding it.
5. Law of Reciprocal Proportions
■ Proposed by: Ritcher.
■ Statement: The ratio of the weights of two elements (X & Y) that separately combine with a fixed weight of a third element (Z) is a simple multiple of the ratio in which X and Y combine directly.
Example:
H & O separately combine with S:
H₂S: 2 g H + 32 g S
SO₂: 32 g O + 32 g S
Ratio of H:O (with S) = 1:16
H₂O formation: H:O ratio = 1:8 (a simple multiple).
Importance: Helped in understanding chemical equivalents and equi valent weights
Solved Examples
9. Carbon dioxide contains 27.27% of carbon, carbon disulphide contains 15.79% of carbon and sulphur dioxide contains 50% of sulphur. Are these figures in agreement with the law of reciprocal proportions?
Sol. Yes,
1 g C will combine with S = 84.21 5.33g 15.79 =
1 g C will combine with O = 72.73 2.67g 27.27 =
∴ Ratio of masses of S and O, which combine with fixed mass of carbon (via 1 g) = 5.33 : 2.67 = 2 : 1. Ratio of masses of S and O which combine directly with each other = 50 : 50 = 1 : 1. Thus, the two ratios are simple multiples of each other.
Try yourself:
6. What is the difference between the law of multiple proportions and the law of reciprocal proportions?
Ans: In multiple proportions, two elements combine in more than one ratio of masses to form different compounds. In reciprocal proportions, two different elements combine separately with the same weight of the third element.
TEST YOURSELF
1. Choose the correct statement(s) among the following.
(1) Chemical equation is balanced according to the law of conservation of mass.
(2) Law of conservation is not applicable for nuclear reactions.
(3) Law of combining volumes was proposed by Gay-Lussac.
(4) All of the above
2. Two gaseous samples were analysed. One contained 1.2 g of carbon and 3.2 g of oxygen. The other contained 27.3% carbon and 72.7% oxygen. The experimental data is in accordance with the
(1) law of conservation of mass
(2) law of definite proportions (3) law of reciprocal proportions (4) law of multiple proportions
3. Percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This proves the law of (1) constant proportions (2) reciprocal proportions (3) multiple proportions (4) conservation of mass
4. “The total mass of reactants is always equal to the total mass of products in a chemical reaction.” This statement is known as (1) the law of conservation of mass (2) the law of definite proportions (3) the law of equivalent weights (4) the law of combining masses
5. The reaction hydrogen (g) + oxygen (g) gives water vapour; the ratio of volumes is 2 : 1 : 2. This illustrates the law of (1) conservation of mass (2) combining weights (3) combining volumes (4) all of the above
6. Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates the (1) law of reciprocal proportions (2) law of constant proportions (3) law of multiple proportions (4) law of equivalent proportions
7. One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combine together, the ratio of their masses will be governed by
(1) the law of definite proportions (2) the law of multiple proportions
(3) the law of reciprocal proportions (4) the law of conservation of mass
8. 14 g of element X combines with 16 g of oxygen. On the basis of this information, which of the followings is a correct statement?
(1) The element X could have an atomic weight of 7 and its oxide is XO.
(2) The element X could have an atomic weight of 14 and its oxide formula is X 2O.
(3) The element X could have an atomic weight of 7 and its oxide is X 2O.
(4) The element X could have an atomic weight of 14 and its oxide is XO 2.
9. Carbon and oxygen combine to form two oxides, carbon monoxide and carbon dioxide, in which the ratio of the weights of carbon and oxygen are, respectively, 12 : 16 and 12 : 32. These figures illustrate the (1) law of multiple proportions (2) law of reciprocal proportions (3) law of conservation of mass (4) law of constant proportions
10. Two elements X (at. mass 16) and Y (at. mass 14) combine to form compounds A, B, and C, in which the ratio of weights of Y combining with a fixed mass of X is 1 : 3 : 5, respectively. If 32 parts by mass of X combine with 84 parts by mass of Y in B, then in C, 16 parts by mass of X will combine with
(1) 14 parts by mass of Y (2) 42 parts by mass of Y
(3) 70 parts by mass of Y (4) 84 parts by mass of Y
11. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at STP. The data illustrates the (1) law of conservation of mass (2) law of constant proportions (3) law of multiple proportions (4) law of reciprocal proportions
12. Which of the following pairs can be cited as an example to illustrate the law of multiple proportions?
(1) Na2O, K2O (2) CO, CO2 (3) SO2, SO3 (4) Both 2 and 3
13. In compound A, 1.00 g nitrogen unites with 0.57 g oxygen. In compound B, 2.0 g nitrogen combines with 2.28 g oxygen. In compound C, 3.00 g nitrogen combines with 5.13 g oxygen. These results obey which of the following laws?
(1) Law of constant proportion (2) Law of multiple proportions
(3) Law of reciprocal proportions (4) Dalton’s law of partial pressure
14. A sample of pure carbon dioxide, irrespective of its source, contains 27.27% carbon and 72.73% oxygen. The data supports the (1) law of constant composition (2) law of conservation of mass (3) law of reciprocal proportions (4) law of multiple proportions
15. Hydrogen combines with oxygen to form H 2O, in which 16 g of oxygen combines with 2 g of hydrogen. Hydrogen also combines with carbon to form CH 4, in which 2 g of hydrogen combines with 6 g of carbon. If carbon and oxygen combine together, then they will show in the ratio of (1) 6 : 16 (2) 6 : 18 (3) 1 : 2 (4) 12 : 24
16. Two elements X and Y have atomic weights of 14 and 16. They form a series of compounds A, B, C, D, and E, in which the same amount of element X combines with Y in the ratio 1 : 2 : 3 : 4 : 5, respectively. If the compound A has 28 parts by weight of X and 16 parts by weight of Y, then the compoud C will have 28 parts by weight of X and (1) 32 parts by weight of Y (2) 48 parts by weight of Y (3) 64 parts by weight of Y (4) 80 parts by weight of Y
17. n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as, X + Y= R + S. The relation that can be established by the amounts of the reactants and the products will be (1) n–m=p–q (2) n+m=p+q (3) n=m (4) p=q
18. A sample of calcium carbonate (CaCO 3) has the following percentage composition: Ca = 40%; C = 12%; O = 48%, If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate from another source will be (1) 0.016 g (2) 0.16 g (3) 1.6 g (4) 16 g
Answer Key
1.6 DALTON’S ATOMIC THEORY
■ Matter consists of indivisible atoms.
■ All the atoms of a given element have identical properties, including identical mass. Atoms of different elements differ in mass.
■ Compounds are formed when atoms of differ ent elements combine in a fixed ratio.
■ Chemical reactions involve reorganisation of atom s. These are neither created nor destroyed in a chemical reaction.
■ Dalton’s theory could explain the laws of chemical combination.
1.7 ATOMIC MASS AND MOLECULAR MASS
■ Mass of one H-atom = 1.66 × 10 –24 g.
■ Atomic Mass and Mass Number
■ Reference Element for Atomic Mass: Initially H (taken as 1 amu), later O, and finally C-12 (since 1961).
■ Standard Definition: 1 amu = 1 12 × mass of one C–12 atom (also called Dalton, Da).
■ Symbol: ‘u’ replaces ‘amu’ on the C-12 scale (unified scale).
■ Example: H: 1.007825 amu (≈1)
■ O: 15.995 amu (≈16)
■ Atomic Mass: The relative weight of an atom compared to 1 amu.
Massofoneatomofelement Massof ×
■ Atomic mass of element = 112 12C
■ Average Atomic Mass: Depends on isotopes & their abundance.
Average atomic mass of xAyA12 E xy + = +
■ Mass Number (A): Whole number sum of protons + neutrons in an atom
■ Molecular Mass (Molecular Weight)
■ Definition: The number of times a molecule is heavier than 1 amu.
■ Molecular weight = 12 1 12 Weight of one molecule of substance ×WeightofC
■ Formula: Molecular Mass = Sum of atomic masses of all atoms in the molecule.
■ Definition: Ionic compounds do not have independent molecules; they consist of ions and are represented by a formula.
■ Formula Mass is used instead of Molecular Mass.
■ Example: Sodium chloride (NaCl)
Formula Mass = 23 (Na) + 35.5 (Cl) = 58.5 amu
1.8 MOLE CONCEPT AND MOLAR MASS
■ Definition of mole: One mole is the amount of a substance that contains as many particles (or entities) as there are atoms in exactly 12 g (or 0.012 kg) of the C-12 isotope.
No. of carbon atoms in one mole of carbon-12 23 23 12g/mol 6.02210 1.99264810g/atom ==× × atoms/mol.
This number is called Avogadro’s number (N A or NO).
■ Gram Atom : one mole of atoms is gram atom
Number of gram atoms = Weightingramsofelement GAW
Gram Atomic Weight & Gram Molecular Weight
■ Gram Atomic Weight: The weight (in grams) of one mole of atoms of an element.
Example: Atomic weight of carbon = 12 amu, so one gram atom of carbon = 12 g.
■ Gram Molecular Weight (Molar Mass): The weight (in grams) of a substance numerically equal to its molecular mass.
Example: If molecular mass of a substance is 18 amu, its molar mass = 18 g
Examples of gram molecules are given in Table.
■ Gram Molecular Weight: The mass of one mole of a molecular compound is called its gram molecular weight.
■ Molar Volume: The volume occupied by one mole of any gas at STP (273 K, 1 atm) is 22.4 L. 28g
Chemical Entities & Avogadro’s Number
■ Avogadro’s Number represents the number of elementary units in one mole of a substance.
■ These units can be molecules, atoms, ions, electrons, bonds, or other specified particles.
Number of moles = Weightingramsof compound GMW (or) Weightingramsofelement GAW
Number of molecules = A Weightingramsofsubstance N Grammolecularweight ×
moles
Number of atoms =
Weightingramsof anelement
Mass and the number of particles
× 1023 molecules
× 1023 molecules
■ Number of chemical units denoted by molar mass of substances is given in Table 1.6 and some more relationships are illust rated in Table.
Examples of mole relationships
× 1024 atoms
The Importance of Mole
■ Mole concept is important and useful as it accounts for the following:
One mole of an elements represents one gram atomic weight of the element. It is called gram atom.
CHAPTER 1: Some Basic Concepts of Chemistry
One mole of a substance represents one gram molecular weight of substance. It is called gram mole.
One mole of an element means the mass of 6.022 × 10 23 atoms of the element.
One mole of a covalent substance means the mass of 6.022 × 1023 molecules of the substance.
One mole of an ionic substance means the mass of 6.022 × 10 23 formula units of the substance.
■ The formation of ammonia from its elements in terms of mole concept is summarised in Table.
Formation of ammonia - mole concept
One mole of Nitrogen Three moles of hydrogen Two moles of ammonia 28 g of Nitrogen 6 g of hydrogen 34 g of ammonia 6.022 × 1023 N2 molecules
× 1023 H2 molecules
22.4 L of N2 at STP 67.2 L of H2 at STP
× 1024 NH3 molecules
L of NH3 at STP
One mole of a gas or vapour occupies a volume called GMV. The GMV at STP conditions is 22.4 L.
The charge of one mole of electrons is called faraday. One faraday is given as 96,500 couloumbs.
The concept of mole represents Avogadro cons tant.
Density of a gas in g L -1 GMW = 22.4
Loschmidt number is the number of molecules present in one cc. of a gas at STP.
Loschmidt number NA 22400 =
Number of molecules in a molecular compound = Number of gram molecules × N A
Number of atoms in a pure substance = Number of molecules × atomicity
Number of moles in a c ompound = Mass GMW
Mole is extremely useful in calculations because it simplifies the work of a chemist.
TEST YOURSELF
1. Number of molecules in one litre of oxygen at STP conditions is (1) 23 6.0210 32 × (2) 6.02×1023
2. Which of the following has the least number of atoms?
(1) 0.5 g atom of Zn (2) 0.645 g of Zn (3) 0.25 mole of Zn (4) 6.45×1020 amu of Zn
3. Air contains nitrogen and oxygen in the volume ratio of 4 : 1. The average molecular weight of air is (1) 30 (2) 28.8 (3) 28 (4) 14.4
4. The number of gram atoms of sulphur present in 3 moles of hydrogen sulphide is (1) 3 (2) 2 (3) 1.5 (4) 1.0
5. 1 mole of water vapour is condensed to liquid at 25 °C. Now, this water contains
I) 3 moles of atoms
II) 1 mole of hydrogen molecules
III) 10 moles of electrons
IV) 6 g of oxygen
The correct combination is (1) (I) and (II) (2) (I) and (III) (3) (I) and (IV) (4) All are correct
6. One mole of CH4 contains (1) 6.02 × 1023 atoms of hydrogen (2) 4 gram atoms of hydrogen (3) 3 g of carbon (4) 1.81 × 1023 molecules of CH4
7. The number of molecules in one litre of water is (density of water = 1 g/mL) (1) 6 × 1023 / 22.4 (2) 3.33 × 1025 (3) 3.33 × 1023 (4) 3.33 × 1024
8. The ratio between the number of molecules in equal masses of CH 4 and SO2 is (1) 1 : 1 (2) 4 : 1 (3) 1 : 4 (4) 2 : 1
9. The number of sulphur atoms present in 0.2 moles of sodium thiosulphate is (N = Avogadro number) (1) 4N (2) 0.2N (3) 0.4N (4) 0.1N
10. The number of nitrogen molecules present in 1 cc of gas at NTP is (1) 2.67×1022 (2) 2.67×1021 (3) 2.67×1020 (4) 2.67×1019
11. A gaseous mixture contains oxygen and nitrogen in the ratio 1 : 4 by weight. The ratio of their number of molecules is (1) 1 : 4 (2) 4 : 1 (3) 7 : 32 (4) 3 : 16
12. An a -particle changes into a helium atom. In the course of one year, the volume of helium collected from a sample of radium was found to be 1.12×10–2 mL at STP. The number of particles emitted by the sample of radium in the same time is (1) 6 × 1017 (2) 3 × 1017 (3) 1.5 × 1017 (4) 1.2 × 1018
13. The mixture containing the same number of molecules as that of 14 g of CO is
(1) 14 g of nitrogen + 16 g of oxygen
(2) 7 g of nitrogen + 16 g of oxygen
(3) 14 g of nitrogen + 8 g of oxygen
(4) 7 g of nitrogen + 8 g of oxygen
14. The number of atoms of hydrogen present in 1.5 moles of H 2O is (1) 1 N (2) 2 N (3) 3N (4) 0.5 N
15. Which of the following contains the maximum number of atoms?
(1) 10 g of CaCO3 (2) 4 g of hydrogen (3) 9 g of NH4NO3 (4) 1.8 g of C6H12O6
16. Which contains more number of molecules?
(1) 1 mole of carbon dioxide (2) 4 g of hydrogen (3) 6 g of helium (4) 33.6 litres of oxygen at STP
17. Which of the following is the heaviest?
(1) 50 g of iron (2) 5 moles of nitrogen (3) 0.1 gram atom of silver (4) 1023 atoms of carbon
18. The density of a gas is 2, relative to nitrogen, under the same conditions. The molecular weight of the gas is (1) 5.6 (2) 28 (3) 56 (4) 14
19. The density of a gas at STP is 1.5 g/L. Its molecular weight is (1) 22.4 (2) 33.6 g (3) 33.6 (4) 44.8
20. A copper plate of 20 cm × 10 cm is to be plated with silver of 1 mm thickness. The number of silver atoms required for plating is (density of silver = 10.8 g/cc) (1) 1.2×1024 (2) 2.4×1024 (3) 1.2×1013 (4) 2.4×1023
21. Which of the following has the number of molecules present equal to those present in 16 grams of oxygen?
(1) 16 g O3 (2) 32 g SO2 (3) 16 g SO2 (4) All the above
22. What is the mole percentage of O 2 in a mixture of 7 g of N 2 and 8 g of O2? (1) 25% (2) 75% (3) 50% (4) 40%
23. 7.5 g of a gas occupies 5.6 litres as STP. The gas is (1) NO (2) N2O (3) CO (4) CO2
24. Total number of sulphate ions present in 3.92 g of chromic sulphate is (Cr = 52, S = 32, O = 16) (1) 1.8 × 1022 (2) 1.8 × 1023 (3) 1.2 × 1021 (4) 6 × 1023
25. Number of atoms in 558.5 grams of Fe (at.wt. of Fe = 55.85 g mol –1) is (1) twice that in 60 g of carbon (2) 6.023 × 1022 (3) Half that in 8 g of He (4) 558.6 × 6.023 × 1023
26. The mass of 1.5 × 1026 molecules of a substance is 16 kg. Molecular mass of the substance is (1) 64 g (2) 64 amu (3) 16 amu (4) 32 amu
Answer Key
1.9 PERC ENTAGE COMPOSITION
■ A chemical formula represents the elements in a compound along with their relative proportions.
■ The composition of a pure compound remains fixed, as per the law of definite proportions.
■ Weight percent of an element is the mass of the element present in 100 g of the comp ound
■ Weight percent of an element in a compound =
Weight of element in one mole of the compound Gram molecular weight of compound 100 ×
■ Methods and equations for the estimation of different elements are listed in Table
Methods of estimating elements in a given substance (w)
Carbon Liebig’s method
Hydrogen Liebig’s method
Nitrogen Duma’s method
Nitrogen Kjeldahl’s method
Sulphur Carius tube method
Chlorine Carius tube method
of ammonia (w3)
of barium sulphate (w4)
Carius tube method
Carius tube method
Carius tube method
Types of Chemical Formulas
■ Empirical Formula – The simplest formula showing the relative number of atoms of different elements in a compound.
■ Molecular Formula – Represents the actual number of atoms of each element in a molecule.
For some compounds, both formulas are the same, but in general: Molecular Formula = Empirical Formula × n where n is a whole number (1, 2, 3, etc.).
= Molecularweight n Empiricalformulaweight
■ Empirical Formula Determination Steps
Determine the weight percentage of each element in the compound.
Divide each percentage by the atomic weight to find the relative number of atoms.
Obtain the simplest whole number ratio by dividing all values by the smallest one.
If necessary, multiply by a suitable integer to get whole numbers.
The final atomic ratio represents the empirical formula.
■ Empirical formulas are the same for compounds with identical elemental compositions.
■ Empirical and molecul ar formulae of some substances
Method for Determining the Molecular Formula of a Gaseous Hydrocarbon
■ The combustion of hydrocarbon is given by ()()()() 22242 1 4 excess
++→+
+
Mix a known volume of the hydrocarbon with an excess volume of oxygen or air in a eudiometer over mercury.
Ignite the mixture using an electric spark to cause combustion.
Cool the system, allowing water vapor to condense into liquid (its volume is negligible).
Measure the remaining gas volume to determine the amounts of CO₂ and unreacted oxygen.
Use volume ratios to deduce the molecular formula of the hydrocarbon
+
4 x cc
CO₂ Absorption: KOH is added, which absorbs CO₂.
Decrease in volume = Volume of CO₂ produced.
O₂ Absorption: The remaining gas is passed through alkaline pyrogallol, which absorbs O₂.
Further decrease in volume = Volume of unused O₂.
■ The molecular formula is determined using these volume measurements. Values of x and y are then calculated from the following data:
Volume of O2 used per cc of hydrocarbon y xcc
Volume of CO2 produced = x cc
Contraction on explosion and cooling
TEST YOURSELF
1. The weight percentage of oxygen in NaOH is (1) 40 (2) 6 (3) 8 (4) 20
2. Caffeine contains 28.9% by mass of nitrogen. If molecular mass of caffeine is 194, then the number of N atoms present in one molecule of caffeine is (1) 3 (2) 4 (3) 5 (4) 6
3. 10 g of hydrofluoric acid gas occupies 5.6 litres of volume at STP. If the empirical formula of the gas is HF, then its molecular formula will be (Atomic mass of F = 19) (1) HF (2) H 3 F 3 (3) H2F2 (4) H 4 F 4
4. An organic compound made of C, H, and N contains 20% nitrogen. What will be its molecular mass if it contains only one nitrogen atom in it? (1) 70 (2) 140 (3) 100 (4) 65
5. A peroxidase enzyme contains 2% selenium (Se = 80). The minimum molecular weight of the enzyme is (1) 1000 (2) 2000 (3) 4000 (4) 800
6. Haemoglobin contains 0.33% iron (Fe = 56). The molecular weight of haemoglobin is 68000. The number of iron atoms in one molecule of haemoglobin is (1) 2 (2) 3 (3) 4 (4) 5
7. A gaseous alkane requires five times its volume of oxygen under the same conditions for complete combustion. The molecular formula of the alkane is (1) C2H6 (2) C4H10 (3) C3H8 (4) CH4
8. A dibasic acid containing C, H, and O was found to contain C = 26.7% and H = 2.2%. The vapour density of its diethyl ester was found to be 73. The molecular formula of the acid is (1) CH2O2 (2) C2H2O4 (3) C3H3O4 (4) C2H4O4
9. 0.36 g of an organic compound, on combustion, gave 1.1 g of CO 2 and 0.54 g of H2O. The percentages of carbon and hydrogen in the compound are (1) 75, 25 (2) 60, 40 (3) 83.33, 16.67 (4) 77.8, 22.2
10. 40 ml of a hydrocarbon undergoes combustion in 260 ml of oxygen and gives 160 ml of CO 2. If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is (1) C3H8 (2) C4H8 (3) C6H14 (4) C4H10
Answer Key
1.10 STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
■ Stoichiometry deals with the relative quantities of the substances taking part in a reaction.
CH4(g) + 2O2 → CO2(g) + 2H2O
■ The above chemical reaction gives the following information.
The coefficients, one each for CH 4 and CO2 and the coefficients, 2 each for O 2 and H2O, are called stoichiometric coeffic ients.
The qualities of the substances involved are as follows:
1. Relative number of moles involved
2. Relative number of molecules involved
3. Relative masses of the substances involved
4. Relative volumes at STP
■ Thus, one mole of CH4 reacts with 2 moles of O2 to give one mole of CO2 and 2 moles of H2O.
Limiting Reagent
■ In a reaction, the reactant present in the least quantity determines the amount of product formed.
■ The reactant that is completely consumed is called the limiting reagent; others are excess reagents.
■ Example: Mg + 2HCl → MgCl₂ + H₂
Given: 6 moles of Mg and 6 moles of HCl.
Mg needs 12 moles of HCl, but only 6 moles are available.
HCl is the limiting reagent; Mg is in excess.
■ Finding the Limiting Reagent: Divide the given moles of each reactant by its stoichiometric coefficient.
The reactant with the smallest ratio is the limiting reagent. Example: Consider the reaction, 3BaCl2 + 2Na3PO4 → 6NaCl + Ba3(PO4)2
■ If the number of moles of BaCl2 and Na3PO4 taken is 10 each, ratio of moles of 2 10 BaCl3.33 3 ==→ lowest value.
Ratio of moles of 34 10 NaPO5 2 == . Hence, the limiting reagent is BaCl 2.
Volume – Volume Relationship
■ If gases are used as reactants and gases are produced in the reaction, Gay-Lussac’s law of combining volumes may be used for calculations based on equations.
■ The mole coefficients of gaseous substances give the volumes or volume ratio of the substances directly.
TEST YOURSELF
1. What is the weight of oxygen that is required for the complete combustion of 2.8 kg of ethylene? (1) 6.8 kg (2) 6.4 kg (3) 96.0 kg (4) 9.6 kg
2. 30 g of marble stone, on heating, produced 11 g of CO 2. The percentage of CaCO3 in marble is (1) 75% (2) 80% (3) 83.3% (4) 86.6%
3. The number of moles of CO 2 produced when 3 moles of HCl react with excess of CaCO 3 is (1) 1 (2) 1.5 (3) 2 (4) 2.5
4. The volume of CO2 obtained by the complete decomposition of one mole of NaHCO 3 at STP is (1) 22.4 L (2) 11.2 L (3) 44.8 L (4) 4.48 L
5. 5 g of a sample of magnesium carbonate, on treatment with excess of dilute hydrochloric acid, gave 1.12 L of CO2 at STP. The percentage of magnesium carbonate in the mixture is (1) 42 (2) 40 (3) 84 (4) 80
6. The volume of CO2 formed when 1 litre of O2 is reacted with 2 L of CO under the same condition is (1) 1 L (2) 2 L (3) 3 L (4) 1.5 L
7. Air contains 20% by volume of oxygen. The volume of air required for the complete combustion of 1 L of methane under the same conditions is (1) 2 L (2) 4 L (3) 10 L (4) 0.4 L
8. When 20 ml of methane and 20 ml of oxygen are exploded together and the reaction mixture is cooled to laboratory temperature, the resulting volume of the mixture is (1) 40 ml (2) 20 ml (3) 30 ml (4) 10 ml
9. For the reaction A + 2B → C, 5 moles of A and 8 moles of B will produce (1) 5 moles of C (2) 4 moles of C (3) 8 moles of C (4) 13 moles of C
10 How many litres of CO2 at STP will be formed when 100 ml of 0.1M H 2SO4 reacts with excess of Na2CO3? (1) 22.4 (2) 2.24 (3) 0.224 (4) 5.6
11. When a sample of baking soda is strongly ignited in a crucible, it suffers a loss in weight of 3.1 g. The mass of baking soda is (1) 16.8 g (2) 8.4 g (3) 11.6 g (4) 4.2 g
12. 7 g of a sample of NaCl, on treatment with excess of silver nitrate, gave 14.35 g of AgCl. NaCl in the sample is (1) 80% (2) 50% (3) 65.8% (4) 83.5%
13. 18.4 g of a mixture of CaCO3 and MgCO3, on heating, gives 4 g of magnesium oxide. The volume of CO2 produced at STP in this process is (1) 1.12 L (2) 4.48 L (3) 2.24 L (4) 3.36 L
14. 8 g of sulphur is burnt to form SO2, which is oxidised by chlorine water. The solution is treated with BaCl2 solution. The number of moles of BaSO 4 precipitated is (1) 1 (2) 0.5 (3) 0.25 (4) 0.125
15. One mole of mixture of CO and CO2 requires exactly 20 g of NaOH to convert all the CO2 into Na2CO3. How many more grams of NaOH would it require for conversion into Na 2CO3, if the mixture is completely oxidised into CO 2? (1) 80 g (2) 60 g (3) 40 g (4) 20 g
16. The weight of H2O2 to be decomposed to liberate 5.6 litres of O 2 at STP is ______ g. (1) 34 (2) 3.4 (3) 17 (4) 1.7
17. A gas mixture contains acetylene and carbon dioxide. 20 litres of this mixture requires 20 litres of oxygen under the same conditions for complete combustion. The percentage by volume of acetylene in the mixture is (1) 50% (2) 40% (3) 60% (4) 75%
18. Acetylene can be prepared from calcium carbonate by a series of reactions. The mass of 80% calcium carbonate required to prepare 2 moles of acetylene is (1) 200 g (2) 160 g (3) 250 g (4) 320 g
19. 4.9 g of H2SO4 decomposes x g of NaCl to give 6 g of sodium hydrogen sulphate and 1.825 g of hydrochloric acid. The value of x is (1) 6.92 (2) 4.65 (3) 2.925 (4) 1.41
20. One litre of a mixture of CO and CO2 is passed over red hot coke when the volume is increased to 1.6 L under the same conditions of temperature and pressure. The volume of CO in the original mixture is (1) 400 ml (2) 600 ml (3) 500 ml (4) 800 ml
Answer Key
Concentration Methods
■ Mass per cent is obtained by using the following relation: Mass per cent = Mass of one component 100 Total mass of all components ×
■ Mole fraction (X) is the ratio of number of moles of a particular component to the total number of moles of all the components in the solution.
Mole fraction of a component = Numberofmolesofthecomponent
Totalnumberofmolesofallcomponentsofsolution
If nA moles of a substance ‘A’ are dissolved in nB moles of a substance ‘B’, then the mole fractions of A and B are given as:
■ Molarity (M) is the most widely used unit and it is denoted by M. It is defined as the number of moles of the solute in one litre of the solution. Thus ,
Molarity (M) = of moles of solute
Volume of solution in litre Number
CHAPTER 1: Some Basic Concepts of Chemistry
Units of M = moles / L–1. 1 M HCl means 1 mole of HCl in 1 litre of solution, i.e., 36.5 g HCl in 1 litre of solution.
Molarity of a solution depends upon temperature because volume of a solution is temperature dependent.
Dilution formula: M1V1 = M2V2, where M1, M2 = initial and final molarities, V1, V2 = initial and final volumes of solution.
Suppose M1 = 0.2, V1 = 1000 ml
M2 = 1.0, V2 = ?
2 0.21000 V200mL 1.0 × ∴==
■ Suppose we have 1M solution of NaOH and we want to prepare a 0.2M solution from it. 1M NaOH means 1 mole of NaOH present in 1 litre of the solution. For 0.2M solution, we require 0.2 moles of NaOH in one litre solution.
■ Example: Calculate the molarity of a solution prepared by dissolving 4 g NaOH in enough water to form 250 mL of the solution.
Sol.
Numberofmolesofsolute(n)
Molarity w n MW
Volumeofsolutionin litres(v) ====
Molarity ====
Weight4g40g10.4molL0.4M
Molecularweight0.250L
■ Molality (m) is defined as the number of moles of solute present in one kg of solvent. It is denoted by m.
Numberofmolesof solute
Molality
()
Massofsolvent in kg = m
Units of m = moles/kg
■ Molality of a solution does not change with temperature since mass remains unaffected with temperature.
■ Molality of a solution does not change with temperature since mass remains unaffected with temperature.
■ Normality (N) is the number of gram equivalents of solute per liter of solution (eq L –¹), and it decreases with increasing temperature due to volume expansion.
Number of gram equivalents of solute = Volume of solution in litres N (or) Weight of solute Gram equivalent weight × 1000 Volume of solution in ml
Dilution Law
■ Normality = Molarity × n f , nf = n–factor of the solute
■ Whenever two substances react, they always react in 1 : 1 ratio of their equivalents.
■ Number of equivalents of one reactant = Number of equivalents of other reactant
■ VANA = VBNB
■ VA(nf of A × MA) = VB (nf of B × MB)
Volume Strength of H2O2 Solution (Advance)
■ The concentration of H 2O2 is usually represented in terms of volume. If a sample of H 2O2 is labelled as ‘x volume’, it means that one volume of H 2O2 solution gives ‘x volumes’ of O 2 gas at STP on complete decomposition.
Volume strength
2 22 Volume of O gas at STP Volume of HO solution =
Consider the decomposition H 2O2 as 22 22
∴ 22400 mL of O2 gas is liberated by 68g of H 2O2 solution.
∴ x mL of O2 gas will be liberated by 68x17x g 224005600 == of H2O2
It means that 17x g 5600 of H2O2 will present in 1 mL of solution.
∴1000 mL of solution constains H 2O2 = 17x17x 1000 56005.6 ×=
Number of grams of H2O2 per litre 17x 5.6
Strength (gL–1) = Normality × Equivalent weight.
17x34 N 5.62 =× ( Q n-factor of H2O2=2)
x = 5.6 × N
∴ Volume strength of H 2O2 = 5.6 × Normality of H 2O2 = 11.2 × Molarity of H 2O2.
Percentage Labeling of Oleum (Advance)
■ Oleum or fuming sulphuric acid contains SO 3 gas dissolved in sulphuric acid. When water is added to oleum, SO3 reacts with H2O to form H2SO4, thus mass of the solution increases.
SO3 + H2O → H2SO4
■ The total mass of H2SO4 obtained by diluting 100 g of sample of oleum with desired amount of water, is equal to the percentage labeling of oleum.
% labeling of oleum = Total mass of H 2SO4 present in oleum after dilution.
= mass of H2SO4 initially present + mass of H 2SO4 produced on dilution.
CHAPTER 1: Some Basic Concepts of Chemistry
Degree of Hardness of Water(Advance)
■ Hard water is having soluble salts of calcium and magnetism which forms precipitates with the carboxylate ion present in soap.
There are two types of hardness
Temporary hardness:
It is due to the presence of bicarbonates of calcium and magnesium in water.
Permanent hardness:
It is due to the presence of chlorides and sulphates of Ca 2+ and Mg2+ .
■ The concentration of the salts responsible for hardness is expressed in terms of degree of hardness.
■ Degree of hardness is defined a s number of parts by mass of CaCO 3 (equivalent to the quantities of salts in the hard water ) present in one million parts of mass of water.
Hardness of water 6 3 Mass of CaCO 10ppm Mass of water
1 2 w c.f100 m =×
c.f in same of equivalent mass of CaCO 3
c.f = conversion factor
w1 = mass of salt responsible for handless
M1 = Molar mass of salt.
TEST YOURSELF
1. The number of millimoles of H 2SO4 present in 5 litres of 0.2 N H2SO4 solution is (1) 500 (2) 1000 (3) 250 (4) 0.5×10–3
2. The number of glucose molecules present in 10 ml of decimolar solution is (1) 6.0 × 1020 (2) 6.0 × 1019 (3) 6.0 × 1021 (4) 6.0 × 1022
3. 0.1 gram mole of urea is dissolved in 100 g of water. The molality of the solution is (1) 1 m (2) 0.01 M (3) 0.01 m (4) 1.0 M
4. Aqueous NaOH solution is labelled as 10% by weight. Mole fraction of the solute in it is (1) 0.05 (2) 0.0476 (3) 0.052 (4) 0.52
5. The density of 3 molal solution of NaOH is 1.110 g/ml. Molarity of the solution is (1) 1.97 M (2) 2.97 M (3) 3.45 M (4) 2.45 M
6. The volume of 0.025 M Ca(OH) 2 solution, which can neutralise 100 ml of 10 –4 M H3PO4, is (1) 10 ml (2) 60 ml (3) 0.6 ml (4) 2.8 ml
7. When 100 ml of 0.06 M Fe(NO 3)3, 50 ml of 0.2M FeCl3, and 100 ml of 0.26 M Mg(NO 3)2 are mixed, the concentration of NO 3 – ions in the final solution is (1) 0.028 M (2) 0.32 M (3) 0.12 M (4) 0.28 M
8. The molality of 2% (W/W) NaCl solution is nearly (1) 0.02 m (2) 0.35 m (3) 0.25 m (4) 0.45 m
9. 100 ml of 2 M HCl solution completely neutralises 10 g of a metal carbonate. Then, the equivalent weight of the metal is (1) 50 (2) 20 (3) 12 (4) 100
10. Molarity of 200 ml of HCl solution, which can neutralise 10.6 g of anhydrous Na 2CO3, is (1) 0.1M (2) 1M (3) 0.6M (4) 0.75M
Answer Key
# Exercises
JEE MAIN
Level I
Importance of Chemistry
Single Option Correct MCQs
1. In which medical treatment are the drugs Cisplatin and Taxol effective?
(1) Cancer therapy
(2) Cardiac therapy
(3) Physiotherapy
(4) Muscular therapy
2. In which medical condition is Azidothymidine (AZT) primarily used?
(1) AIDS
(2) Leprosy
(3) Skin pigment disorders
(4) Muscular disorders
Nature of Matter
Single Option Correct MCQs
3. Which of the following can be classified as an element?
(1) Water (2) Sodium (3) Ammonia (4) Sugar
4. Which of the following is a pure substance?
(1) Air (2) Copper
(3) Salt solution (4) Brass
5. Which of the following is an example of a homogeneous mixture?
(1) Sand and water (2) Sugar solution
(3) Milk (4) oil and water
6. Which of the following is a characteristic of gases?
(1) Definite shape and definite volume
(2) Definite shape but variable volume
(3) Indefinite shape and definite volume
(4) Indefinite shape and indefinite volume
7. Which of the following processes involves a change from a liquid state to a gaseous state?
(1) Condensation
(2) Freezing
(3) Evaporation
(4) Sublimation
Numerical Value Questions
8. There are _____ fundamental physical states.
Properties of Matter and Their Measurement
Single Option Correct MCQs
9. Which of the following properties of a gas does not vary with pressure and temperature?
(1) Density
(2) Volume of a mole
(3) Volume
(4) Vapour density
10. The prefix zepto stands for (1) 10-15 (2) 109 (3) 10-21 (4) 10-12
11. The SI unit for luminous intensity is (1) mole
(2) candela
(3) ampere
(4) kelvin
12. The SI system has _______ base units.
(1) six (2) seven (3) five (4) three
13. Which of the following is incorrect about SI units?
(1) Density in kg m–3
(2) Force in newtons
(3) Pressure in pascals
(4) Amount of the substance in mol L–1
Numerical Value Questions
14. The number of significant figures in 50000.020 × 10−3 is _______.
Uncertainty in Measurement
Single Option Correct MCQs
15. The proper value of significant figures in 38.0 + 0.0035 + 0.00003 is (1) 38
(2) 38.0035
(3) 38.00353
(4) 38.0
16. The actual product of 4.327 and 2.8 is 12.1156. The correctly reported answer will be
(1) 12 (2) 12.1
(3) 12.12 (4) 12.116
17. After rounding off 1.235 and 1.225, their answers will, respectively, be
(1) 1.23, 1.22
(2) 1.24, 1.123
(3) 1.23, 1.23
(4) 1.24, 1.22
18. The number of significant figures in electronic charge 1.602 × 10 –19 C is (1) 1 (2) 2
(3) 3 (4) 4
19. Which of the following statements is correct?
(1) There is no difference between precision and accuracy.
(2) A good precision always means good accuracy
(3) Accuracy means that all measured values of an experiment are close to the actual value.
(4) A measurement may have good accuracy but poor precision.
20. On dividing 0.25 by 22.1176, the actual answer is 0.011303. The correctly reported answer will be
(1) 0.011 (2) 0.01
(3) 0.0113 (4) 0.013
Numerical Value Questions
21. A student performs a titration with different burettes and finds the values 25.20 ml, 25.25 ml, and 25 ml. The average titer value is _______.
22. After rounding off 4.855 to three significant figures, the answer will be ______
23. The number of significant figures should be present in the answer for 4.0 × 4.364 will be ____
24. When 4000 is converted to scientific notation, the number of significant figures is ________.
25. The value in the multiplication of 1.1 with 1.111 is ______.
26. The number of significant figures in the sum of 8.01, 10.11, and 9.123 is _______.
Laws of Chemical Combinations
Single Option Correct MCQs
27. The law of conservation of mass holds good for all the following, except
(1) nuclear reactions
(2) endothermic reactions
(3) all chemical reactions
(4) exothermic reactions
28. The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This proves the law of
(1) constant proportions
(2) reciprocal proportions
(3) multiple proportions
(4) conservation of mass
29. In compound A, 1.00 g of nitrogen combines with 0.57 g of oxygen. In compound B, 2.00 g of nitrogen combines with 2.24 g of oxygen. In compound C, 3.00 g of nitrogen combines with 5.11 g of oxygen. These results obey which of the follo wing laws?
(1) Law of constant proportions
(2) Law of multiple proportions
(3) Law of reciprocal proportions
(4) Dalton’s law of partial pressure
30. One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combine together, the ratio of their masses will be governed by the
(1) law of definite proportions
(2) law of multiple proportions
(3) law of reciprocal proportions
(4) law of conservation of mass
31. Law of combining volumes was proposed by
(1) Lavoisier
(2) Gay Lussac
(3) Avogadro
(4) Dalton
Daltons Atomic Theory
Single Option Correct MCQs
32. Dalton’s atomic theory could not explain the law of
(1) conservation of mass
(2) multiple proportions
(3) constant proportions
(4) gaseous volumes
33. Which of the following statements are incorrect postulates of Dalton’s atomic theory?
A) Atoms of different elements differ in mass.
B) Matter consists of divisible atoms.
C) Compounds are formed when atoms
of different elements combine in a fixed ratio.
D) All the atoms of a given element have different properties, including mass.
E) Chemical reactions involve the reorganisation of atoms.
Choose the correct answer from the options given below.
(1) B, D, and E only
(2) C, D, and E only
(3) A, B, and D only
(4) B and D only
Atomic and Molecular Masses
Single Option Correct MCQs
34. Air contains nitrogen and oxygen in a volume ratio of 4 : 1. The average molecular weight of air is
(1) 30 (2) 28.8
(3) 28 (4) 14.4
35. It is given that the abundance of isotopes 54Fe, 56Fe, and 57Fe are 5%, 90%, and 5%, respectively. The atomic mass of Fe is
(1) 55.85 (2) 55.95
(3) 55.75 (4) 56.05
36. The mass of 1.5 × 10 26 molecules of a substance is 16 kg. The molecular mass of the substance is
(1) 64 g (2) 64 amu
(3) 16 amu (4) 32 amu
Mole Concept and Molar Masses
Single Option Correct MCQs
37. The number of molecules in one litre of water is (density of water = 1 g/mL)
(1) 6 × 1023 / 22.4 (2) 3.33 × 1025
(3) 3.33 × 1023 (4) 3.33 × 1024
38. The mixture containing the same number of molecules as that of 14 g of CO is
(1) 14 g of nitrogen + 16 g of oxygen
(2) 7 g of nitrogen + 16 g of oxygen
(3) 14 g of nitrogen + 8 g of oxygen
(4) 7 g of nitrogen + 8 g of oxygen
39. The charge present on one mole of electrons is
(1) 96500 coulombs (2) 1 coulomb
(3) 1.60 × 10-19 C (4) 0.1 faraday
40. Which of the following has the highest mass?
(1) 1 gram atom of iron
(2) 5 moles of N2
(3) 1024 carbon atoms
(4) 44.8 L of He at STP
41. The ratio between the number of molecules in equal masses of nitrogen and oxygen is
(1) 7 : 8 (2) 1 : 9
(3) 9 : 1 (4) 8 : 7
42. 1 mole of water vapour is condensed to liquid at 25 °C. Now, this water contains i) 3 moles of atoms
ii) 1 mole of hydrogen molecules
iii) 10 moles of electrons
iv) 16 g of oxygen
Choose the correct combination.
(1) (i) and (ii) are correct.
(2) (i) and (iii) are correct. (3) (i) and (iv) are correct. (4) All are correct.
43. If 1 gram of hydrogen contains 6×10 23 atoms, then 20 grams of Ne contains (1) 6 × 1023 atoms (2) 12 × 1023 atoms (3) 24 × 1023 atoms (4) 1.5 × 1023 atoms
44. The charge present on 10 moles of electrons is
45. The weight of 0.1 moles of Na 2CO3 is (1) 106 g (2) 10.6 g
(3) 5.3 g (4) 6.02 x 1022 g
46. Avogadro’s number of helium atoms has a mass of (1) 2 g (2) 4 g
(3) 8 g (4) 4 × 6.02 × 1023 g
47. The ratio between the number of atoms in equal masses of methane and oxygen is (1) 1 : 5 (2) 5 : 9 (3) 9 : 10 (4) 5 : 1
48. The gas that is twice as dense as oxygen under the same conditions is (1) ozone (2) sulphur trioxide
(3) sulphur dioxide (4) carbon dioxide
49. Forty grams of Calcium contains N atoms. Then 24 grams of Mg contains (1) 6 × 1023 atoms (2) 12 × 1023 atoms (3) 24 × 1023 atoms (4) 1.5 × 1023 atoms
50. Number of electrons in 1.8 grams of H 2O is (1) 6.02 × 1023 (2) 3.01 × 1023 (3) 0.602 × 1023 (4) 60.22 × 1023
51. Which of the following gases has the largest number of molecules?
(1) 0.5 g of H2 (2) 16 g of CO2 (3) 16 g of CH4 (4) 0.5 g of He
52. The number of molecules in 16 g of methane is
(1) 3.0 × 1023 (2) 23 16 10 6.02 ×
(3) 6.023 × 1023 (4) 23 16 10 3.0 ×
53 Number of molecules in one litre of oxygen at STP conditions is (1) 23 6.02 x10 32 (2) 23 6.02 x10 22.4
(3) 32 × 22.4 (4) 32 22.4
54. The atomic masses of two elements A and B are 20 and 40, respectively. If x g of A contains Y atoms, how many atoms are present in 2x g of B?
(1) 2y (2) y/2 (3) y (4) 4y
Numerical Value Questions
55. The number of atoms in 8 g of sodium is x × 1023. The value of x is ____. (Nearest integer) [Given: N A = 6.02 × 10 23 mol; Atomic mass of Na = 23 u]
56. The weight of methane, which occupies the same volume at STP as 7.5 g of ethane, is ___ g.
57. 7 g of nitrogen occupies a volume of 5 litres under certain conditions. Under the same conditions, one mole of a gas, having molecular weight 56, occupies a volume of _________ L.
Percentage Composition
Single Option Correct MCQs
58. A peroxidase enzyme contains 2% selenium (Se = 80). The minimum molecular weight of the enzyme is (1) 1000 (2) 2000 (3) 4000 (4) 800
59. Haemoglobin contains 0.33% iron (Fe = 56). The molecular weight of haemoglobin is 68000. The number of iron atoms in one molecule of haemoglobin is (1) 2 (2) 3 (3) 4 (4) 5
60. An element X forms two oxides. The formula of the first oxide is XO 2. The first oxide contains 50% of oxygen. If the second oxide contains 60% oxygen, the formula of the second oxide is (1) XO3 (2) X2O3 (3) X3O2 (4) X2O
61. 0.262 g of a substance gave, on combustion, 0.361g of CO2 and 0.147g of H2O. What is the empirical formula of the substance?
(1) CH2O (2) C3H6O (3) C3H6O2 (4) C2H6O2
62. A compound contains carbon and hydrogen in the mass ratio 3 : 1. The formula of the compound is
(1) CH2 (2) CH3
(3) CH4 (4) C2H6
63. The empirical formula of acetic acid is the same as that of (1) sucrose (2) glucose (3) oxalic acid (4) formic acid
64. A compound contains 90% C and 10% H. The empirical formula of the compound is (1) C8H10 (2) C15H30 (3) C3H4 (4) C15H40
65. 5.6 g of an organic compound, on burning with excess oxygen, gave 17.6 g of CO2 and 7.2 g of H2O. The organic compound is (1) C6H6 (2) C4H8 (3) C3H8 (4) CH3COOH
66. An organic compound contains 40% of C, 13.33% of H, and 46.67% of N. Its empirical formula is
(1) C2H2N (2) C3H7N (3) CH4N (4) CHN
67. The percentage of silica in sodium silicate is approximately (Atomic weight of Si = 28)
(1) 25 (2) 40 (3) 50 (4) 60
68. A hydro carbon contains 80% by mass of carbon. The molecular formula of the hydro carbon is
(1) C2H4 (2) C3H6 (3) C2H6 (4) C4H10
69. Analysis of chlorophyll shows that it contains 2.68% Mg. The number of magnesium atoms present in 2.4 g of chlorophyll is (1) 2.68 × 6 × 1021 (2) 2.68 × 6 × 1023 (3) 2.68 × 6 × 1020 (4) 2.68 × 6 × 1020/24
70. The weight percentage of nitrogen in urea (NH2CONH2) is
(1) 38.4 (2) 46.6 (3) 59.1 (4) 61.3
Numerical Value Questions
71. A n alkaloid contains 17.28% of nitrogen and its molecular mass is 162. The number of nitrogen atoms present in one molecule of the alkaloid is _____.
72. In diammonium hydrogen phosphate (NH4)2HPO4, the weight percentage of P2O5 is ________.(Nearest integer)
73. An e nzyme contains 2% of sulphur. The molecular weight of the enzyme is 6400. How many sulphur atoms are present in that enzyme molecule?
74. A certain metal sulphide MS 2 is used extensively as a high-temperature lubricant. If MS 2 is 40.00% by mass of sulphur, the atomic mass of M is _____.
Stoichiometry and Stoichiometric Calculations
Single Option Correct MCQs
75. The number of molecules of CO2 liberated by the complete combustion of 0.1 gram atoms of graphite in air is
(1) 1.01 × 1022 (2) 6.02 × 1023
(3) 6.02 × 1022 (4) 3.01 × 1023
76. When one mole of ozone completely reacts with SO 2 , the number of moles of SO 3 formed is (1) 1 (2) 2 (3) 3 (4) zero
77. Six grams each of magnesium and oxygen were allowed to react. Assuming that the reaction is complete, the mass of MgO formed in the reaction is
(1) 42 grams (2) 28 grams
(3) 21 grams (4) 10 grams
78. The mass of CO2 obtained when 2 g of pure limestone is calcined is (1) 44 g (2) 0.22 g (3) 0.88 g (4) 8.8 g
79. X g of Ag was dissolved in HNO 3, and the solution was treated with excess NaCl when 2.87 g of AgCl was precipitated. The value of X is
(1) 1.08 g (2) 2.16 g
(3) 2.70 g (4) 1.62 g
80. 1.2 g of Mg (at. mass 24) will produce MgO equal to
(1) 0.05 g (2) 2 g
(3) 40 mg (4) 4 g
81. The number of molecules of KI needed to produce 0.4 moles of K 2Hgl4 will be 4Kl + HgCl2 → K2HgI4 + 2KCl (1) 1 (2) 3 (3) 16 (4) 1.6
82. 8Al + 30HNO3 → 8Al(NO3)3 + 3NH4NO3 + 9H2O
As per the given equation, the number of moles of aluminium metal that can be oxidised by one mole of HNO 3 is
(1) 8/3 (2) 8/30 (3) 3/8 (4) 30/8
83. The weight of oxygen required to completely react with 27 g of Al is
(1) 8 g (2) 16 g
(3) 32 g (4) 24 g
84. The weight of a pure sample of KClO3 to be decomposed in order to get 0.96 g of O2 is
(1) 2.45 g (2) 1.225 g (3) 9.90 g (4) none
85. 6 g of Mg reacts with excess of an acid. The amount of hydrogen produced will be (1) 0.5 g (2) 1 g
(3) 2 g (4) 4 g
86. The number of moles of Fe2O3 formed when 5.6 L of O2 reacts with 5.6 g of Fe is (1) 0.125 (2) 0.01 (3) 0.05 (4) 0.10
87. What volume of H2 at NTP is required to convert 2.8 g of N2 into NH3? (1) 2240 mL (2) 22400 mL (3) 6.72 L (4) 224 L
88. The volume of CO 2 obtained by the complete decomposition of one mole of NaHCO3 at STP is (1) 22.4 L (2) 11.2 L (3) 44.8 L (4) 4.48 L
Numerical Value Questions
89. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, calculate the maximum number of moles of Ba3(PO4)2 that can be formed. (Nearest integer)
90 12 grams of a mixture of sand and calcium carbonate, on strong heating, produced 7.6 grams of residue. How many grams of sand are present in the mixture?
91. The volume of CO 2 obtained by the complete decomposition of one mole of NaHCO3 at STP is ______.
92. 0.1 mole of a hydrocarbon, on complete combustion, produced 17.6 g of CO2. How many carbon atoms are present in each molecule of the hydrocarbon?
93. ‘ x’ g of calcium carbonate was completely burnt in air. The weight of the solid residue formed is 28 g. The value of ‘x’ is _______.
94. How many moles of H2SO4 can be reduced to SO2 by 2 moles of aluminium?
95. Ammonia is oxidised by oxygen to give nitric oxide and water. The weight of water produced per gram of nitric oxide is _____ grams.
96. The mass of ammonia (in grams) produced when 2.8 kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is ____.
97. The minimum number of moles of O 2 required for the complete combustion of a mixture containing 1 mole of propane and 2 moles of butane is ____.
Level II
Importance of Chemistry and Nature of Matter
Single Option Correct MCQs
1. (X) and (Y) are two pure substances. (X) can’t be decomposed into simpler substances by chemical reactions but (Y) can be decomposed into simpler substances. Identify (X) and(Y). X Y
(1) H2 O2
(2) H2 CO
(3) CO NH3
(4) CO2 Fe
Pr operties of Matter - Measurement and Uncertainty in Measurement
Single Option Correct MCQs
2. Which of the following have same number of significant figures?
A. 0.00253 B. 1.0003
C. 15.0 D. 163
Choose the correct answer from the options given below.
(1) A, B, and C only
(2) C and D only
(3) B and C only
(4) A, C, and D only
3. 81.4 g sample of ethyl alcohol contains 0.002 g of water. The amount of pure ethyl alcohol (to the proper number of significant figures) is
(1) 91.398 g
(2) 81.44 g
(3) 81.4 g
(4) 81 g
Laws of Chemical Combinations
Single Option Correct MCQs
4. An unbalanced chemical equation is against (1) the law of gaseous volumes (2) the law of constant proportions (3) the law of mass action (4) the law of conservation of mass
5. Whi ch of the following reactions is not correct according to the law of conservation of mass?
6. The percentage of hydrogen in water and hydrogen peroxide is 11.2% and 5.94%, respectively. This illustrates the law of (1) constant proportions (2) conservation of mass (3) multiple proportions (4) gaseous volume
7. 4.4 g of an oxide of nitrogen gives 2.24 L of nitr ogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at STP. The data illustrates the (1) law of conservation of mass (2) law of constant proportions (3) law of multiple proportions (4) law of reciprocal proportions
8. Which one of the following combinations illustrates the law of reciprocal proportions?
(1) Law of conservation of mass (2) Law of constant composition
(3) Law of multiple proportions (4) All of the above
10. According to Dalton’s atomic theory, an atom can ____ (1) Be created (2) Be destroyed (3) Neither be created nor destroyed (4) Centrifuged
Atomic and Molecular Mass
Single Option Correct MCQs
11. The unified atomic mass unit has a value of
(1) 1.661 × 10–27 g
(2) 1.661 × 10–27 kg
(3) 1.661 × 10–25 g
(4) 1.661 × 10–25 kg
12. The atomic masses of two elements A and B are 20 and 40, respectively. If x g of A contains Y atoms, how many atoms are present in 2x g of B?
(1) 2Y (2) Y/2 (3) Y (4) 4Y
13. 0.2 mol of an alkane, on complete combustion, gave 26.4 g of CO 2 . The molecular weight of alkane is (1) 16 (2) 30 (3) 44 (4) 58
14. The molar mass of a substance is 20 g. The molecular mass of the substance is (1) 20 g (2) 20 amu (3) 10 g (4) 10 amu
15. If m1 is the mass of 2 neutrons + 2 protons + 2 electrons and m 2 is the mass of an α-particle, then
(1) m1 > m2
(2) m1 < m2
(3) m1 = m2
(4) m1 may be > or < m2, depending on its physical state.
16. The vapour density of a volatile chloride of a metal is 59.5 and the equivalent mass of the metal is 24. The atomic mass of the element will be
(1) 96 (2) 48 (3) 24 (4) 12
17. Mass of CO2 molecule in kg [approximately] (1) 44 (2) 44 × 10–3
(3) 73.04 × 10−27 (4) 63.04 × 10−3
18. Gram molecular weight of water is 18 g. The gram molecular weight of glucose is 180 g. Mass of 1 molecule of glucose is (X) and mass of 1 molecule of water is ( Y ). Relation between (X) and (Y) is
(1) X = Y (2) 10 X = Y
(3) X = 5 Y (4) X = 10 Y
19. A compound contains 7 carbon atoms, 2 oxygen atoms, and 1.0×10 −23 g of other elements. The molecular mass of compound is (NA = 6×1023)
(1) 122 (2) 116 (3) 148 (4) 154
20. 30 g of element x contains 18.069 × 10 23 atoms of x. The gram molar mass of x
(1) 20 amu (2) 10 amu
(3) 10 g (4) 20 g
21. A certain compound has the molecular formula X4O6. If 10.0 g of the compound contains 5.62 g of X, the atomic mass of X is
(1) 62.0 amu
(2) 48.0 amu
(3) 32.0 amu
(4) 30.8 amu
Numerical Value Questions
22. An organic compound, on analysis, was found to contain 0.032% of sulphur. If one molecule contains two sulphur atoms, the molecular mass of organic compound is x × 105. Then, x is ______.
Mole Concept and Molar Mass
Single Option Correct MCQs
23. The correct increasing order of volume occupied by the following samples at NTP will be
A. Four gram atoms of nitrogen
B. One mole of methane
C. 14 grams of carbon monoxide
D. 1.5 grams atoms of helium
(1) C, D, B, A
(2) B, C, A, D
(3) B, D, C, A
(4) C, B, D, A
24. Which of the following gases will occupy the same volume that is occupied by 3 moles of CO2 at STP?
(1) 96 g of O2
(2) 2.8 g of N2
(3) 10 g of H2
(4) 72 g of SO3
25. Suppose, the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mol of XY2 weighs 10 g and 0.05 mol of X3Y2 weighs 9 g, the atomic weights of X and Y are
(1) 40, 30 (2) 60, 40
(3) 20, 30 (4) 30, 20
26. Which of the following is heaviest?
(1) 50 g of iron
(2) 5 mol of nitrogen
(3) 0.1 g atom of silver
(4) 1023 atoms of carbon
27. Which of the following has the least number of atoms?
(1) 0.5 g atom of Zn
(2) 0.645 g of Zn
(3) 0.25 mol of Zn
(4) 6.45 × 1020 amu of Zn
28. Consider the following compounds:
A. 1 g of CO2 (mol. wt. = 44)
B. 1 g of CH3CHO (mol. wt. = 44)
C. 1 g of C3H8(mol. wt. = 44)
The correct order for the total atoms present in these compounds is
(1) A = B = C (2) A > B > C
(3) C > B > A (4) B > A > C
29. Study the following table:
Which two compounds have least weight of oxygen? (molecular weights of compounds are given in brackets)
(1) II and IV (2) I and III
(3) I and II (4) III and IV
30. Number of moles of gas (A) is less than number of moles of gas (B) when equal masses of the gases are taken seperately under same conditions. Gas (A) and (B) are respectively,
(1) CH4 and SO2 (2) H2S and SO2
(3) NH3 and H2S (4) SO3 and CH4
31. The sample(s) containing the same number of Na atoms as there are Na atoms in 5.3 g of Na2CO3 is/are
(1) 4 g of NaOH
(2) 6.85 g of NaCl
(3) 0.25 mol of Na2SO4
(4) 5.6 g of Na3PO4
32. The mass of 1.5 × 1020 atoms of an element is 15 mg. The atomic mass of the element is
(1) 60 g (2) 60 mg
(3) 60 kg (4) 60 amu
33. 1 mol of water vapour is condensed to liquid at 25 °C. Now, this water contains
(i) 3 mol of atoms
(ii) 1 mol of hydrogen molecule
(iii) 10 mol of electrons
(iv) 16 g of oxygen
The correct combination is (1) (i) and (ii)
(2) (i) and (iii)
(3) (i) and (iv)
(4) (i), (ii), (iii) and (iv)
34. One mole of any gas
A. occupies 22.4 L at STP
B. contains 3.05 × 1023 molecules
C. contains 6.023 × 1023 molecules
D. contains same number of molecules as 22 g of CO2
Correct statements among the given are
(1) B, D (2) A, C
(3) B, C (4) A, D
35. Avogadro’s number is
(a) the number of atoms in gram atomic weight of substance
(b) the number of molecules in gram molecular weight of substance
(c) the number of atoms in 0.012 kg of C-12
(1) (a) and (b) only (2) (b) and (c) only
(3) (c) only (4) (a), (b), and (c)
36. I) 4.4 g of CO2
II) 2.4 g of CH4
III) 10 g of SO3
Correct increasing order of number of atoms present in the above three samples is
(1) III, II, I
(2) I, III, II
(3) II, III, I
(4) II, I, III
37. Correct the statement among the following.
A) Mass of one gram molecule oxygen is 32 g.
B) Mass of one gram atom oxygen is 16 g
C) Mass of one molecule oxygen is 32 amu.
D) Mass of one oxygen atom is 16 amu.
(1) A, B, C, and D (2) A, B, and C only
(3) B and C only (4) A and B only
Numerical Value Questions
38. Henry thinks that a mole contains 6.023 × 1024 molecules. Hence, the mass of Henry’s mole of nitrogen (in grams) is ___
Percentage Composition
Single Option Correct MCQ
39. An organic compound having C, H, and S elements contains 4% sulphur. The minimum molecular weight of the compound is
(1) 800 (2) 400 (3) 200 (4) 600
40. 60 g of a compound, on analysis, gave 24 g C, 4 g H, and 32 g O. The empirical formula of the compound is
(1) C2H4O2 (2) C2H2O2 (3) CH2O2 (4) CH2O
41. The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of the compound will be (1) C2H2 (2) C3H3 (3) C4H4 (4) C6H6
42. A compound contains 5 g sulphur and 5 g oxygen atom. The empirical formula of the compound is
(1) SO (2) SO2
(3) S2O (4) SO3
43. E ach 9.4 g of a compound contains 7.2 g carbon, 0.6 g hydrogen, and the rest is oxygen. The empirical formula of the compound is
(1) C3H3O (2) C6H3O
(3) C6H6O (4) C3H6O2
44. A gaseous hydrocarbon gives, upon combustion, 0.72 g of water and 3.08 g of CO 2 . The empirical formula of the hydrocarbon is
(1) C3H4 (2) C6H5
(3) C6H8 (4) C2H4
45. A carbon compound contains 12.8% of carbon, 2.1% of hydrogen, and 85.1% of bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula of the compound.
(1) CH3Br
(2) CH2Br2
(3) C2H2Br2
(4) C2H3Br3
46. The relative number of atoms of different elements in a compound are as follows:
A = 1.33, B = 1 and C = 1.5. The empirical formula of the compound is
(1) A2B2C3 (2) ABC
(3) A 8B6C9 (4) A 3 B3C4
47. 0.14 g of an element, on combustion, gives 0.28 g of its oxide. What is the element?
(1) Nitrogen (2) Carbon
(3) Fluorine (4) Sulphur
48. A compound contains carbon and hydrogen in the mass ratio 3 : 1. The formula of the compound is
(1) CH2 (2) CH3
(3) CH4 (4) C2H6
49. A sample of oleum is labelled as 118%. The number of moles of NaOH needed for complete neutralisation of 100 g oleum is
(1) 2.0 (2) 20 49 (3) 118 49 (4) 59 49
Numerical Value Questions
50. Calculate the percentage composition of oxygen in calcium nitrate.
Stoichiometry and Stoichiometric Calculations
Single Option Correct MCQs
51. The molality of 2% (W/W) NaCl solution is nearly
(1) 0.02 m (2) 0.35 m
(3) 0.25 m (4) 0.45 m
52. Assuming that sea water is a 3.5 wt% aqueous solution of NaCl, what is the molality of sea water?
(1) 2 molal (2) 2.5 molal
(3) 0.62 molal (4) 1.5 molal
53. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution?
(1) 0.190 (2) 0.086
(3) 0.050 (4) 0.100
54. 6 g of urea is dissolved in 90 g of water. The mole fraction of solute is
(1) 1/5 (2) 1/50 (3) 1/51 (4) 1/501
55. Mole fraction of solute in an aqueous solution is 0.2, the molality of the solution is (1) 13.88 (2) 1.388
(3) 0.138 (4) 0.0138
56. 1 kg of 2 m urea solution is mixed with 2 kg of 4 m urea solution. The molality of the resulting solution is (1) 3.33 m (2) 10 m (3) 1.67 m (4) 5m
57. Molarity and molality of a solution of a liquid (mol. mass = 50) in aqueous solution is 9 and 10, respectively. What is the density of the solution?
58. A solution of acetic acid has molarity equal to 1.35 M and molality equal to 1.45 mol kg−1. The density of solution will be (1) 1.251 g L–1 (2) 1.125 g L–1 (3) 1.012 g L–1 (4) 0.994 g L–1
59. The number of ion s present in 1 mL of 0.1 M CaCl2 solution is (1) 1.8 × 1020 (2) 6.0 × 1020 (3) 1.8 × 1019 (4) 1.8 × 1021
60. At 25 °C, for a given solution, M = m. Then, the correct relation between them if the temperature is increased to 50 °C, is (1) M = m (2) M > m (3) M = 2m (4) M < m
61. Normality of 250 mL of acidified solution containing 3.16 g of KMnO 4 (M = 158 g/mol) is (1) 0.4 N (2) 1 N (3) 0.5 N (4) 0.2 N
62. Normality of 2% (W/V) of H2SO4 solution is nearly (1) 2 (2) 4 (3) 0.2 (4) 0.4
63. 500 mL of 1M H2SO4 completely neutralises 42 g of metal carbonate. Equivalent weight of metal is (1) 24 (2) 12 (3) 30 (4) 23
64. 1 g Ca was burnt in excess of O2 and the oxide was dissolved in water to make up one litre of solution. The normality of the solution is
(1) 0.04 N (2) 0.4 N
(3) 0.05 N (4) 0.5 N
65. Density of a solution is 0.5 g/cc. (w/v)% of 40% (w/w) of the solution will be (1) 20% (2) 80%
(3) 15% (4) 60%
66. Sulphuric acid and orthophosphoric acid have the same molar mass. The ratio of the masses of these acids needed to neutralise the same amount of NaOH, if the sulphate and dihydrogen orthophosphate were formed in their separate reactions, is (1) 1 : 2 (2) 2 : 1
(3) 1 : 3 (4) 1 : 1
67. T he minimum amount of O 2 (g) consumed per gram of reactant is for the reaction (Given atomic mass: Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
(1) 2Mg(s)+O2→ 2MgO (s)
(2) 4 Fe (s) +3 O2(g)→ 2Fe2 O3 (s)
(3) P4(s)+5 O2(g)→ P4O10 (s)
(4) C3H8(g)+5 O2(g)→ 3CO2(g)+4H2O(l)
68. The number of moles of KI required to produce 0.4 mol K2HgI4 is 4KI + HgCl2→ K2HgI4 + 2KCl (1) 1 (2) 3 (3) 16 (4) 1.6
69. Acetylene, C 2 H 2 , reacts with oxygen according to the unbalanced equation:
C2H2(g)+O2(g) → CO2(g)+H2O(g)
What is the O 2 /C 2 H 2 ratio when this
equation is correctly balanced?
(1) 2 1 (2) 3 1
(3) 4 1 (4) 5 2
70. 20 mL of nitric oxide combines with 10 mL of oxygen at STP to give NO 2. The final volume will be
(1) 30 mL (2) 20 mL
(3) 10 mL (4) 40 mL
71. 2 9810 3 × g of tribasic acid is completely neutralised by 100 mL of 0.05M Ca(OH) 2 solution. The molecular weight of the acid is
(1) 49 (2) 74
(3) 37 (4) 98
Numerical Value Questions
72. Calculate the milliequivalents of HCl in its 3.65 g (at. wt. of H = 1, Cl = 35.5) is __.
73. Haemoglobin contains 0.34% of iron by mass. The number of Fe atoms in 3.3 g of haemoglobin is _______ ×10 19 .
Multiple Concept Questions
Single Option Correct MCQs
74. A compound has 40% of carbon by weight. If the molecular weight of the compound is 90, the number of carbon atoms present in 1 molecule of the compound is (1) 3 (2) 2 (3) 1 (4) 5
75. Mass percentage of nitrogen in uracil is O NH N H uracil O
(1) 70 (2) 50
(3) 80 (4) 25
76. What is the percentage of free SO 3 in a sample of oleum labelled a s 104.5%?
(1) 20% (2) 40%
(3) 60% (4) 80%
77. If 20 g water is added to 50 g of 109% oleum, the final composition of the sample is
(1) 50 g H2SO4
(2) 54.5 g H2SO4
(3) 54.5 g H2SO4 + 15.5 g H2O
(4) 54.5 g H2SO4 + 4.5 g SO3
78. The specific gravity of 98% H 2 SO 4 is 1.8 g/cc. 50 mL of this solution is mixed with 1750 mL of pure water. Molarity of the resulting solution is
(1) 0.2 M (2) 0.5 M
(3) 0.1 M (4) 1 M
79. The molarity of a sulphuric acid solution is 2.32 mol dm−3. If the density of solution is 1.14 g cm−3, the molality of the solution will be
(1) 2.54 mol kg–1
(2) 2.25 mol kg–1
(3) 2.62 mol kg–1
(4) 1.98 mol kg–1
80. What volume of 75% alcohol by weight ( d = 0.8 g/cm3) must be used to prepare 159 cm 3 of 30% alcohol by mass ( d = 0.90 g/ cm3)
(1) 67.5 mL
(2) 56.25 mL
(3) 44.44 mL
(4) 4.44 mL
81. The amount of wet NaOH containing 15% water required to prepare 70 L of 0.5 N solution is
(1) 1.65 kg (2) 1.4 kg
(3) 16.5 kg (4) 140 kg
Level III
Single Option Correct MCQs
1. 20.8 g of barium chloride is dissolved in 100 g of water. 14.2 g of sodium sulphate is dissolved in 100 g of water. Both solutions are mixed and 23.3 g of barium sulphate and 11.7 g of sodium chloride are formed.
(At. wt. Ba = 137, Cl = 35.5, S = 32, Na = 23, O = 16). This data explains
(1) law of conservation of mass
(2) law of definite proportions
(3) law of constant proportions
(4) law of multiple proportions
2. n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as X + Y → R + S. The relation that can be established in the amounts of the reactants and the products will be
(1) n – m = p – q
(2) n + m = p + q
(3) n = m
(4) p = q
3. A sample of calcium carbonate (CaCO 3) has the following percentage composition: Ca = 40%; C = 12%; O = 48%. If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate from another sourc e will be
(1) 0.016 g (2) 0.16 g
(3) 1.6 g (4) 16 g
4. Percentage composition of a natural sample of CuCO3 is as follows:
% Cu % O % C x y z
If the law of definite proportions is obeyed, then the correct composition of same elements in synthetic sample of CuCO 3 is
% Cu % O % C
(1) x y z
(2) >x >y >z
(3) <x <y <z
(4) >x <y >z
5. Which of the following statements indicates that the law of multiple proportions is being followed?
(1) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1 : 2.
(2) Carbon forms two oxides, namely CO2 and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2 : 1.
(3) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(4) At constant temperature and pressure, 200 mL of hydrogen will combine with 100 mL of oxygen to produce 200 mL of water vapour.
6. Hydrogen combines with oxygen to form H2O, in which 16 g of oxygen combines with 2 g of hydrogen. Hydrogen also combines with carbon to form CH4, in which 2 g of hydrogen combines with 6 g of carbon. If carbon and oxygen combine together, then they will show in the ratio of
(1) 6 : 16 (2) 6 : 18
(3) 1 : 2 (4) 12 : 24
7. An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67%, and N = 46.67%, while the rest is oxygen. On heating, it gives NH 3 along with a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is
(1) CH3NCO (2) CH3CONH2
(3) (NH2)2CO (4) CH3CH2CONH2
8. 40 mL of a hydrocarbon undergoes combustion in 260 mL of oxygen and gives 160 mL of CO 2. If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is
(1) C3H8 (2) C4H8
(3) C6H14 (4) C4H10
9. 10 mL of a gaseous hydrocarbon, on combustion, gives 40 mL of CO2 and 50 mL of H2O vapour under the same conditions. The hydrocarbon is
(1) C4H6 (2) C6H10
(3) C4H8 (4) C4H10
10 For the reaction, 2x + 3y + 4z → 5 w, initially, if 1 mole of x, 3 mol of y, and 4 mol of z is taken. 1.25 mol of w is obtained. Then, the percentage yield of this reaction is
(1) 50% (2) 60%
(3) 12% (4) 25%
11. In which of the following solutions mole fraction of solute is highest? (density of water = 1 g/ml)
(1) 220 g of aqueous solution containing 1 mol of NaOH (mol. wt. = 40)
(2) 282 g of aqueous solution of urea containing 162 g of water
(3) 540 g of glucose added to 8 moles of water
(4) 4 mol of KOH dissolved in 126 g of water
12. Wood’s metal contains 50.0% bismuth, 25.0% lead, 12.5 % tin, and 12.5% cadmium by mass. What is the mole fraction of tin?
(atomic mass: Bi = 209, Pb = 207, Sn = 119, Cd = 112)
(1) 0.202 (2) 0.158
(3) 0.182 (4) 0.221
13. The expression converting molality (m ) into molarity (M) is
(where M2 is the molar mass of solute and r is the density of the solution)
(1) M =(1+ mM2) m r
(2) M = m r (1+mM2)
(3) M = mM2/ r
(4) M = m r/ M2
14. Relation between mol arity (M) and mole fraction is
(1) A AABA .1000d M .mm χ× = χ+χ
(2) B AABB 1000d M .mm χ×× = χ+χ
(3) A m M V100 ×χ = ×
(4) A M d 1 V χ =
15 Thr ee solutions X, Y, and Z of HCl are mixed to produce 100 mL of 0.1 M solution. The molarities of X, Y, and Z are 0.07 M, 0.12 M, and 0.15 M, respectively. What respective volumes of X, Y, and Z should be mixed?
(1) 50 mL, 25 mL, 25 mL
(2) 20 mL, 60 mL, 20 mL
(3) 40 mL, 30 mL, 30 mL
(4) 55 mL, 20 mL 25 mL
16. One mole of KClO3 is heated in the presence of MnO2 (catalyst). The produced oxygen is used in burning of Al. Then, oxide of Al that will be formed is
(1) 2 mol (2) 1 mol
(3) 4 mol (4) 3 mol
17. One litre of a mixture of CO and CO 2 is passed over red hot coke, when the volume increased to 1.6 L under the same conditions of temperature and pressure. The volume of CO in the original mixture is
(1) 600 (2) 200
(3) 800 (4) 400
18. Iron has a density of 7.86 g cm −3 and an atomic mass of 55.85 amu. The volume occupied by 1 mol of Fe is
(1) 0.141cm3 mol–1
(2) 7.11 cm3 mol–1
(3) 4.28 × 104 cm3 mol–1
(4) 22.8 cm3 mol–1
19. The percentage of water of crystallisation in CaCl2·6H2O is (1) 39.4 (2) 43.9 (3) 49.3 (4) 46. 5
20. 0.607 g of a silver salt of tribasic organic acid was quantitatively reduced to 0.37 g of pure Ag. What is the molecular mass of the acid?
(1) 207 (2) 210 (3) 531 (4) 324
21. An organic base is tetra-acidic. If, from every 10 g of the chloroplatinate salt of the base, 3.9 g of residue of platinum is obtained, then what will be the molecular mass of the base? [Pt = 195]
(1) 180 (2) 360 (3) 90 (4) 270
22. The shape of tobacco mosaic virus (TMV) is cylindrical, having length and diameter 3000 A o and 170 A o , respectively. The density of the virus is 0.08 g/mL. The molecular weight of TMC is (1) 3.28
(2) 5.44 × 10–24
(3) 5.44 × 10–18
(4) 3.28 × 106
23. Manganese forms non-stoichiometric oxides having the general formula MnO x The value of x for the compound, which on analysis gave 64% by mass of Mn, is (1) 1.16 (2) 1.83
(3) 2 (4) 1.93
24. An organic compound contains C, H, and N. When 9 volumes of a gaseous mixture containing the organic compound and just sufficient quantity of O 2 was exploded, 2 volumes of N2, 6 volumes of water vapour, and 4 volumes of CO 2 were produced. If all measurements are done under similar conditions of temperature and pressure, what will be the formula of the organic compound?
(1) C2H6N (2) C2H6N2
(3) C3H6N2 (4) C2H6N3
25. A hydrocarbon CnH2n yields CnH2n+2 by reduction. The molar mass of the product is 2.38% more than the reactant. The value of n is
(1) 8 (2) 4
(3) 6 (4) 5
26. A granulated sample of aircraft alloy (Al, Mg, Cu) weighing 8.72 g, was first treated with alkali and then with very dilute HCl, leaving a residue. The residue, after alkali boiling, weighed 2.10 g and the acid insoluble residue weighed 0.69 g. What is the composition of the alloy?
(1) Al = 75.9%, Mg = 16.2%, Cu = 7.9%
(2) Al = 16.2%, Mg = 75.9%, Cu = 7.9%
(3) Al = 7.9%, Mg = 16.2%, Cu = 75.9%
(4) Al = 75.9%, Mg = 19.2%, Cu = 4.9%
27. A mixture of ethane and ethene occupies 16.42 L at 1 atm and at 400 K. The mixture reacts completely with 51.2 g of O 2 to produce CO 2 and H 2 O. Assuming ideal gas behaviour, calculate the mole fractions of C2H6 and C2H4 in the mixture.
(1) 0.3, 0.2 (2) 0.66, 0.33
(3) 0.75, 0.25 (4) 0.4, 0.6
28. x g of KHC2O4 requires 100 mL of 0.02 M KMnO 4 in acidic medium. In another experiment, y g of KHC2O4 requires 100 mL of 0.05 M Ca(OH)2. The ratio of x and y is
(1) 1 : 1 (2) 1 : 2
(3) 2 : 1 (4) 5 : 4
29. What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?
(1) 330 (2) 316 (3) 320 (4) 325
30. For the following reaction, A+2 B → C, if equal mass of A and B are taken, which of the following is correct?
[MA and MB are the molar masses of A and B, respectively]
(1) If MA = 2MB, then none of the reactant is left.
(2) If MA = MB, then A will be limiting reagent.
(3) If MB = MA, then B will be limiting reagent
(4) If A B 2 > M M then A wi ll be limiting reagent
Numerical Value Questions
31. A 300 mL bottle of soft drink has 0.2M CO2 dissolved in it. Assuming that CO2 behaves as an ideal gas, the volume of the dissolved CO2 at STP is ______ mL (nearest integer).
(Given: At STP molar volume of an ideal gas is 22.7 L mol–1
32. 2 mol of carbon reacts with oxygen in air or oxygen forming two oxides, CO and CO2, depending on the amount of oxygen available. If 19.2 g of CO is formed, then the percentage of carbon converted into CO is (Atomic weights of C = 12, O =16) ______.
33. 2 mol of a mixture of O2 and O3 is reacted with excess of acidified solution of KI. The iodine liberated required 1L of 2 M hypo solution for complete reaction. The weight percentage of O3 in the initial sample is x Find x
34. A gaseous mixture contains SO 3(g) and CH4(g) in 5 : 1 ratio by mass. Calculate Q, where Q = 200 × ratio of total number of atoms present in SO3(g) to total number of atoms present in CH4(g).
35. Lithium nitride hydrolyses in water to produce a basic solution. What is the mass of HCl required to neutralise the 1 litre of solution, in which 2 mol of lithium nitride is dissolved?
36. One litre of 2M Na3PO4 and 1 litre of 1M BaCl 2 solutions are mixed. What is the normality of phosphate ions in the filtrate?
THEORY-BASED QUESTIONS
Statement Type Questions
(Q.No. 1-10)
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
1. S–I : Sugar syrup is heterogeneous mixture.
S–II : Air is homogeneous mixture.
2. S-I : During a chemical reaction (at STP), the total volume remains constant in gaseous reactions.
S-II : During a chemical reaction, the total mass remains constant.
3. S–I : Formation of oxides of nitrogen explains the law of definite proportions.
S–II : Equivalent weight of nitrogen in oxides of nitrogen is always constant.
4. S-I : The number of atoms in a given mass of dioxygen (oxygen) and trioxygen (ozone) gases is same.
S-II : The number of atoms depends on atomic mass, not on molecular mass.
5. S-I : 12 C has atomic mass of exactly 12 atomic mass units (amu).
S-II : One amu is defined as a mass exactly equal to 1/12 of the mass of one carbon-12 atom.
6. S–I : Fructose and ethanoic acid have the same percentage composition of elements by weight.
S–II : Fructose and ethanoic acid have the same empirical formula
7. S–I : 16 g each of ozone and oxygen contain the same number of atoms.
S–II : Nitrogen cannot exhibit +5 oxidation state in its compounds.
8. S–I : 1 cc of nitrogen at STP contains 2.69 × 1019 molecules.
S–II : Molar volume of an ideal gas at STP contains Avogadro’s number of molecules.
9. S–I : 0.5 mol of P4O10 contains five gram atoms of oxygen.
S–II : 6.023 ×10 22 atoms of hydrogen can make 0.1 g atom of hydrogen.
10. Consider the given reaction and statements.
2A+3B → C
S–I : 4 3 mol of ‘C’ is always produced when 3 mol of ‘A’ and 4 mol of ‘B’ are added.
S-II : ‘B’ is the liming reactant for the given data.
Assertion and Reason Questions
(Q.No. 1-29)
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as (1) if both (A) and (R) are true and (R) is the correct explanation of (A), (2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
11. (A) : One mole of CO 2 is formed when 1 mol of O2 is reacted with 1 mol of CO, according to the reaction 2CO+O2→ 2CO2.
(R) : CO is the limiting reagent in the above reaction.
12. (A) : 3.4 g of NH 3 (g), on complete decomposition into N 2 and H 2 (g), produces 0.6 g of H2(g).
(R) : Law of conservation of mass is followed by the chemical reaction.
13. (A) : During the conversion of Fe 2O3 : Fe, the stoichiometric coefficients of iron are in the ratio of 1 : 2.
(R) : During a chemical reaction, atoms can neither be created nor destroyed.
14. (A) : Formation of C2H5OH and CH3OCH3 from respective elements obeys the law of definite proportions.
(R): Both compounds have the same molecular formula and the elements C : H : O are combined in the ratio of 12 : 3 : 8 by mass.
15. (A) : A certain element X forms three binary compounds with chlorine, containing 59.68 %, 68.95%, and 74.75% chlorine, respectively. This data illustrates the law of multiple proportions.
CHAPTER 1: Some Basic Concepts of Chemistry
(R) : According to the law of multiple proportions, the relative amounts of an element combining with some fixed amount of a second element in a series of compounds is in the ratio of small whole numbers.
16. (A) : In Haber’s process, N2 and H2 combine in 1 : 3 volume ratio.
(R) : Gases combine in a simple volume ratio.
17. (A) : For reaction 2A(g)+ 3B (g) → 4C (g) + D (g), the vapour density remains constant throughout the course of the reaction.
(R) : In all gaseous chemical reactions, vapour density remains constant.
18. (A) : Atomic weight of an atom can never be in fraction.
(R) : Average atomic weight of chlorine is 35.5.
19. (A) : Atomic mass of Mg is 24.
(R) : An atom of magnesium is 24 times heavier than 1 th 12 of the mass of a carbon atom (C12).
20. (A) : Boron has relative atomic mass of 10.81.
(R) : Boron has two isotopes 10 5 B and 11 5 B, and their relative abundances are 19% and 81%, respectively.
21. (A) : 18 g of water vapour and 18 g of ice will not contain the same number of molecules.
(R) : Number of molecules is independent of temperature and pressure.
22. (A) : The empirical mass of ethene is half of its molecular mass.
(R) : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
23. (A) : Molality of a solution increases with temperature.
(R) : Molality expression does not involve any volume term.
24. (A) : 3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.
(R) : Molar mass of hydrated oxalic acid is 126 g.
25. (A) : Mass of a solution of 1 litre of 2 M H2SO4 [density of solution =1.5 g/mL] is greater than the mass of solution containing 400 g MgO which is labeled as 40% (w/w) MgO.
(R) : Mass of H2SO4 in 1 litre 2 M H2SO4 [density of solution =1.5 g/mL] is greater than the mass of MgO in 1 litre 40% (w/w)[density of solution =2 g/mL] solution.
26. (A) : The number of glucose molecules present in 10 mL of decimolar solution is 6.023×1020
(R) : Avagadro’s number of molecules present in 1 g molar volume of solution.
27. (A) : 22.4 L of N2 at NTP and 5.6 L O2 at NTP contain equal number of molecules.
(R) : Under similar conditions of temperature and pressure all gases contain equal number of molecules.
28. (A) : Combustion of 16 g of methane gives 36 g of water and 44 g of CO 2.
(R) : In the combustion of methane, water and CO2, are products.
29. (A) : A reactant that is entirely consumed when a reaction goes to completion is known as limiting reactant.
(R) : The amount of limiting reactant limits the amount of product formed.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Choose the correct statement(s).
(1) The mole is the amount of a substance containing the same number of chemical units as there are atoms in exactly 12 g of 12C.
(2) Avogadro’s number is the number of units in a mole.
(3) The weight of one gram atom of an element means its atomic weight in g.
(4) One gram atom of each element contains the same number of atoms.
2. A+2B + 3C → AB2C3. Reaction of 6.0 g of atoms of B, and 0.036 mol of C yields 4.8 g of compound AB2C3. If the atomic masses of A and C are 60 and 80 amu, respectively, the atomic mass B of is (Avogadro’s constant = 6 × 1023)
(1) 70 amu (2) 60 amu
(3) 50 amu (4) 40 amu
3. ‘ x’ grams of 50% pure potassium chlorate liberates 0.3 mol of oxygen on thermal decomposition. x grams of pure H 2 SO 4 requires g of NaOH for complete neutralisation
(mol wt. of potassium chlorate = 122.5)
(1) 20 g (2) 40 g
(3) 60 g (4) 50 g
4. 40 g of a sample of carbon, on combustion, left 10% of it unreacted. The volume of oxygen required at STP for this combustion reaction is
(1) 11.2 L (2) 22.4 L
(3) 44.8 L (4) 67.2 L
5. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3–Na+ (molar mass: 206). What would be the maximum uptake of Ca+2 ions by the resin, when expressed in mole per gram resin?
(1) 2 309 (2) 1 412
(3) 1 103 (4) 1 206
6. If the vapour density of a mixture of NO 2 and N 2 O 4 is 34.5, then the percentage abundance (by mol) of NO2 in the mixture is
(1) 50% (2) 25%
(3) 40% (4) 60%
7. Phosphine, o n decomposition, produces phosphorus and hydrogen. When 100 mL of phosphine is decomposed, the change in volume under laboratory conditions is
(1) 50 mL increase
(2) 50 mL decrease
(3) 900 mL decrease
(4) 75 mL increase
8. 1.878 g of MBrx, when heated in a stream of HCl gas, was completely converted to chloride MCl x, which weighed 1.0 g. The specific heat of metal is 0.14 cal/g. Then, the molecular weight of metal bromide is (atomic weight of Br = 80)
(1) 285.5
(2) 185.7
(3) 85.7
(4) 216
9. Which of the following can explain the law of reciprocal proportions?
(1) H2O, CH4, CO2
(2) CO, CO2, C3O2
(3) HF, CF4, CH4
(4) N2O, NO2, N2O4
10. “Compounds are formed when atoms of different elements combine in a fixed ratio” is one of the statements of Dalton’s atomic theory. Which of the following laws is not related to this statement?
(1) Law of conservation of mass
(2) Law of definite proportions
(3) Law of multiple proportions
(4) Avogadro’s law
11. Acetic acid and glucose have the same
(1) empirical formula
(2) weight composition of elements
(3) ratio of masses of individual elements
(4) number of gram atoms of each element per mole
12. The pair of species having different percentage (mass) of carbon is
(1) CH3COOH and C6H12O6
(2) CH3COOH and C2H5OH
(3) HCOOCH3 and HCOOH
(4) C2H5OH and CH3OCH3
13. 80% carbon is present in an alkane, by weight. The possible conclusions are
(1) the empirical formula of the compound is CH3
(2) the minimum number of carbons in the molecule is 2
(3) the compound has gram atoms of C and H in 4 : 1 ratio
(4) this composition suits all alkanes
14. 10% w/v NaOH is same as
(1) 2.5 M
(2) 2.5 N
(3) 1.5 m
(4) 10% w/w NaOH
15. 4.1 g of phosphorous acid H3PO3 is present in 250 mL of a solution. The strength of the solution is
(1) 0.2 M (2) 0.4 N
(3) 16.4% w/v (4) 8.2% w/v
16. Solution (s) containing 23 g HCOOH is/ are
(1) W 46gof70%HCOOH V
(dsolution = 1.40 g/mL)
(2) 50 g of 10 M HCOOH
(dsolution=1g/mL)
(3) W 50g of 25%HCOOH V
(4) 46 g of 5MHCOOH
(dsolution =1g} /mL)
17. If 400 mL of 0.1 M HCl, 500 mL of 0.3 M H2SO4 and 100 mL of 0.1 M KNO3 are mixed, then what is correct for the resultant solution?
(1) [H+] = 0.19 M
(2) [Cl ] = 0.04 M
(3) [NO3–] = 0.01 M
(4) [K+] = 0.01 M
18. 100 mL of a solution contains 12 mg MgSO4 The concentration of the solution is (1) 10–3 M
(2) 2 × 10–3 N
(3) 120 ppm (4) 10–3 m
19. If 3.65% w/v HCl solution has density 1.0365 g/cc, then its concentration is (1) 1 M (2) 1 m
(3) 1 N (4) 0.5 M
20. If 1 mL of 1 M solution is mixed with 999 mL of pure water, then (1) 10–3 M solution is formed
(2) the mass of solute per mL decreases by 1000 times
(3) the quantity of solute decreases in the solution
(4) 10 mL of resultant solution contains 10–5 moles of solute
21. An aqueous solution of ammonia has molarity to 2 M. If the density of the solution is 1.534 g/mL, then identify the options in which the correct concentration terms are mentioned.
(1) Molality 4 3 = m
(2) () 34 %w/w 15.34 =
(3) %(w/v) = 6.8
(4) mole fraction of NH3 = 3/128
22. 1 g atom of nitrogen represents (1) 6.02 × 1023 of N2 molecules
(2) 22.4 L of N2 at 1 atm and 273 K
(3) 11.2 L of N2 at 1atm and 273 K
(4) 14 g of nitrogen
23. Which of the following are correct statements?
(1) Unbalanced equations are against the law of conservation of mass.
(2) 1 mL of any gas at STP contains 2.69 × 10 19 molecules.
(3) The number of atoms present in 8 g of sulphur is same as that in 10 g of calcium.
(4) The oxidation number of ‘Cr’ in CrO5 is +10.
24. 8 g of O2 has the same number of molecules as in
(1) 11 g of CO2 (2) 22 g of CO2
(3) 7 g of CO (4) 14 g of CO
25. 6.023×1022 atoms of hydrogen can make (1) 0.05 mol of H2 molecules
(2) 0.1 g of hydrogen atoms
(3) 0.1 g molecules of hydrogen
(4) 0.1 g atoms of hydrogen
26. 1 mol of 143 7 N ions contains
(1) 7 × 6.023 × 1023 electrons
(2) 7 × 6.023 × 1023 protons
(3) 7 × 6.023 × 1023 neutrons
(4) 14 × 6.023 × 1023 protons
27. We have 1.6 g CH 4, 1.7g NH 3, and 1.8 g H2O. Select the correct statement(s).
(1) There are equal number of moles of each reactant.
(2) Total number of atoms in CH4 > NH3 > H2O.
(3) Total number of H-atoms are in the ratio of 4 : 3 : 2.
(4) Total number of C-atoms in CH4 < that of N-atoms in NH3 < that of O-atoms in H2O.
28. 18 grams of glucose contains
(1) 0.6 gram atom of carbon
(2) 0.6 gram molecule of hydrogen
(3) 0.6 gram molecule of CO2
(4) 1.2 gram atom of hydrogen
29. 3200 g sulphur is present in
(1) 9800 g H2SO4
(2) 20 mol H2SO4
(3) 100 mol H2SO4
(4) 6400 g of SO2
30. The molar mass of haemoglobin is about 65000 g mol−1. Every haemoglobin contains 4 iron atoms. Thus,
(1) iron content in haemoglobin is 0.35% by mass
(2) 1 mol of haemoglobin contains 56 g iron
CHAPTER 1: Some Basic Concepts of Chemistry
(3) 1 mol of haemoglobin contains 224 g iron
(4) if iron content is increased to 0.56%, molar mass of haemoglobin would be higher than 65000 g mol−1
31. 1021 molecules are removed from 440 mg of CO2. It becomes
(1) 366 mg (2) 8.3 millimol
(3) 200 mg (4) 4.1 millimol
32. One takes 0.62 g of P 4 and 4.00 g of O 2 separately for each of the following reactions. (Atomic mass of P = 31; O = 16)
I. P4+5O2 → P4O10
II. P4+3 O2 → P 4 O6
Select the correct statements about I and II.
(1) P 4 is the limiting reagent in 1 and O2 is the limiting reactant in II.
(2) P 4 is the limiting reactant in II and O 2 is the limiting reactant in I.
(3) 1.42 g of P4O10 in I and 1.10 g of P4O6 in II are formed.
(4) P 4 is the limiting reactant in both I and II.
33. If 100 mL of O2 is subjected to silent electric discharge till volume of O 2 left and O 3 formed are equal, then
(1) decrease in volume is 20 mL
(2) volume of O3 formed is 40 mL
(3) volume of O2 reacted is 60 mL
(4) volume of mixture is 80 mL
34. Consider the following reactions:
I. (NH4)2SO4 +2 NaOH 40% → Na2SO3 + 2H2O +2 NH3
II. 80% 34 NHHClNHCl +→
If 4 g of NaOH is taken, then
(1) produced mole of NH4Cl (in reaction II) is 1.6 times of produced moles of Na2SO4 (in reaction I)
(2) reacting mole of HCl (in reaction II) is 20% lesser than original (NH 4)2SO4 moles
(3) reacting mole of HCl (in reaction II) is lesser than reacting moles of NaOH (in reaction I)
(4) produced mass of NH4Cl is 2.71 g
35. When taken in an eudiometer tube operating at room temperature and pressure and subjected to complete reaction, in which of the following options, the contraction in volume is greater than or equal to 30% of the original volume? [Note: Produced H2O is in liquid state]
(1) CO(g) and O2(g) taken in a molar ratio of 2 : 1
(2) 10 mL of CH4(g) and 30 mL of O2(g)
(3) N2(g) and H2(g) taken in a molar ratio of 3 : 1
(4) N2 and H2 taken in a molar ratio of 1 : 3
36. Which of the following requires six times their volume of oxygen for complete combustion?
(1) n-butane (2) 1-butene
(3) 2-butene (4) cyclobutane
37. Combustion of 2.24 L ethane at STP requires
(1) 7.84 L of O2 (2) 0.35 mol of O2
(3) 11.2 g of O2 (4) 5.6 L of O2 at STP
38. 1 mol of iron reacts completely with 0.65 mol of O2 to give a mixture of only FeO and Fe2O3. Choose the correct statement(s).
(1) Ratio of number of moles of FeO to Fe2O3 is 3 : 2.
(2) Ratio of number of moles of FeO to Fe2O3 is 4 : 3.
(3) Higher mass of Fe2O3 is formed.
(4) The sum of the masses of Fe and O2 equals the sum of the masses of FeO and Fe2O3
39. What volume of 0.2 M Ba(MnO4)2 solution is required for the complete oxidation of 25 g of 89.6% pure FeC2O4 in acidic medium, according to the reaction MnO4– + FeCr2O4 → Fe+3+Cr2O7-2+ Mn+2?
(1) 700 mL (2) 175 mL
(3) 350 mL (4) 200 mL
Numerical Value Questions
40. O n heating two moles each of Li 2 CO 3 , K2CO3 and NaHCO3, how many grams of CO2 are evolved?
41. If the relative atomic mass of oxygen is 64 units, the molecular mass of CO becomes .
42. The empirical formula of an organic compound is CH2O. Its vapour density is 45. The molecular formula of the compound is x. How many atoms are present in x?
43. A gas mixture contains C2H4 and CO2 and 40 L of this mixture require 30 L of O2 for combustion. The percentage of CO2 in the original mixture is
44. A certain metal sulphide MS 2 is used extensively as a high temperature lubricant. If MS2 is 40.00% by mass of sulphur, what is the atomic mass of M?
45. The molality of a 10% ( v / v ) solution of di-bromine solution in CCl 4 (carbon tetrachloride) is ‘x’.
x = × 10–2 M (Nearest integer)
[Given: Molar mass of Br 2 = 160 g mol−1
Atomic mass of Cl = 35.5 g mol –1
density of dibromine = 3.2 g cm –3
density of CCl4 = 1.6 g cm−3]
46. The mole fraction of glucose (C 6H12O6) in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is .
47. When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid
solution is × 10–2 M. (Nearest integer) (Molar mass of nitric acid is 63 g mol –1)
48. Density of methanol is 0.0008 kg L–1. What volume of methanol in mL is required to prepare 2.5 L of a 0.25 M solution of methanol?
49. The mole fraction of a solute in a 100 molal aqueous solution is × 10 –2 (Round off to the nearest integer).
50. An element X has the following isotopic composition. 200
The weighted average atomic mass of the naturally occurring element ‘X’ is closest to
51. A gaseous mixture contains 40% O 2 , 40% N 2, 10% CO 2,10% CH 4 by volume. Calculate the vapour density of the gaseous mixture.
52. A mixture of O2 and gas ‘Y’ (mol. wt. 80) in the mole ratio a : b has a mean molecular weight of 40. What would be the mean molecular weight, if the gases are mixed in the ratio b : a under identical conditions? (Gases are non-reacting)
53. Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL −1 at 257 °C and 100 mm Hg. The molar mass of geraniol is g/mol (Nearest integer) [Given: R = 0.082 Latm K–1mol–1]
54. What is the percentage of free SO 3 in an oleum that is labelled as 104.5% H 2SO4?
55. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g/mL.
CHAPTER 1: Some Basic Concepts of Chemistry
56. The molarity of HNO3 in a sample, which has density 1.4 g/mL and mass percentage of 63%, is . (Molecular weight of HNO3 = 63)
57. The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4.2H2O) in 250 mL of water in mol L −1 is x × 10 −2. The value of ‘ x ’ is (Nearest integer) [Atomic mass: H:1.0, C:12.0, O: 16.0].
58. At NTP, how much volume in mL of NH 3 must be passed into 30 mL of 1 N H 2 SO 4 so that its normality becomes 0.2 N?
59. The number of nitrogen atoms in 681 g of C7H5N3O6 is x × 1021. The value of x is . NA = 6.02 × 1023 mol–1) (Nearest integer)
60. What is the amount of Mg in grams to be dissolved in dilute H 2 SO 4 to liberate H2,which is just sufficient to reduce 160 g of ferric oxide?
61. 10 g impure NaOH is completely neutralised by 1000 mL of 1 N 10 HCl. Calculate the percentage purity of the impure NaOH.
Integer Value Question
62. Three isotopes of an element have mass number m , m +1, and m +2. If the ratio of abundance is 4 : 1 : 1 and mean mass is 1 m x + , then x is
63. Two elements A and B form 0.15 mol of A2B and AB3 type compounds. If both A2B and AB 3 weigh equally, then the atomic weight of A is times of the atomic weight of B.
64. Calculate the total moles of atoms of each element present in 122.5 g of KClO 3
65. An organic compound containing carbon, hydrogen, and nitrogen have the percentage 40, 13.33, and 46.67, respectively. If its empirical formula is Cx H y N z, then x + y + z is
66. An element X forms two oxides. Formula of the first oxide is XO 2. The first oxide contains 50% oxygen. If the second oxide contains 60% oxygen, the atomicity of the second oxide is
67. The ratio of the mass percentages of ‘C and H’ and ‘C and O’ of a saturated acyclic organic compound ‘X’ are 4 : 1 and 3 : 4, respectively. Then, the moles of oxygen gas required for complete combustion of two moles of organic compound ‘X’ is
68. 0.01 mol of a gaseous compound C2H2Ox was treated with 224 mL of O2 at NTP. After complete combustion, the total volume of gases is 560 mL at NTP. On treatment with KOH solution the volume decreases to 112 mL. The value of x is
69. A sample of CaCO3 has 40% has Ca, 12% C and 40% O 2 . The mass of Ca in 5 g of CaCO3 from another source will be g.
70. 1 M HCl and 2 M HCl are mixed in volume ratio of 4 : 1. What is the final molarity of HCl solution?
71. How many blood cells of 5 mL, each having [K +] = 0.1M, should burst into 25 mL of blood plasma having [K +]= 0.02M, so as to give final [K+]= 0.06 M?
72. One litre each of 1M Al 2 (SO 4 ) 3 and 1M BaCl2 are mixed. What is the molarity of free sulphate ions in the resultant solution?
73. The mole fraction of a solute in a solution is 0.1 at 298 K. Molarity of this solution is the same as its molality. Density of this
solution is 2 g/cc. The ratio of molecular weight of solute and solvent w w Msolute Msolvent is .
74. Number of moles of valence electrons present in 6.022 × 1023 NH4+ ion is
75. Total 10 mol of KClO3 is decomposed by the following two parallel reactions. Calculate the moles of KClO4 produced if 3 mol of O2 is produced.
32 2KClO2KCl3O (1) ∆+→→
34 4KClO3KClOKCl (2) ∆+→→
76. Due to partial corrosion of a piece of copper into cuprous sulphide, Cu 2 S, it gains weight. If the percentage of total copper that has undergone corrosion is 31.75 %, then the percentage gain in weight of piece of copper is ______.
10 mL of the mixture containing CO and N2 required 7 mL oxygen to form CO2 and NO, on combustion. The volume of N 2 in the mixture will be _____mL.
Passage-based Questions
Q(78-79)
Isotopes are the atoms of the same element; they have same atomic number but different mass numbers. Isotopes have different numbers of neutrons in their nucleus. If an element exists in two isotopes having atomic masses a and b in the ratio m : n, then the average atomic mass will be manb mn + +
Different isotopes of same element have same positi on in the periodic table. The elements that have single isotope are called monoisotopic elements. Greater the percentage composition of an isotope, more will be its abundance in nature.
78. The isotopes of chlorine with mass number 35 and 37 exist in the ratio (if its average atomic mass is 35.5)
(1) 1 : 1 (2) 2 : 1
(3) 3 : 1 (4) 3 : 2
79. Atomic mass of boron is 10.81. It has two isotopes, namely 5 B 11 and 5 B x with their relative abundance of 80% and 20% respectively. The value of x is
(1) 10.05 (2) 10
(3) 10.01 (4) 10.02
Q(80-81)
Vapour density of a compound is defined on the ratio of mass of a certain volume of gas to the mass of the same volume of hydrogen gas under identical conditions of temperature and pressure.
Vapour density
Mass of certain volume of gas (22.4L) at STP
2
Mass of same volume of H gas (22.4L) at STP = mw
2 =
i.e., molecular mass of gas = vapour density × 2
Vapour density is a unitless quantity. It is unaffected by variation of temperature and pressure.
Vapour density may also be defined in terms of density.
Density of gas
Vapour density = Density of hydrogen
Mass of molecules of gas = Mass of molecules of hydrogen gas n n
80. At STP, 5.6 L of a gas weights 60 g. The vapour density of the gas is ______.
81. The average density of the universe on a whole is estimated at 3 × 10 –29 g/mL. If we assume that the entire mass is only H atoms, then what is the average volume (in litres) of space that contains one H-atom?
(N = 6 × 1023)
Q(82-83)
A hydrocarbon gas (made up of only C-12 and H-1 isotopes) has 90% C.
CHAPTER 1: Some Basic Concepts of Chemistry
82. The number of C-atoms present in the empirical formula of the gas is ____.
83. If empirical formula and molecular formula of the gas are same, then the sum of protons and neutrons per molecule of the gas is x then 10 x is ______.
Q(84-85)
A gaseous hydrocarbon consumed 5 times its volume of oxygen for combustion. The volume of CO 2 produced in the reaction is thrice the volume of hydrocarbon under the same conditions.
84. How much of water is produced by the combustion of 0.1 mol of the given hydrocarbon?
(1) 7.2 g (2) 3.6 g
(3) 14.4 g (4) 1.8 g
85. What is the ratio of molecular weight to empirical formula weight of the hydrocarbon?
(1) 1 (2) 2
(3) 3 (4) 4
Q(86-87)
Empiri cal formula is the simplest formula of the compound, which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula = (Empirical formula) × n.
86. Two metallic oxides contain 27.6% and 30% oxygen, respectively. If the formula of the first oxide is M3O4, then the formula of the second one will be
(1) MO (2) MO2
(3) M2O5 (4) M2O3
87. Which of the following formulae is the one odd out?
(1) C6H12O6 (2) C12H22O11
(3) HCOH (4) C5H10O5
Q(88-89)
Empirical formula is the simplest formula of the compound that gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula = (Empirical formula) × n. n = Molecular mass/Empirical formula mass. A compound may have the same empirical and molecular formula. Both these formulae are calculated using percentage composition of constituent elements.
88. Which of the following represents the formula of a substance that contains 50% oxygen?
(1) N2O (2) CO2
(3) NO2 (4) CH3OH
89. An oxide of iodine (at. wt. of iodine = 127) contains 25.4 g of iodine and 8 g of oxygen. Its formula could be
(1) I2O3 (2) I2O (3) I2O5 (4) I2O7
Q(90-91)
2 litre of 9.8% (w/w) H2SO4 (d = 1.5 g/mL) solution is mixed with 3 litres of 1 M KOH solution.
90. The number of moles of H 2SO4 added is (1) 1 (2) 2 (3) 3 (4) 0.5
91. The concentration of H+ if solution is acidic or concentration of OH– if solution is basic in the final solution is (1) 0 (2) 3/10 (3) 3/5 (4) 2/5
Q(92-93)
Avogadro’s law states that, under conditions of constant temperature and pressure, equal volume of gases contain equal number of
particles. Experimental investigation shows that, at one atmosphere pressure and a temperature of 273 K, one mole of any gas occupies a volume which is very close to 22.4 L. Therefore, the number of moles in any gas sample can be found by comparing its volume at STP with 22.4 L.
92. At STP, 40 L of CO2 contains ___ mole.
93. Number of gram atoms of oxygen present in 0.3 gram moles of H2C2O4.2H2O is ___.
Q(94-96)
When we assume that one mole of a substance contains the same number of elementary entities as one mole of any other substance, we don’t actually need to know what that number is. Sometimes, however, we will need to work with the actual number of elementary entities in a mole of a substance. This number is called Avogadro’s number. N A = 6.022137 × 10 23 molecules. The unit mol–1, which we read as per mole, signifies that a collection of NA molecular level entities is equivalent to one mole at the macroscopic level. For example, a mole of carbon contains 6.02 × 1023 atoms of C. A mole of oxygen gas contains 6.02 × 1023 molecules of O2.
94. The number of atoms present in 8 g of ozone is
(1) NA (2) 3NA
(3) NA/6 (4) NA/2
95. Which of the following is a reasonable value for the number of atoms in 1.00 g of helium?
(1) 0.25 (2) 4.0 (3) 4.1 × 10–23 (4) 1.5 × 1023
96. The dot at the end of a sentence has a mass of about one microgram. Assuming that the black stuff is carbon, the number of carbon atoms present in it is
(1) 5 × 1016 (2) 5 × 1017
(3) 8 × 1023 (4) 12 × 1023
Q(97-98)
Chlorine is a very strong oxidising agent and can oxidise several metals, non-metals, and compounds.
(at. wt. Fe = 56, Cl = 35.5, S = 32, H = 1)
97. A compound x is formed by adding 3.2 g of sulphur to sodium suphite in alkaline medium. What is the weight of chlorine required to oxidise the compound x?
98. What is the weight of the product formed when 3.55 g of chlorine reacts with iron?
Q(99-101)
In a reaction vessel, 100 g H2 and 100 g Cl2 are mixed and suitable conditions are provided for the reaction:
H2(g) +Cl2(g)→ 2HCl(g)
99. The amount of HCl formed in this reaction (at 100% yield) will be
(1) 102.8 g (2) 73 g
(3) 36.5 g (4) 142 g
100. The amount of excess reactant remaining is
(1) 50 g (2) 97.2 g (3) 46 g (4) 64 g
101. The amount of HCl formed (at 90% yield) will be
(1) 36.8 g (2) 62.5 g (3) 80 g (4) 92.53 g
Q(102-103)
102. How many litres of NH3 at STP is required to produce 2.25 g of HNO3? (Round-off to nearest integer)
103. What volume of oxygen is required to oxidise 20 litres of ammonia (at STP) oxidised to HNO 3 , if yield is 100 %? (Round-off to nearest integer)
Q(104-106)
10 mol of SO2 and 4 mol of O2 are mixed in a closed vessel of volume 2 litres. The mixture is heated in the presence of Pt catalyst. Following reaction takes place: 2SO2(g) + O2(g) → 2SO3(g)
Assume that the reaction proceeds to completion.
CHAPTER 1: Some Basic Concepts of Chemistry
104. Select the correct statement.
(1) SO2 is the limiting reagent.
(2) O2 is the limiting reagent.
(3) Both SO2 and O2 are limiting. (4) This cannot be predicted.
105. Number of moles of SO 3 formed in the reaction will be (1) 10 (2) 4 (3) 8 (4) 14
106. Number of moles of excess reactant remaining is (1) 4 (2) 2 (3) 3 (4) 8
Matrix Matching Questions
107. Match Column I (molecule) with columnII (formula).
Column I Column II
A. NH3 I) EF = MF
B. N2H4 II) MF = (EF)2
C. N3H III) Maximum percentage of nitrogen by mass
D. C2N2 IV) Least percentage of nitrogen by mass
(A) (B) (C) (D)
(1) I II III IV
(2) II I III IV
(3) III I IV II
(4) IV III II I
108. Match the Column I (molecules) with column- II (composition).
Column I Column II
A. N2 I) 40% carbon by mass
B. CO II) empirical formula CH2O
C. C6H12O6 III) vapour density : 14
D. CH3COOH IV) 14 NA electrons in a mole
(A) (B) (C) (D)
(1) III, IV III, IV I, II I, II
(2) IV, III II, I I, III IV
(3) I II III IV
(4) II, I III IV,I IV
109. Match the Column I (concentration term) with Column II (Units)
Column I Column II
A. Molarity I) mol
B. Mole fraction II) Unitless
C. Mole III) mol L–1
D. Molality IV) mol Kg–1 V) Temperature dependent
(A) (B) (C) (D)
(1) III II IV I
(2) III I II IV
(3) III, V II I IV
(4) II III I IV
110. Match Column I (solution) with Column II (amount of solute present in the solution).
Column I
Column II
A. 0.5 L of 1M H2SO4 I) 1 equivalent of solute
B. 6% w/v aqueous solution of urea II) 1 mol of solute
C. 100 mL of 0.2N aqueous Na2CO3 solution III) 1.06 g of solute
D. 518 g of 0.2 m aqueous solution of glucose
IV) 0.1 mol of solute
V) 180 g of solute
(A) (B) (C) (D)
(1) I II III IV
(2) II I III IV
(3) I II III V
(4) I III II IV
111. Match Column I (solution) with Column II (concentration).
Column I Column II
A. 100 mL of solution containing 0.98 g of H2SO4 I) Decinormal
B. 0.5 L solution containing 50 milli equivalents of H3PO4 II) Semimolal
C. 109 g of solution containing 9 g of glucose III) Decimolar IV) Semimolar
(A) (B) (C)
(1) III I IV
(2) I II III
(3) II I III
(4) III I II
112. When 200 mL 0.2 M NaCl and 200 mL 0.4 M BaCl 2 and 100 mL 0.2 M KCl are mixed, match Column I (the ion) with Column II (molarity)
Column I Column II
(A) molarity of Na+ ion I) 0.04 M
(B) molarity of Ba+2 ion II) 0.08 M
(C) molarity of Cl– ion III) 0.16 M
(D) molarity of K+ ion IV) 0.44 M
(A) (B) (C) (D)
(1) II III IV I
(2) I II III IV
(3) II I III IV
(4) V II III I
113. Match Column I (equation)with Column II (equivalent weight).
Column I Column II
A. BrO3– → Br- I) 3 M
B. P → PH 3 II) 6 M
C. S–2 → SO4-2 III) 8 M
D. Cr2O7–2 → Cr+3 IV) 5 M
(A) (B) (C) (D)
(1) II I III II
(2) IV I III II
(3) II I III IV
(4) IV I III IV
114. Match Column I (mole) with Column II (quantity)
Column I Column II
A. 0.5 mol of SO2(g) I) Occupy 11.2 L at 1 atm and 273 K
B. 1 g of H2(g) II) Weighs 24 g
C. 0.5 mol of O3(g) III) Total number of atoms =1.5×NA
D. 1 g molecule of O2(g) IV) Weigh 32 g
(A) (B) (C) (D)
(1) I, II, III, IV I, II I, II, III, IV IV
(2) III, IV I I, III IV
(3) I, III I, II I, II, III I, IV
(4) I, III, IV I I, II, III IV
115. Match Column I (compound) with Column II (equivalent quantity)
Column I Column II
A. One gram molecule of oxygen gas I) One mole of H2
CHAPTER 1: Some Basic Concepts of Chemistry
B. Gram molar volume of H2 II) One mole of O2
C. 44 g of CO2 III) 22.4 L at STP
D. 18 g water IV) 3 NA atoms
(A) (B) (C) (D)
(1) II, III I, III II, III, IV I, IV
(2) II I III IV
(3) I, II II, III III, IV IV, II
(4) I, II, IV III, IV, II II, I, IV II, IV
116. Match Column I (compound) with Column II (percent carbon).
Column I Column II
A. CH4 I) 90% of carbon
B. C2H6 II) 75% of carbon
C. C2H4 III) 80% of carbon
D. C3H4 IV) 85.7% of carbon V) 60% of carbon
(A) (B) (C) (D)
(1) V I II IV
(2) II III IV I
(3) III II IV I
(4) II I V III
117. Match Column I (reaction) with Column II (limiting reagent)
C. () 22 1 g 22.4L at NTP C+OCO → (III) 6.25 × 10 –2 mol
D. 222 1 g1 g 2H+O2HO → (IV) 6.67 × 10 –2 mol
(A) (B) (C) (D)
(1) II III IV I
(2) I II III IV
(3) III I IV II
(4) IV I II III
118. Match Column I (reaction) with Column II (limiting reagent).
Column I
A.
Column II
223 0.2 mole0.7 mole N+3H2NH → (I) H2
B. 2 22 24 g 1 g H+2CCH → (II) C
C. 2 0.5 g 2.24 lit C+OCO → (III) O2
D. 222 2 g2 g 2H+O2HO → (IV) N2
(A) (B) (C) (D)
(1) IV I II III
(2) II III IV III
(3) III I II IV
(4) I II IV III
119. Match Column I with Column II.
Column I
Column II
A. 222 3 g22.68 g 2H + O2HO → (I) 25.5 g product is formed
B. 223 24.5g5.5g N + 3H2NH → (II) 0.25 g of a reactant is left
C. 22 1.4g43g H + Cl2HCl → (III) H2 is the limiting reagent
D. 24 20g 6.375g C + 2HCH → (IV) 41.12 g product
(A) (B) (C) (D)
(1) I II IV III
(2) II I IV III
(3) I IV II III
(4) I III IV II
FLASHBACK ( Previous JEE Questions )
JEE Main
1. Mass of methane required to produce 22 g of CO2 after complete combustion is ______g. (C = 12.0, H = 1.0, O = 16.0) (2024)
2. 1 mole of PbS is oxidised by “X” moles of O3 to get “Y” moles of O2. Then, the value X + Y = __________. (2024)
3. Number of moles of methane required to produce 22 g of CO2(g), after combustion is x × 10–2 moles. The value of x is (2024)
4. A sample of CaCO3 and MgCO3. Weighing 2.21 g is ignited to a constant weight of 1.152 g. The composition of the mixture is: (Given molar mass in g mol–1 CaCO3.100, MgCO3:84) (2024)
(1) 1.187 g CaCO3 + 1.023 g MgCO3
(2) 1.023 g CaCO3 + 1.023 g MgCO3
(3) 1.187 g CaCO3 + 1.187 g MgCO3
(4) 1.023 g CaCO3 + 1.187 g MgCO3
5. A protein ‘A’ contains 0.30% of glycine (molecular weight 75). The minimum molar mass of the protein ‘A’ is ______________ ×103 g mol–1[nearest integer]. (2022)
6. CNG is an important transportation fuel. When 100 g CNG is mixed with 208 g oxygen in vehicles, it leads to the formation of CO2 and H2O and produces large quantity of heat during this combustion. Then, the amount of carbon dioxide produced (in grams) is _______ (nearest integer). Assume CNG to be methane.
(2022 )
7. The moles of methane required to produce 81 g of water after complete combustion is ____ × 10–2 mol. (Nearest integer)
(2022)
8. 4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is x × 10–1.The value of x is ____. (Rounded off to the nearest integer)
(2021)
9. Complete combustion of 750 g of an organic compound provides 420 g of CO2 and 210 g of H2O. Then, percentage composition of carbon and hydrogen in organic compound is 15.3 and ______ respectively. (Round off to nearest integer)
(2021)
10. A 6.50 molal solution of KOH (aq.) has a density of 1.89g cm–3. The molarity of the solution is_____ mol dm –3. (Round off to the nearest Integer). [Atomic masses: K: 39.0 u; O : 16.0 u; H : 1.0 u]
(2021)
11. 10.30 mg of O 2 is dissolved in a litre of sea water of density 1.03 g/mL. The concentration of O2 in ppm is ___________. (2020)
CHA PTER TEST – JEE MAIN
Section A
1. Which of the following reactions does not obey the law of conservation of mass?
(A) C2H4(g)+O2(g) → CO2(g) + H2O(g)
(B) KHCO3 + 2HCl → KCl + H2O + CO2(g)
(C) BaCl 2(aq) +H 2 SO 4(aq) → B a SO4(ppt)+2HCl(aq)
(D) PbO(s)+C(s) → Pb(s) + CO 2(g)
(1) A, B, and C (2) B, C, and D
(3) A, C, and D (4) A, B, and D
2. The number of millimoles of solute present in 0.5 L of 0.02 M aqueous solution is (1) 0.01 (2) 10
(3) 100 (4) 1
3. The density of an aqueous 49% H 2 SO 4 solution is 1.02 g/cm3. Identify the correct relation between molarity (M) and molality (m) for the same solution at 25 °C.
(1) M = m (2) M > m
(3) M < m (4) M = 2 m
4. While determining the empirical formula for a given substance, the percentage by mass of each element is divided by its atomic mass. This ratio gives (1) relative number of moles
(2) simplest molar ratio
(3) simplest whole number ratio
(4) simplest whole number molar ratio
5. 1 g of CaCO3 is dissolved in 99 g of H2O. The mass percentage of CaCO3 in solution is
(1) 10 (2) 0.1
(3) 100 (4) 1
6. The number of hydrogen atoms present in 22 u of C3 H 8 is (1) 8 (2) 4
(3) 2 (4) 1
7. The weight of a mixture of 2 ×1023 molecules of A and 3 ×1023 molecules of B is 130 g. If the molecular mass of B is 180, then the molecular mass of A is (Avogadro number = 6×1023)
(1) 120 u (2) 60 u
(3) 90 u (4) 180 u
8. Match Column I (solution) with Column II (concentration).
Column I
A. 200 mL of aqueous solution containing 0.02 mol of solute.
B. 222 g of aqueous solution containing 60 g of urea.
C. 20 mL of aqueous solution containing 0.5 g of CaCO3
D. 1.18 kg aqueous solution containing 180 g of glucose
Column II
I) Mole fraction of solute = 0.1
II) Seminormal solution
III) Mole fraction of solute = 0.9
IV) Decimolar V) 1 molal solution
(A) (B) (C) (D)
(1) IV I II V
(2) III I II IV
(3) IV II I III
(4) IV III II I
9. The cost of 9 g of glucose is Rs 0.9. What is the cost of 9 mol of glucose?
(1) Rs 1620 (2) Rs 162
(3) Rs 16.2 (4) Rs 1.62
10. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
(A) : Both 100 g of calcium carbonate and 44 g of carbon dioxide contain the same number of carbon atoms.
(R) : Both contain 1 g-atom of carbon, which contains 6.023 ×10 23 carbon atoms.
In light of the given statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A),
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) (A) is true but (R) is false, (4) (A) is false but (R) is true.
11. Given below are two statements
S–I : A pair of compounds, C2H5OH and CH3OCH3, does not obey the law of constant proportions.
S–II : The combining ratio of C : H : O in these compounds by mass is 12 : 3 : 8.
In light of the given statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
12. A compound is made up of two elements, ‘X’ and ‘Y’. The percent by mass composition of these two elements in the compound is 60% ‘X’ and 36% ‘Y’. The relative number of moles of these elements in the compound is 1.25 and 1.8 respectively. Atomic masses of ‘X’ and ‘Y’, respectively, are
(1) 48 and 20 (2) 20 and 48
(3) 2 and 48 (4) 48 and 2
13. A mixture contains 1 mol each of calcium nitride and sodium phosphide. The mixture, on complete hydrolysis, liberates gases with the molar ratio of
(1) 1 : 1 (2) 2 : 1
(3) 3 : 2 (4) 1 : 3
14. 100 cm3 of acetylene and 100 cm3 of oxygen are mixed and subjected to combustion in a closed vessel under identical conditions. Identify the correct statement from the following.
(1) 40 mL of C2H2 is a limiting reagent.
(2) 120 mL of CO2 is formed.
(3) 60 mL of C2H2 is burnt in oxygen.
(4) 80 mL of CO2 is formed.
15. Given below are two statements.
S–I : Mass of 6.02 ×1022 molecules of CO2 is heavier than the mass of 6.02 ×1022 molecules of H2O.
S–II : 1 mole of any covalent compound contains 6.02 ×1023 molecules.
In light of the given statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
16. 8 mol of N2 and 5 moles of O2 are allowed to react in a vessel according to the reaction 2N2 + 5O2 → 2N2O5. The reaction proceeds until 6 moles of N2 are left unreacted. Then, the number of moles of product formed is (1) 1 (2) 2 (3) 6 (4) 5
17. Consider the following statements about a compound:
(A) A molecule of a compound has atoms of different elements.
(B) A compound can’t be separated into
CHAPTER 1: Some Basic Concepts of Chemistry
its constituent elements by physical methods of separation.
(C) A compound retains the physical properties of its constituent elements.
(D) The ratio of atoms of different elements in a compound is fixed.
The correct statements are
(1) A and B only
(2) C only
(3) A, B, and C only
(4) A, B, and D only
18. Given below are two statements.
S–I : 2 mol of water and 3 mol of CO 2 are produced when 30 g of C 2 H 6 completely burns in oxygen.
S–II : 30 g of ethane requires 112 g of pure oxygen for complete combustion.
In light of the given statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
19. Which of the following is the semi-normal solution?
(1) 0.5 M H2SO4 (2) 2.5 M H2SO4
(3) 0.25 M H2SO4 (4) 0.05 M HCl
20. The mass of one molecule of a compound (containing C, H, and O) is 15 times heavier than mass of the C-12 isotope atom. Empirical formula of the compound is CH2O. Number of carbon atoms present in one molecule of the compound is
(1) 2 (2) 6
(3) 4 (4) 8
Section B
21. Caffeine contains 28.9% of nitrogen. The molecular weight of caffeine is 194. The number of nitrogen atoms present in one molecule of caffeine is _____.
22. 200 mL of pure oxygen is subjected to electric discharge. 15% of oxygen is converted into ozone. The volume of ozonised oxygen is 10(10+x). Find the value of x.
23. A mixture of Na2CO3 and NaHCO3, having a total weight of 100 g on heating, produced 11.2 L of CO2 under STP conditions. The percentage of Na2CO3 in the mixture is (x). Find the value of x?
24. Aspirin has the formula C 9 H 8 O 4 . How many atoms of oxygen are there in a tablet weighing 360 mg is x × 10 21. Then value is_____ (NA = 6.02 × 1023).
25. The measured density at NTP of He is 0.1784 g/L what is the weight (in g) of one mole of He?
C HAPTER TEST – JEE ADVANCED
2020 P1 Model
Section-A
[Single Option Correct MCQs]
1. In compound A, 1.00 g nitrogen combines with 0.57 g oxygen. In compound B, 2.00 g nitrogen combines with 2.24 g oxygen. In compound C, 3.00 g nitrogen combines with 5.11 g oxygen. These results obey which of the following laws?
(1) Law of constant proportions
(2) Law of multiple proportions
(3) Law of reciprocal proportions
(4) Dalton’s law of partial pressure
2. One mole of an element contains 4.8×10 24 electrons. If the elements exists as diatomic, then choose the correct option about the element.
(1) Atomic number = 16
(2) Gram atomic weight = 8 g
(3) Gram molecular weight = 16 g
(4) Gram molecular weight = 32 g
3. 8 mol of SO2 and 14 mol of O2 were passed over the catalyst to produce 6 mol of SO 3. The ratio of SO 2 and O 2 moles in the mixture is
(1) 11 : 2 (2) 2 : 11
(3) 2 : 6 (4) 11 : 6
4. On heating 2.44 g of hydrated barium chloride to dryness, 2.08 g of anhydrous salt remained. Formula of hydrated salt is (1) BaCl2.3H2O (2) BaCl2.2H2O (3) BaCl2. H2O (4) BaCl2.6H2O
Section-B
[Multiple Correct Option MCQs]
5. To 50 L of 0.2 N NaOH, 5 L of 1 N HCl and 15 L of 0.1 N FeCl3 solutions are added and ignited. Identify the correct statement from the following.
(1) Mass of Fe2O3 precipitated = 40 g
(2) Normality of NaOH left unreacted = 0.05
(3) Equivalents of NaOH reacting with HCl = 5
(4) Equivalents of NaOH used for FeCl3 = 1.5
6. 124 g of P4 contains (1) 4NA atoms of phosphorus
(2) 4NA molecules of phosphorus
(3) NA molecules of phosphorus
(4) 4 atoms of phosphorus
7. For the reaction aA + bB → cC+ dD, initially ‘X’ moles of reactant ‘A’ and ‘Y’ moles of reactant ‘B’ are taken. Identify the correct statement(s) from the following.
■ Three basic sub-atomic particles are electrons, protons, and neutrons.
■ Other sub-atomic particles: positron, neutrino, antiproton, pions, and mesons.
Discovery of Electron
■ Thomson discovered cathode rays in Crookes’ discharge tube.
■ They are invisible, but cause fluorescence in materials like ZnS (e.g., TV screens).
■ Travel in a straight line unless deflected by electric or magnetic fields.
■ Deflected towards the south pole in a magnetic field.
■ Independent of the cathode material and gas in the tube. Named ‘electron’ by Stoney.
■ Produce X-rays with a heavy metal target.
Charge to Mass Ratio of Electron
■ J.J. Thomson determined the charge-to-mass ratio (e/m) of the electron using a cathode ray tube.
■ Particles with higher charge and lower mass led to greater deflection.
■ Thomson determined the value of charge to mass ratio as =× 111 e e 1.75882010 C kg m
Charge on the Electron
■ Determined by Millikan oil drop method experiment.
■ q = ne, where n is the number of electrons ( n is 1, 2, 3, ....) and e is charge on electron, i.e., 1.6022 × 10–19 coulombs.
CHAPTER 2: Structure of Atom
■ From the values of e/m e and e of electron, it is possible to calculate the mass of electron,
■ mass is termed as the rest mass of electron.
■ It has been found that the mass of an electron is approximately (1/1837)th the mass of a hydrogen atom.
■ The mass of moving electron = ()2
■ If velocity of electron increases, its mass increases and specific charge (e/me) of electron decreases.
■ When v becomes equal to c, mass of the moving electron becomes infinity.
Solved Examples
1. When an oil drop has the charge equal to 1.93 × 10 5 coloumbs, due to only electrons, how many electrons are expected to be in the drop to make the charge?
Sol. 1.6022 × 10–19 C = charge of one electron
2. Mass of electron is not always constant. Is it true? Explain.
Sol. Yes, the mass of electron depends upon its velocity. When the velocity of electron is increased, its mass increases.
Discovery of Proton and Neutron
■ Proton is named by Rutherford.
■ Goldstein discovered positive (canal) rays.
■ Charge-to-mass ratio (e/m) of positive rays depends on the gas in the tube, unlike cathode rays.
■ Thomson determined the e/m ratio, which is much smaller than that of cathode rays.
■ The maximum e/m value (9.58 × 10⁷ C kg –1) occurs when hydrogen gas is used.
■ Proton charge = +1.6022 × 10 –19 C.
■ Rutherford produced proton by bombarding N-14 w ith alpha particles
■ Proton is 1836 times the mass of an electron), nearly equal to a hydrogen atom.
■ Neutron is a neutral subatomic particle.
■ James Chadwick discovered the neutron
■ Neutrons are produced when beryllium-9 is bombarded with alpha particles.
■ A neutron is 1838 times heavier than an electron, and slightly heavier than proton.
■ 1 1 H or protium is the only atom which will not have a neutron in its nucleus.
■ The properties of fundamental particles are listed in Table 2.12.
Solved Examples
3. Sodium atom has 11 protons, 11 electrons, and 12 neutrons. If 12 more electrons are added to the atom, by how many times does its mass increase, compared to its initial mass?
Sol. There is no change in its mass, because the mass of electron is negligible. Thus, number of electrons are added to the atom does not increase its mass.
4. Arrange electron, proton, and neutron particles in increasing order of their specific charge.
Sol. Charge of neutron = Zero
Hence, its e/m = 0
The mass of electron is much less than that of proton, hence, e/m of proton is less than that of electron.
Therefore, the order is Neutron < Proton < Electron.
TEST YOURSELF
1. The increasing order of masses of electron (e–), proton (p), neutron (n), and α-particle is (1) a < n < p < e– (2) e – < p < n < a (3) e – < n < p < a (4) p < n < e– < a
2. Which of the following statements is wrong about cathode rays?
(1) They travel in a straight line towards cathode.
(2) They produce heating effect.
(3) They carry negative charge.
(4) They produce X-rays when they strike with a material having high atomic mass.
3. The increasing order of e/m values for electron, proton, neutron, and alpha particle is (1) e, p, n, α (2) n, p, e, α (3) n, p, α, e (4) n, α, p, e
4. Which one of the following particles has max imum value of specific charge?
(1) Proton (2) Electron (3) Neutron (4) Alpha particle
5. A proton is heavier than an electron by how many times? (1) 2 times (2) 1837 times (3) 1/1837 times (4) 1/3 times
Answer Key (1) 2 (2) 1 (3) 3 (4) 2 (5) 2
2.2 ATOMIC MODELS
Thomson’s Atomic Model
■ Atom is a sphere of positive charge with embedded electrons (watermelon model).
■ Explained electrical neutrality.
CHAPTER 2: Structure of Atom
Rutherford’s Experiment:
■ 99% of α-particles passed through without deflection. Few deflected at small and large angles.
■ 1 in 20,000 bounced back at nearly 180°.
■ Led to the discovery of the nucleus.
Rutherford’s experiment conclusions:
■ Atoms have mostly empty space, as most alpha particles passed through undeflected.
■ A dense, positively charged nucleus is present at the center, causing deflections.
■ The nucleus repels positively charged alpha particles, explaining their scattering in different directions
Solved Examples
5. When a–rays hit a thin gold foil, very few a–particles are deflected back. What conclusion one can make from this observation?
Sol. There is a very small massive body present at the centre of the atom, called nucleus.
6. In Rutherford’s experiment, why do most of the a–particles pass through the gold foil without making any deflection?
Sol. Atom has larger empty space, as compared to the positive portion of the atom. Hence, a –particles went straight, without making any deflection.
Rutherford’s Nuclear Model
■ An atom is spherical and mostly empty .
■ Electrons are held around the nucleus by electrostatic attraction.
■ Resembles the solar system, known as the planetary model.
■ The nucleus has a diameter of about 10−13 cm (fermi), while the atom's diameter is about 10−8 cm (angstrom).
■ The atom is 105 times larger than its nucleus
Properties of fundamental particles
Nature
3. Relative charge –1 unit +1 unit 0 4. Absolute mass 9.1095 × 10–31 kg 1.673 × 10–27 kg 1.675 × 10–27 kg 5. Relative mass 0.0005486 amu
7. Discoverer Thomson Rutherford Chadwick
■ The nucleus is smaller in size than that of an atom. The volume of an atom is 10 12 to 1015 times greater than that of the nucleus.
■ The nucleus is extremely dense, with a density of 10 14 g/cc or 1011 kg/cc.
■ Electrons occupy the extranuclear space, which determines the atom’s volume, while the nucleus accounts for its mass.
Limitations:
■ Failed to explain electron distribution and electron energy levels.
■ Maxwell’s theory states that moving electrons should emit radiation, lose energy, and collapse into the nucleus, contradicting atomic stability.
■ If electrons lost energy continuously, the atomic spectrum should be continuous.
Atomic Number (Z)
■ Determined by Moseley by X-ray spectra.
■ the a metal target is called anticathode.
Mosley’s Experiment
■ The results obtained led to the suggestion that υ must be directly proportional to the atomic number of an element (Z) Fig.2.1
To give accurate results, Moseley modified this equation as () Zbυ∝− where ‘b’ is the screening constant, for K a and K b lines, b = 1. Here, ()aZbυ=−
■ where ‘a’ is the proportionality constant.‘a’ stands for slope = tan q and ‘b’ stands for intercept, where u is the frequency of the characteristic X-rays emitted.
Fig. 2.1 Mosley’s Curve
CHAPTER 2: Structure of Atom
■ a and b are constants, having definite values for that element.
■ This equation is very useful for the calculation of Z if the frequency of K a and K b lines are known.
■ Atomic number (Z) = Number of protons in the nucleus of an atom or ion = Number of electrons in a neutral atom.
Mass Number (A)
■ Also know as nucleons.
■ Mass number (A) = Number of protons (Z) + Number of neutrons (n)
■ The number of neutrons (n) in an atom is n = A – Z
Isotopes, Isobars, and Isotones
■ Isotopes: Atoms of the element with the same Z but different A are called isotopes.
Examples: 1 1 H 2 1 H and 3 1 H
■ isotopes have same chemical properties but different physical properties,
■ Atomic weight of naturally occurring elements fractional due to the existence of isotopes.
■ Isobars: The atoms of different elements which have the same mass number but different atomic numbers are called isobars.
Examples: 14 6 C and 14 7 N ; 4040 1819Ar,K and 40 20 Ca
■ Isobars have same number of nucleons. Isobars are not alike chemically
■ Isotones: Isotones are the atoms of different elements which have the same number of neutrons.
Examples: 3031
Solved Examples
7. 40 18Ar is an isobar of A ZX. If X is calcium, find the number of neutrons in the metal.
Sol Since Ar-40 is isobar of X, the mass number of X is also 40. Z of Ca = 20. Hence, its number of neutrons = A – Z = 40 – 20 = 20.
8. The atoms x and y are isobars. They contain 37 nucleons. The x – contains 11.1% more neutrons than the electrons. What is the atomic number of x?
Sol. Since x and y are isobars, the number of nucleons in x = 37. If the number of electrons in the x – ions = e
The number of neutrons 11.1 eee0.111e1.111e 100 +=+=
37 = p + n = (e–1) + 1.111e
2.111e – 1 = 37 e = 38/2.111e = 18.
Hence, proton number or atomic number of
9. The wavelengths of the characteristic K a X-rays of iron and potassium are 1.93 × 10 –8 and 3.737 × 10–8 cm, respectively. What is the atomic number of an element for which the characteristic K α wavelength is 2.289 × 10–8 cm?
Sol. From Moseley’s law
Try yourself:
1. What is the property of gold metal that is responsible for its use in Rutherford’s experiment? Answer: The property is its high malleability. Rutherford needed a metal which could be as thin as possible. Hence, gold, being the most malleable metal, was selected for the experiment.
Test Yourself
1 Which one of the following is an isobar of 14 6 C? (1) 13 6 C (2) 12 6 C (3) 14
2. In which of the following, protium, deuterium, and tritium atoms do not differ?
3. Read the following and choose the option with the correct statement.
(A) Atom whose nucleus contains 20 protons and 15 neutrons (B) Atom whose nucleus contains 18 protons and 21 neutrons (C) Atom whose nucleus contains 18 protons and 17 neutrons (D) Atom whose nucleus contains 19 protons and 20 neutrons
(1) B and C are isobars. (2) B and D are not isobars. (3) A and C are isobars. (4) A and D are isotones.
4. Choose the correct state ment about the given pairs.
Pair (A): 3032 1416SiandS
Pair (B): 7676 3234GeandSe (1) ‘A’ represents isotopes and ‘B’ represents isobars (2) ‘A’ represents isobars and ‘B’ represents isotopes (3) ‘A’ represents isobars and ‘B’ represents isotones (4) ‘A’ represents isotones and ‘B’ represents isobars
5. How many different molecular forms of hydrogen can exist with all the isotopes of hydrogen? (1) 10 (2) 6 (3) 4 (4) 7
Answer Key
(1) 3 (2) 2 (3) 3 (4) 4 (5) 2
2.3 DEVELOPMENTS LEADING TO BOHR’S MODEL OF ATOM
■ James Maxwell proposed that an accelerating charged particle emits electromagnetic waves with alternating electric and magnetic fields.
Wave Theory
■ Maxwell proposed that oscillating charges produce perpendicular electric and magnetic fields, which propagate as waves as shown in Fig.2.2.
Fig. 2.2 Electric and magnetic components of electromagnetic radiation
■ Vertical component is electric field, horizontal component is magnetic field.
■ Wave length (λ)
■ distance between two consecutive crests or troughs,
■ The SI unit of wavelength is metres (m)
1A0 = 10–10 m or 10–8 cm
1nm = 10–9 m or 10–7 cm = 10A°
1pm = 10–12m or 10–10 cm =10-2A°
10–6m is denoted as ‘micron’.
Frequency (ν)
The number of waves passing a given point per second, measured in cps, s -1, or Hertz (Hz), which is its SI unit. 1 cps = 1Hz = 1s –1
Velocity
■ Distance traveled by a wave in one second.
■ The speed of all electromagnetic waves in a vacuum is constant, c is 3 × 10 8 ms–1.
Wavelength × Frequency = Speed
This can be written as, λ= ν c or ν= λ c
Wave Number ν
■ The number of waves that can be present in unit length is called wave number.
IL ACHIEVER SERIES FOR JEE CHEMISTRY
■ It is the reciprocal of wavelength. 1 wavelength ν=()ν=λ= λν 11 or
■ it is expressed in cm –1 or m–1. 1 cm–1 = 100 m–1
■ l , ν , n , and c are mathematically =λ×νν=ν corc
Energy (E)
■ Energy of light is directly proportional to its frequency ( n ). ∝ν E E = hv, where ‘h’ is Planck’s constant
ν==ν
■ Unit of E: kcal/mol or kJ/mol.
Amplitude (A)
■ The height of the crest or depth of the trough of a wave,
■ A helps in determining the intensity, or brightness of radiation.
Electromagnetic Spectrum
■ Arrangement of different E.M.radiations in the order of increasing wavelength or decreasing frequency.
Applications of E.M.waves
■ Radiofrequency – broadcasting.
■ Microwave – radar.
■ Infrared – heating.
■ Visible Light – visible region (380–760 nm).
■ UV Rays - high penetration power.
■ X-rays – to study the interior of objects.
■ Gamma Rays: cancer treatment
■ Radio Waves: communication
Particle Nature
■ Newton proposed that light travels as tiny particles called corpuscles. This idea was later discarded, by wave theory of light.
■ Electromagnetic wave theory explains reflection, refraction, polarization, diffraction, and interference
CHAPTER 2: Structure of Atom
■ E M theory failed to explain,Radiation emitted by black body radiation, photoelectric effect.
■ Variation of heat capacity of solids with temperature.
Fig. 2.3 Differences between line and band spectra
Solved Examples
10. The wavelength of a light wave having frequency, 5 × 10 9 s–1 is _______ m.
Sol. 81 2 91 C310ms 610m. 510s × λ===× ν×
11. Yellow light emitted from a sodium lamp has a wavelength of 580 nm. Its wave number is _____ m–1
Sol. 61 9 11 1.7210m 58010m ν===× λ×
Black Body Radiation
It is perfect absorber and perfect emitter.
Hollow sphere with platinum black coating.
A graph is obtained by plotting the intensity of radiation against wavelength, as shown in Fig.2.4.
A study of the curve reveals the following:
The nature of radiation depends upon the temperature of the black body.
The radiations are due to vibrations of charged particles (electrons) of atom.
At a given temperature, the intensity of radiation increases with the wavelength, reaches maximum, and then decreases.
Fig. 2.4 Wavelength (nm)
The intensity of radiation is greatest at the middle wavelengths and least at the highest and lowest wavelengths. The l, where intensity is maximum, is known as ‘ l max’.
As the temperature inc reases, the maximum intensity of l shifts towards the shorter wavelength.
■ Wave theory fails to explain these results as it predicts constant wavelength with only intensity changes.
■ If energy emission wer e continuous, the intensity curve would follow the dotted line.
solved example
12. A black body is taken at temperatures, T1 and T2. At these temperatures, the maximum intensities are I1 and I2, and the corresponding wavelengths are l1 and l2, respectively. Compare their magnitudes.
Sol. When T1 < T2, I1 < I2. Then l 1 > l 2
Try yourself:
2. At the temperature 5900 K, if the radiation of sun shows the maximum intensity, what will be its corresponding wavelength? [Assuming the radiation obeys the equation, T l max = 2.9×10–3 m-K]
Answer: 4.9 × –710 m
Planck’s Quantum Theory
To explain black body radiation.
Postulates:
■ Radiation from a hot body is due to vibrating charged particles, emitting energy in discrete packets called quanta.
CHAPTER 2: Structure of Atom
■ Energy is proportional to frequency: E a ν or E = hν, where h=6.6256×10 −34 Js.
■ Total energy emitted or absorbed is an integral multiple of E=nhν (n = 1,2,3…).
■ This is called energy quantisation. The emitted energy propagates as waves.
Photoelectric Effect (Discovered by Hertz)
■ The ejection of electrons from a metal surface when exposed to light of suitable frequency. The emitted electrons are called photoelectrons..
Fig. 2.5 Photoelectric effect
■ Light of a given frequency strikes metal surface in side the vacuum chamber.
Explanation of Photoelectric Effect
■ Threshold frequency is the minimum frequency required for the effect; light below this frequency cannot eject electrons.
■ The number of ejected electrons depends on light intensity but not on frequency.
■ Einstein explained this using photons—light behaves as a stream of particles, each carrying energy E=hν.
■ If a photon's energy exceeds the metal's work function, the excess energy appears as the kinetic energy of the emitted electron.
E = E 0 or w + KE h n = w + KE
■ If the threshold frequency is n 0, the threshold energy or work function, w = h n 0 hv = h n 0 + KE or h n = h n 0 + ν 2 1 m 2 or KE = hv – h n 0
■ If (v > v0), the excess energy, is imparted to the ejected electrons as kinetic energy.
■ As frequency increases, the photoelectron's kinetic energy also increases.
■ If the incident radiation frequency (above threshold) doubles, the kinetic energy increases more than twice, as the work function remains unchanged.
■ The metal with a lower work functio n is widely used in photoelectric effect. (Table 2.3)
IL ACHIEVER SERIES FOR JEE CHEMISTRY
Work function of some metals
■ Each photon ejects one electron.
■ Increasing light intensity raises the number of photons striking the surface, increasing electron ejection, but kinetic energy remains unchanged.
Stopping Potential
■ The minimum potential required to stop electron which has maximum kinetic energy is called stopping potential.
According to law of conservation of energy,
2 SS 1 me,stopping potential 2 e = charge of electron ν=νν=
Solved Examples
13. Energy of a photon is 3 × 10 –12 ergs. What is its wavelength in nm? (h = 6.626 × 10 –34 Js)
14. Calculate the kinetic energy of the electron ejected in joules, when yellow light of frequency 5.2 × 1014 s–1 falls on the surface of potassium metal. (Threshold frequency of potassium is 5 × 10 14 s–1)
3. Among the metals, which one can be best used in photoelectric cells?
4. How many photons of light of wavelength 7000 A° can make an energy of one joule? Answers: 3. Caesium (because of its low ionisation enthalpy) 4. nhcE hcE,n λ == λ = 3.52×1018
Spectra and Atomic Models
■ The image recorded when radiant energy is passed through a prism is called a spectrum, by using spectrograph or spectrometer.
■ Spectra are classified into absorption and emission spectra.
Emission Spectrum:
■ spectrum of light emitted by a substance when its atoms or molecules are excited and then return to a lower energy state, producing distinct wavelengths of light.
■ emission spectrum is of two types continuos spectrum and discontinuous spectrum
Continuous Spectrum:
■ White light, such as from the Sun or an incandescent lamp, consists of seven colors (VIBGYOR), as shown in Fig. 2.6.
Fig. 2.6 Continuous spectrum of white light
■ This type of spectrum is without any gap hence, known as a continuous spectrum.
■ A heated solid emits white light, producing a continuous spectrum when passed through a prism, whereas gases give discontinuous spectra.
Discontinuous Spectrum :
■ emission spectrum which have gaps in its pattern is called as discontinuous spectrum
■ it is mainly of two types line spectrum and band spectrum
■ Each element has a unique spectrum, like sodium’s yellow lines at 5890 Å and 5896 Å
■ No two elements share the same spectrum, making it useful for identification.
■ Helium was found in the Sun using spectroscopy.
■ Band spectrum is due to vibrations and rotations of atoms within a molecule.
■ Absorption Spectrum
■ When white light passes through unexcited substance, certain wavelengths are absorbed, appearing as dark lines on a bright background.
■ These dark lines correspond to radiation absorbed by atoms or molecules in the gas.
■ An absorption spectrum is essentially the photographic negative of the emission spectrum of a substance.
Hydrogen Emission Spectrum
■ The hydrogen emission spectrum is the simplest atomic spectrum, consisting of multiple series.
■ The Balmer series, discovered in the visible region, has four prominent lines (Hα, Hβ, Hγ, Hδ).
■ Balmer derived an empirical equation to express the wave number of these spectral lines.
Wave number =
where n = 3,4,5,6 ....
R is Rydberg constant. The value of Rydberg constant for hydrogen,
RH = 1,09,677 cm–1 or 1.09677 × 107 m–1
■ The wave numbers of different lines in the series are obtained by substituting n values.
When n = 3, the spectral line in the Balmer s eries is called H a line. For H a line,
■ The limiting wavelength (series limit) of the Balmer series is obtained by taking n as infinity ( ∞)
The limiting wavelength, l∞ = 3648A°.
■ Lyman observed some more spectral lines of the hydrogen emission in ultraviolet region. Paschen, Brackett, and Pfund observed separately, three different series of lines in the infrared region.
■ Different spectral lines of hydrogen emission are shown diagrammatically in Fig. 2.7 . and summarised in Table 2.4.
Fig. 2.7 Transition lines of hydrogen spectrum
CHAPTER 2: Structure of Atom
The wave numbers of all the lines in all the series can be calculated by the Rydberg equation.
where n1 and n2 are whole numbers, n2 > n1.
For one electron species, like He+, Li2+ and Be3+ , the value of R is 109677 cm –1 × z2, where z is the atomic number of the species.
1. The line spectrum has sharp, distinct, well defined lines.
2. The line spectrum is characteristic of atoms and is also called atomic spectrum.
3. The line spectrum is due to transition of electrons in an atom.
4. The line spectrum is given by inert gases, metal vapours, and atomised non-metals.
1. The band spectrum has many closed lines.
2. The band spectrum is characteristic of molecules and is also called molecular spectrum.
3. The band spectrum is due to vibrations and rotations in a molecule.
4. The band spectrum is given by hot metals and molecular non-metals.
The wave number for any single electron species, like He+, Li2+ and Be3+, can be calculated from the equation
15. When an electron in H atoms jumps from 3 rd orbit to ground state, Balmer spectral line of x cm–1 of wave number is emitted. The value of x is ________.
Sol. It is Balmer spectral line, hence, n 1 = 2.
16. What is the wave number of limiting line emission spectral line of hydrogen spectrum in Paschen series?
Sol. For Paschen series, n1 = 3, for limiting line in the series, n2 = ∝
Try yourself
5. The shortest wavelength transition in the Paschen series of Li +2 spectrum occurs at 91.2 nm. At what wavelength does it occur in hydrogen spectrum?
Answer: 9 × 91.2 = 820.8 nm
Test Yourself
1. Which of the following properties of a wave is independent of the other three? (1) Wave number (2) Wavelength (3) Frequency (4) Amplitude
2. The energy of a photon is 3 × 10 –12 ergs. What is its wavelength in nm? (h = 6.62 × 10-34 J-s) (1) 662 (2) 1324 (3) 66.2 (4) 6.62
3. If the threshold wavelength (λ 0) for ejection of electrons from a metal is 330 nm, then work function for the photoelectric emission is (1) 1.2 × 10–18 J (2) 6 × 10–10 J (3) 6 × 10–19 J (4) 3 × 10–19 J
4. Kinetic energy of photoelectrons is independent of ________ incident radiation. (1) wavelength (2) wave number (3) frequency (4) intensity
5. The wavelength of electromagnetic radiation is 400 nm. Thus, the wave number will be (1) 2.5 × 104 cm–1 (2) 2.5 × 103 cm–1 (3) 2.5 × 105 cm–1 (4) 2.5 × 102 cm–1
6. The frequency of an EMR is 6000 kHz. Number of waves formed in 100 km distance is (1) 2 (2) 1 (3) 20 (4) 10
7. Consider the electromagnetic radiations: (a) Visible light (b) IR light (c) UV light (d) Micro waves The correct order of their increasing energy is (1) a > b > c > d (2) a < b < c < d (3) d < b < a < c (4) b < c < d < a
8. The correct relation about the velocity of any photon is (1) dependent on its wavelength (2) dependent on its intensity (3) equal to cube of its amplitude (4) independent of its wavelength
9. The correct order of wavelength is (1) IR >Visible > UV (2) UV > IR > Visible (3) Visible > IR > UV (4) Visible > UV > IR
10. The ratio of energies of two photons of wavelength 2000 and 4000 A° is (1) 1 : 4 (2) 4 : 1 (3) 1 : 2 (4) 2 : 1
11. The energy of one mole of photons of a radiation of wavelength 300 nm is (h = 6.63 × 10–34 Js; NA = 6.02 × 1023) (1) 321 kJ/mol (2) 325 kJ/mol (3) 399 kJ/mol (4) 435 kJ/mol
■ Bohr’s model is based on the application of quantum theory of radiation.
Postulates of Bohr’s Theory
■ Electrons revolve around the nucleus in fixed circular orbits (K, L, M, N) with definite velocity.
■ Bohr stated that an electron can occupy only those orbits where its angular momentum is an integral multiple of h/2π.
ν= π h mr 2 n ,
angular momentum = I × w
Since I = m e r2 and w= v/r, where v is the linear velocity.
∴ Angular momentum = me r2 × v/r = m e vr
■ Electrons in stationary orbits have fixed energy and do not lose or gain energy while revolving.
■ Energy is absorbed when an electron moves to a higher orbit and emitted when it jumps to a lower orbit.
■ The energy emitted or absorbed in a transition is equal to the difference between the energies of the two orbits
E2 – E1 = ∆E = h v
Here, E1 and E2 are the energies of the lower and higher allowed ene rgy states.
υ== 21EEE hh is called Bohr’s frequency rule.
Radius of Hydrogen Atom
■ Radius for ‘nth’ orbit, 22 n 22 nh r 4mez = π
Substituting the values of constants in cgs we get radius of nth orbit 2 8 n 0.52910cm z =×× n r or 2 n = 0.529 A z ×° n r
■ 0.529 A0 or 0.0529 nm or 52.9 pm is known as Bohr’s radius.
■ Similarly, for hydrogen-like ions (He+, Li2+ and Be3+), the radii may be given by the expression
Radius 0.5292 A Z × = ° n ,
■ where z is the atomic number of the species. Distance between two successive orbits x and y in hydrogen-like species
0.529 A Z = ° xy +
Energy of Electron in Hydrogen Atom
■ The energy of an electron at an infinite distance, where it is not attracted by a nucleus of an atom, is considered zero.
■ When it approaches to the nucleus, its energy decreases and, hence, the energy of an elect ron always has a negative sign.
Potential energy (PE) = 2 e r
■ The total energy is negative and kinetic energy is positive and have same magnitude.
■ Total energy of electron E n, E n = KE + PE
■ Substituting the value of r, we get
■ Except n, all are constants in hydrogen atom.
Hence, 2 K E = n n
Here, K is a constant,
■ Substituting the values of m, e, h, and p in the equation,
■ Energy expression for hydrogen like ions (He +, Li2+ and Be3+) is
CHAPTER 2: Structure of Atom
■ Radius and energy of electron in nth orbit of hydrogen atom can also be calculated in SI units.
■ where e 0 is permittivity of air and it is equal to 8.854 × 10–12 Farad metre–1 K = 9 × 109 N–m2 c–2 (In CGS units K=1 dyne cm 2 esu–2)
■ The energy of electron has certain discrete restricted values . This is called quantisation of energy of electron.
Rydberg Equation
This equation is similar to Rydberg equation.
Rydberg constant R should be equal to
Substituting the values, we get RH= 1,09,681 cm–1 .
■ This value is almost equal to Rydberg’s constant 1,09,677 cm –1 .
■ The frequencies of the spectral lines in the hydrogen spectrum calcula ted by using
Velocity of Electron
■ The velocity (v) of electron in nth orbit of hydrogen atom is given by the expression,
v ∴= n h22 2 4me 2m / π × π n h2 2 2e h π = n
■ Substituting the values of p , e, and h in the above expression,
vcms × = n n
8 2.18101
■ Thus, the velocity of electron in the first orbit of hydrogen atom is 2.18 × 10 8 cm s–1.
■ The velocity of electron in one electron species He+, Li2+, and Be3+ is given as
vZcms × =× n n
8 2.18101
■ Number of revolutions or orbital frequency of an electr on in hydrogen like speci es = Velocityofanelectron Circumferenceoftheorbit
Time period of revolution (Tn) = n
Explanation of Hydrogen Spectrum
■ Bohr’s theory explains the hydrogen emission spectrum, where electrons absorb energy, get excited, and return to lower levels, emitting radiation.
■ Though hydrogen has one electron, its spectrum has multiple lines due to different transitions, with intensity depending on photon count.
■ Lyman series (UV) occurs when electrons return to n=1, Balmer (visible) to n=2, Paschen (near IR) to n=3, Brack ett (mid IR) to n=4, and Pfund (far IR) to n=5
■ The maximum number of lines produced when electrons jump from n th level to ground level is equal to, ()() 21 1 or 2 ∑− nn nn
The Possible lines produced when electrons jump from n2 to n1 orbit = () () 21211 2 −−+nnnn
No. of Lyman lines = n – 1
No. of Balmer lines = n – 2
No. of Paschen lines = n – 3
No. of Brackett lines = n – 4
No. of Pfund lines = n – 5
CHAPTER 2: Structure of Atom
■ The different spectral lines in the spectra of atoms correspond to different transitions of electrons from higher energy levels to lower energy levels. This is shown in Fig. 2.8.
Fig.2.8 Formation of different lines in the hydrogen emission spectrum
Advantages of Bohr’s Model
Bohr’s model explains the stability of the atom. The stationary orbits does not lose energy.
Bohr’s theory explains the spectrum of hydrogen and also for unielectron species He+, Li2+ and Be3+, etc.
Experimental frequencies of lines are close to those of Bohr’s theory.
Experimental Rydberg constant value of H is close to that of Bohr's theory
Limitations of Bohr’s Model
failed to explain the spectra of multielectron atoms.
failed to explain fine spectrum
failed to explain Zeeman effect and Stark effect.
failed to explain wave nature of electron.
Bohr’s theory is not in agreement with Heisenberg’s uncertainty principle. Sommerfeld’s Extension
■ Sommerfeld explained the fine structure of the hydrogen spectrum.
■ Sommerfeld proposed that the moving electron might describe elliptical orbits in addition to circular orbits.
■ mvr = h k, 2 π where k is an integer and is called azimuthal quantum number.
The ratio of n to k is equal to the ratio of the length of major axis to the length of minor axis.
When, n = 1, k = 1 and n r = 0
n = 2, k = 1, 2 and n r = 0, 1
klength of minor axis = n
length of major axis
n = 3, k = 1, 2, 3 and n r = 0, 1, 2
■ The n determines the major axis of the elliptical orbit. It is equal to the diameter of the Bohr’s circular orbit.
■ The k determines the minor axis of the elliptical orbit and its eccentricity. k cannot be zero.
■ is related to k by the expression = k – 1.
Fig. 2.9 Transitions-fine structure
solved examples
17. In an electron transition, in hydrogen atom, electron experiences change of distance from nucleus is 8r1 (r1 is radius of first orbit of H-atom). Between which two orbits may this transition occur?
Sol. From Bohr’s theory, radius of nth orbit in H-atom, r n = r1 × n2 21 2222 1nn1211121 8()()
22 21 8, −=nn
Hence, transition will be from 3 rd to 1st orbit.
18. The velocity of electron moving in 3rd orbit of He+ is v. What is the velocity of electron moving in 2nd orbit of Li+2?
Sol. Velocity of electron in nth orbit is given by 1 =× n z vv n (here, v1 = velocity of electron in 1st orbit of H-atom.
Velocity of electron in 3rd orbit of He+, () 31 2 He 3 vvv + =×= (given)
Hence, 1 3 2 vv = Velocity of electron in 2nd
Try yourself
6. Why does the energy level of an electronic shell, that is closer to the nucleus, have a lower energy compared to one that is father away?
When a free electron approaches closer to the nucleus, it has to loose some energy due to the electrostatic interactive forces. More the closeness of the electron to the nucleus more is the loss of its energy. Thus, the one osercl to the nucleus has the lower energy compared to the one that is farther away from it.
Answer: At an infinite distance from the nucleus, the energy of a free electron is zero, i.e., infinityE = 0.
Test Yourself
1. The ratio of wavelengths of the first line of Lyman series of H-spectrum to second line of Balmer series of H-spectrum is (1) 1 : 2 (2) 2 : 1 (3) 1 : 4 (4) 4 : 1
2. The ratio of radii of first orbits of H, He +, and Li+2 is (1) 2 : 3 : 5 (2) 1 : 2 : 3 (3) 6
3. An electronic transition in hydrogen atom occurs from sixth level to first excited state in all possible ways. The number of spectral lines observed in Paschen series is (1) 5 (2) 4 (3) 3 (4) 6
4. If the radius of the first orbit of H-atom is x , circumference of 4th orbit in H-atom would be (1) 32 p x (2) 64 p x (3) 16 p x (4) 128 p x
5. If the shortest wavelength of H atom in Lyman series is x, then the longest wavelength in Balmer series of He+ is (1) 9x 5 (2) 36x 5 (3) x 4 (4) 5x 9
6. If the first ionisation energy of hydrogen is E, the ionisation energy of He + would be (1) E (2) 2 E (3) 0.5 E (4) 4 E
7. The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of Be3+ is (1) 1 : 4 (2) 1 : 8 (3) 1 : 16 (4) 16 : 1
Answer Key (1) 3 (2) 3 (3) 3 (4) 1 (5) 1
2.5 TOWARDS QUANTUM MECHANICAL MODEL OF ATOM
■ There are two important developments in the formulation of the new model of atom. de Broglie’s Wave Theory
■ Wave nature explains diffraction and interference, while particle nature explains photoelectric and Compton effects, proving light’s dual nature.
■ De Broglie proposed that matter also exhibits dual behavior like radiation.
■ Moving microparticles (electrons, protons, atoms, molecules) have an associated wave nature. This is de Broglie’s equation,
■ For wavelength of a particle, h/p m == λλ ν
■ Momentum (p) is given as product of mass and velocity.
■ Here, l signifies wave nature and P signifies particle nature.
de Broglie Equation – Derivation
■ From Einstein’s generalisation of Planck’s theory, Ehhc =ν= λ
■ From Einstein’s mass–energy relationship, E = mc2 (here, m is mass equivalent to energy).
From the above,
2 hc h Emc mc ==∴λ= λ
■ The wavelength of macrobodies, quite small in magnitude and they may be ignored.
■ The electron moving with high speed possesses both the particle nature and the wave nature.
■ The waves associated with material particles are known as matter waves or particle waves.
■ The wavelength of a particle in motion is inversely proportional to its momentum.
■ The wave property of electron was proved experimentally by Davission and Germer.
■ The wave property of electron is utilised in the co nstruction of electron microscope, which is a powerful tool in modern scientific research.
de Broglie’s concept and Bohr’s theory: Two types of waves are possible for an electron moving around the nuclues in the circular path:
A standing or stationary or non-energy radiating wave
A non–stationary or energy radiating wave
■ The electron wave is in-phase when the circumference of orbit is equal to the integral (whole number) multiple of the wavelength of the electron wave.
i.e., 2 p r = n l
2r π λ=∴= n n
() integer or whole number
We know that, h2rhh mvr mvmv2 π λ=⇒=⇒= π n n
■ de-Broglie wavelength of an electron in a given orbit for hydrogen like atoms is
o 3.33Aλ=× n z
Circumference of orbit = 2 o A 3.33 × n Z
de Broglie’s wavelength = λ= h 2mE where m = mass of particle
■ For a charged particle like electron acclerated with potential, V.
■ E = kinetic energy of particle = qV (for charged particles)
∴λ= h 2mqV
where q = charge of particle and V= accelerating potential
For electron 12.27 A V λ=°
■ de Broglie’s concept supports the postulate of Bohr that the angular momentum of an electron in an orbit is quantised.
■ In case the circumference of the electron orbit (2 pr) is bigger or smaller than nl, the electron wave is said to be out of phase.
■ Then, there occurs destructive interference of electron waves, causing radiation of energy. Such an orbit cannot exist, as shown in Fig.2.10.
Fig. 2.10 de Broglie’s electron waves in an orbit
19. What is the de Broglie wavelength of a particle with mass 1 g and velocity 100 m/s ?
Sol. h
Heisenberg’s Uncertainty Principle
■ The uncertainty principle is one of the basic principles of wave mechanics.
■ It is the consequence of dual behaviour of matter and radiation.
■ It states “it is impossible to determine simultaneously and accurately the exact position and the exact momentum or velocity of microscopic particles.”
■ The high certainty in one factor introduces the high uncertainty in another factor.
■ mathematically
■ This principle says product of uncertainty in position and momentum of moving microparticles cannot be less than h/4 p
■ () () ∆∆ν≥ π h x 4m
■ Other forms of Heisenberg’s uncertainty principle are ∆×∆≥∆∆λ≥λ ππ h2 Et;x. 44
(E= energy, t = time)
■ If the position of the particle is known exactly (∆x = 0), ∆v becomes infinity (∞) and vice versa.
■ The product of uncertainties is inversely proportional to mass of the particle
() () 1 xv m ∆∆∝
If A and B are two particles, () () AB A B xv m xvm ∆∆ = ∆∆
■ Heisenberg’s uncertainty principle is not applicable to those objects which cannot change their position by themselves when light falls on them.
Significance of Uncertainty Principle
Uncertainty principle rules out the existence of definite paths or trajectories of electrons and other similar particles.
This principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.
In dealing with milligram size or heavier objects, the associated uncertainties are hardly of any real consequence.
■ In view of Heisenberg’s principle, Bohr’s theory of definite orbits for electron has no meaning at all. It is not possible to know the exact position of an electron, as implied by Bohr’s theory.
■ Case-I: Consider a particle of mass 1mg. The product of uncertainty in position and velocity can be given as
■ Therefore, if one tries to find the exact location of an electron, say only 1 × 10 –8 m, then the uncertainty in velocity is 5.8 × 10 3 ms–1 = 1.92 × 10–3 m.
CHAPTER 2: Structure of Atom
Solved Example
20. In an atom, an electron is moving with a speed of 600 ms–1 with an accuracy of 0.005%. Certainty with which the position of the electron can be located is ___ ____.
Sol. h xp 4 ∆⋅∆= π
Try yourself
7. If uncertainty in position and velocity are equal, then what is the uncertainty in momentum? Answer: 1mh 2 π
Test Yourself
1. A hydrogen molecule and helium atom are moving with the same velocity. Then, the ratio of their de Broglie wavelength is (1) 2 : 1 (2) 2 : 3
2. For an electron to have the same de Broglie wavelength as that of a deuteron, its velocity should be x times that of deuteron.
The value of x is
3. Which of the following is responsible for ruling out the existence of definite paths or trajectories of electrons?
(1) Paul’s exclusion principle (2) Heisenberg’s uncertainty principle (3) Hund’s rule of maximum multiplicity (4) Aufbau principle
4. The graph between momentum p and deBroglie wavelength l of proton is
5. Decreasing order of uncertainty in velocity, when ∆ x is same, is
Answer Key
2.6 QUANTUM MECHANICAL MODEL OF ATOM
■ Classical mechanics ignores the dual behaviour for microscopic particles.
■ The branch of science that takes into account this dual behaviour of matter is called quantum mechanics.
■ It deals with both wave–like and particle–like properties of microscopic objects.
Schrodinger’s Wave Equation
■ The wave equation describes the electron in motion as a three dimensional wave is called Schrodinger wave equation.
■ Here x, y, z are Cartesian coordinates of the electron.
m = Mass of electron; h = Planck’s constant
E = Total energy of the electron (KE + PE)
V = potential energy of the electron (PE)
y = wave function of the electron
■ Using the symbol ∇ instead of the three partial differentials, the wave equation is shortened to
■ For an atom or molecule, whose energy does not change with time, Schrodinger’s equation is
HE ,
H is Hamiltonian operator.
CHAPTER 2: Structure of Atom
■ As the space around the nucleus is spherically symmetric the behaviour of an electron should not only be viewed in radial distance terms but also in angular parameters.
■ Hence, the Cartesian coordinates are to be taken in terms of polar coordinates that describe the angular placement of electron
x = r sin 0 cos ϕ,
y = r sin 0 sin ϕ,
z = r cos 0
■ Schrodinger equation converts polar coordinates with imposing the boundary conditions to obtain wave functions of electrons as
()()()() r,,Rr.. ψθφ=ΘθΦφ
■ R(r) represents the radial part that describes the variation of wave function only on the basis of radial distance from nucleus.
■ ()() Θθ⋅Φφ (Angular wave function) represents the dependence of Ψ on the angular orientation of electron in the spherical space.
Boundary Conditions
■ Schrodinger equation has several solutions for Ψ but most of them are imaginary, hence are not valid. Values of Ψ are valid if the following conditions are obeyed.
Wave function must be finite and continuous.
It must be single valued, i.e., at a given point, there cannot be more than one value for Ψ.
Ψ must become shall at large distances of r from the nucleus in order to depict bound stationary state, where the electron is held close to the proton.
It is normalised, i.e., 2 dr1 +∞
∫ψ=
■ Values of ‘ Ψ ’ which are valid are called eigen functions and values of E corresponding to valid ‘ Ψ ’ values are called eigen values.
Specification of the Wave Functions (Orbitals)
■ The wave functions depend on all three quantum numbers.
()() n.l.m n,l r,,Rrψθϕ=
■ The radial function Rn,1(r) is a polynomial in r of degree n-1 multiplied by an exponential factor 0 r na e, , where a 0 is the Bohr’s radius. This exponential factor ensures that R(r ) will become small as r gets large, in order to conform to the boundary condition for a bound state.
■ The angular function consists of products of polynomials of sin 0 and cos 0 multiplied by a complex exponential function e imφ, where i1 =−
Significance of yy is a wave function. It gives the amplitude of the electron wave. y is expressed in terms of coordinates x, y and z
■ The wave function may have positive or negative value, depending upon the values of coordinates.
■ The value of amplitude increases and reaches the maximum, which is indicated by the peak in the curve. The value of amplitude decreases after reaching the maximum.
■ y is +ve in the region above the x-axis. y is zero along the x-axis. y is –ve in the region below the x-axis.
The variation of amplitude y or the field strength is shown in Fig. 2.11
Fig. 2.11 Wave function
■ The acceptable solutions of y are called eigen wave functions.
■ The probability of finding an electron at a point in space is given by y 2 (x,y,z), where x, y, and z are Cartesian coordinates.
■ y2 in case of electron wave denotes the probability of finding an electron in the space around the nucleus or electron density around the nucleus.
■ y 2 is directly proportional to electron density. y 2 is a probability function.
■ If y 2 is maximum, the probability of finding the electron is also maximum (95%).
Important Features of Quantum Mechanics
■ Electrons in an atom have quantised energy due to their wave-like nature.
■ Schrödinger’s wave equation defines allowed energy levels and orbital wave functions (ψ), which lack physical meaning.
■ Electron paths are uncertain, but ψ² gives the probability of finding an electron at a point.
■ Wave functions satisfying boundary conditions are eigenfunctions, with corresponding energy values called eigenvalues.
■ Orbital is different from the orbit. The distinction between an orbit and the orbital is represented in Table 2.7.
Shapes of Orbitals
■ The shape of an atomic orbital is a 3D surface enclosing the electron’s probable space.
■ Orbitals differ in size, shape, and orientation, affecting electron probability in certain directions.
■ Radial probability (D-function) represents the likelihood of finding an electron at a certain distance from the nucleus.
■ Radial probability = D-function = 4 p r2.dr. y 2, where dr is the thickness of subshell.
■ Radial probability distribution curves show how probability varies with electron distance from the nucleus.
■ The point or the space around the nucleus where the probability of finding the electron is zero is called a node or radial node or nodal region, ( y 2 = 0)
CHAPTER 2: Structure of Atom
■ The number of nodal points or number of radial nodes present in an orbital = ( n – l – 1). It is (n–1) for ‘s’ orbitals, ( n–2) for p-orbitals, ( n–3) for d-orbitals and ( n–4) for f-orbitals.
Example: Number of radial nodes in
1s - orbital = 1 – 0 – 1 = 0
2s - orbital = 2 – 0 – 1 = 1
3p - orbital = 3 – 1 – 1 = 1
3d - orbital = 3 – 2 – 1 = 0
■ The plane in which the probability of finding the electron is zero is called nodal plane or angular node.
■ Number of nodal planes present in an orbital = l.
■ s, p, d and f-orbitals of any shell contain 0, 1, 2 and 3 nodal planes, respectively.
■ In the radial probability distribution curves, the number of minimas in the curve is equal to the number of radial nodes.
■ Total nodes = n – l – 1 + 1 = (n – l)
s-orbital
■ s-orbitals (l=0) exist in all orbits starting from n=1and are spherical in shape. Their size increases with n, maintaining spherical symmetry.
■ Radial probability is zero at the nucleus (r=0), and the most probable electron distance (r 0) increases with n (r0 for 1s = 0.53 Å, 2s = 2.6 Å, 3s > 2s).
■ Orbital size follows 1s < 2s < 3s < 4s. The number of peaks in radial probability curves equals n.
An orbit represents the movement of electron in one plane.
3. In an orbit, the position as well as the velocity of the electron can be known with certainty.
2. An orbital represents the movement of electron in three dimensional space.
3. In an orbital, it is not possible to find the position as well as velocity of electron with certainty.
4. Orbits are circular or elliptical in shape. 4. Orbitals have different shapes. s-orbital is spherical and p-orbital is dumbbell shaped.
5. Orbits do not have directional characteristics. 5. Except s-orbitals, all other orbitals have directional characteristics.
6. An orbit can have a maximum number of 2n2 electrons.
6. An orbital can accommodate a maximum of only two electrons.
■ For 2s orbital, probability is low at 0.53 Å and high at 2.6 Å, with a radial node at 1.1 Å.
Differences between orbit and orbital
1. An orbit is a well defined circular path around the nucleus in which the electron revolves.
1. An orbital is the region of space around the nucleus where the probability of finding the electron is maximum (95%).
2.
■ the probability of finding the electron around the nucleus is same in all directions, as shown in Fig. 2.12.
Fig.2.12 Probability density plots of s orbital
■ Radial probability distribution curves for 1s, 2s, and 3s orbitals are shown in Fig. 2.13. 2s
distance 3s
distance
Fig. 2.13 Radial distribution curves for 1s, 2s and 3s orbitals
The curve for 2s orbital contains 2 peaks and one minima, 3s orbital contains 3 peaks and 2 minimas.
Variation of orbital wave function y against the distance of an electron from the nucleus (r) for 1s and 2s orbitals is shown in Fig. 2.14.
Fig. 2.14 Variation of orbital wave function
■ Variation of probability density y 2 against the distance of electron from the nucleus ( r) for 1s and 2s orbitals is shown in Fig. 2.15.
Fig. 2.15 Variation of probability density
■ For 1s orbital, the probability density decreases sharply as we move away from the nucleus.
CHAPTER 2: Structure of Atom
■ For 2s orbital, the probability density first decreases sharply to zero, and then starts increasing. After reaching a small peak (maxima), it decreases again as ‘ r’ value increases.
■ The region where the probability density function reduces to zero is called ‘nodal surface’ or radical node, where the probability of finding an electron becomes zero.
■ The electron probability density is represented by the density of dots in Fig.2.16 and no dots for nodes.
p-sublevel
■ The p-sublevel starts from the 2nd orbit.
■ For the p-sublevel, l = 1, meaning it has three orbitals with m values: -1, 0, +1.
■ The three p-orbitals are px, py, and pz, based on electron density orientation.
■ In the px-orbital, electron density is along the x-axis, and p-orbitals have a dumbbell shape.
■ Each p-orbital has two lobes separated by a nodal plane where the probability density is zero.
■ The nodal planes for px, py, and pz are yz, zx, and xy planes, respectively.
■ The three p-orbitals have the same shape, size, and energy but different orientations (different m values).
■ The m values assigned are: ±1 for px, ±1 1 for py, and 0 for pz.
■ These three orbitals are perpendicular to each other and the angle between any two p-orbitals is 90o.
■ Radial probability distribution curves for 2p, 3p, 3d, 4s, 4p, 4d and 4f orbitals is shown in Fig.2.17.
Fig.2.17 Penetrating power of orbitals
■ This diagram illustrates the order of penetrating power of the orbitals: 4s > 4p > 4d > 4f.
Fig.2.16 Radial distribution curves
d-sublevel
■ The d-sublevel begins from the 3rd orbit (n = 3).
■ For the d-sublevel, l = 2, meaning it has five orbitals with m values: -2, -1, 0, +1, +2.
■ The five d orbitals are dxy, dyz, dzx, dx²-y², and dz².
■ Conventionally, the dz² orbital is assigned m = 0.
■ All d orbitals, except dz², have a double dumbbell shape.
■ Each d orbital has four lobes separated by two nodal planes (angular nodes).
■ The dz² orbital has a dumbbell shape along the z-axis with a small ring (collar) in the xy plane.
■ In dxy, dyz, and dzx orbitals, electron density is present between the corresponding axes.
■ In dx²-y² and dz² orbitals, electron density is along the axes.
■ The dxy orbital has yz and zx as nodal planes.
■ The dyz and dzx orbitals have xy, zx, and xy, yz planes as nodal planes, respectively.
■ The dx²-y² orbital has two nodal planes at 45° to the x- and y-axes.
■ The dz² orbital has no nodal planes.
■ The number of radial nodes depends on the principal quantum number.
■ radial and nodal planes of orbitals are given in the table below.
Nodes and nodal planes of orbitals
■ The five d-orbitals have the same energy in the ground state, but their energy and size increase with n value
■ Radial probability distribution curve for 3d orbital is shown in the Fig.2.18. f-sublevel
■ The f-sublevel begins from the 4th orbit (n = 4) and has seven orbitals with m values -3 to +3, having complex shapes.
Fig.2.18 Boundary surfaces of various orbitals
■ The order of energies of different sublevels o f a shell is s < p < d < f.
■ The boundary surfaces of 1s, 2p, 3d orbitals are given in the Fig.2.18.
Quantum Numbers
■ An electron's behavior in an atom is described by a wave function or orbital. Quantum numbers precisely define atomic orbitals, with the first three (principal, azimuthal, and magnetic) derived from Schrödinger’s equation, while the spin quantum number is not.
■ Principal Quantum Number (n)
Introduced by Niels Bohr,
n is a positive integer (1, 2, 3...∞) representing the main energy level or shell of an electron.
It determines the shell's size, energy, and maximum electron capacity (2n²), increasing with higher n values.
■ Azimuthal Quantum Number ( ) or Subsidiary Quantum Number or Orbital Quantum Number
introduced by Sommerfeld. It is denoted by ‘ ’.
In an energy level, n, the possible values of = 0, 1, 2 . . . . ( n–1), totally, n values.
It designates the subshells to which the electron belongs.
■ The energy of subshells increases with increase in value of (except for hydrogen atom). The energy order of subshells present in nth shell: s < p < d < f
■ It governs the shape of orbital.
■ It determines angular momentum of an electron.
Names of Subshells
Spherical
1 p Dumbbell
2 d Double dumbbell (except 2 z d )
3 f Complex
Magnetic Quantum Number (m)
■ This quantum number was proposed by Lande to explain Zeeman effect and Stark effect. It is denoted by m or m .
■ This quantum number describes the behaviour of electron in magnetic field.
■ Under the influence of external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus.
■ The number of m values gives the number of preferred orientations of the electron in a subshell or the number orbitals in the subshells.
■ The values of m in a subshell, = – to + , including zero. Thus total number of orbitals in ‘ ’ subshell = 2 + 1
■ The subshells and the corresponding m-values are as follows:
Subshells and Orbitals
CHAPTER 2: Structure of Atom
Spin Quantum Number (m s)
■ Proposed by Goudsmit and Uhlenbeck to explain doublets and triplets of multi-electron atoms.
It is denoted by ms or s.
The m s values +− 11 and 22 .
■ The electrons that have different m s values are said to have opposite spins; they are clockwise spin and anti-clockwise spin.
■ The two values of s are equal in magnitude but opposite in direction. They are represented by ↑ (upward) and ↓ (downward) arrows.
The total spin of an atom = × 1 n 2
■ The spin angular momentum of an electron is given by ()() h ss1ss1 2 +=+ π
■ Spin-only magnetic moment of an atom or ion is given by ()µ=+ s nn2 B.M, n = number of unpaired electrons with parallel spin.
Solved Examples
21. In 4th orbit of an atom, a maximum of how many electrons can exist?
Sol. In an nth orbit, the maximum number of electrons is given by 2 n2. Therefore, in 4th orbit, the maximum number of electrons = 2 × 42 = 32.
22. Among 4s, 4p, 4d, and 4f subshells of multi electronic atoms, which one has the highest energy?
Sol. For 4s, 4p, 4d, and 4f subshells, n value is 4. Their l values are 0, 1, 2, 3 and 4, respectively. Since energy increases with the increase in the value of l , the 4f subshell has the highest energy among the given subshells.
Try yourself:
8. For which of the orbitals of p and d subshells, value of ml = 0?
Answer: For zp and d z 2 orbitals, the value of ml = 0.
Electronic Configuration of Multi-electron Atoms
■ The electronic configuration describes the arrangement of electrons in shells, subshells, and orbitals of an atom.
Representation Methods
1. nℓx Method – Uses n (principal quantum number), ℓ (azimuthal quantum number as s, p, d, f), and x (number of electrons in that subshell).
2. Orbital Diagram Method – Represents electron distribution using arrows in boxes.
Aufbau Principle
■ Electrons fill orbitals in order of increasing energy. The orbital with the lowest (n + ℓ) value is filled first. If two orbitals have the same (n + ℓ) value, the one with the lower n value is filled first.
Order of increasing energy of orbitals:
■ 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s and so on.
Pauli’s Exclusion Principle
■ No two electrons in an atom can have the same set of four quantum numbers. Each orbital can hold a maximum of two electrons with opposite spins.
Maximum Electron Capacity
Main energy shell: 2n² electrons
Subshell: (4ℓ + 2) electrons
Orbital: Max 2 electrons with opposite spins
Subshells and Electrons
Hund’s Rule
■ Electrons occupy degenerate orbitals singly before pairing, following Hund’s rule of maximum multiplicity:
■ "Pairing of electrons in orbitals of the same subshell (p, d, or f) does not occur until each orbital has one electron with the same spin."
■ In p (3 orbitals), d (5 orbitals), and f (7 orbitals) subshells, pairing starts with the 4th, 6th, and 8th electrons, respectively.
■ The stability of nitrogen, noble gases, chromium (Cr), and copper (Cu) is based on Hund’s rule.
Stability of Atoms
■ Extra stability occurs when degenerate orbitals are half-filled or completely filled due to symmetrical electron distribution and exchange energy.
■ Exchange energy (Eex) is the energy released when two electrons with the same spin exchange positions without changing the electronic configuration.
■ Greater exchange energy leads to greater stability, with maximum exchange pairs in halffilled or completely filled subs hells.
■ Cr ([Ar] 4s¹ 3d⁵) and Cu ([Ar] 4s¹ 3d¹⁰) have exceptional configurations due to increased stability from exchange energy.
CHAPTER 2: Structure of Atom
■ Electrons with parallel spins remain farther apart, reducing Coulombic repulsion, lowering energy, and increasing stability.
■ The extra stability of parall el spin configurations is attributed to exchange energy (E ex)
■ Here, N is the number of electrons having identical spins. As the number of identical spins increases, the exchange energy also increases.
■ The exchange energy of chromium if the configuration is 3d4 4s2, is 10 K. If the configuration is 3d5 4s1, the exchange energy is 15 K.
Configuration of Ions
■ Electronic configuration of anions is written in the same method of writing configuration of neutral atoms. Oxide ion is O 2– (10 electrons) is 1s 2 2s2 2p6.
■ Chloride ion is Cl– (18 electrons) is 1s 2 2s2 2p6 3s2 3p6.
■ Electronic configuration of cations is also written by electrons removed from ultimate shell .
■ Electronic configuration of Fe 2+ is 1s2 2s2 2p6 3s2 3p6 3d6.
■ A set of atoms or ions with same electronic configuration is called isoelectronic species.
An isoelectronic group of 18 electrons is P 3–, S2–, Cl–, Ar, K+, Ca2+, and Sc3+
An isoelectronic group of 28 electrons is Cu + and Zn2+ .
Magnetic Properties of Species
Paramagnetic nature: Any species with unpaired electrons show paramagnetic nature. G reater the number of unpaired electron greater will be the paramagnetic. Examples: Cl, O2, Na . . . . etc
Diamagnetic nature: Species with pairs of electron will be diamagnetic in nature. Examples: Na+, Cl–, Ne . . . . etc
Solved Examples
23. An atom has 18 electrons in M shell and one electron in N shell (outermost shell) in its ground state. In which shell does it have symmetrical electron distribution?
Sol. Its electron configuration is 3s 2 3p6 3d10 4s1. In this way, its 3rd shell or penultimate shell has symetrical electron distribution.
24. What is the maximum number of electrons in an atom that can have the quantum numbers, n = 4, m = +1?
Sol. 4s electron cannot have m = +1 two 4p, two 4d and two 4f electrons can have m = +1. Thus, totally, six electrons can have these values.
25. Corresponding to n = 3, l = 2, m = 0, what is the maximum number of electons possible?
Sol. When n = 3, l = 2, m = 0, the possible atomic orbital is 3d z 2. It is with maximum two electrons
TEST YOURSELF
1. For which of the orbitals, m = 0 for all orbitals?
(1) s, px, and d xy
(3) P y , d xy , and dzx
(2) s, pz, and dz2
(4) pz, d xy , and dzx
2. Which quantum number is different for the unpaired electrons of carbon atom?
(1) n value
(3) l value
(2) ml value
(4) s value
3. For which orbital, there are two radial nodes and sum of radial and angular nodes is five?
(1) 5d (2) 6d
(3) 6f (4) 4f
4. Which of the following sets of quantum numbers is not valid?
(1) 1 3,2,2, 2 ====+ nlms
(2) 1 2,0,0, 2 ====− nlms
(3) 1 4,2,1, 2 ===−=+ nlms
(4) 1 4,3,4, 2 ====− nlms
5. Number of nodal planes present in s, p x, py, and pz, respectively, are (1) 0, 1, 1, 1 (2) 0, 2, 1, 1 (3) 0, 2, 2, 2 (4) 0, 0, 0, 0
6. Which of the following orbitals is not possible?
(1) 4d (2) 5g
(3) 3f (4) 2s
7. How many electrons are possible with n = 3, l = 1, s = + 1/2?
(1) 18 (2) 9 (3) 6 (4) 3
Answer Key
(1) 2 (2) 2 (3) 3 (4) 4 (5) 1 (6) 3 (7) 4
CHAPTER 2: Structure of Atom
# Exercises
JEE MAIN LEVEL
Level I
Discovery of Sub-atomic Particles
Single Option Correct MCQs
1. A discharge tube is filled with He gas. The e/m of the cathode ray particles is
(1) equal to the e/m of anode ray particle
(2) nearly 7295 times the e/m of the anode ray particles
(3) nearly 1837 times e/m of the anode ray particles
(4) nearly 1 1837 times e/m of the anode ray particles
2. Which of the following represents a beta particle?
(1) Proton
(2) Neutron
(3) Electron
(4) Dipositive helium ion
3. The study of discharge of electricity through gases led to the discovery of
(1) structure of the atom
(2) nucleus
(3) spectral lines
(4) electron
4. Which of the following species has larger specific charge
e m value?
(1) Electron (2) He+
(3) Na+ (4) Proton
5. Which of the following statements is incorrect regarding the oil drop experiment?
(1) Oil droplets in the form of mist, produced by the atomiser, were allowed
to enter through a tiny hole in the upper plate of the electrical condenser.
(2) The air inside the chamber was ionised by passing a beam of X-ray through it.
(3) The fall of these charged oil droplets can be retarded.
(4) Magnitude of electric charge ‘q’ on the droplets is always a fractional multiple of the electric charge e
6. Formula for calculating the mass of the electron (me) is
(1) ()eee mme·e/m =
(2) () e e e m e/m =
(3) e e e me m
(4) e e e me m
7. The mass of neutron is nearly (1) 10–23 kg (2) 10–24 kg
(3) 10–26 kg (4) 10–27 kg
8. Which of the following statements regarding the characteristics of positively charged canal rays is incorrect?
(1) Unlike cathode rays, mass of positively charged particles depends upon the nature of gas present in the cathode ray tube.
(2) The charge to mass ratio of these particles depends on the gas from which they originate.
(3) The behaviour of these particles in electric or magnetic field is same as that observed for electron or cathode rays.
(4) Some of the positively charged particles carry a multiple of the fundamental unit of fundamental charge.
9. The heaviest particle is (1) Beta
(2) neutron
(3) proton
(4) electron
Numerical Value Questions
10. A certain particle carries 2.5 × 10 –16 C of static electric charge. Calculate the number of electrons present in it.
11. The e/m value of electron is 1.75 × 10x C kg−1. The value of x is
12. The mass of neutron is (nearest whole number) __________ amu.
13. The charge of neutron is ____________.
Atomic Models
Single Option Correct MCQs
14. What was Thomson’s model of the atom called?
(1) Plum pudding model
(2) Solar system model
(3) Spherical model
(4) Model of atomic theory
15. According to J.J. Thomson, an atom is a positively charged sphere. Which subatomic particle is embedded in it?
(1) Neutron
(2) Neutrino
(3) Electron
(4) Meson
16. When alpha particles are sent through a thin metal foil, most of them go straight through the foil because i) alpha particles are much heavier than electrons ii) alpha particles are positively charged iii) most part of the atom is empty space
iv) alpha particles move with high velocity
(1) i, ii, iii, iv (2) i, iii (3) ii, iii (4) iii, iv
17. Rutherford’s experiment on the scattering of alpha particles showed for the first time that an atom has (1) electrons
(2) protons
(3) nucleus
(4) neutrons
18. Alpha particle consists of (1) 2 electrons, 2 protons, and 2 neutrons (2) 2 protons and 2 neutrons only (3) 2 electrons and 4 protons (4) 2 protons only
19. The radius of 27 13Al nucleus will be (1) 1.2 × 10−15 m
(2) 27 × 10−15 m
(3) 10.8 × 10−15 m (4) 3.6 × 10−15 m
20. The radius of the atom is of the order of (1) 10–10 cm (2) 10–13 cm
(3) 10–15 cm (4) 10–8 cm
Numerical Value Questions
21. The diameter of the atom is 10 m times the diameter of the nucleus. What is the value of m?
22. The ratio of the volume of the atom to the volume of the nucleus is in the order of 10x Find the value of x.
Developments Leading to Bohr’s Model of Atom
Single Option Correct MCQs
23. The frequency associated with a photon of radiation having a wavelength of 6000 A° is (1) 5 × 1014 Hz (2) 5 × 1010 Hz
(3) 2 × 1014 Hz (4) 5 × 1015 Hz
24. Wavelength of light is not measured in (1) Cps (2) A° (3) Pm (4) nm
25. The energy of an electromagnetic radiation is 19.875 × 10 –13 erg. What is the wave number in cm–1?
h = 6.625 × 10−27 erg s; c = 3 × 1010 cm s−1
(1) 1000
(2) 106
(3) 100
(4) 10000
26. Time period of a wave is 2.0 × 10 −10 s.
Frequencyofwaveis
(1) 2.0 × 10−10 s−1
(2) 50.0 × 108 s
(3) 50.0 × 109 s−1
(4) 500.0 × 107 s−1
27. If the wave number of two electromagnetic radiations A and B are 100 m-1 and 100 cm-1, then λA : λB is (1) 1 : 1
(2) 10 : 1
(3) 1 : 100
(4) 100 : 1
28. With respect to the figure given below, choose the incorrect statement.
29. The value of Planck’s constant is
(1) 6.625 × 10−27 erg/s
(2) 6.625 × 10−34 joule/s
(3) 6.625 × 10−27g cm2s−1
(4) 6.625 × 10−34 erg-s
30. According to Planck’s quantum theory, which of the following statements are correct?
a) The vibrating particle in the black body does not emit energy continuously.
b) Radiation is emitted in the form of small packets called quanta.
c) Energy associated with emitted radiations is inversely proportional to frequency.
d) The emitted radiant energy is propagated in the form of waves.
(1) a, b, c
(2) b, c
(3) a, b, d
(4) b, d, c
31. Which of the following is not a characteristic property of Plank’s quantum theory?
(1) Energy is neither absorbed nor emitted in simple multiples of quanta.
(2) Radiation is related to energy.
(3) Energy is not radiated continuously but in small packages of energy called quantum.
(4) The energy of a quantum is proportional to its frequency.
(1) As wavelength increases, intensity first increases and then decreases.
(2) T 3 > T2 > T1
(3) As temperature increases, the wavelength corresponding to maximum intensity increases.
(4) At very large wavelength, intensity is less.
32. A black body is an object that is capable of (1) only emitting all frequencies of radiation uniformly
(2) only absorbing all frequencies of radiation uniformly
(3) only emitting and absorbing all frequencies of radiation uniformly
(4) only emitting and absorbing all frequencies of radiation at random
33. Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from a metal surface?
(1) K.E. of electrons Energy of light O
(2) Number of electrons Frequency of light O
(3) K.E. of electrons Frequency of light O
(4) K.E. of electrons Intensity of light O
34. Which of the following elements requires radiation of the highest wavelength to show a photoelectric effect, and that the photoelectron emits with zero velocity?
(1) Li (2) Na
(3) K (4) Cs
35. In photoelectric effect, the number of photoelectrons emitted is proportional to (1) intensity of incident beam
(2) frequency of incident beam
(3) wavelength of incident beam
(4) all of the above
36. Identify the incorrect statement regarding the photoelectric effect.
(1) The number of electrons ejected is proportional to the intensity of light.
(2) The kinetic energy of photoelectrons increases with the increase in frequency of the light used.
(3) The difference in energy [ hν0 hν] is transferred as the kinetic energy of the photoelectron (where hv = energy of striking photon, hv0 = work function).
(4) The number of ejected photoelectrons does not depend on the energy of incident radiation.
37. Kinetic energy of a photoelectron emitted on the y-axis is plotted against the frequency of incident radiation on the x-axis. Identify the correct statement.
(1) slope y–intercept h h ϑ o
(2) slope y–intercept h ϑ o h ϑ
(3) slope y–intercept –h ϑ –h ϑ o
(4) slope y–intercept h –h ϑ o
38. Light of wavelength λ falls on a metal having work function of 0 hc λ . The phenomenon pho toelectric effect will take place only when
(1) 0 2 λ λ≥ (2) λ ⩾ λ0
(3) λ ⩾ 2λ0 (4) λ ⩽ λ0
39. Kinetic energy of photoelectrons is independent of of incident radiation.
(1) wavelength (2) wave number
(3) frequency
(4) intensity
40. The energy required to remove an electron a nd bring it to the surface of a metal is called as
(1) activation energy
(2) threshold energy (3) critical energy
(4) kinetic energy
Numerical Value Questions
41. 1 mole of photons, each of frequency 250 s−1, would have a total energy of ___erg.
42. How many metals among the given will show photoelectric effect (nearest whole number) with given threshold energies. By using incident radiation of 248 nm falling on their metal surface?
Metals: X Y Z W P Q R
E 0 (ev): 2.4 2.8 4.6 5.9 3.8 4.2 3.6
43. If the work function of a metal is 6.63 × 10−19 J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ________ nm. [Nearest integer]
[Given: h = 6.63 × 10−34 Js and c = 3 × 108 ms−1]
Bohr's model Hydrogen Atom
Single Option Correct MCQs
44. Bohr’s theory is not applicable to (1) H (2) He+ (3) Li2+ (4) H+
45. If the radius of the first Bohr’s orbit is a0, the de Broglie’s wavelength of an electron revolving in the second Bohr’s orbit will be (1) 6πa0 (2) 4πa0
(3) 2πa0 (4) πa0
46. The ratio of the radii of the first three orbits in an atom of hydrogen is (1) 1 : 4 : 9 (2) 9 : 4 : 1 (3) 1 : 2 : 3 (4) 3 : 2 :1
2: Structure of Atom
47. The radius of the second Bohr’s orbit of a hydrogen atom is r2. The radius of the third ohr’s orbit will be
(1) 2 4 r 9 (2) 4r2
(3) 2 9 r 4 (4) 9r2
48. In H-atom, if an electron falls from n = 3 to n = 2, then emitted energy is
(1) 10.2 eV (2)12.09 eV (3) 1.9 eV (4)0.65 eV
49. The velocity of an electron in the fourth Bohr’s orbit of hydrogen is ‘u’. The velocity of the electron in the first orbit would be (1) 4u (2) 16u
(3) u 4 (4) u 16
50. If the first ionisation energy of hydrogen is E, the ionisation energy of He+ wouldbe (1) E (2) 2E
(3) 0.5E (4) 4E
51. Velocity of the electron in the second orbit of hydrogen is x. What is the velocity of the electron in the first orbit of Li 2+ ion?
(1) 2 x 3 (2) 3 x 2
(3) 4x (4) 6x
52. Kinetic and potential energies (in eV) of an electron present in the second Bohr’s orbit of a hydrogen atom are, respectively, [Energy of first Bohr’s orbit = −13.6 eV]
(1) −13.6, −27.2 (2) −1.51, −3.02
(3) 3.4, −6.8 (4) 3.02, −1.51
53. As the electron moves away from the nucleus, its potential energy ________ and kinetic energy _________.
54. The energy difference between which of the following two successive levels is minimum?
(1) 1 and 2
(2) 2 and 3
(3) 3 and 4
(4) 4 and 5
55. The wave number of the 4 th line in the Lyman series of ‘H’ spectrum is (R = Rydberg constant)
(1) 1 24 cm 25R
(2) 1 25 cm 24R
(3) 1 25R cm 24
(4) 1 24R cm 25
56. What is the wavelength of Hβ line in Balmer series of hydrogen spectrum? (R = Rydberg constant)
(1) 36/5R
(2) 5R/36
(3) 3R/16
(4) 16/3R
57. The ratio of the shortest wavelength of two spectral series of a hydrogen atom is 9 : 1. The spectral series are
(1) Paschen and Lyman
(2) Balmer and Brackett
(3) Paschen and Pfund
(4) Lyman and Brackett
58. The wave number of the first emission line in the Balmer series of He + spectrum is (R = Rydberg constant)
(1) 5 R 9 (2) 9 R 400
(3) 7 R 6 (4) 3 R 4
59. The shortest wavelength of a hydrogen atom in the Lyman series is x. The longest wavelength of He+ in the Balmer series is
(1) 5 9x (2) 9x 5
(3) 5x 9 (4) 36x 5
Numerical Value Questions
60. Ra dius of the tenth Bohr’s orbit of the hydrogen atom is ____________ A 0 .
61. An electron in a hydrogen atom (ground state) jumps to a higher energy level x, such that the potential energy of the electron becomes half of its total energy at the ground state. What is the value of x?
62. The energy of an electron, when it is present at an infinite distance from the nucleus, is _______.
63. The velocity of an electron in a certain Bohr’s orbit of a hydrogen atom bears in the ratio 1 : 275 to the velocity of light. The quantum number (n) of the orbit is ________.
64. How many waves are there in the sixth Bohr’s orbit of hydrogen?
65. The electronic velocity of the 4th Bohr’s orbit in H-atom is V. The velocity of the electron in the first Bohr’s orbit would be _____ times V.
66. The wavelength of a spectral line emitted by a hydrogen atom in the Lyman series is 16/15R cm. The value of n2 [R = Rydberg constant] is ____.
67. The wavelength of Hδ line of Balmer series of a hydrogen atom is nearly ______ × 103 A° (R = 1.08 × 107 m−1).
Towards Quantum Mechanical Model of Atom
Single Option Correct MCQs
68. If travelling at the same speed, which of the following matter waves has the shortest wavelength?
(1) Electron
(2) Alpha particle (He2+)
(3) Neutron
(4) Proton
69. If the radius of the first Bohr’s orbit of H atom is x, then the de Broglie wavelength of an electron in the third orbit is nearly (1) 2πx (2) 6πx (3) 9x (4) x/3
70. An electron is moving in Bohr’s fourth orbit. Its de Broglie wavelength is λ. Then, what is the circumference of the third orbi t?
(1) λ 2 (2) λ 6 4
(3) λ 9 4 (4) λ 4
71. The momentum of a particle (in g cm s –1) of wavelength 1 A° is
(1) 6.625 × 10−27
(2) 6.625 × 10−19
(3) 6.625 × 10−16
(4) 6.625 × 10−23
72. Mass of electron = 9.1 × 10–31 kg. The kinetic energy of an electron is 4.55 × 10–25 J. What is its wavelength (in nm)?
(1) 728 (2) 850 (3) 896 (4) 786
73. If uncertainty in position is zero, the uncertainty in momentum of an electron will be
(1) zero (2) infinity (3) unity (4) negative
CHAPTER 2: Structure of Atom
74. If the uncertainties in position and momentum are equal, then the uncertainty in velocity is (1) π 1h m (2) π h (3) π 1h 2m (4) π h 2
75. Heisenberg uncertainty principle can be explained as (1) ∆× ∆≥ π h 4 P x (2) ∆×∆≥ π 4 h xP (3) ∆×∆≥ π h xP (4) ∆≥π ∆ h P x
76. Which among the following is an incorrect expression of Heisenberg’s uncertainty principle?
(1) ∆×∆≥ π h xP 4 (2) ∆×∆≥ π h tE 4
(3) ∆×∆≥ π hm xV 4 (4) ∆×∆λ≥λ π 2 x 4
77. Decreasing order of uncertainty in velocity, when ∆x is the same, is (1) n > e– > p (2) e – > p > n (3) p > e– > n (4) p > e– > n–
Numerical Value Questions
78. The uncertainty in the momentum of an electron is 10–5 kg ms–1. The uncertainty in its position will be __________ 10 -30m.
79. If th e u ncertainties of position and momentum of a particle of mass m are equal, then the uncertainty of velocity is π 1 h xm . Find the value of x.
80. The value of de Broglie wavelength of ‘He’ gas at 100 K is 5 times that of a gas at 1250 K. What is the atomic mass of the gas?
81. The de Broglie’s wavelength of an electron in the fourth orbit is ________ πa 0. (a 0 = Bohr’s radius)
82. If the radius of the first orbit of a hydrogen atom is a0, then the de Broglie’s wavelength
of an electron in the third orbit is _______
Quantum Mechanical Model of Atom
Single Option Correct MCQs
83. Select the correct statement(s).
a) Solutions of Schrodinger’s wave equation give an idea about E (energy) and ψ.
b) Schrodinger’s wave equation for a multi-electron system can be easily solved with accuracy.
c) Solution of Schrodinger’s wave equation gives an idea about n, l, and m.
(1) All are correct.
(2) Only a and b are correct.
(3) Only a and c are correct.
(4) Only b is correct.
84. Which of the following correctly represents Schrodinger’s wave equation for a system whose energy does not change with time?
(1) ψ=ψ ˆ HE
(2) ψ=ψ ˆ E
(3) = ˆ HE
(4) ψ= ˆ HE
85. Based on quantum mechanical model of atom, ψ (2, 1, 0) is represented as (1) 2px (2) 2py (3) 2s (4) 2pz
86. ψ 2 at any point in an atom, is directly proportional to (1) energy of the electron
(2) radius of the atom
(3) the probability of finding the electron at that point
(4) 1 r of the atom (r is the radius of the atom)
87. In Schrodinger’s wave equation, ψ represents (1) orbitals
(2) wave function
(3) amplitude function (4) 2 and 3
88. The following graph is plotted for nsorbitals.
4prR2
r(A)
The value of ‘n’ is (1) 1 (2) 2 (3) 3 (4) 4
89. Graphs for 2s orbital are shown below. (R = ψ, R2 = ψ2) X
X, Y, and Z, respectively, are (1) R, R2, and 4πr2R2 (2) R2, R, and 4πr2R2
(3) 4πr2, R2, R2, and R (4) R2, 4πr2R2, and R
90. Which of the following graphs represents the radial probability function of a 3d electron?
91. Hamilitonian operator ˆ H represents the total energy of the system, which does not include
(1) kinetic energies of all the sub-atomic particles
(2) repulsive potential between electrons and protons
(3) repulsive potential among electrons and nuclei individually
(4) attractive potential between electrons and nuclei
92. The number of radial nodes of 3s and the number of angular nodes of 3p orbitals, respectively, are
(1) 2, 2
(2) 2, 0
(3) 0, 1
(4) 2, 1
93. Which of the following have the same number of radial nodes?
(1) 4s, 5p
(2) 5s, 5p
(3) 1s, 2s
(4) 4s, 6p
94. Which of the following orbitals doesn’t have both nodal planes and radial nodes?
(1) 1s
(2) 3dxy
(3) 3dz2
(4) Both 1 and 3
95. Which of the following orbitals has two spherical nodes?
(1) 2s (2) 4s
(3) 3d (4) 6f
96. The electron density of 3dxy orbital in yz plane is
(1) 95% (2) 50%
(3) 33.33% (4) zero
97. The energy of the ‘2s’ orbital is highest in (1) K (2) Na
(3) Li (4) H
98. Which of the following statements is incorrect?
(1) In a hydrogen atom, the shapes of 2s and 2p orbitals are different but the energy is the same.
(2) In a hydrogen atom, the order of energies of orbitals is 1s < 2s = 2p < 3s = 3p = 3d
(3) ‘+’ and ‘–’ signs of the lobes of the orbital represent geometric signs of wave functions.
(4) Energy of E2s(H) = E2s(Li) = E2s(Na)
99. For which of the following sets of quantum numbers will an electron have the highest energy?
(1) + 1 3,2,1, 2
(2) −+ 1 4,2,1, 2
(3) + 1 4,1,0, 2
(4) + 1 5,0,0, 2
100. The maximum number of electrons with s = +1/2 in an orbital, for which l = 2, is
(1) 1 (2) 3
(3) 5 (4) 7
101.For a given value of azimuthal quantum number (l ), the total number of possible magnetic quantum numbers ( m) will be
(1) + = 1 2 lm
(2) = 1 2 lm
(3) + = 21 2 lm
(4) + = 21 2 l n
102. Which of the following quantum numbers are not possible?
(1) n = 2, l = 1, m = −1, s = −1/2
(2) n = 3, l = 2, m = −3, s = +1/2
(3) n = 2, l = 0, m = 0, s = +1/2
(4) n = 3, l = 2, m = −2, s = +1/2
103. The maximum number of electrons having the same spin, present in an atom for n + = 4, is
(1) 6 (2) 2
(3) 8 (4) 4
104. ‘P’ subshell of any energy level has three orientations. This is indicated by the values of
(1) n (2)
(3) m (4) s
105. If Aufbau rule is not followed, K-19 will be placed in
(1) s-block (2) p-block
(3) d-block (4) f-block
106. The orbital diagram in which the Aufbau principle is violated is (1) 2s 2p (2) 2s 2p (3) 2s 2p (4) 2s 2p
107. Which orbital fills completely immediately before the 4f?
(1) 6s (2) 5p (3) 5d (4) 4d
108. Which of the following is a violation of Pauli’s exclusion principle?
(1) (2) (3) (4)
109. In the absence of Pauli’s exclusion principle, the configuration of Li would have been
(1) 2s 1s 2p (2) 2s 1s 2p
(3) 1s (4) 2s 1s 2p
110. Which orbital representation follows Hund’s rule?
(1)
1s 2p (2)
(3)
(4)
111. Hund’s rule states that the most stable arrangement of electrons (for a ground state electron configuration)
(1) has three electrons per orbital, each with an identical spin
(2) has m 1 values greater than or equal to +l
(3) has the maximum number of unpaired electrons, all with the same spin in degenerate orbital
(4) has two electrons per orbital, each with opposing spin
112. For a d 5 configuration, Hund’s rule is followed for maximum multiplicity and not followe d minimum multiplicity then their values will be ______.
(1) 6, 2 (2) 5, 1
(3) 51 , 22 (4) 6, 1
113. For which of the following species is the electronic distribution spherically symmetrical?
(1) C (2) Na
(3) Cl– (4) B
114. The electronic configuration of Cr (Z = 24) is taken as 1s22s22p63s23p63d54s1 instead of 1s22s22p63s23p63d44s2 due to which of the following in the first case?
(1) Higher exchange energy
(2) Higher symmetry of electronic distribution
(3) Smaller Coulombic repulsion energy
(4) All of the above
Numerical Value Questions
115.The number of nodes possible in the radial probability distribution curve of 3d orbital is _______.
116.The wave equation represents the motion of an wave in ________ dimensions.
117 .The number of orbitals with n = 5 and m = +2 is _______. (Round off to the nearest integer)
118. The number of electrons in Cr having ‘m’ value zero is _______. (Atomic number: Cr = 24)
119. The orbital angular momentum of an electron in 3s orbital is π xh 2 . The value of x is_______.
120. For how many electrons does an argon atom have l = 1?
121. The wave function of an orbital is represented as Ψ 4,2,0 . The azimuthal quantum number of that orbital is ________.
Level II
Discovery of Sub-Atomic Particles
Single Option Correct MCQs
1. The specific charge value of neutron is
(1) 1.78 × 108 C/kg
(2) 1.75 × 108 C/kg
(3) 1.79 × 108 C/ g
(4) 0
2. Select the correct statements from the following.
A. Atoms of all elements are composed of two fundamental particles.
B. The mass of the electron is 9.10939 × 10–31 kg.
C. All the isotopes of a given element show same chemical properties.
D. Protons and electrons are collectively known as nucleons.
E. Dalton’s atomic theory regarded the atom as an ultimate particle of matter.
Choose the correct answer from the options given below.
(1) B, C, and E only
(2) A and E only
(3) C, D, and E only
(4) A, B, and C only
3. Charge of an electron in e.s.u is
(1) –4.8 × 10–10
(2) + 4.8 × 10–10
(3) –1.602 × 10–19 C
(4) +1.602 × 10–19 C
Atomic Models
Single Option Correct MCQs
4. Rutherford’s alpha particle scattering experiment led to the conclusion that (1) the point of impact with matter can be precisely determined
(2) electrons occupy space around the nucleus
(3) mass and energy are related
(4) neutrons are buried deep in the nucleus
5. The fraction of volume occupied by the nucleus with respect to the total volume of an atom is
(1) 10–15 (2) 10–5
(3) 10–30 (4) 10–10
6. The nucleus and an atom can be assumed to be spherical. The radius of the nucleus is given by 1.25 × 10–13 × A1/3 cm. The atomic radius of atom is 1 Å. If the mass number is 64, the fraction of the atomic volume that is occupied by nucleus is
(1) 1.0 × 10–3 (2) 5.0 × 10–5
(3) 2.5 × 10–2 (4) 1.25 × 10–13
Development Leading to Bohr's Model of Atom
Single Option Correct MCQs
7. The energy of one mole of photons of radiations of wavelength 300 nm is (Given: h = 6.63 × 10–34 J s, NA = 6.02 × 1023 mol–1, c = 3×108 ms–1)
8. The emission spectrum of hydrogen in the visible region consists of (1) a continuous band of light (2) a series of equally spaced lines (3) a series of lines that are closer at low energies (4) a series of lines that are closer at high energies
9. If the electron in hydrogen atom jumps from the third orbit to the second orbit, the wavelength of the emitted radiation is given by (1) 36 5R λ= (2) 5R 36 λ= (3) 5 R λ= (4) R 6 λ=
10. Order of wavelength in electromagnetic spectrum is
(1) Cosmic rays >γ-rays > X-rays > UV > Visible > IR > Micro waves > Radio waves
(2) Cosmic rays > X-rays > γ-rays > UV > IR > Visible > Micro waves > Radio waves
(3) Cosmic rays < γ-rays < X-rays < UV < Visible < IR < Micro waves < Radio waves
(4) Cosmic rays < X-rays < γ-rays < Visible < UV < Radio waves < IR < Micro waves
Numerical Value Questions
11. If there are 3 atoms of a hydrogen-like species in the second, third, and fourth excited state respectively, then how many maximum different Balmer and Paschen lines can be produced?
Bohr's Model for Hydrogen Atom
Single Option Correct MCQs
12. In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given: Bohr radius a0 = 52.9 pm]
(1) 105.8 pm
(2) 2116. pm
(3) 211.6 p pm
(4) 52.9 p pm
13. If the radius of the 3 rd Bohr’s orbit of hydrogen atom is ‘r3’ and the radius of 4th Bohr’s orbit is ‘r4’ then
(1) 43 9 16 = rr (2) 43 16 9 = rr
(3) 43 3 4 = rr (4) 43 4 3 = rr
14. What is the ratio of time periods (T 1/ T 2) in second orbit of hydrogen atom to third orbit of He+ ion?
(1) 8/27 (2) 32/27
(3) 27/32 (4) 27/8
15. With respect to an electron moving from first orbit to second orbit, choose the incorrect statement.
(1) Velocity is halved.
(2) Energy is increased by 4 times.
(3) Radius is increased by 4 times.
(4) Energy is decreased by 4 times.
16. For a 3d–orbital, the orbital angular momentum is
(1) 2h 2 π (2) zero
(3) h 6 2 π (4) h 2 2 π
17. The angular velocity of an electron occupying t he second Bohr orbit of He + ion is (in s−1)
(1) 2.067 × 1016
(2) 3.067 × 1016
(3) 1.067 × 1018
(4) 2.067 × 1017
18. The ratio of the radii of the first three orbits in an atom of hydrogen is
(1) 1 : 4 : 9 (2) 9 : 4 : 1
(3) 1 : 2 : 3 (4) 3 : 2 : 1
19. According to Bohr’s theory, the angular momentum of electron in 5th orbit is
(1) h 2.5 π (2) h 25 π
(3) h 1.0 π (4) h 10 π
20. Energy of which orbital is maximum for hydrogen atom?
(1) 6s (2) 5p
(3) 4d (4) 5f
Numerical Value Questions
21. If radius of second stationary orbit (in Bohr’s atom) is R, then radius of third orbit will be 9 R X . Calculate the value of x.
22. If an electron in H–atom has an energy of –78.4 kcal/mol, the electron is present in _____ orbit.
Towards Quantum Mechanical Model of Atom
Single Option Correct MCQs
23. The mathematical expression for the uncertainty principle is
(1) h xp 4m ∆∆≥ π (2) h Et ∆∆≤ π
(3) h xp p ∆∆≥ (4) h Et 4 ∆∆≥ π
24 Choose the correct decreasing order of energy, for the orbitals having the following set of quantum numbers:
A. n = 3, l = 0, m = 0
B. n = 4, l = 0, m = 0
C. n = 3, l = 1, m = 0
D. n = 3, l = 2, m = 1
(1) D > B > C > A (2) B > D > C > A
(3) C > B > D > A (4) B > C > D > A
25. Correct expression for Heisenberg uncertainly principle is
(1) 4 h xV π ∆×∆≥
(2) 4 h xV mπ ∆×∆≥ (3) 2 h xV π ∆×∆≥ (4) 2 h xV mπ ∆×∆≥
26. de Broglie wavelength of 150 g ball moving with a velocity of 50 ms −1 is
(1) 6.626 × 1034 cm
(2) 6.626 × 10–34 cm
(3) 8.83 × 10–33 cm
(4) 2.26 × 10–34cm
27. Which of the following sets of quantum numbers is not possible?
(1) 1 3,2,2, 2 ===−=+ nlms
(2) 1 4,0,0, 2 ====− nlms
(3) 1 2,1,0, 2 ==−==+ nlms
(4) 1 5,3,2, 2 ===−=− nlms
28. xy plane is nodal plane for (1) d xy (2) 22 xy d (3) 2 z d (4) d xz
29. Which quantum number is different for the unpaired electrons of carbon atom?
(1) Principal quantum number
(2) Azimuthal quantum number
(3) Magnetic quantum number
(4) Spin quantum number
30. Which of the following is an incorrect statement?
(1) Electrons in atom can have all four quantum numbers same.
(2) Orbital quantum number tells us what energy level the electron has.
(3) Higher orbits have more energy.
(4) An orbital can be described by n, l, m quantum numbers.
31. According to Aufbau principle, which orbital takes place just after 5s in a multielectron atom?
(1) 5p (2) 5d
(3) 4f (4) 4d
32. Which orbital representation is following Hund’s rule?
1S 2S 2P (1)
1S 2S 2P (2)
1S 2S 2P (3)
1S 2S 2P (3)
Numerical Value Questions
33. The value of de Broglie wavelength of He gas at –73° is ‘ x ’ times that of de Broglie wavelength of Ne at 727 °C. What is the value of ‘x’?
34. Number of orbitals present in L-shell is.
35. The number of unpaired electrons present in chromium atom is _______.
36. Sum of the total number of radial and angular nodes of 4 d orbital is ______.
37. The atomic number of iron is 26. The maximum number of electron(s) of iron atom having 1 s 2 =+ is
38. How many of the following ions have the same magnetic moment? Fe2+, Mn2+, Cr2+, Ni2+
39. If uncertainty in position is 25 . 4π The uncertainty in measurement of de Broglie wavelength (in A°) is x. The value of x/5 is ______.
Multiple Concept Questions
Single Option Correct MCQs
40. 1 mole of diatomic element X 2 contains 34 and 40 mol of electrons and neutrons, respectively. The isotopic formula of the element is
(1) 34 X74
(2) 17 X37
(3) 34 X40
(4) 20 X40
41. Which of the following shown an increasing value of e m ?
(1) n < a < p < e
(2) n < p < a < e
(3) n < p < e < a
(4) p < n < a < e
42. What will be the difference in the mass number if the number of neutrons is halved and the number of electrons is doubled in 8O16?
(1) 25% decrease
(2) 50% increase
(3) 150% increase
(4) No difference
43. The wavelength of the first line of Lyman series of hydrogen, He+, and Li2+ ions are λ1, λ2, and λ3, respectively. The ratio of these wavelengths is
(1) 1 : 4 : 9 (2) 9 : 4 : 1
(3) 36 : 9 : 4 (4) 6 : 3 : 2
44. Which of the following transitions represents limiting line of Balmer series?
(1) ∞ → 1 (2) ∞ → 2
(3) ∞ → 3 (4) ∞ → 4
45. According to Einstein's photoelectric equation, the correct graph between kinetic energy of photoelectrons ejected and the frequency of the incident radiation is
(1) KE
46. The ratio of wavelength values of series limit lines (n 2 = ∞) of Blamer series and Paschen series are
(1) 4 : 9 (2) 9 : 4
(3) 2 : 3 (4) 3 : 2
47. The true relation about the velocity of photon is (1) dependent on its wavelength (2) dependent on its intensity (3) equal to cube of its amplitude (4) independent of its wavelength
48. If uncertainty in position of particle is equal to the uncertainty in momentum, then its uncertainty in velocity is
(1) 4 h mπ (2) 4 h mπ
(3) 1 2 × h m π (4) 2 h m
49. Which of the following has correct matching of curve and orbital? I)
I) I(2p)II(1s)III(4p)
II) I(3p) II(3d) III (3s)
III) I(4d) II(2p) III(5d)
IV) I(2s) II(4f) III(3d)
(1) II and III (2) I and III
(3) III and IV (4) I, II, III, and IV
50. Magnetic moments of the following isoelectronic species (24 electrons) Mn + , Cr, Fe2+, and Co3+, are in the order (1) Fe2+= Co3+ < Mn+ = Cr
(2) Fe2+ = Cr < Co3+ = Mn+
(3) Cr < Mn+< Fe2+ < CO3+
(4) Fe2+< Co3+ < Mn+ < Cr
Numerical Value Questions
51. The work function of sodium metal is 4.41 ×10–19 J. If the photons of wavelength 300nm are incident on the metal, the kinetic energy of the ejected electrons will be ____ × 10–21 J. (h = 6.63 × 10–34 Js; c = 3 × 108 m/s)
52. Number of photons of light with a length of 4000 pm, necessary to provide one joule of energy, is approximately x × 1016 . Then, x is ____.
53. The wavelength of an electron of kinetic energy 4.50 × 10 –29 J is______ × 10 –5 m (nearest integer). Given: Mass of electron is 9 × 10–31 kg, h = 6.6 × 10–34 Js
54. The de Broglie wavelength of a neutron at 27 °C is l . The wavelength at 927 °C will be x λ . The value of x is _____.
55. The number of radial nodes of 3s and 2p orbitals are x and y, respectively. x + y is________.
56. What is the maximum number of electrons that can have n + l = 4 in an atom?
Level III
Single Option Correct MCQs
1. Electrons in a cathode ray tube have been emitted with a velocity of 1000 m s–1. Which of the following statements are true about the emitted radiation?
Given: h = 6 × 10–34 J s, me = 9 × 10–31 kg
1) The de Broglie wavelength of the electron emitted is 666.67 nm.
2) The characteristic of electrons emitted depends upon the material of the electrodes of the cathode ray tube.
3) The cathode rays start from cathode and move towards anode.
4) The nature of the emitted electrons depends on the nature of the gas present in cathode.
(1) 1 and 3 (2) 2 and 4
(3) 1 and 4 (4) 1, 2, 3 and 4
2. A graph is plotted for an element by putting its mass on x–axis and the corresponding number of atoms on y -axis. What is the atomic mass of the element for which the graph is plotted? (NA = 6.0 × 1023)
number of atoms q= tan –1 (7.5 × 1021)
Mass in (grams)
(1) 80 (2) 40 (3) 0.025 (4) 20
3. From the observations and conclusions one can draw from Rutherford α-particle scattering experiment on thin gold foil, select the incorrect statement.
(1) Unlike Thomson’s model, atoms are highly non-uniform, where mass and charge are non–uniformly distributed. While electrons can occupy volume of radius 10−8 cm, positive charge is more concentrated in a very tiny space.
(2) From Rutherford experiment, neutrons were discovered.
(3) α-particle passing close to nucleus has a hyperbolic path.
(4) As the kinetic energy of α particles is decreased, more α particles will be deflected by relatively larger angles.
4. Which of the following expressions represents the spectrum of Balmer series (if n is the principal quantum number of higher energy level) in hydrogen atom?
(1) () () 2 11−+ = Rnn v n
(2) () () 2 22 4 −+ = Rnn v n
(3) () () 2 22−+ = Rnn v n
(4) () () 2 11 4 −+ = Rnn v n
5. A dye absorbs a photon of wavelength l and re-emits the same energy into two photons of wavelengths l 1 and l 2 respectively. The wavelength l is related with λ1 and l 2 as:
(1) 12 12 λ+λ λ= λλ
(2) 12 12 λλ λ= λ+λ
(3) 22 12 12 λλ λ= λ+λ
(4) () 12 2 12 λλ λ= λ+λ
6. In a photoelectric experiment, the stopping potential is plotted against 1 λ of incident radiation for two different metals, and the curve is as shown in the figure. Predict which of the following statements is incorrect
v0 (volts)
Metal I Metal II
1 ( ( l m–1
work function of metal I = 0.24 eV
(1) Slope of the curves for both the metals is 1.242 ×10−6 Jm coulomb–1 .
(2) When an electromagnetic radiation of wavelength 100 nm strikes the two metals separately, the stopping potential for metal I is 12.18 V.
(3) The stopping potential for metal I is more than that of metal II, if both the metals are exposed to electromagnetic radiation of wavelength 200 nm separately.
(4) The stopping potential for metal II is more than that of metal I, if both the metals are exposed to electromagnetic radiation of wavelength 200 nm separately.
7. Kinetic energy of photoelectron emitted on y-axis is plotted against frequency of incident radiation on x-axis. Identify the correct statement.
Slope y-intercept
(1) h hϑ o
(2) hϑ o hϑ
(3) –hϑ –hϑ o
(4) h –hϑ o
8. The de Broglie wavelength (λ) associated with a photoelectron varies with the frequency (v) of the incident radiation as [v0 is threshold frequency]
(1) () 0 1 vv
(2) () 1 4 0 1 vv
9. When n = 6, the correct sequence for filling of electrons will be
(1) ns → (n−2)f → (n–1)d → np
(2) ns → (n–1) d → (n–2) f → np
(3) ns → (n–2) f → np → (n–1)d
(4) ns → np(n–1)d → (n–2)f
10. For a uni-electronic species, compare the energy of the following orbitals (represented by S).
S 1 = A spherically symmetrical orbital having two spherical nodes
S 2 = An orbital that is double-dumb bell shaped and has no radial nodes
S 3 = An orbital with orbital angular momentum zero and three radial nodes
S4 = An orbital having one planar and one radial node
(1) S1 = S2 = S3 = S4 (2) S1 = S2 = S4 < S3
(3) S1 > S2 > S3 > S4 (4) S1 < S2 < S3 < S4
Numerical Value Questions
11. At what minimum atomic number would a transition from n = 2 to n = 1 energy level result in the emission of X-rays with λ = 3 × 10–8 m?
12. A light of wavelength 200 nm falls upon a surface and two different wavelength photons, λ = 800 nm and λ = 400 nm, are emitted from the surface. 80% of the energy absorbed is re-emitted in the form of photon. Number of photons emitted as λ = 800 nm is three times than that of number of photons emitted as λ = 400 nm. If the ratio of the total absorbed photon to total emitted photon is x, then find the numerical value of 64x.
13. The minimum number of waves made by an electron moving in an orbit having maximum magnetic quantum number + 3 is ______.
14. In the plots of radial distribution function for the hydrogen 3s orbital versus ‘r’, the number of peaks is ______.
15. When a certain metal was irradiated with light of frequency 4.0 × 1016 s–1 , the photoelectrons emitted had three times the kinetic energy as the kinetic energy of photoelectrons emitted when the metal was irradiated with light of frequency 2.0 × 1016 s–1. The critical frequency (v0) of the metal is 1×10x s−1, what is ‘x ’ ?
16. The dissociation energy of H 2(g) is 435.0 kJ. mol−1. If a radiation of wavelength 2640 A° is used for atomisation of hydrogen gas, what percentage of energy of radiation will be converted into kinetic energy of hydrogen atoms?
17. The wavelength of an electron and a neutron will become equ al when the velocity of
the electron is ‘x’ times the velocity of the neutron. The value of ‘x’ is (nearest integer) (mass of electron is 9.1 × 10–31 kg and mass of neutron is 1.6×10–27 kg)
18. A tube emits light of wavelength 5296 A o . The bulb is rated as 200 watt and 20% of the energy is converted into light. If the whole light emitted is allowed to fall on a metal surface having work function 3.00 × 10–19 J, the number of electrons ejected in 10s is 106.58 × 10x. The value of ‘x’ is ____.
THEORY-BASED QUESTIONS
Statement Type Questions
(Q.No. 1-10)
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
1. S-I : Atoms of elements are composed of two fundamental particles.
S-II : Specific charge of cathode rays are constant.
2. S–I : Cathode rays consist of negatively charged particles, called electrons.
S–II : In the presence of electrical/magnetic field, the behaviour of cathode rays is similar to the negatively charged particles.
3. S–I : The kinetic energy of the photoe lectron ejected increases with increase in intensity of incident light.
S–II : Increase in intensity of incident light always increases the photoelectric current.
4. S–I : According to Bohr’s model of an atom, the velocity of electron increases with decrease in hold on the electron by the nucleus.
S–II : According to Bohr’s model of an atom, the velocity of electron increases with decrease in principal quantum number.
5. S-I : In the line spectra observed in the case of multielectron atoms, some lines actually occur in doublets, triplets, etc.
S-II : Energies of the orbitals in the same subshell decrease with increase in the atomic number (Zeff).
6. S-I : Wavelength of limiting line of Lyman series is less than wavelength of limiting line of Balmer series.
S-II : Rydberg constant value is same for all elements.
7. S-I : According to Bohr’s model of hydrogen atom, the angular momentum of an electron in a given stationary state is quantised.
S-II : The concept of electron in Bohr’s orbit violates the Heisenberg uncertainty principle.
8. S-I : The kinetic energy of an electron is greater than α-particle having same velocity.
S-II : Mass of α-particle is greater than the mass of electron.
9. S–I : If the potential difference applied to an electron is made 4 times, the de Broglie wavelength associated is halved. (Initial kinetic energy of the electron was zero.)
S–II : On m aking potential difference 4 times, velocity is doubled and hence,
according to de Broglie hypothesis,λ is halved.
10. S–I : Principal quantum number of the outermost electron in Fe is 4.
S–II : The last electron is filled in 3d.
Assertion and Reason Questions
(Q.No. 11-21)
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
11. (A) : For a 2p-orbtial, the probability density function is zero on the plane where the two lobes touch each other.
(R) : Such a plane is called nodal plane.
12. (A) : A photon of energy 12 eV can break three molecules of A2 into atoms. A2 has bond dissociation energy of 4 e V/molecule.
(R) : Total energy is conserved and interaction is always one to one between photon and molecule.
13. (A) : The energy of ultraviolet radiation is greater than the energy of infrared radiation.
(R) : The velocity of ultraviolet radiation is greater than the velocity of infrared radiation.
14. (A) : KE of photoelectrons increases with increase in frequency of incident light (v > v0).
(R) : Whenever intensity of light is increased, the magnitude of photocurrent always increases.
15. (A) : A spectral line will not be seen for a 2px –2py transition.
(R) : Only Balmer lines are observed in the visible region.
16. (A) : As the intensity of the incident photon increases, photo current increases.
(R) : Intensity corresponds to number of photons emitted per unit time.
17. (A) : Humphrey series discovered in H–atomic spectra has lowest energy radiations among all series.
(R) : Lowest state for this series is n1 = 6.
18. (A) : Black body emits radiations in a discontinuous manner.
(R) : Black body gives a continuous spectrum.
19. (A) : Bohr’s orbits are called stationary orbits.
(R) : Electrons remain stationary in these orbits for some time.
20. (A) : Hydrogen has only one electron in its orbit, but it produces several spectral lines.
(R) : There are many excited energy levels available.
21. (A) : The five d-orbitals are degenerate in the absence of magnetic field only.
(R) : d–d splitting is the reason for magnetic properties of a speci es.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. Which of the following species will be isoelectronic with Ne?
(1) N–3 (2) Na+
(3) Al+3 (4) F–
2. During Rutherford’s α particle scattering experiment,z
(1) distance of closest approach is of the order of 10–14 m
(2) at the distance of closest approach, the kinetic energy of the α− particle is transformed into electrostatic potential
(3) the α− particles moving towards the nucleus will stop and start retracing their path
(4) maximum number of α particles go straight through the metal foil
3. Which of the following statements are incorrect?
(1) In absorption spectrum, the bright lines are formed on dark background.
(2) In emission spectrum, dark spectral lines are formed on bright background.
(3) Hydrogen spectrum is an example of absorption spectrum.
(4) The intensity of spectral lines depends on population of energy levels, between which electronic transition takes place.
4. A photon of wavelength 4× 10–7 m strikes a metal surface, the work function of the metal being 3.4 × 10–19 J . Select the correct statements.
(1) The energy of photon is 4.97×10 −19 J.
(2) The kinetic energy of the emission is 1.57×10−19 J.
(3) The kinetic energy of the emission is 0.98 eV.
(4) The velocity of photoelectron is 5.87 × 105 ms−1.
5. In H-atom, an electron jumps from 5th orbit to first orbit. Then,
(1) total spectral lines emitted are 10
(2) number of Lyman lines is 4
(3) number of Bracket lines is 1
(4) number of Balmer lines is 5
6. n = 3 3 1 2 n = 2
n = 1
For the above transitions in hydrogen-like atoms, select the incorrect relation(s).
(1) ϑ3 = ϑ1+ ϑ2 (2) 12 3 12 ϑϑ ϑ ϑϑ = +
(3) λ3 = λ1 + λ2 (4) 12 3 12 λλ λ λλ = +
7. According to Bohr’s model o f atom,
(1) the radius of nth orbit ∝ n2
(2) the total energy of electron in n th orbit ∝ 1 n
(3) the angular momentum of electron is integral multiple of h/2π
(4) the magnitude of potential energy of an electron in an orbit is greater than kinetic energy
8. Calculate the λ of electron having kinetic energy of 3 eV
(1) o 50A (2) o 60A
(3) 7.07Å (4) 8.07Å
9. de Broglie wavelength associated with charged particles for electron and proton is (1) o 12.27 A = v λ (2) o 0.286 A = v λ (3) o 0.101 A = v λ (4) 12.27 pm =
10. Which is correct in case of p-orbitals?
(1) They are spherically symmetrical. (2) They have strong directional character. (3) They are three-fold degenerate orbitals.
(4) Their charge density along x, y and z-axes is zero.
11. If an electron is accelerated through V volt, its de Broglie wavelength is
(1) 2. h meV (2) 12.27 Å V (3) 1 2 150
Å V (4)
12. Which of the following is/are correct about the radial probability curves?
(1) 3d z 2 has 3 angular nodes.
(2) The number of angular nodes = l
(3) The number of radial nodes is equal to n- l -1.
(4) The number of maxima in 2s orbital is two.
Numerical Value Questions
13. Total number of protons, electrons, and neutrons in 35 Br80 is __.
14. In Millikan’s experiment, static electric charge on the oil drops was obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18 C, calculate the number of electrons prese nt on it.
15. In the Rutherford scattering experiment, the number of alpha particles scattered at an angle θ = 60° is 36 per minute. The number of alpha particles per minute scatte red at angle θ = 90° is (Assume all other conditions to be identical) ______.
16. Maximum velocity of photoelectrons emitted by a photo meter is 1.8 × 106 m/s.
Taking 11 e 1.810 m =× C/kg for electrons, the stopping potential of emittor is _____
17. Ratio of energies of two photons of 3000 Å to 2400 nm is x : 1. Give the value of x
18. A gas absorbs a photon of 355 nm and emits two wavelengths. If one of the emissions is at 680 nm, the other is at ___ nm.
19. The amount of energy required to remove the electron from a Li+2 ion in its ground state is x times that of the amount of energy required to remove the electron from an H-atom in its ground state. Find ‘ x’.
20. The speed of electron revolving around H-nucleus is 0.547 × 106 m/s. The distance of electron from the nucleus is _____.
21. Radius of nth orbit 2 11 10cm =×× n n ry z . What is ‘y’?
22. Find the number of waves made by a Bohr electron in one complete revolution in it’s 3rd orbit.
23. The atomic masses of gaseous elements A and B are 4 and 20 amu, respectively. The value of the de Broglie wavelength of A gas at −73 °C is ‘M’ times that of the de Broglie wavelength of B at 727 °C. The value of ‘M’ is ____.
24. Find the total number of orbitals present in N shell of an atom.
25. The wave function of an orbital is represented as Ψ 4, 2, 0 . The azimuthal quantum number of that orbital is ____.
26. The ma gnitude of an orbital angular momentum vector of an electron is 6 2 h π . Into how many components will the vector split, if an external field is applied to it?
27. A c ompound of vanadium has magnetic moment of 1.73 BM. It suggests the vanadium oxidation state in the compound is + X. Then, the value of ‘X’ is _____________.
Integer Value Questions
28. The velocity of an electron in a certain Bohr’s orbit of H-atom bears the ratio 1 : 550 to the velocity of light. The quantum number n of the orbit is _____.
29. For which orbit in He + ion, the circumference is 26.5 A 0?
30. The ratio of wave number of the first line of Lyman series in H-atom to wave number of the first line in Balmer series of Li 2+ ion is 3 : x. The value of ‘x’ is ____.
31. How many of the following statements is/ are true?
(I) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.
(II) For a given orbital on the boundary, surface value of |ψ|2 remains constant.
(III) In the ground state, angular momentum of an electron is equal to h 2 π .
(IV) The plot of |ψ|2 versus r gives two radial nodes for 3P orbital.
32. The value of n (i.e., principal quantum number) for the valence shell of palladium is (atomic number = 46) ______.
33. The uncertainty in the position of an electron is equal to its de Broglie wavelength. The minimum percent error in its measurement of velocity under these circumstances will be approximately _________.
34. Calculate the minimum value of uncertainty in the position of a particle whose de Broglie wavelength is π A° and uncertainty in de Broglie wavelength is 0.05 A o . (Express your answer in Ao.)
35. For 3s orbital of hydrogen atom, the normalised wave function is given by
The above mentioned orbital (3s) has two spherical nodes at 2 a 0 and xa 0. Find the value of x/4. [use: 335 = ]
36. A hydrogen-like species is in a spherical symmetrical orbital S 1 , having 3 radial nodes. It gets de-excited to another level S2, having no radial node. Energy of S2 orbital is 2.25 times the energy of 1st Bohr orbit of hydrogen atom. What is the combined total number of nodes (radial + angular) in S 2?
37. Electrons, each with =1, are separated into three groups in an atom (17 th group and 4th period of modern periodic table). The number of electrons in the group which experience the lowest effective nuclear charge than any other element, is____. (Each electron of same group experiences same effective nuclear charge than any other group)
Passage-based Questions
Q (38-39)
A metal is irradiated with a radiation of wavelength 300 nm. Then, photoelectrons are found to have velocity of 3.7×10 5 m/s. When a graph is drawn between KE of photoelectrons and frequency of radiation, straight line is obtained cutting x-axis.
38. The KE of photoelectrons is y × 10−20 J. y is (1) 6.23 (2) 6.4 (3) 6.625 (4) 6.956
39. The value of x-intercept is c ×1014. c is (1) 6 (2) 10.2 (3) 10.9 (4) 8.9
Q (40-41)
The electron in the H-atom resides under ordinary conditions in the first orbit. When energy is supplied, the e moves to higher energy shells, depending upon the amount
of energy absorbed. When an e emits energy, Lyman, Balmer, Paschen, Bracket and Pfund series are formed.
40. If the shortest wavelength of H atom in Lyman series is x, then the longest wavelength in Balmer series of He+ is how many times of x?
(1) 7.2 (2) 0.25 (3) 1.8 (4) 0.55
41. The ratio of number of spectral lines obtained when an electron jumps from 7th to 1st orbit and from 6 th to 3rd orbit is (1) 7 (2) 3.5 (3) 10 (4) 2.5
Q (42-43)
A gas of identical H-like atom has some atoms in the lowest energy level, A, and some atoms in a particular excited state, B, and there are no atoms in any other energy level. The atoms of the gas make transition to higher level by absorbing photon having energy 2.7 eV. Subsequently, the atoms emit radiation of only six different photon energies. Some of the emitted photons have energy 2.7 eV. Some have more and some have less than 2.7 eV.
42. The principal quantum number of initially excited level ‘B’ is __________.
43. Ionisation energy for gas atoms in eV/atom is ____________.
Q (44-45)
According to de Broglie, the wavelength λ of an electron is inversely proportional to the momentum P of a material particle.
44. What is the ratio of the velocity of CH4 and O 2 molecules so that they are associated with de Broglie waves of equal wavelength?
45. An electron is continuously accelerated in a vacuum tube by applying potential difference. If its de Broglie wavelength is decreased by 1%, the change in the kinetic energy of the electron is nearly ____.
Q (46-47)
The behaviour of an electron in an atom is described mathematically by a wave function or orbital. It turns out that each wave function contains three variables, called quantum numbers, which are represented as n, l, and ml. These quantum numbers describe the energy level of an orbital and define the shape and orientation of the region in space where the electron will be found.
46. Radial nodes are maximum in (1) 4s (2) 4p (3) 3d (4) 5f
47. The numerical value ψ4, 3, 0 denotes (1) 3d-orbital (2) 4f-orbital (3) 2s-orbital (4) 4d-orbital
Q (48-49)
The wave functions for any one-electron system, such as hydrogen atom, can be expressed as ψ=k ye −kt,where k and y are constants. For hydrogen atom, we have
48. If the nodes at infinity are not neglected, then what is the total number of radial and angular nodes of 3px orbitals?
(1) 4 (2) 3
(3) 5 (4) infinity
49. Which of the following is the correct representation of plot of radial function on y -axis vs distance from the nucleus in x -axis for an electron of 3p atomic orbital?
(1) r x y r.f
(2) r x y r.f
(3) r x y r.f
(4) r x y r.f
Matrix Matching Questions
50. Match column I (nuclides) with column II (number of nucleons).
Column I Column II
The number of angular nodes is given by the value of angular quantum number and angular node is directional in nature. The total number of nodes is nothing but the sum of radial nodes (n – l – 1) and angular nodes (l).
A. Isotopes I) Same number of neutrons
B. Isobars II) Same isotopic number
C. Isotones III) Same mass number
D. Isodiaphers IV) Same number of protons
(A) (B) (C) (D)
(1) IV III I II
(2) III IV I II
(3) IV III II I
(4) I II III IV
51. Match column-I (electronic configuration) with column-II (rule).
Column I
Column II
A. 1s 2s 2p I) Hund’s only
B. 1s 2s 2p II) Aufbau only
C. 1s 2s 2p III) Pauli only
D. 1s 2s 2p IV) Hund’s and Pauli’ only
E. 1s 2s 2p V) Hund’s and Aufbau only
VI) Aufbau, Hund’s and Pauli
VII) No rule violated
(A) (B) (C) (D) (E)
(1) III V II VI VII
(2) IV I II VI VII
(3) II V II I VI
(4) III I II IV VI
52. Match column I (electron state) with column II (radial probability).
Column I Column II
A. An orbital with two radial nodes and two nonplanar angular nodes I) r y(r)
B. An orbital with one radial node and one angular node
II) Radial wave function has two maxima in Ψ(r) vs r plot
Column I Column II
C. Orbital with = 4, m = −4, n = 5 III) r y2(r)
D. Orbital with maximum z-component of orbital angular momentum as h π and one radial node
IV) Has more energy than other three orbitals of coloum-I
V) Can have minimum number of electrons
(a) (b) (c) (d)
(1) V II,V III,V I,V
(2) III,V III,V I,V III,V
(3) V III,V I,V III,V
(4) III II V,IV I,V
53. Match the entries in Column-I with correctly related quantum number (s) in Column-II.
Column-I Column-II
A. Orbital angular momentum of the electron in a hydrogen-like atomic orbital I) Principal quantum number (n)
B. A hydrogenlike one electron wave function obeying pauli’s exclusion principle
II) Azimuthal quantum number (l)
C. Shape, size and orientation of hydrogen like atomic orbitals
D. Probability function of electron at the nucleus in hydrogen like atom
III) Magnetic quantum number (m)
IV) Electron spin quantum number(s)
(A) (B) (C) (D)
(1) I,III IV I,II I
(2) II I,II,III,IV I,II,III I,II,III
(3)I,III,IV I,IV II,III,IV III,IV
(4) I,II I,III III,IV I,IV
FLASHBACK (Previous JEE Questions)
JEE Main
1. The number of electrons present in all the completely filled subshells having n =4 and 1 s 2 =+ is _______ (2024)
(W here n = principal quantum number and s = spin quantum number)
2. Total number of ions from the following with noble gas configuration is _________
Sr2+(Z = 38), Cs+ (Z = 55), La2+ (Z = 57) Pb2+ (Z = 82), Yb2+ (Z = 70) and Fe2+ (Z = 26) (2024)
3. The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is: (2024)
(1) 1
5,0,0, 2 +
(2) 1 5,0,1, 2 +
(3) 1 5,1,0, 2 +
(4) 1 5,1,1, 2 +
CHAPTER 2: Structure of Atom
4. Number of spectral lines obtained in He + spectra, when an electron makes transition from fifth excited state to first excited state will be _____ (2024)
5. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R). (2023)
(A) : Isotopes of hydrogen have almost same chemical properties, but difference in their rates of reaction.
(R) : Isotopes of hydrogen have different enthalpy of bond dissociation.
In the light of the above statements, choose correct answer from the options given below.
(1) (A) is not correct but (R) is correct.
(2) (A) is correct but (R) is not correct.
(3) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(4) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
6. Which one of the following sets of ions represents a collection of isoelectronic species? (Given : Atomic Number: F:9, Cl:17, Na = 11, Mg = 12, Al= 13, K=19, Ca=20, Sc=21)
(1) N3-, O2-,F-, S2-
(2) Ba2+,Sr2+, K+, Ca2+
(3) K+, Cl–, Ca2+, Sc3+
(4) Li+, Na+, Mg2+, Ca2+
7. The number of following statement/s which is/are incorrect is (2023)
A) Line emission spectra are used to study the electronic structure.
B) The emission spectra of atoms in the gas phase show a continuous spread of wavelength from red to violet.
C) An absorption spectrum is like the photographic negative of an emission spectrum.
D) The element helium was discovered in the sun by spectroscopic method.
8. T he number of incorrect statement/s about the black body from the following is _______
A) Emit or absorb energy in the form of electromagnetic radiation.
B) Frequency distribution of the emitted radiation depends on temperature.
C) At a given temperature, intensity vs frequency curve passes through a maximum value.
D) The maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature. (2023)
9. Electrons in a cathode ray tube have been emitted with a velocity of 1000 ms–1. The number of following statements which is/are true about the emitted radiation is _____. Given: h = 6 × 10 –34 J s, m e = 9 × 10–31kg.
A) The de Broglie wavelength of the electron emitted is 666.67 nm.
B) The characteristic of electrons emitted depends upon the material of the electrodes of the cathode ray tube.
C) The cathode rays start from cathode and move toward anode.
D) The nature of the emitted electrons depends on the nature of the gas present in cathode ray tube. (2023)
10. The electron in the nth orbit of Li2+ is excited to (n+1) orbit using the radiation of energy 1.47 × 10 –17 J (as shown in the diagram). The value of n is
Given: RH = 2.18× 10–18 J n + 1
1.47 × 10–17J n
11. If the radius of the first orbit of hydrogen atom is a0, then de Broglie’s wavelength of electron in 3rd orbit is (2023)
(1) 3πa0 (2) πa0/6 (3) πa0/3 (4) 6πa0
12. The energy of an electron in the first Bohr orbit of hydrogen atom is −2.18 × 10−18J. Its energy in the third Bohr orbit is ______. (2023)
(1) Three times of this value
(2) 1 27 of this value
(3) 1 9 th of this value
(4) One-third of this value
13. Which of the following sets of quantum numbers is not allowed? (2022)
(1) 1 3, 2, 0, 2 l nlms ====+
(2) 1 3, 2, 2, 2 l nlms ===−=+
(3) 1 3, 3, 3, 2 l nlms ===−=−
(4) 1 1 3, 0, 0, 2 nlms ====−
14. The number of radial and angular nodes in 4d orbital are, respectively, (2022)
(1) 1 and 2 (2) 3 and 2
(3) 1 and 0 (4) 2 and 1
15. Outermost electronic configurations of four elements A, B, C, D are given below:
(A) 3s2 (B) 3s2 3p1
(C) 3s2 3p3 (D) 3s2 3p4
The correct order of first ionisation enthalpy for them is (2022)
(1) (A) < (B) < (C) < (D)
(2) (B) < (A) < (D) < (C)
(3) (B) < (D) < (A) < (C)
(4) (B) < (A) < (C) < (D)
16. The energy of one mole of photons of radiation of wavelength 300 nm is (Given: h = 6.63 × 10–34 J s, NA = 6.02 × 1023 mol–1, c =3 × 108 ms–1) (2022)
(1) 235 kJ mol–1 (2) 325 kJ mol–1
(3) 399 kJ mol–1 (4) 435 kJ mol–1
17. The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is (2022)
[Given: The threshold frequency of platinum is 1.3 × 1015 s-1 and h = 6.6 × 10–34Js.]
(1) 3.21 × 10–14J (2) 6.24 × 10–16J
(3) 8.58 × 10–19J (4) 9.76 × 10–20J
JEE Advanced
18. For He+, a transition takes place from the obit of radius 105.8 pm to the orbit of radius 26.45 pm. The wavelength (in nm) of the emitted photon during the transition is
[Use:
Bohr radius, a = 52.9 pm
Rydberg constant, RH = 2.2 × 10–18 J
Planck’s constant, h = 6.6 × 10 –34 Js
Speed of light, c = 3 × 108 ms–1] (2023)
19. Consider the Bohr’s model of a one-electron atom, where the electron moves around the nucleus. In the following, List I contains some quantities for the nth orbit of the atom and List II contains options showing how they depend on n. (2019)
List I List II
I) Radius of the nth orbit (P) ∝ n−2
II) Angular momentum of the electron in the nth orbit (Q) ∝ n–1
III) Kinetic energy of the electron in the nth orbit (R) ∝ n0
IV) Potential energy of the electron in the nth orbit (S) ∝ n1 (T) ∝ n2 (U) ∝ n1/2
Which of the following options has the correct combination considering List I and List II?
(1) (III), (P) (2) (I), (S)
(3) (IV), (Q) (4) (II), (U)
CHAPTER TEST – JEE MAINS
Section - A
This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3), and (4) for its answer, of which ONLY ONE option is correct. Marking scheme: +4 for correct answer, 0 if not attempted, and –1 in all other cases.
1. A certain positive ion A +2 has 22 neutrons and 18 electrons. What is the mass number of most abundant isotope of A?
(1) 42 (2) 38
(3) 40 (4) 44
2. Based on Bohr’s theory, when ‘n’ value increases, choose the incorrect statements.
A. Velocity of electron increases.
B. Total energy of electron increases.
C. Radial distance between successive orbits decreases.
D. Energy difference between successive orbits decreases.
(1) A and D only
(2) B and C only
(3) B and D only
(4) A and C only
3. According to Rutherford’s theory,
(A) : A few α-particles were deflected when passed through gold foil.
(R) : Positively charged α-particles are repelled by nucleus of an atom.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
4. The shortest wavelength of H-atom in the Lyman series is λ1. The longest wavelength in the Balmer series of He + is
(1) 1 9 5 λ (2) 1 5 9 λ (3) 1 27 5 λ (4) 1 36 5 λ
5. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
(A) : In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than the threshold frequency strikes the surface.
(R) : When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place.
In light of the above statements, choose the most appropriate answer from the options given below:
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) Both (A) and (R) are false.
6. The magnitude of product of wavelength and velocity of a particle is 12.57. The product of uncertainty in velocity and uncertainty in position of the particle would be
(1) 10 (2) 4 (3) 2 (4) 1
7. The maximum number of electrons possible in subshell is equal to (1) 2l + 1 (2) 2n2 (3) 2l2 (4) 4l + 2
8. In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic field and electric fields?
(a) n = 1, l = 0, m = 0
(b) n = 2, l = 0, m = 0
(c) n = 2, l = 1, m = 1
(d) n = 3, l = 2, m = 1
(e) n = 3, l = 2, m = 0
(1) (a) and (b) (2) (b) and (c) (3) (c) and (d) (4) (d) and (e)
9. The ratio of energies of two photons of wavelengths of 2000 and 4000 A0
(1) 1 : 4 (2) 4 : 1
(3) 1 : 2 (4) 2 : 1
10. Identify the incorrect statement from the following.
A. Shell for which n > 3, does not have a d-subshell.
B. The different wave functions are characterized by n and l only.
C. For 2pz orbital in the xy plane, the wave function is zero.
D. The number of peaks appearing in the radial distribution graph for orbitals is equal to 2.
E. Quantum mechanics locates the electron with certainty around the nucleus.
(1) A, B, and C (2) C and D
(3) A, B, and E (4) C, D, and E
11. The graph between |ψ| 2 and r (radial distance) is shown below. This represents
r
(1) 3s orbital (2) 2p orbital (3) 1s orbital (4) 2s orbital
12. Identify the incorrect statement from the following.
(1) A circular path around the nucleus in which an electron moves is proposed as Bohr’s orbit.
(2) An orbital is the one-electron wave function (ψ) in an atom.
(3) The existence of Bohr’s orbits is supported by hydrogen spectrum.
(4) Atomic orbital is characterised by the quantum numbers n and l only.
13. Column-I (Configuration) Column II (Rule violated)
A. I) No rule violated
B. II) Hund’s and Aufbau only
C. III) Hund’s only
D. IV) Pauli only V) Hund’s and Pauli only VI) Aufbau only
(A) (B) (C) (D)
(1) V III V V
(2) VI I V IV
(3) IV I I V
(4) II III V IV
14. The correct order of energies of electron in ‘H’ atom in the ground state is
(1) 1s < 2s < 2p < 3s
(2) 1s < 2s = 2p = 3s
(3) 1s > 2s = 2p > 3s
(4) 1s < 2s = 2p < 3s
15. Degenerate orbitals are those orbitals having (1) same angular momentum values and different energies
(2) different angular momentum values and different energies
(3) different angular momentum values and same energies
(4) same angular momentum values and same energies
16. Consider the following statements and identify the correct statements.
(A) XY plane is the nodal plane of pz orbital.
(B) XY plane and YZ plane are the nodal planes for dXZ orbital.
(C) According to quantum mechanical model of atom, electron behaves like a particle but does not show wave nature.
(D) Both 1s orbital and 2s orbital have same shape and size.
(1) A, B, and D only
(2) B, C, and D only
(3) A and B only
(4) A and D only
17. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
(A) : 1s, 2s, 3s atomic orbitals have the same shape but different size.
(R) : For 3P z atomic orbital, the number of nodal planes is 3.
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false .
(4) Both (A) and (R) are false.
18. In which of the following changes both the electrons are removed from sam e orbital?
(1) 2 CuCu2e+−→+
(2) 2 CrCr2e+−→+
(3) 2 GeGe2e+−→+
(4) 2 ZnZn2e+−→+
19. Suppose that a hypothetical atom gives a red, green, blue, and violet line spectrum. Which jump, according to the figure, would give off the red spectral line?
n = 4
n = 3
n = 2
n = 1
(1) 3 → 1 (2) 2 →1 (3)4→1(4)3→2
20. For the Balmer series in the spectrum of H-atom,
23. A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 × 10 14 Hz. The velocity of ejected electron is _____ × 10 5 ms –1 . (Nearest integer)
[Use h = 6.63 × 10–34 Js, me = 9.0 × 10–31 kg]
24. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to 2 2 0 h xma
The value of 10x is _______.
(a 0 is radius of Bohr’s orbit) (Nearest integer) [Given: π = 3.14]
Choose the correct statements among (I) to (IV).
(I) As wavelength decreases, the lines in the series converge.
(II) The integer n1 is equal to 2.
(III) The lines of longest wavelength corresponds to n2 = 3.
(IV) The ionisation energy of helium atom can be calculated.
(1) (II), (III), (IV) (2) (I), (II), (III)
(3) (I), (II), (IV) (4) (I), (III), (IV)
Section - B
Attempt any 5 questions only. If the answer is in decimal, then round off to the nearest integer. Marking scheme: +4 for correct answer, 0 if not attempted, and –1 in all other cases.
21. The azimuthal quantum number for the valence electrons of Ga + ion is ______ (Atomic number of Ga = 31)
22. The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x × 10–12 m. The value of x is _____.Nearest integer)
Given : Mass of electron = 9.1 × 10 –31 kg
Charge on an electron = 1.6 × 10 –19 C
Planck’s constant = 6.63 × 10–34 Js
25. When the excited electrons of H-atom from n = 5 drops to the ground state, the maximum number of emission lines observed is ____.
CHAPTER TEST – JEE ADVANCED
2023 P1 Model Section-A [Single Option correct MCQ's]
1. The IP 1 of the H-atom is 13.6 eV. When irradiated with photons of monochromatic radiation, 12.09 eV is absorbed and gives out induced radiation. Identify the correct statement(s) from the following?
(1) Transition from 3 →1 emits the lowest energetic photon.
(2) The wavelength of one of the induced radiations is 6566.4 A0 .
(3) The wave number of one of the induced radiation is 3R 4 .
(4) The total number of spectral lines possible in the Lyman series is 2.
2. According to Bohr’s atomic theory, which of the following are correct?
(1) Kinetic energy of electron 2 2 Z n ∝
(2) Frequency of revolution of electron in an orbit 2 3 Z n ∝
(3) Coulombic force of attraction of electron ∝ 3 4 Z n
(4) Velocity of electron in an orbit Z n ∝
3. Which among the following is correct about chromium (Cr)?
(1) Its outer electronic configuration is 3d54s1
(2) Total spin of Cr = 3
(3) Magnetic moment of Cr=48 BM
(4) Spin multiplicity of 3d-subshell of Cr = 6
4. A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 A°. What is the uncertainty involved in the measurement of its velocity?
(1) 5 × 106 m s–1
(2) 4.5 × 106 m s–1
(3) 5.7 × 106 m s–1
(4) 7 × 107 m s–1
5. The graph between momentum ‘ p’ and de Broglie wavelength λ of photon is (1) (2) (3) (4)
6. Match Column I (orbital) with Column II (nodal properties).
Column I Orbital Column II Nodal properties
(A) 5s (p) Number of radial nodes = 1
(B) 3d yz (q) Number of angular nodes = 0
(C) 4d xy (r) Sum of radial and angular nodes = 3
(D) 3p (s) Number of radial nodes = 0
(1) A → p , B → q, C → r, D → s
(2) A → s, B → r, C → q, D → p
(3) A → q, B → s, C → r, D → p
(4) A → s, B → q, C → r, D → p
7. In the radial probability distribution curve for the 2s orbital of the hydrogen atom, the minor maximum, the node, and the major maximum occur at what distances from the nucleus, respectively?
(1) 1.1 A°, 0.53 A°, 2.6 A°
(2) 0.53 A°, 1.1 A°, 0.53 A°
(3) 2.6 A°, 1.1 A°, 0.53 A°
(4) 0.53 A°, 2.11 A°, 2.6 A°
Section-B
[Numerical Value Questions]
8 The number of spectral lines that are possible when electrons in 7 th shell in different hydrogen atoms return to the 2nd shell is ____.
9. The minimum energy required to remove the electron from a metal surface is 7.50 × 10–19 J. What will be the stopping potential required when this metal surface is exposed to U.V. light of l = 375 Ao ?
10. How many orbitals are required for the electric configuration of iron?
11. The atomic mass of an element is 19. The second shell of its atom contains 7 electrons. The number of neutrons in its nucleus is 2x. The value of x is _______.
12. For 3s orbital of hydrogen atom, the normalised wave function is as follows
15. Match column-I(orbital) with columnII(radial probability).
Column I Column II
(A) 4s (P)
Distance from nucleus 4πr 2 R 2 (B) 3p (Q)
If distance between the radial nodes is d, calculate the value of 0 d 1.73a .
13. The energy of the election in the first orbit of He+ is –871.6 × 10–20 J. The energy of the electron in the first orbit of hydrogen is –x × 10–20 J. Then, the value of ‘x’ is _____.
Section-C
[Matrix Value Questions]
14. Match column I with column II.
Column I
Column II
(P) Decreasing order of masses (I) e – > p > n
(Q) Decreasing order of e/m values (II) p > e– > n
(R) Decreasing order of de Broglie’s wavelength with same velocities (III) n > p > e–
(S) Decreasing order of uncertainty in velocity when ∆ x is the same (IV) n > e– > p (V) e – > n > p
(1) P-III, Q-I, R-I, S-I
(2) P-I, Q-II, R-III, S-IV
(3) P-IV, Q-III, R-V, S-I
(4) P-V, Q-III, R-II, S-V
Distance from nucleus 4πr 2 R 2 (C) 3d (R)
Distance from nucleus 4πr 2 R 2
(D) 2p (S)
Distance from nucleus 4πr 2 R 2
(1) A-Q, B-R, C-S, D-P
(2) A-P, B-S, C-Q, D-R
(3) A-S, B-R, C-Q, D-P
(4) A-R, B-Q, C-P, D-S
16. Match column I (ratio of wavelengths) with column II (ratio).
Column I (Ratio of wavelengths)
Column II (Ratio)
(A) Series limit of Lyman series of H-atom to series limit of Balmer series of He+ (I) 1 : 1
(B) Series limit of Balmer series of H-atom to series limit of Paschen series of H-atom (II) 4 : 9
Column I (Ratio of wavelengths)
Column II (Ratio)
(C) Series limit of Balmer series of H-atom to series limit of Paschen series of Li+2 (III) 4 : 3
(D) H– a line of Lyman series of H-atom to series limit of Lyman series of H-atom (IV) 1 : 4 (V) 4 : 1
(1) A-I, B-II, C-V, D-III
(2) A-I, B-III, C-V, D-II
(3) A-I, B-IV, C-V, D-III
(4) A-I, B-II, C-V, D-IV
ANSWER KEY
JEE Main
Level I
5
15
(41) 1 (42) 6 (43) 300 (44) 4
CHAPTER 2: Structure of Atom
17. Match column I (quantum numbers) with column II (orbital).
3.1 Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
3.2 Redox reactions in Terms of Electron Transfer Reactions
3.3 Oxidation Number
3.4 Redox Reactions and Electrode Process
3.1 CLASSICAL IDEA OF REDOX REACTIONS
oxidation
■ The process in which oxygen is added to an element or a compound.
S+O2→ SO2
Na2SO3 + H2O2 → Na2SO4 + H2O
■ Removal of hydrogen C2H6O → C2H4O + H2
■ Addition of an electronegative element
Mg+Cl2 → MgCl2
■ removal of an electro positive element is also known as oxidation.
K4[Fe(CN)6] → K3[Fe(CN)6]
■ Loss of electron or electrons is oxidation.i.e oxidation is de-electronation.
Zn → Zn+2 + 2e–
2S2O32– → S4O62– + 2e–
■ A substance that undergoes reduction acts as an oxidant. An oxidant is an electron acceptor.
■ The strongest oxidant is F₂ . The other important oxidizing agents are: KMnO₄,. O₃ , MnO₂, K₂Cr₂O₇, Cl₂, O₂, NaOCl, H₂O₂, and CuO., etc.
Reduction
■ Removal of oxygen is reduction . CuO + C → Cu + CO
■ Addition of hydrogen is called reduction.
C2H4 + H2 → C2H6
CHAPTER 3: Redox Reaction
■ Removal of an electronegative element 2FeCl 3 + SnCl2 → 2FeCl2 + SnCl4
■ addition of an electropositive element.
CuCl2 + Cu → Cu2Cl2
■ Gain of electron or electrons is reduction, reduction is electronation.
Al3+ + 3e– → Al
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
■ A substance which undergoes oxidation acts as reductant. A reductant is an electron donor. An oxidant is an electron acceptor.
■ Alkali metals are strong reducing agents. Li is the strongest reducing agent. The other important reducing agents are: Na, Mg, Al, C, H₂, S, H₂S, KI, Na₂C₂O₄, LiAlH₄., etc.
■ Zn + CuSO4 → Cu +ZnSO4
In the above reaction, zinc is reductant and reduces copper ion acts as an oxidant.
■ Cu + 2AgNO3 → 2Ag + Cu(NO3)2
copper acts as a reductant and silver ion acts as oxidant
Try yourself:
1. In the reaction, Na2O2+SO2 → Na2SO4, which one is oxidant?
TEST YOURSELF
Ans: 2O2Na
1. MnO2+4HCl → MnCl2+Cl2+2H2O, MnO2 acts as (1) oxidant (2) reductant (3) both 1 and 2 (4) can’t be predicted
2. In the reaction Mg + N2 → Mg 3N2, the correct option is (1) magnesium is reduced (2) magnesium is oxidised (3) nitrogen is oxidised (4) both magnesium and nitrogen is oxidised
3. In C + H2O → CO + H2, H2O acts as (1) oxidising agent (2) reducing agent (3) both (1) and (2) (4) dehydrating agent
4. In a reaction between zinc and iodine in which zinc iodide is formed, what is being oxidised? (1) Zinc ions. (2) Iodide ions. (3) Zinc atom. (4) Iodine.
5. In the reaction 2Ag+2H2SO4→ Ag2SO4+2H2O+SO2, sulphuric acid acts as (1) oxidising agent (2) reducing agent (3) oxidising as well as reducing agent (4) catalyst
6 In the reaction 2Al+N2→ 2 AlN, Al is (1) reduced (2) oxidised (3) oxidising agent (4) catalyst
7. In which of the following reactions, hydrogen acts as an oxidising agent? (1) With iodine to give hydrogen iodide. (2) With lithium to give lithium hydride. . (3) With nitrogen to give ammonia. (4) With sulphur to give hydrogen sulphide.
8. Reduction does not involve, the (1) removal of an electronegative element (2) addition of an electropositive element (3) removal of an electropositive element (4) decrease in oxidation number
3.2 REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS
■ In the following reactions sodium is oxidised due to the addition of oxygen or chlorine or sulphur.
■ Simultaneously, chlorine, oxygen and sulphur are reduced because, the electropositive element is added. loss of 2e–loss of 2e–loss of 2e–
2Na(s) + Cl2(g)
2Na+ Cl– (s)
2Na(s) + O2(g)
2Na(s) + S(s) (Na+)2O2–(s) (Na+)2S2–(s) gain of 2e–gain of 2e–gain of 2e–
■ For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and the other the gain of electrons.
■ Sodium involves loss of electron and chlorine involves gain of electron 2 Na(s) → 2 Na+(g) + 2e–Cl2(g) + 2e– → 2 Cl–(g)
■ Each of the above steps is a half reaction, Sum of the half reactions gives the overall reaction : 2Na(s) + Cl2 (g) → 2NaCl (s)
■ Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
■ Na acts as a reducing agent by donating electrons, while Cl₂, O₂, and S act as oxidizing agents by accepting electrons .
3.2.1 Competative Electron Transfer Reactions.
■ An electrochemical cell converts chemical energy into electrical energy.
■ Zinc in copper sulfate solution turns red due to copper deposition: Zn(s) + Cu²+(aq) → Zn²+(aq) + Cu(s).
■ A galvanic (voltaic) cell generates electricity through a redox reaction.
■ So, chemical energy is converted into electrical energy in a galvanic cell. A galvanic cell can be reversed to function as an electrolytic ce ll.
Zinc anode
Copper cathode
Zinc sulphate solution Coper sulphate solution
Electrochemical series or Activity series
■ The arrangement of electrodes in the order of increasing standard reduction potentials is called electrochemical series. This is also called activity series.
3.3 OXIDATION NUMBER
■ Oxidation number is the residual charge an atom has or a ppears to have in a molecule.
■ It is often used interchangeably with oxidation state.
■ Stock notation of oxidation numbers is based on electronegativity.
■ An atom's oxidation number can be positive, negative, zero, or even fractional depending on its environment
The rules useful for the oxidation number are stated below:
■ The oxidation number of an atoms in its elementary state (e.g., H₂, Cl₂, S₈, Au, He, Ne) is zero.
■ Alkali metals always have +1, and alkaline earth metals always have +2 oxidation states in compounds.
■ Fluorine always has an oxidation state of –1 due to its highest electronegativity.
■ Hydrogen has an oxidation state of +1 in molecular hydrides and –1 in metal hydrides (e.g., LiH, NaH, CaH₂).
■ Oxygen is usually –2, but in peroxides (H₂O₂, Na₂O₂), it is –1, and in superoxides (KO₂, RbO₂), it is –1/2. It has +1 in O₂F₂ and +2 in OF₂.
■ The sum of oxidation numbers in a neutral molecule is zero; in an ion, it equals the charge of the ion.
■ The oxidation number of a monoatomic ion is equal to its charge (e.g., Cl –1 = –1, O-²= –2, P³- = –3, Fe²+ = +2).
■ Halogens (Cl₂, Br₂, I₂) have positive oxidation numbers in oxoacids and oxoanions.
■ The maximum oxidation number of an element equals its group number in the periodic table.
■ Metals generally have positive oxidation numbers, while nonmetals can have positive or negative oxidation states.
■ The highest oxidation state known is +8 (e.g., OsO₄), and the lowest is –4 (e.g., CH₄).
■ If an element appears multiple times in a molecule, its oxidation number may be calculated as an average.
■ Nitrogen exhibits oxidation states from –3 to +5 in its compounds.in Table 3.1.
Stock Notation
■ The oxidation number is expressed using Roman numerals in parentheses after the metal symbol in the molecular formula. Cuprous chloride and cupric chloride are written as Cu(I) Cl and Cu(II)Cl₂
Solved Examples
1 2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g) is a redox reaction. Identify the species oxidised. What acts as an oxidant? Sol. ()()() 2s2ss2
In this redox reaction, Cu is reduced from +1 to 0, and S is oxidized from –2 to +4. Cu₂O acts as an oxidant, helping S in Cu₂S increase its oxidation number. S in Cu₂S acts as a reductant, helping Cu in Cu₂S and Cu₂O decrease its oxidation number.
3.3.1 Oxidation and Reduction in Terms of Oxidation Number
■ Oxidation increases the oxidation number.
■ Reduction decreases the oxidation number.
■ Oxidizing agent (oxidant) increases the oxidation number of another element and itself gets reduced.
Examples of different oxidation number of atoms of same element
Nitrogen N3H N N || – H N 0, 0, –1 –1/3
Chlorine CaOCl2 Cl–Ca–OCl –1, +1 0
Iodine Kl 3 KI+KI2 –1, 0 ,0 –1/3
Sulphur FeS2 FeS+S –2, 0 –1
Sulphur H2S2O3 S=S(OH)2= 0 –2, +6 +2
Sulphur H2S4O6 HOSO2–S2–SO2OH +5, 0, 0, +5 +5/2
■ Reducing agent (reductant) decreases the oxidation number of another element and itself gets oxidized
■ In this reaction, oxidising agent is Cu2O and reducing agent is Cu 2S.
TEST YOURSELF
1. In which of the following compounds does sulphur exhibit least oxidation state? (1) S2Cl2 (2) H2S2O7 (3) SOCl2 (4) Na2S2O3
2. The oxidation state(s) of Cl in CaOCl 2 (bleaching power) is/are (1) +1 only (2) –1 only (3) +1and –1 (4) –1 and –1
3. The oxidation state of nickel in [Ni(CO) 4] is (1) +2 (2) +4 (3) +5 (4) zero
4. In S4O62–, the oxidation numbers of four sulphur atoms are (1) +4, 0, 0, and +6 (2) +4, –2, –2 and +6 (3) +5, 0, 0, and +5 (4) +5, –2, –2 and +5
5. A compound contains atoms of three elements A, B, and C. If the oxidation number of A is +2, B is +5, and that of C is –2, the possible formula of the compound is (1) A3(BC4)2 (2) A3(B4C)2 (3) ABC2 (4) A2(BC3)2
6. Oxidation numbers of Cr in CrO 5 and K2Cr2O7, respectively, are (1) 5 and 6 (2) 10 and 6 (3) 6 and 6 (4) 6 and 5
7. Oxidation state of iron in Fe 3O4 cannot be (1) +3 (2) +2 (3) both (1) and (2) (4) +5
8. The sum of oxidation numbers of nitrogen in ammonium nitrite (NH 4NO2) is (1) +5 (2) 0 (3) 1 (4) –3
9. The oxidation number of P is + 3 in (1) H3PO3 (2) H3PO4 (3) HPO3 (4) H2P2O7
10. Oxidation number of C in CH 2Cl2 is (1) +2 (2) +4 (3) –4 (4) 0
Answer Key (1) 1
3.3.2 Type of redox reactions
■ Redox reactions involve electron transfer. Types of redox reactions are
Combination Reactions
■ Reactions in which elements combine to form compounds are called as chemical combination reactions.
2Mg+O2→ 2MgO
■ Combustion of graphite, formation of alumina from its elements, rusting of iron, oxidation of sulphur, etc., are other familiar chemical combination reactions.
C + O2 → CO2
4Al + 3O2 → 2Al2O3
Decomposition Reaction
■ A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances. AB→A+B.
■ Most decomposition reactions require an input of energy in the form of heat, light, or electricity.
MnO2 32 2KClO2KCl3O →+
2HgO → 2Hg + O2
2(I) 2HO2HO2(s)2(g) ∆ →+ (S) 2NaH2NaH2(s)2(g) ∆ →+
Displacement Reactions
■ Metal displacement: An active metal displaces, relatively less active metal from its salt solution.
■ By metal displacement reactions pure metals are obtained from their compounds.
TiCl4 + 2 Mg → Ti + 2MgCl2
■ Displacement reactions are useful in predicting the reactivity of metals based on electrode potentials.
■ The electron releasing tendency of the metals, Zn, Ag, Cu is in the order Zn > Cu > Ag.
■ Non-metal displacement: All alkali metals and some alkaline earth metals (Ca, Sr, Ba) displace hydrogen from cold water.
Example: 2Na(s)+2H2O(I) → NaOH(aq)+H2(g)
■ Less active metals, such as Fe, Mg, and Al, react with steam to produce H2 gas.
2Fe3HOFeO3H ∆ +→+
2 232
■ Metals like Cd and Sn do not react with cold water and steam but liberate H 2 gas with dilute acids.
Sn + 2HCl → SnCl2+H2
■ Fluorine is the most powerful oxidising agent. It oxidises water to oxygen and potassium bromide to bromine. Chlorine, bromine, and iodine cannot oxidise water.
2F2 + 2H2O → O2 + 4HF
F2 + 2KBr → Br2 + 2KF
■ A lighter halogen can displace heavier halogen from its halide compound.
Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l)
Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s)
■ The above reactions are called non-metal displacement reactions.
■ As F2 is the strongest oxidant, it cannot be prepared from F – by chemical oxidation, F 2 can be prepared from F– by electrolysis only.
Disproportionation Reaction
■ Disproportionation reactions occur when the same element undergoes both oxidation and reduction simultaneously.
■ These reactions are possible if the element has at least three oxidation states. The reactant's oxidation number must be intermediate, not at an extreme oxidation state.
■ Some Examples of Disproportionation Reaction:
P4+3NaOH+3H2O → PH3+3NaH2PO2
6KOH + 3Br2 → 5KBr + KBrO3 + 3H2O
2S2Cl2 + 2H2O → 4HCl + SO2 + 3S
2NO2+2NaOH → NaNO2 + NaNO3 + H2O
S8+12NaOH → 4Na2S+2Na2S2O3+6H2O
Cl2(g)+ 2NaOH(g)→ NaClO(aq)+ NaCl(aq)+ H2O(l)
Comproportionation Reactions
■ It is a chemical reaction in which two reactants, containing same element in different oxidation state, will form a product in which the elements reach the same oxidation state. Some examples of comproportionation reaction:
■ AgSO4+Ag → Ag2SO4
■ C + CO2 →2CO
■ CuCl2 +Cu →Cu2Cl2
3.3.3 Balancing of Redox Reaction
■ Redox reactions are balanced by following methods.
Solved Example
2. Reaction between hydrogen sulphide and sulphurdioxide gives sulphur. Which type of redox reaction is this?
Sol. The reaction between hydrogen sulphide and sulphur dioxide is given as,
2H2S + SO2 → 3S + 2H2O
Oxidation state of S in H2S is –2 and is oxidised to ‘0’. Oxidation state of S in SO2 is +4 and is reduced to ‘0’. In the product, sulphur shows zero oxidation number, which is in between –2 and +4 of the sulphur in the reactants. Therefore, this reaction is an example of comproportionation.
Try yourself:
2. 4KClO3 → 3KClO4+KCl What type of reaction is this?
TEST YOURSELF
Ans: Disproportionation
1. Which of the following is only a redox reaction but not a disproportionation reaction? (1) 4H3PO3 → 3H3PO4+PH3 (2) 2H2O2 → 2H2O+O2 (3) P4+3NaOH+3H2O → 3NaH2PO2+PH3 (4) P4+8SOCl2→ 4PCl3+2S2Cl2+4SO2
2. Which one of the following does not undergo disproportionation in alkaline medium? (1) F2 (2) S (3) Cl2 (4) P 4
3. Which of the following is not an example of a decomposition reaction? (1) 2HgO → 2Hg+O2 (2) 2H2O → 2H2+O2 (3) 2KClO3→ 2KCl+3O2 (4) CH4+2O2→ CO2+2H2O
4. Which of the following is a decomposition reaction? (1) 2HgO → 2Hg+O2 (2) CH4+2O2→ CO2+2H2O (3) S+O2→ SO2 (4) Cl2+2KBr → 2KCl+Br2
5. 2CuI → Cu+CuI2, the reaction is (1) disproportionation (2) neutralisation (3) oxidation (4) reduction
6. Which of the following is not a comproportionation reaction? (1) Ag+2 + Ag → Ag+1 (2) SO2+H2S → S (3) IO3–+I–→I 2 (4) H2O2 →H2O + O2
Answer Key (1) 4 (2) 1 (3) 4 (4) 1 (5) 1 (6) 4
Oxidation Number Method
■ Electron transfer method is used to balance ionic and molecular reactions.
Steps for balancing:
1. Assign oxidation numbers to all atoms.
2. Identify atoms undergoing oxidation and reduction.
3. Calculate the increase and decrease in oxidation number.
4. If multiple atoms are involved, find the total change by multiplying accordingly.
5. Equalize the total oxidation and reduction change using suitable multiples.
6. Balance all atoms except H and O first.
7. Balance O by adding H₂O molecules as needed.
8. For ionic equations, balance O with H₂O and H with H + .
9. In basic medium, add equal OH – to both sides, forming H₂O.
10. Remove duplications and simplify coefficients to whole numbers.
Half Reaction Method
■ Half-reaction method (ion-electron method) is used to balance ionic equations, excluding spectator ions.
■ Steps for balancing:
Write the ionic equation and assign oxidation numbers.
Identify atoms undergoing oxidation and reduction.
Split into oxidation and reduction half-reactions and balance each separately:
¾ Balance atoms except H and O first.
¾ Balance O by adding H₂O.
¾ Balance H: In acidic medium, add H + .
¾ In basic medium, balance hydrogens by using OH – and H₂O.
¾ Remove duplications and simplify coefficients.
Add electrons to balance charges.
Equalize electrons in both half-reaction s by multiplying appropriately.
Combine both reactions to get a balanced final equation with no free electrons.
■ The following are some examples: Permanganate oxidises sulphite to sulphate in acidic solutions.
The ionic skeleton equation is written as H 222 43 4 MnOSOMnSO + −−+− +→+
■ White phosphorous reacts with aqueous caustic soda to give hypophosphite and phosphine.
The ionic skeleton equation is OH 4 322 PPHHPO →+
Writing oxidation numbers 2 03111
4322 PPHHPO −+++ →+
Locating atoms, undergoing change in oxidation numbers
031
4322 PPHHPO −+ →+
Divide the reaction into two half reactions and balance in basic medium, separately
■ Oxidation half-Reaction: P 4→ H2PO2–
Step 1: Balance the phosphorous atoms
P 4→4 H2PO2–
Step 2: Balance the oxygen atoms
P 4 + 8H2O →4 H2PO2–
Step 3: Balance the hydrogen atoms
P4+8H2O+8OH–→4 H2PO2–+8H++8(OH-)
Step 4: Balance the charge
P4+8OH- →4 H2PO2–+4e–.....(a)
■ Reduction Half reaction: P 4→ PH 3
Step 1: Balance the phosphorous atoms
P 4→ 4PH 3
Step 2: Balance the hydrogen atoms
P 4 +12H++12OH- → 4PH3+12OH–12H2O+ P4 → 4PH3+12OH–
Step 3: Balance the charge
P 4 + 12H2O + 12e– → 4PH 3 + 12OH–
■ Equalising the electrons and adding the two half reactions, eq. (a) × 3 + eq. (b) × 1, we get
3P24HO12HPO12e
P12HO12e4PH12OH
4222 42 3 4 2322
4P12OH12HO4PH12HPO +→+ ++→+ ++→+
P4+3OH–+3H2O → PH3+3H2PO2–
This is the balanced equation.
■ Acetylene is oxidised by permanganate in acidic solutions to liberate carbon dioxide. The ionic skeleton equation is written as
Writing oxidation numbers,
Locate atoms, undergoing change in oxidation numbers
Divide the reaction into two half reactions and balance in acidic medium, separately.
■ Oxidation Half-Reaction: C2H2→ CO2
Step 1: Balance the carbon atoms.
C2H2→ 2CO2
Step 2: Balance the oxygen atoms.
C2H2 + 4H2O →2 CO2
Step 3: Balance the hydrogen atoms.
C2H2 + 4H2O → 2CO2 + 10H+
Step4: Balance the charge.
C2H2 + 4H2O → 2CO2 + 10H+ + 10e–---(a)
■ Reduction Half reaction: MnO4–→ Mn2+
Step 1: Balance the oxygen atoms
MnO4–→ Mn2++4H2O
Step 2: Balance the hydrogen atoms
MnO4–+8H++5e– → Mn2++4H2O.....(b)
■ Equalising the electrons and adding the two half reactions. eq. (a) × 1 + eq. (b) × 2, we get
3.3.4 Equivalent Weight
■ Equivalent weight is the weight of a substance that reacts with or displaces 1.008 g of hydrogen, 8 g of oxygen, 35.5 g of chlorine, or 108 g of silver.
■ Chemical substances do not necessarily react in a 1:1 ratio of moles, masses, or volumes.
CHAPTER 3: Redox Reaction
■ Reactants always combine in equal ratios of their equivalents, and products form in the same equivalent ratio.
■ Equivalent weight of an element = Atomic weight / Valency.
■ The same element can have different equivalent weights due to variable valency (e.g., carbon in CO and CO₂ has equivalent weights of 6 and 3, respectively). The equivalent weights of some elements are given in Table 1.11.
Equivalent weights of some elements
■ Equivalent weight of an ion is the number of grams of the ion which can be discharged by the loss or gain of one mole of electrons.
■ Equivalent weight is the ratio of ionic weight and charge magnitude of ion.
Equivalent weight of ion = ionic weight charge
■ Equivalent weights of important cations and important anions are respectively given in table below .
■ Equivalent weight of an acid is the number of grams of the acid which gives one mole of protons in water. It is the ratio of molecular weight of acid and basicity of acid.
■ (Number of H+ ions furnished by one molecule of an acid is called basicity).
Equivalent weight of acid = Molar mass of acid Basicity of acid
Equivalent weights of some cations
some anions
■ Equivalent weight of a base:
■ The number of OH– ions furnished by one molecule of base is called acidity.
Eq.Wt of base = Molecularweightofbase Acidityofbase
CHAPTER 3: Redox Reaction
■ Equivalent weight of a salt:
Eq.wt of salt
Formulaweightofsalt
Totalnumberofpositive(ornegative)chargeunits =
■ Eq.wt of salt = Eq.wt of cation + Eq. wt of anion
■ Equivalent weights of important acids, bases, and salts are given in Table 1.14, Table 1.15, and Table 1.16, respectively.
Equivalent Weights of Oxidant and Reductant
■ The equivalent weight of an oxidant is the weight that gains one mole of electrons.
■ The equivalent weight of a reductant is the weight that loses one mole of electrons.
■ Equivalent weight of oxidant or reduc tant = Molecular weight No.of electrons gained or lost per molecule
Equivalent weight of oxidant or reductant=
Equivalent weights of some acids
Equivalent weights of some bases
Numberofelectronsgainedorlostpermolecule
Equivalent weights of some salts
The following examples explain calculation of equivalent weights.
■ EW of a substance undergoing disproportion:
a) EW of Cl2 with cold and dil. NaOH
Cl2+OH–→ Cl–+ClO–
ECl2 = E1+E2 = MM 71 22 +=
b) EW of Cl2 with hot and conc. NaOH:
3Cl2+6OH–→ 5Cl–+ClO3–+3H2O
ECl2 = E1 +E2 = 12 MMM3M371
■ EW of a substance undergoing comporportionation:
a) EW of sulphur in the following reaction: S2–+SO2→ S, s12 MM3M332 EEE 24 2444 × =+=+===
■ EW of a substance undergoing thermal decomposition:
a) 2KClO3→ 2KCl+3O2, n- factor=6
b) PCl5→ PCl3+Cl2, n–factor=2
■ EW of a substance in which two elements either both oxidised (or) reduced:
a) FeC2O4 in acidic medium:
Fe2+ oxidises to Fe3+ and C2O42– to CO2
FeC2O4→ Fe3+→ +2CO2, n-factors = 1+2 = 3
b) [Fe(CN)6]4–→ Fe3++NO3–+CO32–, n-factor = 1+12+48 =61
3.3.5 Redox Reaction as the Basis for Titrations
■ Analysis of chemical substances is generally of two types: qualitative analysis and quantitative analysis.
■ Quantitative analysis of solutions in volumes is useful in the determination of concentration, volume of the solution, and also amount of a solute.
■ Volumetric analysis is method of measuring the volume of a known solution required to bring about the completion of the reaction with a measured volume of unknown solution. The method is also known as titrimetric analysis and is termed as titration or titrimetry.
Solved Examples
3. How many electrons and protons are present in the balanced half equation? NO2 → NO
Sol. 3 2 2 NONO; ++ →
22 NO2HNOHO −++→+
22 NO2HeNOHO −+−++→+
Number of electrons = 1;
Number of protons = 2
4 How many moles of acidified permanganate are required to oxidise one mole of ferrous oxalate?.
Sol. Ferrous oxalate FeC2O4 is a dual reductant
e 23 l FeFe++ →
2e 2 24 CO2CO2 → One mole of FeC2O4 is involved in 3 electron change.
5e 2 4 MnOMn−+ →
One mole of permanganate in acidic medium gains 5 moles of electrons.
5 moles of FeC2O4 = 3 moles of MnO4-, 1 mole of FeC2O4 = ?
Number of moles of permanganate that can be ox idised by one mole of ferrous oxalate = × 1 = 0.6
Try yourself:
3. How many electrons are transfered in the reaction, IO 3–+5I–+6H+→3I 2+3H 2O?
Ans: Five electrons
TEST YOURSELF
1. In the redox reaction, XKMnO4+YNH3→ KNO3+MnO2+KOH+H2O; X and Y are (1) X = 4, Y = 6 (2) X = 3, Y= 8 (3) X = 8, Y = 6 (4) X = 8, Y = 3
2. Find the value of n in the following equation, if balanced. 2 233 27 2 CrO14HnFe2CrnFe7HO −++++ ++→++ (1) 2 (2) 3 (3) 7 (4) 6
3. Consider the redox reaction: 22 424 22 MnOCOHMnCOHO −−++ ++→++ The correct coefficients of the reactants, 2 MnO,CO424 and H+ for the balanced reaction, respectively, are (1) 2, 5, 16 (2) 16, 3, 12 (3) 15, 16, 12 (4) 2, 16, 5
4. 2 23 42 MnOFeHMnFeHO. ++++ ++→++ The stoichiometric coefficient associated with H + in the stoichiometric equation is (1) 6 (2) 7 (3) 8 (4) 10
5. How many of OH ions are appearing in the given balanced equation? () 2 22 2 3 OH 4 CrOHHOHOCrO ++ → (1) 2 (2) 3 (3) 4 (4) 6
6. Consider the following reaction: 22 424 22 z xMnOyCOzHxMn2yCOHO 2 −−++ ++→++ The values of x, y, and z in the reaction are, respectively, (1) 2, 5, and 8 (2) 2, 5, and 16 (3) 5, 2, and 8 (4) 5, 2, and 16
7. 222 43 4 . MnOSOHMnSO −−++− ++→+
The number of H+ ions involved is (1) 2 (2) 6 (3) 8 (4) 16
8. The value of n in 2 42 MnO8HeMn4HO −++ ++→+ n is (1) 5 (2) 4 (3) 3 (4) 2
9. The number of moles of acidified KMnO 4 required that can oxidise 0.5 mole of Mohr salt is (1) 1 (2) 2 (3) 0.1 (4) 0.2
10. The number of electrons involved in the reduction of Cr 2O72- ion in acidic solution to Cr 3+is (1) 3 (2) 4 (3) 2 (4) 6
11. In the equation, 223 NOHONO2Hne −+−+→++ , n stands for (1) 4 (2) 1 (3) 2 (4) 3
12. The number of electrons lost or gained during the change Fe+H 2O → Fe3O4+H2 is (1) 2 (2) 4 (3) 6 (4) 8
13. Molecular weight of orthophosphoric acid is M. Its equivalent weight is (1) 3M (2) M (3) M 2 (4) M 3
14. 0.5 g of a metal, on oxidation, gave 0.79 g of its oxide. The equivalent mass of the metal is (1) 10 (2) 14 (3) 20 (4) 40
15. The equivalent weight of Bayer’s reagent, is (alkaline KMnO 4) (1) 31.6 (2) 52.6 (3) 79 (4) 158
16. Molecular weight of KMnO4 is ‘M’. In a reaction KMnO4 is reduced to K2MnO4. The equivalent weight of KMnO4 is (1) M (2) 2 Μ (3) M 3 (4) 5 Μ
17. When ferrous sulphate acts as reductant, its equiv alent weight is (1) twice that of its molecular weight (2) equal to its molecular weight (3) one-half of its molecular weight (4) one-third of its molecular weight
18. 2H2O → 4e– + O2 + 4H+.
The equivalent weight of oxygen is (1) 32 (2) 16 (3) 8 (4) 4
19. In acidic medium dichromate ion oxidises ferrous ion to ferric ion. If the gram-molecular weight of potassium dichromate is 294 g, its equivalent weight is (1) 294 (2) 147 (3) 49 (4) 24.5
20. The equivalent weight of hypo in the reaction [M = molecular weight] 2Na2S2O3 + I2 → 2NaI + Na2S4O6 is (1) M (2) M 2 (3) M 3 (4) M 4
21. The equivalent weight of CuSO 4 when it is converted to Cu 2I2 is [M=mol.wt] (1) M 1 (2) M 2 (3) M 3 (4) 2 M
22. The atomic weight of a metal (M) is 27 and its equivalent weight is 9. The formula of its chloride will be (1) MCl (2) MCl9 (3) M3Cl4 (4) MCl3
23. Molecular weight of Mohr’s salt is 392. Its equivalent weight when it is oxidised by KMnO 4 in acidic medium is (1) 392 (2) 196 (3) 130.6 (4) 78.5
24. H3PO4 + 2KOH → K2HPO4 + 2H2O
Based on the above reaction, equivalent weight of H 3PO4 is (1) 196 (2) 98 (3) 49 (4) 32.67
25. In the standardisation of Na2S2O3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is (1) (molecular weight)/2 (2) (molecular weight)/6 (3) (molecular weight)/3 (4) same as molecular weight
■ The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titrations.
I2 + 2e– → 2I– (reduction)
2I– → I2 + 2e– (oxidation)
These are divided into two types.
Iodimetric titrations:
■ These are the titrations in which free iodine is used. As it is difficult to prepare the solution of iodine (volatile and less soluble in water), it is dissolved in potassium iodide solution.
KI + I2 → KI 3 (Potassium tri-iodide)
■ This solution is first standardised before use. With the standard solution of I 2, substances, such as sulphite, thiosulphate, arsenite, etc., are estimated.
■ Following examples explain iodimetric titrations.
■ I2–Na2S2O3 .5H2O. This is iodimetry titration, since standard solution of iodine is used.
Reaction: 2Na2S2O3 + I2 → Na2S4O6 + 2Nal
Indicator: Freshly prepared starch solution.
End point: Disappearance of blue colour
■ As2O3 – I2 titration
■ Iodometry: Iodine is liberated during chemical reactions.
■ Following examples explain iodometric titrations
■ CuSO4 – Na2S2O3 . 5H2O
2CuSO4 + 4KI → Cu2I2 + K2SO4 + I2
2Na2S2O3 + I2 → Na2S4O6 + 2Nal
Indicator: Freshly prepared starch solution
End point: Disappearance of blue colour or violet
■ K2Cr2O7 – Na2S2O3 . 5H2O :
Reaction : K2Cr2O7 + 7H2SO4 + 6KI →
4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2
2Na2S2O3 + I2 → 2NaI + Na2S4O6
■ Indicator: Freshly prepared starch solution.
End point: Disappearance of blue colour.
Solved Examples
5. Determine the volume of M/8 KMnO 4 solution required to react completely with 25.0 cm 3 of M/4 FeSO4 solution in acidic medium.
Sol. The balanced ionic equation for the reaction is
MnO4–+5Fe2++8H+ → Mn2++5Fe3++4H2O
From the balanced equation, it is evident that 1 mole of KMnO 4 ≡ 5 moles of FeSO4
Applying molarity equation to the balanced redox equation, we have, 11 22 44 12 MV MV (KMnO)(FeSO) nn = 1V1 125 8145 × =× × or 3 1 1258 10.0 45 ×× == × V cm
Thus, the volume of M/8 KMnO4 solution required = 10.0 mL.
Try yourself:
4. A solution of ferrous oxalate has been prepared by dissolving 3.6 g L–1. Calculate the volume of 0.01 M KMnO4 solution required for complete oxidation of 100 mL of ferrous oxalate solution in acidic medium.
Ans: 150 mL
TEST YOURSELF
1. In the titration of KMnO 4 versus Mohr's salt, the indicator used is (1) KMnO4 (self) (2) diphenyl amine (3) starch (4) methyl red
2. When KMnO 4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because (1) CO2 is formed as the product (2) reaction is exothermic (3) MnO4 catalyses the reaction (4) Mn2+ acts as an autocatalyst
3. 50 mL of 0.02 M KMnO4 oxidises 50 mL of potassium iodide solution in faint alkaline medium. Molarity of potassium iodide solution is (1) 0.1 M (2) 0.01 M (3) 0.0166 M (4) 0.06 M
4. The weight of I2 (MW=254) required to completely oxidise 1 mole of Na2S2O3 according to the following equation (unbalanced), is I2+Na2S2O3→ NaI+Na2S4O6 (1) 127g (2) 254 g (3) 80 g (4) 98 g
5. The amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl (1) gets oxidised by oxalic acid to chlorine (2) furnishes H+ ions in addition to those from oxalic acid (3) reduces permanganate to Mn 2+. (4) oxidises oxalic acid to carbon dioxide and water
6. Consider the titration of potassium dichromate solution with acidified Mohr’s salt solution using dimethylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is (1) 3 (2) 4 (3) 5 (4) 6
7. The number of moles of ferrous oxalate oxidised by one mole of KMnO 4 in acidic medium is (1) 5 2 (2) 2 5 (3) 3 5 (4) 5 3
8. 0.8 M FeSO4 solution requires 160 mL of 0.2 M Al2(Cr2O7)3 in acidic medium. Calculate the volume of FeSO4 consumed. (1) 480 mL (2) 240 mL (3) 720 mL (4) 40 mL
Answer Key
3.4 REDOX REACTIONS AND ELECTRODE PROCESSES
Redox Reaction and Daniel Cell
■ When a zinc rod is dipped in CuSO₄ solution, Zn is oxidized to Zn² +, and Cu²+ is reduced to Cu, with direct electron transfer and heat evolution.
■ To enable indirect electron transfer, Zn metal and CuSO₄ solution are separated into different beakers.
■ Each beaker contains a metal strip (electrode) dipped in its metal salt solution (Zn in ZnSO₄, Cu in CuSO₄), forming redox couples: Zn² +/Zn and Cu²+/Cu.
Daniell Cell Setup
■ The two beakers are connected by a salt bridge (KCl/NH₄NO₃ in agar gel) to maintain electric contact without mixing solutions.
■ Zinc and copper rods are connected by a metallic wire, allowing electron flow when the switch is on.
Observations when the circuit is closed:
1. Electrons flow from Zn to Cu² + through the wire, producing electricity.
2. Ions migrate through the salt bridge to balance the charge.
3. A potential difference exists between Zn and Cu rods, known as electrode potential.
■ Electrode Potential and Standard Electrode Potential (E°)
■ Electrode potential depends on the tendency of a species to remain in oxidized/reduced form.
■ Standard Electrode Potential (E°) is measured under standard conditions:
■ Concentration = 1 M, Pressure = 1 atm, Temperat ure = 298 K
■ The Standard Hydrogen Electrode (SHE) is assigned E° = 0.00 V.
■ A negative E° means the redox couple is a stronger reducing agent than H +/H₂.
■ A positive E° means the redox couple is a weaker reducing agent.
■ Electrochemical Series and Reactivity
■ Electrodes arranged by increasing E° form the electrochemical series (or activity series).
■ Fluorine (F₂) has the highest E°, making it the strongest oxidizing agent.
■ Lithium (Li) has the lowest E°, making Li metal the strongest reducing agent in aqueous solutions.
■ It may be seen that as we go from top to bottom in table , the standard electrode potential increase.
■ Electrochemical Series and Its Applications
■ Tendency for Reduction and Reducing Power
■ The tendency to undergo reduction decreases down the electrochemical series.
■ The reducing power of elements also decreases in the same order.
■ Applications of Electrochemical Series
■ Predicts spontaneity of reactions and metal-salt solution interactions.
■ Helps estimate the oxidizing and reducing abilities of elements.
■ Chemical Reactivity of Metals
■ Metals with highly negative standard reduction potential lose electrons easily, forming cations, making them highly reactive.
■ Reactivity decreases down the activity series.
■ Relative Strength of Oxidizing and Reducing Agents
■ Reducing power decreases down the series.
Table Comparison between the two types of cells
4. Na(aq)eNa(s) +−+→
– 2.71
5. 2 Mg(aq)2eMg(s) +−+→ Mg2+/Mg – 2.37
6. 3 Al(aq)3eAl(s) +−+→ Al3+/Al – 1.66
7. 2 Zn(aq)2eZn(s) +−+→ Zn2+/Zn – 0.76
8. 2 Fe(aq)2eFe(s) +−+→ Fe2+/Fe – 0.44
9. 2 Co(aq)2eCo(s) +−+→ Co2+/Co – 0.28
10. 2 Ni(aq)2eNi(s) +−+→ Ni2+/Ni – 0.25
11. 2 Sn(aq)2eSn(s) +−+→ Sn2+/Sn – 0.14
12. 2 2H(aq)2eH(g) +−+→
H+/H2 0.00
13. 2 Cu(aq)2eCu(s) +−+→ Cu2+/Cu + 0.34
14. 2 I(s)2e2I(aq) +→
I2/I– + 0.54
15. Ag(aq)eAg(s) +−+→ Ag+/Ag + 0.80
16. 2 2 Hg(aq)2e2Hg(l) +−+→ Hg+/Hg + 0.85
17. 2 Br(l)2e2Br(aq) +→
18. 2 Cl(g)2e2Cl(aq) +→
19. 3 22 O(g)2H(aq)2eO(g)HO(l) +− ++→+
20. 2 F(g)2e2F(aq) +→
Br2/Br– + 1.07
Cl2/Cl– + 1.36
O3/O2 + 2.07
F2/F– +2.87
■ Oxidizing power increases with higher reduction potential.
■ Oxidizing power a reduction potential
■ Reducing power a oxidation potential
■ Fluorine (highest reduction potential) is the strongest oxidizing agent.
■ Lithium (lowest reduction potential) is the strongest reducing agent in aqueous medium.
■ A metal can reduce metal ions below it in the series.
■ A metal ion can oxidize metals above it.
■ A metal can reduce non-metals below it in the series.
■ For storing a salt solution a metal container with smaller SRP value should not be used. Instead a metal container with higher SRP can be used.
Examples:
■ FeSO4 solution () 2 o Fe/Fe E0.44v + =− cannot be stored in Zn container () 2 o Zn/Zn E0.76v + =− but it can be stored in a Cu container () 2 o Cu/Cu E0.34v +=+
■ A non-metal can oxidise all metal atoms present above it in the activity series. It can also oxidise non-metal anions present above it.
3.4.4 Displacement of Hydrogen from Dilute Acids by Metals
■ A metal with negative SRP can displace H + from dilute acids to liberate H 2 gas.
■ A metal with positive SRP like Cu, Ag, Pt etc., cannot displace H + from dilute acids to liberate H2
TEST YOURSELF
1. The standard reduction potentials for the following half-reactions are given against each at 298 K.
()() 2 aq s Zn2eZn;0.762 +−+=−
3 aq s Cr3eCr;0.74V +−+=−
()()
()2g 2H2eH;0.0V +−+= () 32 maq aq FeeFe;0.77V +−++=+
Which is the strongest oxidising agent?
(1) Zn(s) (2) Cr(s) (3) H2(g) (4) Fe3+(aq)
2. The standard electrode potentials M/M E + ο of four metals A, B, C and D are –1.2 V, 0.6 V, 0.85 V, and –0.76 V, respectively. The sequence of deposition of metals on applying potential is
(1) A,B,C,D (2) B,D,C,A (3) C,B,D,A (4) D,A,B,C
3. Which of the following has least tendency to liberate H 2 from mineral acids (1) Cu (2) Mn (3) Ni (4) Zn
4. Which of the following is true for electrochemical cell made with SHE and Cu electrode?
(Given 2 Cu/Cu E0.36V + ο=+ )
(1) H2 is cathode and Cu is anode (2) H2 is anode and Cu is cathode
(3) Reduction occurs at H2 electrode
(4) Oxidation occurs at Cu electrode
5. Regarding the standard hydrogen electrode, the incorrect statement is (1) H2 gas is bubbled at 1 bar
(2) 2 M HCl is electrolyte
(3) SRP is arbitrarily taken as Zero. (4) it is a primary reference electrode.
6. Zn gives H2 gas with H2SO4 and HCl but not with HNO 3 because (1) Zn acts as oxidising agent when reacts with HNO 3
(2) HNO3 is weaker acid than H2SO4 and HCl.
(3) in electrochemical series Zn is above H 2
(4) NO3 is reduced in preference to H 3O+
7. The standard reduction potentials of Cu +2, Ag+, Hg+2, and Mg+2 are +0.34 V, +0.80 V + 0.79 V, and –2.37 V respectively. With increasing voltage, the sequence of deposition of metals on the cathode from a molten mixture containing all those ions is
(1) Ag, Hg, Mg, Cu
(3) Ag, Hg, Cu, Mg
8. The chemical reaction
()()()() s2g aq s 2AgClH2HCl2Ag +→+
(2) Cu, Hg, Ag, Mg
(4) Cu, Hg, Mg, Ag
taking place in a galvanic cell is represented by the notation.
(1)
(2)
(3)
(4)
()()()()() s2g aq PtH1bar1MKClAgClAgss ,
()()()()() s2g aq aqs PtH1bar1MHC1 , lMAgAg +
()()()()() s2g aq PtH1bar1MHClAgClAgss ,
()()()()() s2g aqs s PtH1bar1MHClAgAgCl ,
Answer Key
# Exercises
JEE MAIN
Level I
Classical idea of redox reactions
Single Option Correct MCQs
1. When zinc is added to Cu SO 4 solution, copper is precipitated, it is so because of (1) reduction of Zn (2) reduction of Cu2+ (3) hydrolysis of CuSO4 (4) reduction of SO4 2–
2. PbS + 4H2O2 → PbSO4 + 4H2O. In this reaction PbS undergoes (1) oxidation (2) reduction (3) both (4) none of these
3. The largest oxidation number exhibited by an element depends on outer electronic configuration. With which of the following outer electronic configuration the element will exhibit the largest oxidation number?
(1) 3d14s2 (2) 3d34s2
(3) 3d54s2 (4) 3d54s1
4. Which of the following is an oxidation and reduction reaction?
5. Which one of the following reactions does not involve either oxidation or reduction?
(1) VO2+ → V2O3 (2) Na → Na+
(3) CrO42- → Cr2O72- (4) Zn2+ → Zn
6. The gain of oxygen is known as (1) oxidation (2) reduction (3) halogenation (4)chlorination
7. In a reaction between zinc and iodine, in which zinc iodide is formed, which one is being oxidised?
(1) Zinc ions (2) Iodide ions
(3) Zinc atom (4) Iodine
8. Which of the following is an oxidation and reduction reaction?
(1) BaO2 + H2SO4 → BaSO4 + H2O2
(2) N2O5+H2O2→2HNO3
(3) AgNO3+KI → AgI + KNO3
(4) SnCl2 + HgCl2 → SnCl4 + Hg
9. In which reaction is hydrogen acting as an oxidising agent ?
(1) With iodine to give hydrogen iodide (2) With lithium to give lithium hydride (3) With nitrogen to give ammonia (4) With sulphur to give hydrogen sulphide
10. Which is not an oxidising agent?
(1) KClO3 (2) O2
(3) C6H12O6 (4) K2Cr2O7
11. Reduction does not involve in (1) removal of an electronegative element (2) addition of an electropositive element (3) removal of an electropositive element (4) decrease in oxidation number
Numerical Value Questions
12. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
Choose the correct statements that justifies the above equations
i) Bromine is a stronger oxidant than iodine.
ii) Bromine is a weaker oxidant than iodine.
iii) Thiosulphate undergoes oxidation by bromine
iv) Thiosulphate is reduced by iodine
v) Bromine undergoes oxidation
vi) iodine undergoes reduction
13. Which of the following mainly act as oxidising agents?
O3, SO2, TeO2, PbO2
14. The Oxidation state of Nickel in [Ni (CO)4] is_____
Redox reactions in terms of electron transfer reactions
Single Option Correct MCQs
15. In the reaction: C 2O 4 2– + MnO 4 – + H + → Mn2+ + CO2 + H2O, the oxidizing agent is (1) MnO4– (2) C2O4–(3) Mn2+ (4) H+
16. Which is strongest oxidizing agent? (1) O3 (2) O2 (3) Cl2 (4) F2
17. In acidic solution, the reaction MnO 4 → Mn+2 can occur through (1) oxidation by loss of 3 electrons (2) reduction by gain of 3 electrons (3) oxidation by loss of 5 electrons (4) reduction by gain of 5 electrons
18. In the reaction 2Cu 2 O+ Cu 2 S → 6Cu + SO2 the species which acts as oxidant and reductant, respectively?
(1) Cu+, O–2 (2) Cu+, S–2 (3) Cu, S–2 (4) Cu, O–2
19. The oxyacid which acts both as oxidizing and reducing agent is (1) HNO2 (2) H3PO4 (3) H2SO4 (4) HCIO4
20. When copper is added to a solution of silver nitrate, silver is precipitated. This is due to the reduction of (1) silver (2) copper (3) silver ion (4) cuprous ion
CHAPTER 3: Redox Reaction
21. MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O,
MnO2 will act as (1) oxidant (2) reductant
(3) both (4) can’t predicted
22. The correct option for a redox couple is (1) both are oxidised forms involving same element.
(2) both are reduced forms involving same element.
(3) both the reduced and oxidized forms involve same element.
(4) cathode and anode together.
23. In the following reaction, which is the species being oxidized?
(1) Fe3+ (2) I–(3) I2 (4) Fe2+
24. Which of the following can act as both oxidizing and reducing agent?
i) SO2
ii) HNO3
iii) HNO2
iv) HClO4
(1) SO2, HNO2
(2) HNO2, HNO3
(3) HNO2, HClO4
(4) HClO4, SO2
25. In which of the following reactions hydrogen is acting as an oxidizing agent?
(1) With sodium to give sodium hydride
(2) With iodine to give hydrogen iodide
(3) With nitrogen to give ammonia
(4) With sulphur to give hydrogen sulphide.
26. Which of the following species can function both as oxidising as well as reducing agent?
(1) Cl– (2) ClO4–
(3) ClO– (4) MnO4–
27. Which of the following is a set of reducing agents?
(1) HNO3, Fe2+, F2
(2) F–, Cl–, MnO4–
(3) I–, Na, Fe2+
(4) Cr2O72-, Cr2O42-, Na
28. Which of the following change requires a reducing agent?
(1) CrO42– → Cr2O72–(2) BrO3 → BrO–(3) NH3 → NF3
(4) Al(OH)3 → Al (OH)4 –
29. The compound that cannot act both as oxidising and reducing agent is:
(1) H2O2
(2) HNO2
(3) H3PO4
(4) H2SO3
30. Loss of electron is termed as (1) combustion
(2) oxidation
(3) reduction
(4) neutralisation
31. In equation NO2– + H2O → NO3 – + 2H+ + ne–, the value of n in balanced equation is (1) 1
(2) 0
(3) 2
(4) 3
Numerical Value Questions
32. How many compounds can act as oxidizing agents
KClO3, O3, C6H12O6, K2Cr2O7, KMnO4
33. How many electrons are involved in the following half reaction in acidic medium?
Cr2O7-2 → 2Cr+3
34. The total number of electrons transferred from 3 molecules of reductant to oxidant in the following redox process is _________?
Oxidation Number
Single Option Correct MCQs
35. Highest oxidation number that is exhibited by fluorine is
(1) –1 (2) 0
(3) +1 (4) +7
36. Oxidation state of ‘S’ in S 8 molecule is (1) 0
(2) +2
(3) +4
(4) +6
37. Oxidation state of Fe in K 4[Fe(CN)6]
(1) +6
(2) +4
(3) +2
(4) +5
38. Oxidation number and valency of oxygen in OF2 are
(1) +1, 2
(2) +2, 2
(3) +1, 1
(4) +2, 1
39. In which of the following the oxidation state of chlorine is +5?
(1) HClO4
(2) HClO3
(3) HClO2
(4) HCl
40. All elements commonly exhibit an oxidation state of (1) +1 (2) –1
(3) Zero (4) +2
41. The element that always exhibits a negative oxidation state in its compounds is (1) Nitrogen
(2) Oxygen
(3) Fluorine
(4) Chlorine
42. Oxidation number of iron in Na2[Fe(CN)5NO] is (1) +2 (2) +3
(3) +1 (4) 0
43. The oxidation number of phosphorus in sodium hypophosphite is
(1) +3 (2) +2
(3) +1 (4) –1
44. Which of the following reactions does not involve the change in oxidation state of metal?
(1) VO–2 → V2O3 (2) Na → Na+
(3) CrO42- → Cr2O72- (4) Zn2+ → Zn
45. Oxidation state of oxygen in potassium superoxide is
(1) –1/2 (2) –1
(3) –2 (4) 0
46. Fluorine do not undergo disproportionation, because
(1) Fluorine is always exhibit -1 oxidation state
(2) Fluorine exhibit only two oxidation numbers
(3) Fluorine exhibit three oxidation numbers
(4) Fluorine always exhibit +1 oxidation state
47. 2H 2O 2 → 2H 2O + O 2. This reaction is an example for (1) decomposition
(2) combination
(3) disproportionation
(4) Both 1 and 2
48. K+Cl → KCl. This reaction is an example of (1) oxidation (2) reduction (3) a redox reaction (4) none of these
3: Redox Reaction
49. In the reaction P4 + 3OH– + 3H2O →3H2PO2–+ PH3; phosphorus is undergoing (1) oxidation (2) reduction
(3) disproportionation
(4) hydrolysis
50. Which of the following is not a redox reaction?
(1) 2BaO + O2 → 2BaO2
(2) BaO2 + H2SO4 → BaSO4 + H2O2
(3) 2KClO3 → 2KCl + 3O2
(4) SO2 + 2H2S → 2H2O + 3S
51. This reaction is:
2CuI → Cu + CuI2 (1) disproportionation reaction
(2) neutralisation reaction
(3) oxidation reaction
(4) reduction reaction
52. In a reaction between zinc and iodine, in which zinc iodide is formed, the reductant is
(1) Zinc ions
(2) Iodide ions
(3) Zinc atom
(4) Iodine
53. The reaction 3ClO–(aq) → ClO3–(aq) + 2Cl–(aq) is an example of (1) oxidation reaction (2) reduction reaction
54. In the reaction: Cl2 + H2S →2HCl + S, the oxidation number of S changes from (1) 0 to 2
(2) 2 to zero
(3) - 2 to zero
(4) – 2 to – 1
55. Molecular weight of orthophosphoric acid is M. Its equivalent weight is (1) 3M (2) M
(3) M/2 (4) M/3
56. Which of the following acid has the same molecular weight and equivalent weight?
(1) H3PO2 (2) H3PO3 (3) H3PO4 (4) H2SO4
57. The equivalent mass of CaCO 3 is (1) 100 (2) 50
(3) 33.3 (4) 25
58. 0.5 g of a metal on oxidation gave 0.79 g of its oxide. The equivalent mass of the metal is
(1) 10 (2) 14 (3) 20 (4) 40
59. Which one among the following is not a fixed quantity?
(1) Atomic weight of an element
(2) Equivalent weight of an element
(3) Molecular weight of a compound
(4) Formula weight of a substance
60. The equivalent weight of Bayer’s reagent is (alkaline KMnO4)
(1) 31.6 (2) 52.6
(3) 79 (4) 158
61. Molecular weight of KMnO 4 is “M”. In a reaction KMnO 4 is reduced to K 2MnO 4 The equivalent weight of KMnO 4 is (1) M (2) M/2
(3) M/3 (4) M/5
62. When Ferrous sulphate acts as reductant, its equivalent weight is (1) twice that of its molecular weight (2) equal to its molecular weight (3) one-half of its molecular weight (4) one-third of its molecular weight
63. 2H 2O → 4e – + O 2 + 4H +. The equivalent weight of oxygen is (1) 32 (2) 16 (3) 8 (4) 4
64. In acidic medium dichromate ion oxidises ferrous ion to ferric ion. If the grammolecular weight of potassium dichromate is 294 g, its equivalent weight is (1) 294 (2) 147 (3) 49 (4) 24.5
65. The equivalent weight of hypo in the reaction [M = molecular weight]
2Na2S2O3 + I2 → 2NaI + Na2S4O6 is (1) M (2) M/2
(3) M/3 (4) M/4
66. The minimum oxidation state that nitrogen exhibits is (1) -2 (2) -3 (3) -4 (4) -5
67. In the conversion of K2Cr2O7 to K2CrO4, the oxidation number of the following changes (1) K (2) Cr (3) O (4) None
68. The number of electrons involved in the half-reaction;
81. Number of H + ions involved in the half reaction, H 23 NONO + → is (1) 4 (2) 1 (3) 3 (4) 2
82. 20ml of 0.1M KMnO 4 in acidic medium can completely oxidise 40 ml of Mohr’s salt solution. Molarity of Mohr’s salt solution is
(1) 0.25 M
(2) 0.5 M
(3) 1 M
(4) 0.125 M
83. The strength of an aqueous solution of I 2 can be determined by titrating the solution with standard solution of (1) Oxilic acid
(2) Sodium thiosulphate
(3) Sodium hydroxide
(4) Mohr’s salt
84. The role of starch in iodometric titrations is that, (1) it acts an oxidant (2) it is reducing agent (3) it acts as indicator (4) it is a medium
85. In the standardization of Na2S 2O 3 using K 2 Cr 2 O 7 by iodometry, the equivalent weight of K2Cr2O7 is (1) (molecular weight)/2 (2) (molecular weight)/6 (3) (molecular weight)/3
(4) same as molecular weight
86. The role of permanganate in permanganometric titration’s is (1) it acts an reductant (2) it is a reducing agent
(3) it acts as an self-indicator and oxidant (4) it is a dehydrating agent
87. Titration of oxalic acid Vs KMnO4 is a type of
(1) precipitation titration
(2) redox titration
(3) acid base titration
(4) complexometric titration
88. Titration is carried out by using KMnO4 (taken in Burette) and hot acidified oxalic acid in acidic media. The indicator used to identify the end point is
(1) methyl orange
(2) methyl red
(3) phenolphthalein
(4) KMnO4 is the self-inductor
89. 0.2 moles of KMnO 4 canoxidize_____ moles of oxalic acid in acidic medium
(1) 0.3 (2) 0.4
(3) 0.2 (4) 0.5
90. What are the modes of exchange of gases (O 2 and CO 2) at the level of alveoli and tissues respectively?
(1) Active transport, active transport
(2) Active transport, diffusion
(3) Diffusion, active transport
(4) Diffusion, diffusion
91. In the titration of KMnO4 verses Mohr salt, the indicator used is
(1) KMnO4 (self)
(2) diphenyl amine
(3) starch
(4) methyl red
Numerical Value Questions
92. Number of moles of H+ ions required by 1 mole of MnO4 to oxidise oxalate ion to CO2 is ____.
93. Total number of species from the following which can undergo disproportionation reaction is ______?
H2O2, ClO3 , P4, Cl2, Ag, Cu+1, F2, NO2, K+
94. 2MnO 4 + bI + cH 2O → xI 2 + yMnO 2 + zOH
If the above equation is balanced with integer coefficients, the value of z is
95. In the given combination reaction, what is the value of x+y? [x and y indicate oxidation states] y x 4222
CH2OCO2HO +→+
96. Find the number of moles of K 2 Cr 2 O 7 reduced by three mole of Sn 2+ ions.
97. One mole of Fe 2+ reacts with x moles of KMnO4 is ____, in the balanced equation.
98. In the titration of KMnO 4 and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is_______
99. The oxidation number of Cr in CrO5 is +x then x is________
100. The sum of oxidation number of chlorine in bleaching powder (CaOCl 2) is_____
101. The oxidation state of sulphur in Marshalls acid (H2S2O8) is ___
Redox reactions and Electrode Processes
Single Option Correct MCQs
102. Some statements about the modified Daniel cell are given below:
(A) Cu vessel acts as a cathode.
(B) Porous pot helps to maintain electric neutrality.
(C) ZnSO4 solution is electrolyte.
(D) Anode is a negative electrode
Among these statements, correct statements are
(1) A and B only (2) A and C only
(3) B, C and D only (4) A, B and D only
103. The metal that cannot displace hydrogen from dilute HCl is:
(1) Al (2) Fe (3) Cu (4) Zn
104. The standard electrode potentials of Zn, Ag and Cu are – 0.76, 0.80 and 0.34 volts respectively; then
(1) Ag can oxidize Zn and Cu
(2) Ag can reduce Zn2+ and Cu2+
(3) Zn can reduce Ag+ and Cu2+
(4) Cu can oxidize Zn and Ag
105. Consider the following reduction process:
Zn2+ + 2e- → Zn(s); E° = -0.76V
Ca2+ + 2e- → Ca(s); E° = -2.87V
Mg2+ + 2e- → Mg(s); E° = -2.36V
Ni2+ + 2e- → Ni|(s); E° = -0.25V
The reducing power of the metals increases in the order:
CHAPTER 3: Redox Reaction
(1) Ca < Zn < Mg < Ni
(2) Ni < Zn < Mg < Ca
(3) Zn < Mg < Ni < Ca
(4) Ca < Mg < Zn < Ni
106. Which reaction is not feasible
(1) 2KI + Br2 → 2KBr+I2
(2) 2KBr + I2 → 2KI + Br2
(3) 2KBr + Cl2 → 2KCl + Br2
(4) 2H2O + 2F2 → 4HF + O2
107. In salt bridge normally KCl is used because (1) It is a strong electrolyte
(2) It is a good conductor of electricity (3) K+ and Cl– ions have nearly same ionic mobility (4) It is an ionic compound
108. Using the d ata given below find out the strongest reducing agent 23 27 0 CrO/Cr E1.33V −+ = 2
(1) Cl– (2) Cr (3) Cr3+ (4) Mn2+
109. Standard reduction potential is most positive for
11. Which of the following is not a metal displacement reaction?
(1) Zn+CuSO4 →ZnSO4+Cu
(2) 2Na+H2O→2NaOH+H2
(3) TiCl4+2Mg→Ti+2MgCl2
(4) 3Fe3O4+8Al→4Al2O3+9Fe
12. Which one of the following generally gets displaced by more electropositive metals in non-metal displacement reactions?
(1) H2 (2) N2
(3) F2 (4) Cl2
13. 2CuI→Cu+CuI2, the reaction is (1) disproportionation (2) neutralisation (3) oxidation (4) reduction
14. Which one of the following reaction involve comproportionation?
(1) 2CuI+Cu+CuI2
(2) 2H2S+SO2→3S+2H2O
(3) 2H2O2→O2+2H2O (4) 2KClO3→3O2+2KCl
Numerical Value Questions
15. In the given combination reaction, what is the value of x + y ? [x and y indicades oxidation state]
4222 CH+2OCO+2HO → xy
16. In the given displacement reaction, what is the change in the oxidation number of oxygen?
23 CrO+2Al2Cr+AlO23 ∆ →
Balancing of Redox Reactions
Single Option Correct MCQs
17. For the redox reaction, –2–+ 2+ 424 22 MnO+CO+HMn+CO+HO → the correct coefficients of the reactants for the balanced reaction are respectively –2–+ 424 : MnO,CO,H
18. In balancing the half reaction HCHO → HCOO– taking place in alkaline medium, the number of electrons(e ) to be added are
(1) 2 to the reactants side
(2) 2 to the products side
(3) 3 to the reactants side (4) 3 to the products side
Numerical Value Questions
19. What is the value of ‘ n ’ in the following half equation: ()–– 2– –42 4 CrOH+OHCrO+HO+ne →
Redox Reactions as the Basis for Titration
Single Option Correct MCQs
20. x mmol of KIO3 reacts completely with y mmol of KI to give I2 quantitively If z mmol of hypo are required for complete titration against this I 2, then, which statement is incorrect?
(1) z = 6x (2) 6y = 5z
(3) 5x = y (4) x + y = 2z
Numerical Value Questions
21. In iodimetric titration (where hypo is directly treated with iodine in the presence of starch), What is the average oxidation state of sulphur in the Sulphur containing product.
Multiple Concept Questions
Single Option Correct MCQs
22. The equivalent weight of phosphoric acid (H3PO4) in the reaction, NaOH+H3PO4→NaH4PO4+H2O is (1) 59 (2) 49 (3) 25 (4) 98
23. A compound contains X, Y, Z atoms. The oxidation states of X, Y and Z are +2, +2, -2 respectively. The probable formula of the compound is
(1) Y2(XZ3)2 (2) XYZ2 (3) X3(Y4Z)2 (4) X3(YZ4)2
24. Which of the following acid has the same molecular weight and equivalent weight.
(1) H3PO2 (2) H3PO3 (3) H3PO4 (4) H2SO4
25. The oxidation state of Fe in K4[Fe(CN)6] is (1) +2 (2) +6 (3) +3 (4) +4
Numerical Value Questions
26. Ionisible H atom s in H 3 PO 4 is x and in H3PO2 is y. The ratio of x : y is_____.
Level III
1. Which of the following is not a disproportionation reaction?
2. An example of comproportionation reaction (1) 2H2O+SO2→3S+2H2O (2) 3Cl2+6NaOH→5NaCl+NaClO3+3H2O (3) P4→PH3+H2PO–2 (4) 2CuBr→CuBr2+Cu
3. Among P4(S), S8(S), Cl2(g), and NO2(g), how many species can undergo disproportionation in the alkaline medium?
4. How many are disproportionation reactions? (1) 2Cu+→Cu2++Cu° (2) 2-+4 422 3MnO+4H2MnO+MnO+2HO → (3) 42422 2KMnOKMnO+MnO+O ∆ → (4) -2+ + 422 2MnO+3Mn+2HO5MnO+4H →
5. – –2 +2 424 22 2MnO+ CO+ H Mn+ CO+ HO +→ bcxyz if the above equation is balanced with integer co-efficients. The value of c is __.
6. In a titration, certain amount of H2O 2 is treated with y moles of KMnO 4 in acidic medium. The left out KMnO4 when treated with X+ in basic medium oxidises X+1 to X+6 and 0.2 M, X L of X+ was consumed. The mole of given H2O2 solution is:
(1) 5 yx (2) 5 2 yx
(3) () 5 10 yx (4) () 5 5 yx
7. 25 mL solution containing mixture of Na2CO3 and NaOH required 20 mL of 1N HCl for the phenolphthalein end point and
in a separate titration 25 mL of the same mixture required 25 mL of same HCl for methyl orange end point. Ratio of moles of Na2CO3 and NaOH taken in initial mixture and convert that into decimal form.
8. Concentrated nitric acid is act as oxidising agent. It oxidises I2 to HIO3, the n-factor of the I2 is x, and it oxidises S8 to H2SO4, then n-factor of S8 is y. What is the value of x+y
(1) 38 (2) 48
(3) 58 (4) 26
9. In the reaction the stoichiometry coefficients of 2-272 CrO,NO and H+ respectively are 2--+3+272 CrO+NO+HCrNO32 + +HO →
(1) 1, 3, 8 (2) 1, 4, 8 (3) 1, 3, 12 (4) 1, 5, 12
10. A solution contains mixture of H 2 SO 4 , H 2C 2O 4. 20 mL of this solution requires 40 mL of M/10 NaOH for neutralisation and 20 mL of N/10 KMnO4 for oxidation. The molarity of H2C2O4,H2SO4 are (1) 0.1, 0.4 (2) 0.1, 0.05 (3) 0.05, 0.1 (4) 0.05, 0.05
11. The oxidation states of transition metal atoms in K 2 Cr 2 O 7 , KMnO 4 and K 2 FeO 4 respectively are x, y and z. The sum of x, y and z is ____.
(1) 18 (2) 19
(3) 20 (4) 22
12. KMnO4 reacts with oxalic acid according to the equation - -2+ +2 424 22 2MnO+5CO+16H2Mn+10CO+8HO →
Here 20 mL of 0.1 M KMnO4 is equivalent to
(1) 120 mL of 0.25 M H2C2O4
(2) 150 mL of 0.10 M H2C2O4
(3) 25 mL of 0.20 M H2C2O4
(4) 50 mL of 0.20 M H2C2O4
13. How many of OH are present in the balanced equation? () OH- -2 22 24 3 CrOH+HOHO+CrO →
(1) 2 (2) 3
(3) 4 (4) 6
14. In order to oxidise a mixture having 1 mol each of FeC 2 O 4 ,Fe 2 (C 2 O 4 ) 3 ,FeSO 4 , how many mole of KMnO4 is required in acidic medium _______
15. Calculate number of mole of Na2S2O3 that will react with I2 obtained when 1 mol of K2Cr2O7 reacts with excess of KI in acidic medium as per the reaction
H2SO4 + K2Cr2O7 + KI → K2SO4 + Cr2(SO4)3 + H2O+I2
16. The number of electrons involved in the reaction of permanganate to manganese dioxide in acidic medium is ___.
17. In the balanced chemical reaction, ––+ 3 22 IO+I+HHO+I; abCd →
The sum of the coefficients of a, b, c, and d is 10+y. Then the value of y is _______.
18. How many mL of 0.3 M K2Cr2O7 required for complete oxidation of 5 mL of 0.2 M SnC2O4 solution in acidic medium. (K2Cr2O7 oxidises SnC2O4 to Sn4++CO2)
19. In basic medium, 2 4 CrO oxidises 3 2 2 SO to form 2 SO46 and itself changes into Cr(OH)4 The volume of 0.154 M 2 4 CrO required to react with 40 mL of 0.25 M 2 SO23 is ____ mL. (Rounded off to the nearest integer)
20. 0.01 M KMnO 4 solution was added to 20.0 mL of 0.05 M Mohr’s salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO 4 solution left in the burette after the end point is ____________ mL. (nearest integer)
21. Calculate the equivalent weight of phosphorous acid H3PO3 in the following reaction.
2NaOH + H3PO3 → Na2HPO3 + 2H2O
(at. wt. H = 1, P = 31, O = 16)
22. 0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is
(1) 30.6% (2) 40.6%
(3) 20.6% (4) 50%
23 2KMnO 4 + 3H 2 SO 4 + 5H 2 O 2 → K 2 SO 4 + 2 MnSO4+ 8 H2 O+5O2
Find the normality of H 2 O 2 solution, if 20 mL of it is required to react completely with 16 mL of 0.02 M KMnO 4 solution. (molar mass of KMnO4 = 158 g mol–1 )
(1) 4 × 10–2 N
(2) 2 × 10–2 N
(3) 6 × 10–2 N
(4)8 × 10–2 N
24. The oxide of a metal contains 40% of oxygen. The valency of metal is 2. What is the atomic weight of the metal?
25. 20 ml of 0.1 M FeC2O4 solution is titrated with 0.1 M KMnO 4 in acidic medium. Calculate the volume (mL) of KMnO 4 solution required to oxidise FeC 2 O 4 completely.
THEORY BASED QUESTIONS
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). In light of the given statements, choose the most appropriate answer from the options given below.
(1) if both statement I and statement II are correct.
(2) if both statement I and statement II are incorrect.
(3) if statement I is correct, but statement II is incorrect.
(4) if statement I is incorrect, but statement II is correct.
1. S-I : I n redox titration, the indicators used are sensitive to change in pH of the solution
S-II : In acid base titration, the indicators used are sensitive to change in oxidation potential
2. S-I : 1 mol of H 2 SO 4 is neutralised by 2 mol of NaOH ; However, 1 equivalent of H2SO4 is neutralised by I equivalent of NaOH.
S-II : Equivalent mass of H2SO4 is half of its molecular mass, however, the equivalent mass of NaOH is equal to its molecular mass
3. S-I : 2CuCl→CuCl 2 +Cu is a disproportionation reaction.
S-II : All transition metals show disproportionation reactions.
4. S-I : Aqueous solution of K 2 Cr 2 O 7 is preferred as a primary standard in volumetric analysis over Na2Cr2O7 aqueous solution.
S-II : Na2Cr2O7 has a higher solubility in water than Na2Cr2O7
5. S-I : Equivalent mass of H3PO2 is equal to its molecular mass
S-II : H3PO2 is a monobasic acid
6. S-I : HNO2 acts both as an oxidising as well as a reducing agent.
S-II : The oxidation number of nitrogen can increase above +3 and can also decrease below +3
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statement mark the correct answer.
(1) if both (A) and (R) are true and (R) is the correct explanation of (A).
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) if (A) is true but (R) is false.
(4) if both (A) and (R) are false.
7. (A) : The decomposition of H2O2 to form water and oxygen is an example of disproportionation reaction.
(R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O 2 and –2 oxidation state in H2O.
8. (A) : Photosynthesis is an example of redox reactions.
(R) : In the reaction, both carbon dioxide and water are oxidised to oxygen and carbohydrates respectively.
9. (A) : Neutralisation reactions can never be redox reactions.
(R) : There is no exchange of electrons between the atoms.
10. (A) : Average oxidation state of carbon in glucose is zero.
(R) : Oxidation state of an atom in any compound is zero.
JEE ADVANCED LEVEL
1. In acidic medium dichromate ion oxidises stannous ion as
(1) the value of x : y is 1:3
(2) the value of x+y+z is 18
(3) a : b is 3 : 2
(4) the value of z-c is 17
2. Which of the following may show fixed equivalent weight?
(1) Mg (2) Al
(3) Zn (4) Cu
3. Which statement(s) about oxidation number is (are) correct?
(1) The oxidation number is the number of electrons lost (+ve) or gained (-ve) by an atom during the formation of ionic compounds.
(2) For covalent compounds, the oxidation number is indicated by the charge that an atom of element would have acquired if the substance would have been ionic.
(3) Oxidation number may have fractional values.
(4) Oxidation number is always negative.
4. Which of the following statements(s) is (are) correct?
(1) All reactions are oxidation and reduction reactions.
(2) Oxidising agent is itself reduced.
(3) Oxidation and reduction always go side by side.
(4) Oxidation number during reduction decreases.
5. The oxidation number of Cr is +6 in
(1) FeCr2O4
(2) KCrO3Cl
(3) CrO5
(4) [Cr(OH4)]–
6. The different oxidation state(s) exhibited by oxygen is (are)
(1) –2 (2) –1
(3) 0 (4) –1/2
7. Which of the following reactions involve oxidation–reduction?
(1) 2Rb+2H2O→2RbOH+H2
(2) 2CuI2→2Cul+I2
(3) NH4Cl+NaOH→NaCl+NH3+H2O
(4) 4KCN+Fe(CN)2→K4[Fe(CN)6]
8. Which reactions involves iodimetric titrations?
(1) 2Na2S2O3+I2→N2S4O6+2NaI
(2) –-+ +2 4 22 2 x MnO+2I+16H2Mn+8HO+5I →
(3) Na3AsO3+I2+H2O→Na3AsO4+2HI
(4) -2+- +3 27 CrO+14H+5I2Cr+7HO+3I22 →
9. Which of the following statements are correct?
(1) In the titration of HCl vs NaOH , indicator is phenolphthalein.
(2) In the titration of KMnO4 vs H2C2O4, no external indicator is used.
(3) In the titration of I2 vs Na2S2O3, starch is the indicator.
(4) Equivalent weight of hypo is half of its molecular weight in the reaction I2+2Na2S2O3→2NaI+Na2S4O6.
10. Which of the following is/are disproportionation reaction(s)?
(1) 2H2O2(aq)→2H2O(l)+O2(g)
(2) ()()()()() aq aq 2322 4s l g P+3OH+3HOPH+3HPO →
(3)
()()()() 2 (aq) 2 g aqaq l Cl+2OHClO+Cl+HO →
12. Methane is converted into formaldehyde. What is the ratio of molecular weight to equivalent weight of methane?
13. In the chemical reaction K2Cr2O7+xH2SO4+ ySO2→K2SO4+ySO2→K2SO4+Cr2(SO4)3+zH2O. Here sum of x,y and z is_____.
14. What will be the n–factor of the reactant in the following reaction? ()4272232 2 NHCrON+CrO+4HO ∆ →
15. The oxide of an element contain 67.67% of oxygen. What is its equivalent weight?
16. 1 litre solution of KIO3 unknown molarity is given to titrate with KI in strong acidic medium 50 mL solution of KIO 3 requires 10 mL of 0.1 M KI for complete reduction to I 2. The molarity of KIO 3 solution is x ×10–3 M. Then find the value x is _____.
17. Equivalent weight of an element ‘X’ is 3. If vapour density of volatile chloride of ‘X’ is 77, find out the number of chlorine atoms combine with element ‘X’ to form a molecule.
18. The ion An+ is oxidised to AO–3 by MnO –4 changing to Mn2+ in acidic medium. Given that 2.68×10–3 mol of An+ requires 1.61×10–3 mole of MnO–4. Thus, the value of n is
19. Find magnitude of sum of two different oxidation states of sulphur atoms in Na2S4O6 (Sodium tetra thionate).
20. What is the value of n in the following half equation? ()–- –2 –42 4 CrOH+OHCrO+HO+e → n
(4)
()()()() 2 (aq) 2 g aq2g l 2F+2OH2F+OF+HO →
Numerical Value Questions
11. In acidic medium, dichromate ions () 2–CrO27 oxidize ferrous ions to ferric ions. The molar mass of potassium dichromate is 294 g/mol. The equivalent weight is ____.
21. In order to oxidise a mixture two mole of each of FeC 2O 4,Fe 2(C 2O 4) 3 and FeSO 4 in acidic medium, the number of moles of KMnO4 required is:
22 In the balanced chemical equation, the value of (p+q+r)–(s+t+u) is ____. pCu2O + qMnO–4 + rH2O → sMnO2 + tCu(OH2) + uOH–
23. The equivalent weight of CaCO3 in g eq-1 is
24. 20 mL of 0.1 M FeC2O4 solution is titrated with 0.1 M KMnO 4 in acidic medium. Calculate the volume (mL) of KMnO 4 solution required to oxidise FeC 2 O 4 completely?
Passage-based Questions
Passage I: 'R' is bleaching powder 'S' is one of the component in glass
29. In the reaction
As2S3+HNO3→H3AsO4+H2SO4+NO; the element oxidised is:
(1) As only (2) S only (3) N only (4) As and S both
30. In the equation – –+–223 NO+HONO+2H+e → n n, stands for:
(1) 1 (2) 2 (3) 3 (4) 4
Matrix Matching Questions
25. The average oxidation state of chlorine in 'R' is ___.
26. The oxidation state of silicon in 'S' is ____.
Passage II: The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation number is defined as the residual charge which an atom has or appears to have in a molecule when all other atoms are removed the molecule as ions. Oxidation number is frequently used interchangebly with oxidation state. The stock notations of oxidation numbers are based on the periodic property electronegativity. An atom in a molecule can be assigned positive, negative or zero oxidation number by considering its environment. In few cases, oxidation number can even be fractional.
27. Calculate the oxidation number of Cr in K2Cr2O7?
28. Maximum oxidation state of Os is.
Passage III: A redox reaction involves oxidation of reductant liberations, which are then consumed by an oxidant. The sum of two half reactions give r ise to net redox change. In half reaction charge and atoms are always conserved.
31. Match the Column-I with Column-II Column-I (Acid) Column-II (Nature)
(a) H3PO4 (p) mono basic
(b) H2CO3 (q) dibasic
(c) H3BO3 (r) tribasic
(d) H 4P2O7 (s) tetra basic
(a) (b) (c) (d)
(1) q r p s
(2) s p r q
(3) r q p s
(4) p s q r
32. Match the Column-I with Column-II
Column-I Column-II
(a) – 2–222 2OO+O → (p) redox reaction
(b) 2–+ 4 CrO+H → (q) one of the product has trigonal planar
(c) ––+ 42 MnO+NO+H → (r) dimeric bridge tetrahedral metal ion
(d) – 2+ 224 NO+HSO+Fe → (s) disproportio nation
(a) (b) (c) (d)
(1) rq pq p r
(2) rs qr pq rs
(3) ps r pq p
(4) pq pr q rs
33. Match the Column-I and Column-II
Column–I (Reaction)
Column–II (Eq. Wt)
(a) H+ +2 4 KMnOMn → (p) M 2
(b) +2 24 MgCOMg+CO2 → (q) M 5
(c) +3 227 KCrOCr → (r) M 6
(d) +3 5 CrOCr → (s) M 3
(a) (b) (c) (d)
(1) q r p s
(2) s p r q
(3) r s p q
(4) q p r s
34. Match the redox process in List-I with n-factor for underlined species in List- II
List-I (Redox process)
List-II (n-factor for under lined species)
(a) –2–2334 AsSAsO+SO → (p) 28
(b) 23 II+IO → (q) 4/3
(c) 32333 HPOPH+2HPO → (r) 1
(d) H3PO2 + NaOH → NaH2PO2 + H2O (s) 5/3
(a) (b) (c) (d)
(1) p s q r
(2) q s r p
(3) r q p s
(4) s r p q
35. Match the following.
Column-I Column-II
(a) FeCl3 (p) 32 2KClO2KCl+3O ∆ →
(b) Redox reaction (q) reducing agent
(c) H2S (r) oxidising agent
(d) H2O2 (s) oxidising agent and Reducing agent
(a) (b) (c) (d)
(1) p q r s
(2) s q r p
(3) r p q s
(4) q r s p
36. Match the underlined element in the compound in column-I with its oxidation state in column-II
Column-I (Compound) Column–II (Oxidation state)
(a) H2S2O8 (p) +6
(b) H2SO4 (q) +1
(c) CaOCl2 (r) –1
(d) NO2 (s) +4
(a) (b) (c) (d)
(1) p p qr s
(2) p q sq sr
(3) sq rs sq s
(4) sr qs ps pr
37. Match the Column-I with Column-II: Column-I Column-II
(a) +3 oxidation state (i) nitrogen
(b) +1 oxidation state (ii) nitrous oxide
(c) zero oxidation state (iii) nitrate ion
(d) +5 oxidation state (iv) hydroxylamine
(v) nitrite ion
(a) (b) (c) (d)
(1) i iv iii ii
(2) v ii iv iii
(3) iv v iii i
(4) v ii i iii
38. Match Column-I (Compounds) with Column-II (Oxidation states of Nitrogen).
Column–I Column–II
(a) NaN3 (p) +5
(b) N2H2 (q) +2
(c) NO (r) –1/3
(d) N2O5 (s) –1
(a) (b) (c) (d)
(1) r s q p
(2) p s q r
(3) p q r s
(4) s p r q
FLASHBACK (Previous JEE Questions)
JEE Main
1. Chlorine undergoes disproportionation in alkaline medium as shown below:
CHAPTER 3: Redox Reaction
3. Ir on (III) catalyzes the reaction between iodide and persulphate ions in which
a) Fe+3 oxidizes the iodide ion
b) Fe+3 oxidizes the persulphate ion
c) Fe2+ reduces the iodide ion
d) Fe2+ reduces the persulphate ion
(1) b and c only (2) a only
(3) a and d only (4) b only
4. Thiosulphate reacts differently with iodine and bromine in the reactions given below
Which of the following statement justifies the above dual behaviour of thiosulphate?
(1) Bromine undergoes oxidation and iodine undergoes reduction in these reactions
()()()()() 22 aClgbOHaqcClOaqdClaq HOl +→++ ()()()()() 22 aClgbOHaqcClOaqdClaq HOl +→++
The values of a, b, c and d in a balanced redox reaction are respectively :
(1) 1, 2, 1 and 1 (2) 2, 2, 1 and 3
(3) 3, 4, 4 and 2 (4) 2, 4, 1 and 3
2. Given below are two statements:
Statement I : S 8 Solid undergoes disproportionation reaction under alkaline conditions to form S2− and 2 SO23
Statement II : 4 ClO It can undergo disproportionation reaction under acidic conditions.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is correct but statement II is incorrect.
(2) Statement I is incorrect but statement II is correct
(3) Both statement I and statement II are incorrect
(4) Both statement I and statement II are correct
(2) Thisulphate undergoes oxidation by bromine and reduction by iodine in these reactions
(3) Bromine is a stronger oxidant than iodine
(4) Bromine is a weaker oxidant than iodine
5. An indicator ‘X’ is used for studying the effect of variation in concentration of iodide on the rate of reaction of iodide ion with H 2O 2 at room temp. The indicator ‘X’ forms blue coloured complex with compound ‘A’ present in the solution. The indicator ‘X’ and compound ‘A’ respectively are
(1) starch and iodine
(2) methyl orange and H 2O 2
(3) starch and H 2O 2
(4) methyl orange and iodine
6. In neutral or faintly alkaline medium KMnO 4 being a powerful oxidant can oxidise, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be
(1) 5 (2) 1
(3) 0 (4) 3
7. When 10 mL of an aqueous solution of KMnO 4 was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour.
The strength of KMnO4 in g/L is x × 10–2. Find the value of x.
[Atomic mass of K = 39, Mn = 55, O = 16]
8. The difference in oxidation state of chromium in chromate and dichromate salts is _____.
9. See the following chemical reaction: 2–+2+ 3+3 27 2 CrO+H+6FeCr+6Fe+HO →+ xyz
The sum of x, y, and z is ________
JEE Advanced
10. Consider the following molecules : Br 3O8, F2O, H2S4O6, H2S5O6, and C3O2. Count the number of atoms existing in their zero oxidation state in each molecule. Their sum is ___.
CHAPTER TEST - JEE MAIN
Section-A
Single Option Correct MCQs
1. In the following reaction, which is the species being oxidised?
2. KMnO4 reacts with oxalic acid according to the equation
2MnO+5CO+16H2Mn+10CO+8HO →
20 mL of 0.1 M KMnO4 will react with (1) 120 mL of 0.25 M H2C2O4 (2) 150 mL of 0.10 M H2C2O4 (3) 25 mL of 0.20 M H2C2O4 (4) 50 mL of 0.20 M H2C2O4
3. When –BrO3 ion reacts with Br– in acidic medium, Br2 is liberated. The equivalent weight of Br2 in this reaction is____ (1) 5M 8 (2) 5M 3 3) 3M 5 (4) 4M 6
4. In the reaction –2–+ 2+2–43 42 MnO+SO+HMn+SO+HO → the number of H+ ions involved is (1) 2 (2) 6 (3) 8 (4) 16
5. All elements commonly exhibit an oxidation state of (1) +1 (2) –1 (3) Zero (4) +2
6. Equivalent weight of FeC2O4 in the change: FeC2O4→Fe3++CO2 is (1) M 3 (2) M 6 (3) M 2 (4) M 1
7. The element that always exhibits a negative oxidation state in its compounds is (1) nitrogen (2) oxygen (3) fluorine (4) chlorine
8. Following reaction describes the rusting of iron 4Fe+3O 2 →4Fe 3+ +6O 2–. Which one of the following statement is incorrect?
(1) This is an example of a redox reaction. (2) Metallic iron is reduced to Fe 3+ .
(3) Fe3+ is an oxidising agent.
(4) Metallic iron is a reducing agent.
9. Consider the following reaction: 2–+ +2 424 22 MnO+CO+HMn+2CO+HO 2 → z xyzxy
(1) 2, 5 and 16 (2) 5, 2 and 8 (3) 5, 2 and 16
(4) 2, 5 and 8
10. Equivalent weight of pyrophosphoric acid is
(1) M.wt 1 (2) M.wt 2 (3) M.wt 3 (4) M.wt 4
11. Number of mole of KMnO 4 required to oxidise one mole of Fe(C 2 O 4 ) in acidic medium is
(1) 0.6 (2) 0.167 (3) 0.2 (4) 0.4
12. It is found that element 'X' forms a double salt, isomorphous with Mohr’s salt. The oxidation number of 'X' in this compound is:
(1) +3 (2) +2 (3) +4 (4) –4
13. n -factors for Cu 2 S and CuS, when they react with KMnO 4 in acidic medium are, respectively (neglecting the further oxidation of released SO 2)
(1) 7,7 (2) 6,6 (3) 6,8 (4) 8,6
14. In the reaction, I2+2KClO3→2KIO3+Cl2
i) Iodine is oxidised.
ii) Chlorine is reduced.
iii) Iodine displaces chlorine.
iv) KClO3 is decomposed.
The correct combination is
(1) Only i & iv are correct. (2) Only iii & iv are correct. (3) i, ii, iii are correct. (4) All are correct.
15. Which of the following reactions does not involve the change in oxidation state of metal
(1) VO–2→V2O3 (2) K→K+ (3) Cu2+→CuS (4) Cu2+→Cu
16. The number of mole of –4 MnO and 2–CrO27 s eparately required to oxidise 1 mole of FeC2O4 each in acidic medium, respectively
17. The sum of oxidation numbers of nitrogen in ammonium nitrite (NH 4NO2) is (1) +5 (2) 0 (3) 1 (4) –3
18. The equivalent weight of potassium permanganate (M.W=158)in acid medium is
(1) 75 (2) 52.66
(3) 31.6 (4) 158
19. Which of the following is not a redox reaction?
(1) 2BaO+O2→2BaO2
(2) BaO2+H2SO4→BaSO4+H2O2
(3) 2KClO3→2KCl+3O2
(4) SO2+2H2S→2H2O+3S
20. Which one of the following can function as an oxidising agent?
(1) I– (2) H2S
(3) LiAlH 4 (4) –2CrO27
Section-B
21. For the reaction 4FeS2+11O2→2Fe2O3+8SO2 The equivalent mass of FeS 2 is x (M = Molecular mass of FeS 2). Here the value of x =___.
22. –2 –+ +2 244 2 2 CO+MnO+HCO+Mn+HO → xyz The value of (y+z)–x is ____
23. 1.0 g of metal nitrate gave 0.86 g of metal sulphate. Calculate equivalent weight of metal?
24. Calculate the number of moles of Sn 2+ ion oxidised by 1 mol of K 2 Cr 2 O 7 in acidic medium?
25. In the compound YBa 2Cu 3O 7 that shows superconductivity, the oxidation state of Cu is x/3. (assume that the rare earth element Yttrium is in its usual +3 oxidation state) then x is
CHAPTER TEST - JEE ADVANCED
2023 P1 Model
Section-A
[Multiple Option Correct MCQs]
1. Which is/are incorrect statement?
(1) Equivalent mass of –HPO23 is 40.5
(2) Equivalent mass of –HPO24 may be equal to molar mass or less than molar mass because it depends on the reaction.
(3) KMnO4 has maximum equivalent mass in acidic medium.
(4) Oxidation state of H in MgH2 is greater than in H2O2
2. One gram equivalent of Al 2(SO4)3 contain
(1) 0.33 gram ions of Al+3
(2) 2 gram atoms of oxygen
(3) 0.5 gram atoms of sulphur
(4) 0.5 gram ions of sulphate
3. When a mixture of Cu2S and CuS is titrated with Al(MnO 4 ) 3 in acidic medium the oxidation products of Cu 2S and CuS are Cu +2 and SO 2 if the molecular weight of Cu2S, CuS and Al(MnO4)3 be M1, M2 and M 3 respectively then which of the following statements are correct.
(1) Equivalent weight of Cu2S is M1 8
(2) Equivalent weight of CuS is 2 M 5
(3) Equivalent weight of Al(MnO4)3 is 3 M 5
(4) Equivalent weight of Al(MnO4)3 is 3 M 15
Section-B
[Single Option Correct MCQs]
4. 2 mol N 2 H 4 loses 16 mol of electrons is being converted to a new compound x. Assuming that all of the 'N' appears in the new compound, what is the oxidation state of 'N' in compound x?
(1) –1 (2) –2 (3) +2 (4) +4
5. Oxidation states of the metal in the minerals haematite and magnetite, respectively are (1) II, III in haematite, and III in magnetite (2) II, III in haematite, and II in magnetite (3) II in haematite, and II, III in magnetite (4) III in haematite, and II, III in magnetite
6. 20 mL of 0.2 M MnSO4 are completely oxidised By 16 mL of KMnO 4 of unknown normality, each forming Mn 4+ oxidation state. Find out the normality of KMnO4 solutions. (1) 0.5 N (2) 1 N (3) 1.5 N (4) 2 N
7. 0.59 g of the silver salt of an organic acid (mol.wt.210) on ignition gave 0.36 g of pure silver. The basicity of the acid is [At.wt of Ag=108] (1) 1 (2) 2 (3) 3 (4) 4
Section-C
[Integer Value Questions]
8. How many of the following elements (or) ions (or) compounds will disproportionate in acidic (or) in basic medium at proper temperature?
4822 4332 P,S,F,NO,ClO,MnO,HPO,Br
9. 10 g of Fe3O4 is oxidised completely by 50 mL of 0.1M KMnO4 solution. The mass in grams of Fe2O3 in Fe3O4 is ___.
10. When 5.0 g of a metal is strongly heated, 9.44 g of its oxide is obtained. Then the equivalent mass of the metal is __.
11. In hot alkaline solution, Br 2 disproportionates to Br – and –BrO3 2 32 3Br+6OH5Br+BrO+3HO. →
The equivalent weight of Br 2 is 3M/ x (M=mol. wt.) Then the value of x is ___.
12. How many of the following show disproportionation reaction i) XeF4+H2O→ ii) XeF2+H2O→ iii) P4+NaOH+H2O→ iv) F2+NaOH→ v) Cl2+NaOH→ vi) HOCl ∆ → vii) NO2+NaOH→ viii) Zn+NaOH→ ix) S8+KOH
13. When copper is treated with a certain concentration of nitric acid, nitric oxide, and nitrogen dioxide are liberated in equal volumes according to the following equation: xCu+yHNO3→Cu(NO3)2+NO+NO2+H2O. The coefficient of x is _____.
Section-D
[Matrix Matching Questions]
14. Match the reaction in Column-I with titration method in Column-II and select answer from the answer code
15. Equivalent weight of potassium permanganate in different media.
Column–I (potassium permanganate in different media)
Column–II (Equivalent Weight)
(a) Acidic (p) 15.8
(b) Neutral (q) 52.6
(c) Strong alkali (r) 79
(d) Dilute alkali (s) 31.6
(a) (b) (c) (d)
(1) s q p q
(2) q p q s
(3) r q q p
(4) p q r s
16. Consider the redox reactions in Column-I and molar ratios of oxidising to reducing agents in Column-II, respectively. Match the items in the Columns appropriately.
Column-I (Redox Reaction)
Column-II (Molar Ratio of Oxidising to Reducing Agents)
(a) –2–MnO+CO424 MnO+CO22 → (p) 2 : 1
(b) () –3 ClO+FeOH –2–4 Cl+FeO → (q) 3 : 1
(c) ()––2 3 HO+CrOH 2––4 CrO+HO → (r) 2 : 3
(d) () 24 2 NH+CuOH 2 NO+Cu → (s) 3 : 2
(a) (b) (c) (d)
(1) s q p q
(2) q p q s
(3) r q q p
(4) p q r s
17. Match the conditions if column-I with solution on right column-II.
Column-I (Medium)
Column-II (Equivalent weight)
(a) Has maximum pH at the end point when titrated against KOH (p) CH3COOH (Ka=2×10–5)
(b) Has minimum pH at the end point when titrated with KOH (q) HCN(Ka=5×10–4)
ANSWER KEY
JEE Main Level
(c) Evolve maximum heat when treated with NaOH (r) HF(Ka=5×10–4)
(d) Evolve equal amount of heat when titrated with strong acid or strong base (s) NH3(Ka=2×10–5) (a) (b) (c) (d)
