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IL Ranker Series Chemistry for NEET Grade 11 Module 3
ISBN 978-81-970868-7-8 [FIRST
EDITION]
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Key Features of the Book
Chapter Outline
1.1 Types of Solutions
1.2 Methods of Concentration
1.3 Solubility
This outlines topics or learning outcomes students can gain from studying the chapter. It sets a framework for study and a roadmap for learning.
Specific problems are presented along with their solutions, explaining the application of principles covered in the textbook. Solved Examples
Q. What is the molality of a solution of H2SO4 having 9.8% by mass of the acid?
Sol. 9.8% by mass of H2SO4 contains 9.8 g of H2SO4 per 100 g of solution.
Therefore, if mass of solution = 100 g, mass of solute, H2SO4 = 9.8 g,
Try yourself:
In a solution of H2SO4 and water, mole fraction of H2SO4 is 0.9. How many grams of H2SO4 is present per 100 g of the solution?
Ans: 98
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.
CHAPTER REVIEW
Types of Solutions
■ A solution is a homogeneous mixture of two or more non–reacting components. Formation of solution is a physical process.
TEST YOURSELF
1. The mole fraction of a solvent in aqueous solution of a solute is 0.6. The molality of the aqueous solution is (1) 83.25 (2) 13.88 (3) 37 (4) 73
It offers a concise overview of the chapter’s key points, acting as a quick revision tool before tests.
This is a focused practice with topic-wise questions based on NCERT textbook content. It is designed to enhance students’ success in NEET by aligning with recent exam trends.
Exercises
NEET DRILL FURTHER EXPLORATION
Holding substantial weightage in the NEET Biology paper, these questions improve analytical judgement of statements.
Known for their low scoring rate and high weightage in recent NEET exams, these questions play a crucial role in improving students’ critical thinking skills to assess the logical relationship between the assertion and the reason.
MATCHING TYPE QUESTIONS
STATEMENT TYPE QUESTIONS
ASSERTION AND REASON TYPE QUESTIONS
BRAIN TEASERS
FLASHBACK
CHAPTER TEST
Modelled after the NEET exam format, this test is based on a specific chapter. It serves as a tool for students to evaluate their time management skills and gauge their mastery level in a particular chapter.
This section comprises questions that extend beyond the NCERT content yet remain relevant to NEET, preparing students for additional and pertinent challenges beyond the textbook.
These include questions for practising the correlation of information across different topics. A significant number of matching questions appear in the NEET Biology paper and are easy to score.
These complex questions that combine fun and critical thinking are aimed at fostering higher order thinking skills and encourage analytical reasoning.
Hand-picked questions from previous years NEET offer an insight into the types of questions and the important topics that are probable to appear in NEET.
CLASSIFICATION OF ELEMENTS CHAPTER 3
Chapter Outline
3.1 Genesis of Periodic Classification
3.2 Mendeleef’s Periodic Table
3.3 Modern Periodic Law
3.4 Long Form of Periodic Table
3.5 Nomenclature of Elements
3.6 Classification of Elements
3.7 Periodic Properties and trends
When a large number of elements are known, the study of their chemistry becomes very difficult unless the elements are classified in proper way. classification helps in studying the chemistry of elements and their compounds in a systematic manner. In this chapter, a brief history of Classification of elements and the properties of elements are discussed.
3.1 GENESIS OF PERIODIC CLASSIFICATION
Pure chemicals are mainly of two types: elements and compounds. In 1800, only 31 elements were known. By 1865, the number of identified elements had more than doubled to 63. Among the elements known, 88 are available in the elemental form and about 26 are man made. It is difficult to study individually the chemistry of more than one hundred elements known today and their innumerable compounds. The experimental data regarding elements and their compounds can only be systematised if proper classification
is done. The basic object of classification is to arrange the elements and their compounds in such a way so that we may have greater control over their characteristics with lesser possible efforts. The best classification would be the one which puts together those elements which resemble in most respects and separates the others.
After the proposal of John Dalton’s atomic theory, the scientists took atomic weight as the important property of element and tried to seek the relationship between the properties of the elements and their atomic weights.
3.1.1
Dobereiner's Law of Triads
John Dobereiner pointed out that there were sets of three elements which showed similar chemical properties. Such sets are named as triads. The atomic weight of the middle element of a triad was approximately the average of the atomic weights of the other two members. Examples of triads are: Li, Na and K; Ca, Sr and Ba; Cl, Br and I; etc.
Dobereiner’s triads
3.1.2 Telluric Helix
The next attempt was made by A.E.B de Chancourtois in 1862. He arranged the then known elements in increasing order of atomic weights and made a cylindrical table of elements in a spiral graph form, called telluric helix.
Drawback
This classification was not applicable to all the then known elements.
3.1.3 Newland’s Law of Octaves
John Newlands, arranged the elements in increasing order of their atomic weights and noted that a given element was similar to the eighth element that followed it. The relationship was the same as that existing between a musical note and its octave. This development was called the law of octaves.
Newlands’ octaves Element
At.wt. 7 9 11 12 14 16 19
Element
At.wt. 23 24 27 29 31 32 35.5
At.wt. 39 40
Drawback
This system worked well only for lighter elements, up to calcium, and failed for heavier elements.
3.1.4 Lothar Meyer’s Work
Lothar Meyer presented the classification in the form of a curve. He calculated the atomic volumes of known elements as the ratio of atomic weight and density.
Lothar Meyer plotted the physical properties such as atomic volume, melting point, and boiling point against atomic weight and obtained a periodically repeated pattern, which is shown in Fig.3.1.
Fig. 3.1 Lother Meyer’s curves
Q. What would be the group and period of the element with atomic number 80?
Sol. The element with Z value 80 is mercury (Hg). It is present in period 6 and group II B.
TEST YOURSELF
1. Beryllium follows Newland’s Law of Octaves. What is its eight similar element in the classification?
(1) K (2) Mg
(3) P (4) Si
2. The number of gaseous elements available is
(1) 5 (2) 11
(3) 12 (4) 15
3. Which of the following is a Dobereiner's triad?
(1) Li, Na, K (2) Fe, Co, Ni (3) Ru, Rh, Pd (4) Os, Ir, Pt
Answer Key
(1) 2 (2) 2 (3) 1
3.2 MENDELEEF’S PERIODIC TABLE
Mendeleef relied on the similarities in the empirical formulae and properties of compounds by the elements. Mendeleef put forward the periodic law, stating that, “the physical and chemical properties of the
elements are periodic function of their atomic weights”.
In Mendeleef’s periodic table, elements are arranged in the increasing order of their atomic weights Table 3.1.
In this, elements are arranged in horizontal rows called periods and vertical columns called groups. Each group is divided into two sub groups, namely ‘A’ and ‘B’. The first three periods are called short periods and the remaining are called long periods. Each long period contains two rows of elements, called series.
While arranging the elements in the periodic table, Mendeleef not only followed the increasing order of atomic weights but also considered their properties. If the properties of elements did not correspond to what is expected for that place, he left blank places and proposed the properties of elements to be present in blanks. Accordingly, these elements were discovered later Table 3.2.
The properties shown by these are similar to those predicted by Mendeleef.
Example: Eka boron, eka aluminium, and eka silicon are now known as scandium, gallium, and germanium, respectively. These are called missing elem ents. Mendeleef corrected the atomic weights of elements like Be, In, U, etc., based on their position and equivalent weight (Atomic weight = Equivalent weight × valency). Due to consideration of properties in arranging the elements, at four places, inversion of atomic weight took place.
a) 40Ar, 39K b) 59Co, 58Ni
c)128Te, 127I and d) 232Th, 231Pa.
These are called anomalous pairs.
Mendeleef observed that the elements with similar properties had i) similar atomic weights (or)
Examples:
a) Fe(56), Co(59), Ni(59)
b) Os(191), Ir(193), Pt(195)
ii) constant increase in atomic weights
Examples:
a) K(39), Rb(85), Cs(133)
b) Ca(40), Sr(88), Ba(137).
The elements present in VIII group are called transition triads. These are:
i) Fe, Co, Ni;
ii) Ru, Rh, Pd; and
iii) Os, Ir, Pt.
Initially, zero group elements were not known at the time of Mendeleef.
(EkaAl)2O3 Ga2O3 (EkaSi)O2 GeO2
Formula of chloride (EkaAl)Cl3 GaCl3 (EkaSi)Cl4 GeCl4
Isolation of the elements - - -
The elements with dissimilar properties were kept in different sub groups of the main group, e.g., Cu, Ag, and Au are kept along with K, Rb, and Cs.
The fourteen lanthanide elements (rare earth
Table 3.2 Comparison of properties – Mendeleef’s Eka Elements
elements) with different atomic masses are kept in one place only.
No one had any idea of the structure of atom in the days of Mendeleef. Several decades had to pass before the electron was discovered and atomic structure was developed. The electronic configuration provided fundamental basis for the properties of elements.
Moseley’s Experiments: Moseley bombarded various elements, making them anticathodes, with cathode rays in a discharge tube. This resulted in X-rays with characteristic frequencies. Moseley showed that highest frequency and lines of X-rays are related to the charge present in the atomic nucleus of the element used as anticathode.
()aZbυ=−
Here, υ is th e frequency of X-rays, Z is the atomic number, a and b are constants for a selected type of line. A plot of υ against ‘Z’ gives a straight line, as sh own in Fig.3.2
TEST YOURSELF
1. Considering the chemical properties, atomic weight of the element ‘Be’ was corrected based on
(1) valency (2) configuration
(3) density (4) atomic volume
2. Eka silicon is now known as (1) scandium (2) gallium (3) germanium (4) boron
3. The element ‘Sc’ was known long back as (1) eka-aluminium (2) eka-boron (3) eka-silicon (4) eka-mercury
4. Anomalous pair among the following is (1) Boron – Silicon (2) Beryllium – Indium (3) Aluminium – Gallium (4) Cobalt – Nickel
5. Choose the triad not present in group VIII of Mendeleef’s table. (1) Li, Na, K (2) Fe, Co, Ni (3) Ru, Rh, Pd (4) Os, Ir, Pt
oa Intercept of Y axis is ‘ab’
Fig. 3.2. Plot of υ and atomic number (Z)
No such relationship was obtained using mass number. According to Moseley, atomic number stands for a serial number. As atomic number increases, the frequency of characteristic X-rays increases. Hence, atomic number is concluded as the fundamental quantity of element, which increases in regular steps. Thus, properties of an element are dictated by its atomic number and so, by the number of its electrons. This led to the proposal of modern periodic law.
6. The frequency of the characteristic X-ray of line of metal target ‘M’ is 2500 cm –1 and the graph between ν vs ‘Z’ is as follows The atomic number of M is
(1) 49 (2) 50 (3) 51 (4) 25
Answer Key
(1) 1 (2) 3 (3) 2 (4) 4
(5) 1 (6) 3
3.3 MODERN PERIODIC LAW
The empirical evolution of the periodic table reached its peak in 1913, when Moseley showed that atomic number is a more fundamental property of an element than its atomic weight. The position of an element in the periodic table depends on its atomic number, and the reason for the anomalies in the original periodic table at once becomes clear. Moseley modified Mendeleef’s periodic law and stated that “the physical and chemical properties of the elements are periodic function of their atomic numbers”. The periodic law revealed important analogies among the 94 naturally occurring elements Neptunium and plutonium, like actinium and protactinium, are also found in pitchblende, an ore of uranium).
The atomic number is equal to the nuclear charge or the number of electrons in the neutral atom. Further, it was recognised that the periodic law is essentially the consequence of the periodic variation in electronic configurations. The configurations indeed determine the properties of elements and their compounds, and this is the basis for the modern periodic law.
Th e modern periodic law states that, the physical a nd chemical properties of the elements are periodic functions of their atomic numbers or their electronic configurations.
An expanded modern version and widely used form of the periodic table is shown in Table.3.3. It has the same number of horizontal rows as the series of the Mendeleef’s table.
T hese horizontal rows are now called periods. Elements having similar outer electronic configuration in their atoms are grouped in vertical columns. These columns are called groups or chemical families. 18 groups and 7 periods are in use now-a-days.
Q. How would you justify the presence of 18 elements in the 5 th period of the periodic table?
Sol. When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s, and 5p increases is 5s < 4d < 5p. The total number of orbitals available is 9. The maximum number of electrons that can be accommodated is 18 and, therefore, 18 elements are there in the 5 th period.
Q. Why is there a break in the third period elements of the long form of the periodic table?
Sol. In the third period, 3s and 3p orbitals are only filled successively. Four orbitals together can hold eight electrons.
TEST YOURSELF
1. The basis of modern periodic law is (1) atomic number (2) atomic size (3) atomic volume (4) atomic mass
2. Which of the following pairs of elements are from the same group of the peri odic table? (1) Mg, Cs (2) Mg, Sr (3) Mg, Cl (4) Na, Cl
3. Elements of a vertical group have (1) same atomic number (2) same electronic configuration (3) same number of valency electrons (4) same number of core electrons
4. The starting element of fifth period is (1) K (2) Rb (3) Kr (4) Xe
5. As per the modern periodic law, the physical and chemical properties of elements are periodic functions of their (1) atomic volume (2) electronic configuration (3) atomic weight (4) atomic size
6. T he period that contains only gaseous elements is (1) 1 (2) 2 (3) 3 (4) 4
Periodic Table of the Elements (Long form)
(Representing electronic configurations)
Atomic number Symbol Valence-shell configuration
Main group Elements S-subshell is
Main-Group Elements p-subshell is gradually filled up IA Group 0 (zero) Transition Elements d-subshell is gradually filled up 1 H 1s 1
Inner -Transition Elements f-subshell is gradually filled up
Table 3.3 A modern form of the periodic table (long form)
7. The starting element and last element in the largest period in modern periodic table are (1) Rb and Xe (2) Cs and I (3) Cs and Rn (4) Fr and Kr
8. Which of the following has both members from the same period of the periodic table? (1) Na, F (2) Mg, Ca (3) Na, Cl (4) Be, Al
Answer Key
(1) 1 (2) 2 (3) 3 (4) 2 (5) 2 (6) 1 (7) 3 (8) 3
3.4 LONG FORM OF PERIODIC TABLE
The most convenient version of the periodic table was constructed by Bohr, based on the modern periodic law, and he arranged the elements in the order of their electronic configurations.
3.4.1 Salient Features
1. The long form of the periodic table is based on the electronic configuration of the elements. The elements are arranged in the increasing order of atomic numbers. The long form of periodic table is a graphical representation of Aufbau principle.
2. It contains 7 horizontal rows, called periods, and 18 vertical columns, called groups.
3. In each group, elements with similar valence shell configuration are arranged.
4. In the periodic table, the horizontal rows are called periods, and they correspond to the principal quantum number assigned to the outermost orbit of an atom. Each period starts with an alkali metal and ends with a noble gas element. In each period, the differentiating electron enters the s-orbital in the first element and p-orbital in the last element.
5. Fourteen elements of both 6th and 7th periods of 3rd group are placed in separate rows at the bottom.
The sub-energy levels filled in each period of the periodic table are listed in Table 3.4 and the numbers of elements present in each period are listed in Table 3.5
Table 3.4 Sub-energy levels filled in periods
Table 3.5 Number of elements present in different periods
(20) (Incomplete)
TEST YOURSELF
1. The nu mber of elements present in 2nd, 3rd, 4th, and 5th periods of modern periodic table, respectively are (1) 2, 8, 8 and 18 (2) 8, 8, 18 and 32
(3) 8, 8, 18 and 18 (4) 8, 18, 18 and 32
2. Outer shell octet configuration is observed for the elements of the group
(1) 2 (2) 8 (3) 18 (4) 32
3. The element with Z=117 and Z=120 belong to ____and______family, respectively.
(1) halogen family, alkaline earth metals
(2) nitrogen family, alkali metals
(3) halogen family, alkali metals
(4) chalcogens family, alkali metals
4. The following statements are related to elements in the periodic table. Which of the following is true?
(1) All the elements in Group-17 are gases.
(2) The Group-13 elements are all metals.
(3) Elements of Group-16 have lower ionisation enthalpy values compared to those of Group-15 in the corresponding periods.
(4) For Group-15 elements, the stability of +5 oxidation state increases down the group.
5. In a period, elements are arranged in a strict sequence of
(1) decreasing charges in the nucleus
(2) increasing charges in the nucleus
(3) constant charges in the nucleus
(4) equal charges in the nucleus
6. Which of the following pairs has elements containing the same number of electrons in the outermost orbit?
(1) N, O (2) Na, Cl
(3) Ca, Cl (4) Cl, Br
Answer Key
(1) 3 (2) 3 (3) 1 (4) 3 (5) 2 (6) 4
3.5 NOMENCLATURE OF ELEMENTS
The naming of the new elements had been traditionally the privilege of the discoverer (or discoverers) and the suggested name was ratified by the IUPAC. In recent years, this led
to some controversy. The new elements with very high atomic numbers are so unstable that only minute quantities, sometimes, only a few atoms of them, are obtained. Their synthesis and characterisation, therefore, require highly sophisticated and costly equipment and laboratory. Such work is carried out with competitive spirit only in some laboratories in the world. Scientists, before collecting the reliable data on the new element, at times, get tempted to claim its discovery. For example, both American and the Soviet scientists claimed credit for discovering element 104. The Americans named it Rutherfordium, whereas the Soviets named it Kurchatovium. To avoid such problems, the IUPAC recommended that until a new element’s discovery is proved, and its name is officially recognised, a systematic nomenclature will be derived directly from the atomic number of the element, using the numerical roots for 0 and numbers 1–9. These are shown in Table 3.6
Table 3.6 IUPAC notation for naming elements with Z > 100
The roots are put together in order of digits that make up the atomic number and 'ium' is added at the end. The IUPAC names for elements with Z and above 100 are shown in Table 3.7.
* Official name and symbol yet to be announced by IUPAC
A critical observation of the long form of periodic table suggests the following:
■ The number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled.
■ The period number corresponds to the shell number in which the distant electrons from the nucleus of the atom are present.
■ Each successive period in the periodic table is associated with filling up of the next higher principal energy level.
■ The first period starts with filling of lowest energy level, 1s (K-shell).The period has two elements. Hydrogen has the electronic configuration 1s1 and helium has 1s2
■ The second period starts with lithium. The differentiating electron of lithium enters the 2s (L-shell) and has the electronic configuration 1s2 2s1. The 2p orbitals are filled starting from boron. The second shell is completely filled at neon (2s 2 2p6).
Thus, there are eight elements in the second period.
■ The third period starts with sodium with differentiating electron entering the 3s orbital (M-shell) Successive filling of 3s and 3p sub-shells gives rise to third period of eight elements from sodium to argon.
■ The fourth period starts at potassium (2,8,8,1) with filling up of 4s (N-shell). Filling of 4s is complete with calcium. Before the 4p orbitals, filling up of 3d orbitals (M-shell) becomes energetically favourable. Ten elements, from scandium to zinc, are filled with differentiating electrons in 3d sub-shell. The period ends with krypton, with the filling up of the 4p orbitals. Thus, there are eighteen elements in the fourth period. In this period, exceptionally, Cr and Cu have one electron in the outer shell, i.e., 4s orbital.
■ The fifth period begins with rubidium filling up of 5s orbital. Similar to that, in fourth period, 4d sub-shell is filled, from
yttrium to cadmium. The period ends at xenon, with the filling up of the 5p orbitals. Thus, the fifth period also has eighteen elements.
■ e sixth period starts with caesium. e period has thirty two elements, and electrons are filled successively in 6s, 4f, 5d and 6p orbitals. Filling up of the 4f orbitals begins with cerium and ends at lutetium to give 4f series, also called lanthanide series. ere are thirty two elements in the sixth period (longest period).
■ e seventh period starts with francium. is period is incomplete and is expected to end at the element with atomic number 118. Filling up of the 5f orbitals aer actinium gives 5f series, also called Actinide series.
■ While the typical hydrogen is separately shown above the periodic table, the elements of 4f and 5f series are also placed separately. is is to maintain the structure and preserve the principle of classification by keeping elements with similar properties in a single column.
3.5.2 Electronic Configuration in Groups
e main reason for the classification is to group together elements which are chemically similar. Elements present in the same group have similar electronic configuration and the same number of electrons in the outermost shells.
■ The elements of group 1 or alkali metals have only one electron in the outermost shell of each atom. e general electronic configuration of alkali metals is ns 1 .
■ e elements of group 2 or alkaline earth metals have two electrons in the outermost shell of each atom. e general electronic configuration of alkaline earth metals is ns2
■ The group number in Roman symbol for representative elements directly denotes the number of electrons present in the outermost shell of atom of each element.
■ Atoms of the elements of group IIIA (or group 13) have three electrons in the outermost shell. eir general electronic configuration is ns 2np 1. Similarly group IVA (or group 14) elements have general configuration ns2np4, etc.
■ Noble gas elements (group 18), however, have filled in their outermost shells with general electronic configuration ns 2np 6 Groupwise general electronic configuration of the important groups are listed in Table 3.8
■ Nickel group has an exceptional feature. This group is called pseudo-octet configuration group. The element palladium has the electronic configuration1s 2 2s 2 2p6 3s2 3p6 4s2 3d10 4p6 4d10, with eighteen electrons in the outer most shell. In general, elements of the same group have similar outer shell electronic configuration, though there are few anomalies.
Table 3.8 Configuration of first and last element of each period
TEST YOURSELF
1. Select from the following lists, the elements belonging to the same group.
(1) Z = 12, 38, 4, 88 (2) Z = 9, 16, 3, 35 (3) Z = 5, 11, 27, 19 (4) Z = 24, 47, 42, 55
2. Rare earths are generally (1) actinides (2) all f-block elements (3) all inner transition elements (4) lanthanides
4. In the periodic table, transition elements begin with (1) Scandium (2) Zinc (3) Copper (4) Mercury
5. The general electronic configuration (n-1) d3ns2 indicates that the particular element belongs to the group (1) VB (2) VA (3) IVB (4) IIB
6. In the sixth period, the orbitals being filled with electrons are (1) 5s, 5p, 5d (2) 6s, 6p, 6d, 6f (3) 6s, 5f, 6d, 6p (4) 6s, 4f, 5d, 6p
7. The representative elements get the nearest inert gas configuration (1) by losing electrons (2) by gaining electrons (3) by sharing electrons (4) by losing or gaining or sharing electrons
8. The period number and group number in which maximum number of elements placed are, respectively,
(1) 6th and IA (2) 6th and zero (3) 6th and IIIA (4) 6th and IIIB
9. The formula of the compound formed by the pair of elements Al and S is (1) AlS (2) Al2S3
(3) Al3S2 (4) AlS2
10. An element has 18 electrons in the outer most shell. The element is
(1) transition metal
(2) rare earth metal
(3) alkaline earth metal
(4) alkali metal
11. Match the Columns.
Column–I
Column–II
(A) Po (I) Liquid metal
(B) Mercury (II) Liquid non-metal (C) Bromine (III) Diamond (D) Carbon (IV) VIA group
(1) A-IV, B-I, C-II, D-III
(2) A-IV, B-I, C-III, D-II
(3) A-III, B-II, C-I, D-IV
(4) A-I, B-IV, C-III, D-II
12. The electronic configuration of an element ‘X’, is 1s2 2s2 2p6 3s2 3p3. What is the atomic number of the element which is just below ‘X’ in the periodic table?
(1) 33 (2) 34
(3) 31 (4) 49
Answer Key
(1) 1 (2) 4 (3) 3 (4) 1
(5) 1 (6) 4 (7) 4 (8) 4
(9) 2 (10) 1 (11) 1 (12) 1
3.6 CLASSIFICATION OF ELEMENTS
For a systematic study of elements of modern periodic table, further classification of the tabular form is necessary.
3.6.1 Classification into Blocks
Based on the entry of the differentiating electron into the sub-shells, elements are classified into four blocks. They are:
1) s-block elements,
2) p-block elements,
3) d-block elements and
4) f-block elements
Different blocks of the elements in the long form of the periodic table are given in Fig.3.3
Characteristic Properties of s-block Elements
■ The elements are highly electropositive and are soft metals with lower densities.
■ They are very good reducing agents.
■ They have low melting and boiling points.
■ They are very reactive; reactivity increases down the group.
■ They form ionic substances, except lithium and beryllium.
■ They exhibit an oxidation state of +1 or +2.
■ They impart characteristic colours in the flame (except Be and Mg).
p - Block Elements
The elements in which the differentiating electron enters the p-orbitals of the outermost shell are called p-block elements.
Fig. 3.3 Different blocks of elements in periodic table s-block is present at the left side, p-block is at the right, d-block is at the middle, and f-block at the bottom of the long form of the periodic table.
s - Block Elements
The elements in which the differentiating electron enters the s-orbital of the outermost shell are called s-block elements.
The first two elements of each period belong to the s-block. Group 1 and 2 (alkali and alkaline earth metals) constitute the s-block. These elements are located at the left side of the long form of the periodic table. Helium is placed in zero group due to its inert nature, through it has 1s 2 configuration.
The general electronic configuration of s-block elements is ns 1–2 .
Groups 13, 14, 15, 16, 17, and 18 (IIIA to VIIA and zero groups) constitute the p-block. Group 18 element that is misplaced in p-block is helium. These elements are located at the right side of the periodic table. The p block elements, together with s block elements (except zero group), are referred to as representative elements or main group elements.
The general electronic configuration of p-block elements is ns2np1-6or ns2np x (x = 1 to 6).
Characteristic Properties of p-block Elements
■ The elements include all metalloids, most of the non-metals, and some metals.
■ All gaseous elements (except H 2 and He) are p-block elements.
■ Most of these elements are highly electronegative and have high electron gain enthalpy.
■ Some elements are good oxidising agents. Some of them act as reducing agents.
■ Except group 18, these elements are very reactive.
■ They form mostly covalent compounds
Example: Cl2, O2, HCl though ionic halides, oxides, sulphides, nitrides, etc., are also known.
d - Block Elements
The elements in which the differentiating electron enters the d-orbitals of the penultimate shell are called d-block elements.
Groups 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 constitute the d-block. These elements are located at the middle, in between s-block and p-block of the long form of the periodic table.
They have the properties intermediate to those of the s-block and p-block elements. They are referred to as transition elements. The d-block contains four series of elements. The 3d-, 4d- and 5d- series are completely filled with ten elements each. The 6d- series is incomplete and has only eight elements.
Zinc, cadmium, and mercury, which have the electronic configuration, do not show most of the properties of transition elements.
The general electronic configuration of d-block elements is ( n-1)d1-10 ns1or2.
Characteristic Properties of d-block Elements
1. The elements are all electropositive and metals.
2. They are all solids, except mercury, which is a liquid at room temperature.
3. Most of the elements possess catalytic activity.
Example: In the preparation of NH 3 by Haber’s process, catalyst is Fe, promoter is Mo.
Example: In the preparation of H 2 SO 4 by contact process, catalyst is platinised asbestos.
4. They form cations with different magnitudes of the charge.
5. They form ionic as well as covalent compounds.
6. They form complex compounds, Example: [Cu(NH3)4]SO4, K4[Fe(CN)6].
7. They form alloys and interstitial compounds.
8. They mostly form coloured ions, exhibit variable valency (e.g., : Fe +2 ,Fe +3 ) and paramagnetism. ( 4.90 BM for Fe +2, 5.92 BM for Fe+3)
f - Block Elements
The elements in which the differentiating electron enters the f-orbitals of the antipenultimate shell are called f-block elements.
A part of group 3 constitutes the f-block. These elements are located at the bottom of the periodic table.
There are two series of f-block elements. The first series follows lanthanum (Z = 57) and are called lanthanides [Ce ( Z =58) to Lu (Z = 71)].
The second series follows actinium (Z = 89) and are called actinides [Th (Z = 90) to Lr (Z = 103)].
The general electronic configuration of f-block elements is ( n-2)f1-14 (n-1)d0–1 ns2.
La and Ac are d-block elements but lanthanides and actinides are f-block elements.
Characteristic Properties of f-block Elements
■ These elements are heavy metals with high density, and they form coloured ions, complexes and show paramagnetic property, similar to d-block elements.
■ They are naturally available in very small quantities and are called rare earths.
■ Trans-uranic elements ( Z > 92) are all synthetic.
■ They also form complexes and interstitial compounds.
■ They show a great deal of similarity among themselves in their properties.
■ Actinide elements are radioactive.
■ Actinides show more number of oxidation states compared to lanthanides.
■ Many of actinoid elements have been made only in nano gram quantities by nuclear reactions.
3.6.2 Classification of Elements into Types
All elements are classified into four types, considering their properties based on the configuration in different shells of the elements.
Type I elements are called inert gases.
Type II elements are called representative elements.
Type III elements are called transition elements.
Type IV elements are called inner transition elements.
The difference in the electronic configuration of different shells of these four types of elements determines their chemical behaviour.
Inert Gas Elements
Elements in which the outermost s and p subshells are completely filled are called inert gas elements. Due to their low reactivity, elements of group 18 are called inert gases.
The first compound of xenon was reported with fluorine in 1962. Thereafter, these elements are called noble gases. The natural availability of these elements is very less, and they are also called rare gases.
Representative Elements
Elements in which the outermost s and p subshells ( n th shell) are incompletely filled are called representative elements.
Elements of groups 1, 2, 13, 14, 15, 16 and 17 (all ‘A’ groups) constitute this type. All the s- and p- block elements, except group 18, are known as representative elements. They are also called normal elements.
The general electronic configuration of these elements is ns1–2 np0–5. These elements undergo chemical reactions to acquire the nearest inert gas configuration either by losing or gaining or by sharing electrons, and some of them may even get pseudo inert gas configuration. These elements are so named because they represent most of the chemical reactions known. The most reactive element is fluorine.
Many non-metals, metalloids, and some metals are present in this type of element.
Transition Elements
1. Elements in which the ultimate and penultimate shells n and ( n –1) shells are partially filled are called transition elements.
2. These elements are so named as they represent properties intermediate between most electro-positive s-block elements and most electronegative p-block elements. They are the d-block elements. All the transition elements are metals and solids.
3. The general electronic configuration of transition elements is ( n -1)d 1–10 n s 1or2 They possess incompletely filled ( n – 1)d orbitals in their higher oxidation states. The elements of group 12 (Zn, Cd, and Hg) are not included in the transition elements as they have a completely filled penultimate shell in both atomic and ionic states. The elements of group 11 (Cu, Ag, and Au) also have completely filled
penultimate shell. They are called typical transition elements as some of their ions resemble the transition elements.
4. The transition elements exhibit characteristic properties due to (1) the partially filled d-orbitals in the penultimate shell, (2) small atomic size and (3) high nuclear charge. Some special properties of the transition elements are listed below:
i) Transition elements exhibit variable oxidation states, e.g., : Fe +2 , Fe+3
ii) They form coloured compounds and coloured ions in solution due to d-d transition of electrons by absorbing the visible light.
iii) They are paramagnetic. Iron, cobalt and nickel are ferromagnetic.
iv) They possess catalytic activity. Nickel is used as catalyst in the hydrogenation of oils and iron in Haber’s ammonia synthesis. Molybdenum acts as promoter in Haber’s process.
v) They form alloys (e.g., brass, bronze, German silver) and interstitial compounds. Hydrogen is occluded on palladium.
vi) They form complex compounds and alloys.
vii) Common oxidation state of transition elements is +2.
viii) They have high melting point, boiling point and densities.
Inner Transition Elements
1. Elements in which the ultimate, penultimate and antipenultimate shells, n , ( n – 1), and ( n –2) shells are partially filled are called inner transition elements.
2. These elements are so named as they represent a transition of physical and chemical properties among transition elements. They are the f-block elements.
There are two series of inner transition elements, namely lanthanides and actinides, corresponding to the 4f and 5f series. All the inner transition elements are metals.
3. The general electronic configuration of inner transition elements is (n-2)f1-14 (n-1) d0 or 1 ns2. The elements in each series of inner transition elements closely resemble in their properties. Hence, their isolation and separation by the usual methods is very difficult. For the sake of convenience, the inner transition elements are placed as separate f-block below the main body of the long form of the peri odic table.
4. Inn er transition elements are all heavy metals. Their natural availability is either rare or not at all available. Actinides are synthetic, except Th and U.
5. They exhibit variable oxidation states and magnetism. They are also known to form complex compounds. Common oxidation state of inner transition elements is +3.
3.6.3 Classification into Metals and Non-Metals
1. Elements can be also be divided into metals, non-metals, and metalloids based on their properties. Among the elements in the periodic table, more than 75% are metals. These metals appear on the left side of the periodic table.
2. Metals are usually solids at room temperature. [Mercury is an exception, and galium and caesium have very low melting points (303 K, 302 K)]. They usually have high melting and boiling points. They are good conductors of heat and electricity. Metals are malleable and ductile.
3. About a dozen elements are non-metals. They are placed at the top righthand side of the periodic table. Non-metals are usually gases at room temperature. Some
of them are also solids, with low melting and boiling points (boron and carbon are exceptions). These are poor conductors of heat and electricity (except graphite). Most of the non-metallic solids are brittle and are neither malleable nor ductile.
4. Some other elements exhibit both metallic and non-metallic properties. Such elements are called metalloids. Metalloids are placed in the p-block. Examples include Ge, As, Sb, Se, Te, etc.
Q. The element Z = 117 has not been discovered. In which group would you place this element? Give the electronic configuration.
Sol. The element with Z = 117 would belong to the halogen family (group 17) and the electronic configuration would be.
Q. What would be the IUPAC name and symbol for the element with atomic number 120?
Sol. The roots for 1, 2 and 0 are un, bi and nil, respectively. Hence, the symbol and the name respectively are Ubn and Unbinilium.
Try yourself:
An element E easily forms the ion, E –2. What do you understand by this with respect its group number in modern periodic table and its nature?
Ans: Its group number is 16 and its a nonmetallic element.
TEST YOURSELF
1. The period in which s-block, p-block, and d-block elements are present is
(1) period 1 (2) period 6 (3) period 2 (4) period 3
2. Which of the following configurations corresponds to an inert gas?
The word ‘periodic’ means repeated at a regular interval. Periodicity in properties means that the same properties of elements reappear at definite intervals when the elements are arranged in the order of their increasing atomic numbers. The intervals of atomic numbers are 2,8,8,18, 18, and 32.
Atoms of the elements present in a group of the periodic table have similar outer shell configuration. Repetition of similar valence shell configuration is the cause of periodicity in properties.
In the periodic table, the properties of elements change gradually with a change in their electronic configurations. This trend repeats itself at regular intervals. This repetition of a character is called ‘periodicity’ and such properties are called 'periodic properties'.
Some of the properties which mainly depend on the electronic configuration of elements, such as valency, effective atomic number, screening effect, atomic radius, ionic radius, ionisation potential, electron affinity, electronegativity, metallic nature, oxidation
and reduction ability, acidic or basic nature of the oxides, etc., follow the general trend of periodicity. They are called periodic properties. These properties are especially important in s- and p- block elements.
Properties, like specific heat, refractive index, etc., are not called periodic properties. These properties are not related to the electronic configuration of the elements.
Shielding Effect (or) Screening Effect:
The electrons existing between the nucleus and valences shell are called intervening electrons. These intervening electrons decrease the force of attraction between the nucleus and the outermost shell electrons. This decreasing force of attraction due to presence of the intervening electrons is called ‘shielding effect’.
The actual nuclear charge felt by the electrons in outermost shell is decreased by the quantity, s(sigma), called 'screening constant'. The actual nuclear charge experienced by the electrons in the outermost shell is called effective nuclear charge, Zeff.
Zeff= Z– s
Factors Affecting the Magnitude of Zeff
1. Number of intervening electrons:
More the number of intervening electrons between the nucleus and the outer most shell, more is the magnitude of s and lesser is the magnitude of Zeff.
In a group of modern periodic table, Zeff decreases down the group, due to increase in the number of intervening electrons.
2. Atomic size:
With the increase in the size of atom, the value of Zeff decreases. Thus, in a group of modern periodic table, Zeff decreases down the group. Along a period, Zeff increases from left to right hand size.
Calculation of s and Zeff - Slater’s Rule
Part-I: Calculate of the value of s for an electron in ns (or) np of an atom (or) ion:
1. Each of the remaining electrons in the nth shell makes a contribution of 0.35 to the value of s .
2. Each electron in ( n -1) th shell makes a contribution of 0.85.
3. Each electron in the next inner shells makes a contribution of 1.0.
4. No contribution to the value of s by the electrons in higher energy shells than that of electron for which s is calculated.
5. For 1s electron, the contribution is 0.3 from other single electron.
6. For the electron in ns or np sub-shell
s = 0.35 (remaining number of electrons in nth shell)
+ 0.85 (number of electrons in ( n-1)th shell) + 1.0 (number of electrons in inner shells)
Example:
Calculate the of Zeff of 4s1 electron of K (Z =19). Electron configuration of K = 1s22s22p6 3s23p6 4s1 10-electrons 8-electrons
s value for 4s1 electron of K = (0.35 × zero) + (0.85×8) + (1.0 × 10) = 16.8
Zeff of 4s1 electron = 19–16.8 = 2.2
Part-II: Calculation of the value of s for an electron in (n-1) d:
1. ns-electron does not make any contribution to the value of s.
2. Each remaining electron of (n-1)d makes a contribution of 0.35.
3. Each electron of (n-1) and (n-1) p and all other inner shells makes a contribution of 1.0.
4. For (n-1) d electron,
s = 0.35 (number of remaining electrons in (n-1)d)
+ 1.0 (number of electrons in inner shells)
Example:
Calculation of Zeff of 3d electron of Cu (z=29) Electron of Cu = 1s2 2s2 2p6 3s2 3p6 3d10 4s1 18-electrons 10-electrons
s for 3d-electron = 0.35(9) + 1(18) = 21.5
Zeff for 3d-electron of Cu = 29–21.15 = 7.85
3.7.1 Atomic Radius
Atomic radius is the distance between the centre of the atomic nucleus and the electron cloud of the outermost energy level.
Atomic radius is also commonly referred to atomic size. However, atomic size is truly regarded as the diameter of the atom.
Practically it is not convenient to isolate an individual atom and determine its atomic radius. Atomic radius is influenced by many factors, like
■ nuclear charge,
■ principal quantum number,
■ bond character,
■ multiplicity of bond.
■ oxidation state of atom,
■ coordination number of the atom, etc
The following radii are experimentally significant:
■ crystal radius (or atomic or metallic radius)
■ van der Waals radius, and
■ covalent radius
Metallic Radius (Atomic Radius)
All metal s are crystalline. Hence, metallic radius is also called crystal radius. Metallic
3: Classification of Elements
radius is defined as one-half of the internuclear distance between the centres of the two adjacent metal atoms in the metallic crystal
For example, the internuclear distance between two adjacent atoms of sodium crystal is 3.72A°. Therefore, the metallic radius of sodium is 1.86A°. The radii of some metals are listed in Table.3.9
Table 3.9 Atomic radii of some metals
Van der Waals radius
When atoms or molecules of an element are closely packed in the solid state, they are held together by very weak attractive forces. These forces are called van der Waals forces. Due to these forces, one atom can approach another without forming a covalent bond. Hence, it is also called non-bonded atomic radius. It is measured for molecular substances and inert gases in solid state.
Van der Waals radius is half of the distance between the centres of two atoms of different molecules which are close st to each other in solid state.
For example, the internuclear distance between atoms of two nearest chlorine molecules is 3.6A°. Therefore, the van der
Waals radius of chlorine is 1.8A°. The van der Waals radius of some non-metals is shown in Table.3.10.
Table 3.10 van der Waals radius of some elements
van der Waals
Covalent Radius
Covalent radius is one half of the internuclear distance between atoms of a homonuclear covalent molecule.
It is defined for atoms that are bonded with a similar atom. It is calculated by electron diffraction or x-ray diffraction or other spectroscopic techniques. It is measured for non–metals.
It is measured in Angstrom units, A°.
1A0 = 1 × 10–8 cm = 1 × 10–10 m = 100 pm = 0.1 nm
1nm = 10–9 m = 10–7 cm = 10 A0
For example, the internuclear distance between atoms of chlorine molecule is 1.98 A 0 . The covalent radius of chlorine is 0.99 A 0 .
The covalent radius and van der Waals radius of chlorine are shown diagramatically in Fig.3.4.
Fig.3.4 Radii of chlorine
Covalent radii of some elements are given and compared with their atomic radii in Table.3.11.
Table 3.11 Atomic and covalent radii of some elements
If a molecule contains atoms of two different elements, the sum of the covalent radii is expected to be equal to the internuclear distance. The covalent radius of an atom in a heteronuclear molecule is defined as the distance between the nucleus of an atom and the mean position of one shared pair of electrons between the bonded atoms.
If the bonded atoms have more than one bond present between them, the attraction between the atoms will be more. Due to the increased attraction between the bonded atoms, the internuclear distance decreases and the covalent radius decreases.
Comparison of Covalent Radii with Van der Waals' Radii
C ompared to the theoretical atomic radius, covalent radius of an atom is about 20 percent shorter. The formation of covalent bond involves overlapping of atomic orbitals. As a result of this, the internuclear distance between the bonded atoms is decreased.
The van der Waals radius of an atom is usually 40 percent larger than the covalent radius. This is because the non-bonded atoms of the molecules are held together by weak attractive forces. Usually inert gas elements do not combine themselves or with other elements and are monoatomic.
For elements that exist as monoatomic molecules, the van der Waals radius is taken as the atomic radius.
The covalent and van der Waals radii of some non-metallic eleme nts are co mpared in Table.3.12.
Table 3.12 Covalent and van derwaals radii of some elements
Variation of Atomic Radius in Groups
In a gr oup of modern periodic table, the atomic radius gradually increases from top to bottom.
Reason: On moving down the group, the number of shells increases. The nuclear charge also increases. The effect of increase in the number of shells is more pronounoced than the effect of increase in the nuclear charge. Consequently, the size of atom increases (though there is an increase in shielding effect).
It may be noted that atomic radius (crystal) of a metal is greater than its covalent radius.
Table 3.13 Radii of alkali elements
Atomic radius is least for hydrogen among all elements and highest for caesium among the available elements.
The increase in the atomic radius in groups, with increasing atomic number, is shown diagramatically in Fig.3.5.
(Z)
Fig.3.5 Atomic radii in group
(Z)
Fig.3.6 Atomic radii in period
Variation of Atomic Radius in Period
In a period of modern periodic table, atomic radius decreases from left to right hand side upto noble gas element.
Reasons:
1. On moving across the period, the atomic number increases and, hence, effective nuclear charge increases, while the added electron enters the same shell. Consequently, electrons are increasingly attracted towards the nucleus. Thus, atomic radius decreases.
2. From halogen to noble gas, atomic radius increases in the period. This is because (i) For noble gas, van der waals radius, is taken as its radius (ii) in each noble gas atom, there are completely filled subshells causing inter electronic repulsions.
Radii of second period elements are listed in Table 3.14.
The trend in the decrease of atomic radius is valid in any period from alkali element to halogen.
Table 3.14 Radii of 2nd Period Elements
The decrease in the atomic radius in periods, with increase in atomic number, is shown diagramatically in Fig.3.5
Every period starts with an electron entering s-sub shell of a new orbit. When the next electron enters in the same s-sub shell, the resulting decrease in the atomic radius is significant. But the decrease in the radius with the p, d, and f-sub shells are being filled is normal. Radius of representative and transition elements are compared in Table 3.15
Table 3.15 Radii of representative and transition elements
Variation of Atomic Radius in Transition Elements
In a period of transition elements also, there is d ecrease in atomic size from left hand to right hand side. But there is a slow decrease in atomic size.
Reason: In a period of transition elements, atomic number and, hence, nuclear charge increases gradually from left hand to right hand side. The differentiating electrons are gradually added to (n-1)d-subshell, thus shielding the nucleus from attracting valence electrons, (n-1)d-electrons show poor shielding effect. But their presence makes the atomic radius to decrease slightly.
Variation of Radius in Inner Transition Elements
1. In a series or a period of lanthanides, there is a gradual decrease in atomic size from left hand side to right hand side. This is called lanthanide contraction.
Reason: In a series of elements, atomic number and, hence, nuclear charge increases, while electrons are not added to outermost shell, they are added to (n–2)f-sub-shell. Due to peculiar shape of f-orbitals, they provide poor screening for outer electrons from the nuclear attraction.
2. Due to lanthanide contraction, the increase in the size of an atom from 4d element to 5d element in a group is less significant.
3. In actinide also, atomic radius decreases gradually across this series. This is called actinide contraction.
Reason: The decrease in atomic sizes due to gradual increase in atomic number or nuclear charge and poor shielding by 5f-electrons.
Radii of trivalent ions of lanthanides is shown in Fig.3.7
Ionic Radius
The distance between the nucleus and the point upto which the nucleus shows its influence in an ion is called ionic radius.
Cationic Radius
Cation is a positively charged ion. It is formed by the loss of one or more electrons by a neutral atom. The magnitude of the nuclear charge remains the same upon cation formation, while the number of electrons decreases. The same nuclear charge in a cation acts on less number of electrons. The effective nuclear charge per electron increases and, hence, electrons are more strongly attracted in a cation.
The radius of cation is less than that of the parent atom from which it is formed. This is illustrated by the data in the Table 3.16
Table 3.16 Atomic and cation radii of some elements
Fig. 3.7 Ionic radii (in A°) of lanthanides
Anionic Radius
Anion is a negatively-charged ion. It is formed by the gain of one or more electrons by a neutral atom. The magnitude of the nuclear charge remains the same upon anion formation, while the number of electrons increases. The same nuclear charge in an anion acts on more number of electrons. The effective nuclear charge per electron decreases and, hence, electrons are less strongly attracted in an anion.
The radius of anion is more than that of the parent atom from which it is formed.
Example: Radius of 2 OOO >>
Cation radius or anion radius increases gradually from top to bottom in a group.
Radii of Isoelectronic Species
Ions that have equal number of electrons are called isoelectronic ions. Isoelectronic species have nuclei with different charges. Since the negative charge due to electrons outside the nucleus is the same in isoelectronic ions, the radii depend upon the nuclear charge. The smaller the nuclear charge, the greater is the radius. Radii of an isoelectronic ion series are listed in Table 3.17.
Q. Compare the radii of H atom, H + ion and H– ion
Sol. H+ is the nucleus of H atom. Its radius is very small.
H– ion has number of electrons more than number of protons. Its size is more than that of H atom.
The radius is in the order : H + < < H < H–
Q. Which is a bigger ion among Na+, F–, O2– and Mg2+? Why?
Sol. O 2– is bigger ion among the given four. Among isoelectronic ions, the more the negative charge on the ion, the more is its size.
The order is O–2 > F–1 > Na+ > Mg+2
Try yourself:
If the van der Waals radius of hydrogen is 120 pm, what should be the intermolecular distance in solid Hydrogen?
Ans:240 pm
3.7.2 Ionisation Enthalpy
Electrons in an atom are attracted by the positively charged nucleus. To remove an electron from an atom, energy has to be supplied in order to overcome the attractive forces. This energy is known as ionisation enthalpy or ionisation energy or ionisation potential. Energy is always required to remove electrons from an atom and, hence, ionisation enthalpies [ D H] are always positive.
Table 3.17 Atomic and ionic radii of some isoelectronic series
()() +−+→+ gg MIEMe
The minimum energy required to remove elect ron from an isolated gaseous atom to convert it into a gaseous ion is called ionisation energy.
Here, M is the isolated atom, I is the ionisation, potential and M + is the cation formed by the loss of one electron from M.
The units of ionisation enthalpy = kJmol-1 or
kcal mol–1
1eV/atom= 23.06 k cal mol –1
1eV/atom = 96.43 kJ mol –1
1eV/atom = 1.602 × 10 –19 J atom–1
Successive Ionisation Potentials
Once an electron is removed from a neutral atom, it is possible to remove more electrons one after the other. Therefore, it is essential to define the ionisation energy for each electron removed successively. Theoretically, number of successive ionisation potentials are equal to the atomic number of the element. However, experimentally it is limited to the number of valence electrons present in the outermost shell of the atom.
First Ionisation Potential
It is the energy required to remove electron from an isolated neutral gaseous atom.
M(g) + IE1 → M+(g) + e–
Commonly, first ionisation potential is called ionisation potential. It is denoted by I 1. The utilisation of I1 results in the formation of an unipositive ion.
Second Ionisation Potential
It is the energy required to remove electron from an isolated unipositive gaseous ion. It is denoted by I2. The utilisation of I2 results in the formation of dipositive ion.
()() 2 2 ++− +→+ gg MIEMe
In a similar manner, the fourth ionisation
potential, the fifth ionisation potential etc., are defined.
Third Ionisation Potential
It is the energy required to remove electron from an isolated dipositive gaseous ion. It is denoted by I3. The utilisation of I3 results in the formation of tripositive ion.
()() 23 3 gg MIMe ++− +→+
In a similar manner, the fourth ionisation potential, the fifth ionisation potential, etc., are defined.
IE1, IE2, IE3, IE4, IE5, etc., are collectively known as successive ionisation potential values.
In general, the increasing trend in these values is : IE1 < IE2 < IE 3 ...... IEn
The IE1, IE2, and IE3 values of aluminium are, respectively, 578, 1820, and 2750 kJ/mole–1
When an electron is removed from a neutral gaseous atom M, it becomes a unipositive ion M+. The number of electrons in M+ ion will be less than the number of protons present in the ion. The effective nuclear charge per electron increases. The energy required to remove an electron from a unipositive ion will be more than that of a neutral atom. Hence, IE2 is greater than IE1. Similarly, IE3 is greater than IE2, indicating that it is more difficult to remove successive electrons.
The successive ionisation potential values of the elements of second period are listed in Table 3.18.
Factors Influencing Ionisation Potential
Ionisation potential values of elements depend on several factors. The important factors are listed below.
1. Nuclear charge.
2. Atomic radius.
3. Screening effect.
4. Penetration ability.
5. Electronic configuration.
6. Charge on the ion.
Table 3.18 Successive ionisation potentials (kJ mol-1) of some elements
Nuclear Charge
The ionisation potential is directly proportional to the nuclear charge. As the atomic number increases, the charge on the nucleus increases the number of electron shells remains same. As the nuclear charge increases, the outer electrons are more strongly held by the nucleus due to attractive forces. As a result, the energy required to remove the electron from the atom is more, i.e., ionisation energy will be high. Hence, ionisation energy increases, as the nuclear charge increases. Table. 3.19
Table 3.19 Ionisation potential of second period elements
Atomic Radius
As the number of shells increases, the distance between the outer electron and the nucleus increases. The attractive force between the nucleus and the outer electron decreases. It becomes easier to remove the electron and hence, the ionisation potential decreases.
The ionisation potential is inversely proportional to the atomic radius.
Screening Effect
The electrons present in the inner shells act as a screen between the nucleus and the outer electrons. The attractive force of the nucleus on the outer electrons is decreased due to these screens. This is referred as screening effect.
The screening effect increases with an increase in the filled inner shells and also with an increase in the number of electrons present in the inner shells.
The screening efficiency of the orbitals is in the order of s > p > d > f
The ionisation potential is inversely proportional to the screening effect.
Penetrating Power of Orbitals
Since orbitals hold electrons, they are attracted towards the nucleus. The orbital electron cloud moves towards the nucleus. This is called penetration ability of the orbital. The penetration ability of an electron towards the nucleus depends upon the eccentricity of the orbit containing the orbital. In a given energy state, the eccentricity of the orbit decreases with an increase in the azimuthal quantum number, and the penetration ability decreases.
The order of penetration ability of orbitals in an energy state is given as : s > p > d > f.
If the penetration ability of an orbital is more, the electrons will be closer to the nucleus and will be held firmly. The energy required to remove the electron will be more.
Ionisation potential is directly proportional to penetrating power of orbital.
Electronic Configuration
Sub-shells with exactly half-filled or completely filled with electrons are symmetrical. Atoms with symmetrical orbitals are stable. Greater amount of energy is required to remove an electron from such a stable atom. Hence atoms with stable electronic configuration require more energy for the removal of electrons. Therefore, ionisation potential is high for the atoms with stable electronic configuration.
Atoms of inert gases are stable with outer s- and p-orbitals completely filled. Hence, the energy required to remove the electron is very high.
Net Charge on the Ion
Removal of electron or electrons leads to the formation of ions with positive charge. The energy required for the removal of an electron from an atom is less than that required from unipositive ion. In an unipositive ion, the effective nuclear charge per electron is more. Hence the ionisation potential will be more. Similarly, the energy required for the removal of an electron from unipositive ion is less than that required from a dipositive ion. As the effective nuclear charge per electron in a dipositive ion is much more, the ionisation potential will be further increased.
Thus, with an increase in the positive charge on the ion, the energy required to remove the electron increases.
Ionisation potential is directly proportional to the magnitude of positive charge on the ion.
Variation in Groups
In a group, from top to bottom, the ionisation potential gradually decreases.
Reason : As we move down in a group, the outer electrons occupy new shells. Addition of new shells increases the distance between the nucleus and outermost shell and it increases the screening effect. As a result of these, the attraction between the atomic nucleus and outermost electron decreases and the ionisation potential decreases. The gradual decrease in the ionisation potential can be observed from the values of alkali metals shown in Table 3.20.
Table 3.20 Ionisation potential values of elements of group IA
Fig.3.8. Variation of Ionization Enthalpy with Atomic number
Variation in Periods
In a period, from left to right, the ionisation potential gradually increases.
Reason: As we move across the period, the atomic number of the elements increases continuously. The nuclear charge increases, but the electrons are added into the same shell.
The distance between the nucleus and the outer electron decreases. Therefore, the energy required to remove the electron increases.
The gradual increase in the ionisation potential can be observed from the values of elements of third period in Table 3.21.
■ In any period, the element with least ionisation potential is alkali element and with highest ionisation potential is inert gas element.
■ The elements present in the groups 2, 15 and 18, have higher ionisation potential values than those of the adjacent elements of the same period. The elements present in the groups 1, 3, and 16 have ionisation potential value smaller than that of the adjacent elements of the same period.
■ The ionisation potential is the highest for helium among all elements. The value is the least for caesium among the available elements. When ionisation potential values are plotted versus atomic numbers, the ionisation potential curve is obtained as shown in Fig.3.9 and Fig.3.10.
Table 3.21 Ionisation potential of third period
Fig.3.9 Ionisation potential curve for the first eleven elements
Number (Z)
Fig.3.10 Ionisation potential curve for the first eleven elements
The IE of He is higher among inert gases (or) all elements (2467.2 kJ/mole).
Comparison of Be and B
IE1 of Be is greater than that of B. Electronic configuration of Be is 1s 2 2s2.
Beryllium atom contains completely filled outer s-orbital. The configuration has greater stability. Moreover, the s-orbital has greater penetrating power. Hence, energy required to remove electron from beryllium atom is high.
E lectronic configuration of boron is 1s 2 2s2 2p1.
Boron atom contains just one electron in the outer p-orbitals. It is less stable. The penetrating power and screening effect of p-orbital is relatively less. Hence, energy required to remove electron from boron atom is less.
IE1 of Beryllium is 899 kJ mol–1, while that of Boron is 801 kJ mol –1
Similarly IE1 of Magnesium (738 kJ mol–1) is greater than that of Aluminium (577 kJ mol–1).
Comparision of N and O
IE1 of Nitrogen is greater than that of Oxygen. Electronic configuration of Nitrogen is 1s 2 2s2 2p3.
Nitrogen atom contains half-filled p-subshell. Hence, it has extra stability. Energy required to remove electron from nitrogen atom is high.
Electronic configuration of Oxygen is 1s 2 2s2 2p4.
Oxygen atom has neither completely filled nor exactly half-filled p-subshell. The configuration is less stable. Hence, energy required to remove electron from oxygen atom is less.
IE1 of Nitrogen is 1403 kJ mol–1, while that of Oxygen is 1314 kJ mol –1
Similarly IE 1 of Phosphorus is (1010 kJ mol –1) is greater than that of Sulphur (999 kJ mol–1).
Q. The successive ionisation enthalpies of an element M are 5.98, 18.82, 28.44, 119.96, 153.77....eV/atom. What is the formula of chloride of M?
Sol. Observing the IE1, IE2, IE3, IE4, IE5, .... it is noticed that there is a sudden jump from IE3 and IE4.
This observation gives the idea that the element has 3 electrons in the outer most shell, as there is a great difference between 3rd and 4th ionisation enthalpies.
M3+ state is stable and valency is 3.
Formula of chloride of M is MCl 3.
Q. The ionisation enthalpy of sodium is 5.14 eV. How many kcal of energy is required to ionise all atoms present in one gram of gaseous Na atoms?
Sol. 1 eV atom–1 = 23 kcal mol–1
Energy required to ionise all atoms of 23 g (one mole) of gaseous Na atoms = 23 × 5.14 kcal
Energy required for ionisation of all atoms present in one gram of gaseous Na atoms = = 5.14 kcal.
Try yourself:
The 1 st, 2 nd and 3 rd ionisation enthalpies of Aluminium are 577, 1816 and 2744 kJmol –1 . What is the amount of energy required to convert one gaseous Aluminium atom to gaseous Al3+?
Ans:8.53 × 10–18 J/atom..
3.7.3 Electron Gain Enthalpy
In the process of addition of electron to neutral isolated gaseous atom, a certain amount of enthalpy change is involved. This is called Electron gain enthalpy ( D Heg)
XeX,HH +→∆=∆
()() atom ion eg gg
Units: It is expressed in kJ/mol or kcal/mol.
Negative electron gain enthalpy:
The release of energy in the electron gain process is represented by negative D H eg .
Halogens show negative D H eg
Positive electron gain enthalpy:
When atom is reluctant for the addition of a new electron, energy is to be spent in electron addition to the atom. In that case, D H eg is positive.
As electron affinity and ionisation potential are defined at absolute zero, at any other temperature heat capacities of reactants and products have to be taken into consideration.
5 , 2 ∆=−−
where eg H∆ = electron gain enthalpy
A e = electron affinity
Similarly, 5 HI.PRT 2 ∆=+ , where H∆ is ionisation enthalpy
The ionisation potential of a neutral atom A is equal in magnitude with the electron affinity of A+ ion. However, they have opposite signs.
AIAe:AeAE +−+− +→++→+
I n the above equations, I and E are numerically same but the sign is opposite.
Successive Electron Affinities
S uccessive addition of electrons to atomic species involve energy changes.
First Electron Affinity
First electron affinity is the energy involved when electron is added to isolated gaseous neutral atom.
Commonly, first electron gain enthalpy is called electron gain enthalpy. It is denoted by E1. The utilisation of E1 results in the formation of uninegative ion.
XeXE +→+
()() 1 gg
Second Electron Affinity
It i s the energy absorbed when electron is added to the outermost shell of an isolated uninegative ion. It is denoted by E 2 . The utilisation of E 2 results in the formation of dinegative ion.
()() 2 gg XeEX ++→
In a similar manner, the third electron affinity and the fourth electron affinity are
defined. E1, E2, E3 etc., are collectively known as successive electron affinity values.
11
OeO142kJmole +→−
12 1
OeO702kJmole +→+
Two successive electron affinities of group 16 elements are listed in Table 3.22.
Table 3.22 First and secondelectron affinity values of group 16 elements
8.09
– 2.08 + 6.12
– 2.02 + 4.36
– 1.97 + 3.03
Second electron gain enthalpy of elements is positive because energy is required to overcome the repulsions between the uninegative ion and the added electron.
Fac tors Influencing Electron Affinity
Electron affinity depends on some of the properties of atoms.
Nuclear Charge
With an increase in the atomic number of the element, the nuclear charge increases. The ability of the nucleus to attract the incoming electron increases. Hence, the amount of energy liberated increases.
Therefore, the electron affinity is directly proportional to the nuclear charge.
Atomic Radius
With an increase in the number of shells present in an atom, the atomic radius increases. The distance between the nucleus and the outer shell increases. The attraction between the incoming electron and the nucleus decreases. Hence, the energy which is liberated decreases.
Therefore, the electron affinity is inversely proportional to the atomic radius.
Electronic Configuration
The elements with half-filled and completely filled sub-shells are stable. They show only a little tendency to accept an additional electron. Thus, the energy liberated with the addition of electron to the atoms of these elements will be either zero or very small.
Therefore, elements with stable configuration have low electron gain enthalpies.
Variation in
Groups
In a group, from top to bottom, electron gain enthalpy gradually decreases.
Reason: On going down the group, the atomic radius and nuclear charge increase. With an increase in the radius, attraction of the added electron by the nucleus decreases and the electron affinity decreases.
The effect of increase in the nuclear charge is less predominent than the increase in the atomic radius.The trend in the electron affinity in a group can be observed from the data in Table 3.23
Table 3.23 Electron affinity of halogens
The element with most negative electron gain enthalpy is chlorine, and the one with the least negative electron gain enthalpy is phosphorus.
Anomalous Behaviour Down the Group
Electron gain enthalpy of F is less negative (–328 kJ/mol) than that of Cl (–349 kJ/mol).
Reason : When electron is added to F, the added electron goes to smaller, n = 2 quantum level and suffers significant repulsion from the other electrons of this level. But in Cl, added electron goes to n = 3 quantum level, thus, occupies larger region of space and experiences weaker electron-electron repulsions.
Similarly, electron gain enthalpy of ‘O’ is less negative than that of S.
Variation in periods
In a period, from left to right, electron affinity gradually increases.
Reason : On moving across the period, the atomic number increases, the nuclear charge increases and the radius decreases. Because of these, the attraction between the added electron and the nucleus increases and electron gain enthalpy increases
Anomalous Behaviour Across the Period:
1. The electron affinity of inert gases is large positive charge. This is due to their stable octet configuration. The electron affinity of group 15 elements are smaller compared to the corresponding elements of group 14 and 16. ‘N’ has positive gain enthalpy.
2. The electron gain enthalpies of alkali elements are small, but negative. These values of alkaline earth elements are small. Be and Mg have positive gain enthalpy. These values are regarded as zero for all practical purposes.
The trend in the electron gain enthalpy values in a period can be observed from the data in Table 3.24
Table 3.24 Electron gain enthalpy values of some elements (kJ mol–1)
73
Q. Write the descending order of electron affinity values of chalcogens.
Sol. Decreasing order of electron affinity values of chalcogens: S > Se > Te > O. Electron affinity of oxygen is less because oxygen has small atomic size and the added electron experiences greater repulsion on oxygen atom.
Q. Process (A) : F2(g)+ 2e– → 2F–(g) Process (B) : Cl2(g) + 2e– → 2Cl–(g). Which of these processes is easier? Why?
Sol. F2(g) + 2e– → 2F–(g) is easy. Though electron gain enthalpy of Cl(g) to give Cl–(g) is more than that of F(g) to give F–(g), the bond dissociation of F2(g) is very less, compared to that of Cl2(g)
Try yourself:
The quantity of energy released when one million gaseous iodine atoms are converted into I–(g) ions is 5.0 × 10–16 kJ. In this question: I(g) + e – → I–(g) + Q kJ mol–1 . What is the value of Q?
Ans: 301
3.7.4 Electronegativity
Electronegativity is an important relative property of elements.
A covalent bond is formed by sharing of an electron pair between two atoms. The shared electron pair will be attracted by the nuclei of both the atoms. The sharing will be unequal if the pair of electrons is present between atoms of different elements. The electron pair will be closer to the atom that attracts the pair more. This relative tendency of attracting electrons is termed as electronegativity. It is defined as the tendency of the atom of an element to attract the shared electron pair(s) more towards itself.
Electronegativity is not measurable. So, many quantitative scales are used to calculate electronegativity.
1. Mulliken scale: According to Mulliken,
Electronegativity = I.EE.A 2 + ; here, both
IE and EA are in eV unit.
2. Pauling scale: Linus Pauling determined electronegativities with the help of bond dissociation energies.
The Pauling’s equation for the determi nation of electronegativity is written as,
AB XX0.208−=∆ where D is in k.cals.mol-1
AB XX0.1017−=∆ where D is in kJ.mol-1
Here, XA is electronegativity of element A, XB is electronegativity of element B and D is the bond stabilisation or resonance energy
D = Experimental bond energy - Calculated bond energy
EA–A, EB–B and EA–B are the bond dissociation energy values of the bonds A–A, B–B and A–B, respectively.
Pauling arbitrarily assigned a value of 2.1 to hydrogen and determined the electronegativity of fluorine as 4. Considering this, the electronegativity values of other elements are calculated. Pauling’s values of electronegativity are listed in Table 3.25
Factors Influencing Electronegativity
Electronegativity of an element depends upon various factors, some of which are given below.
Nuclear Charge
Greater the nuclear charge of the bonded atom, greater is its ability to attract the electron and greater is its electronegativity.
Atomic Radius
Greater the radius of the bonded atom lesser is the electronegativit y of the atom.
Table 3.25 Pauling’s electronegativity values of representative elements
Penetration Ability
The penetration ability of bonded orbitals is in the order: s > p > d > f. In the case of hybrid orbitals, the greater the s-character of the orbital, the greater is the electronegativity of the element.
s-character
As s-character of hybrid orbital increases, the electro negativity increases.
Order of electronegativity: sp > sp 2 > sp3
Variation in Groups
In a group, from top to bottom, the electronegativity gradually decreases.
Reason: This decrease in electronegativity is due to the increase in the atomic radius and screening of the nuclear charge by the filled inner shells.
Variation in periods
In a period, from left to right, electronegativity gradually increases up to halogen.
Reason: This increase in electronegativity in a period is due to the decrease in the atomic radius and increase in the nuclear charge. Consequently, the attraction of the nucleus on the bonded electrons increases.
Variation of values of electronegativity with atomic number of elements is shown graphically in Fig.3.11.
Fig. 3.11 Variation of electronegativity
The periodic trends in electronegativity can be observed from the data listed in Table 3.26 . In any period, the element with
highest electronegativity is halogen, while elect ronegativity of inert gases is taken as zero as they are not actively involved in bond formation.
The element with highest electronegativity is fluorine. The element with least electronegativity is caesium. The values in Pauling’s scale for these elements are 4.0 and 0.7, respectively.
Applications of Electronegativity
1. Electronegativity is a direct measure of the non-metallic nature. Elements with electrone-gativity values of 2 or more in Pauling’s scale are generally known as non-metals.
2. Electronegativity may be related to the chemical reactivity of an element. Fluorine is the most reactive element and helium is the least reactive element. Elements with similar electronegativity may possess similar chemical activity.
3. Electronegativity is useful in predicting the nature of the chemical bond. If the difference in the electronegativity of bonded atoms is 1.7 or more, the bond formed is generally an ionic bond.
4. If the difference in electronegativity between the bonded atoms is 1.7, then the chemical bond has 50% ionic character and 50% covalent character.
Hanny and Smith Relationship
Percentage of ionic character
= 16 (XA – XB ) + 3.5(XA – XB)2, where XA and XB are the electro-negativities of two atoms A and B.
Q. How is the nature of covalent bond between two atoms predicted?
Sol. Nature of the covalent bond between two atoms is predicted based on the difference in electro-negativity.
If there is no difference in the electronegativity of bonded atoms, the bond is pure covalent. If electronegativity difference arises between bonded atoms, the bond is polar covalent, and further, increase of difference of electronegativity leads to the formation of ionic bond.
Q. Bond energies of H 2 , Cl 2 , and HCl are, respectively, 104, 58, and 100 kcal mol –1 Calculate Pauling’s electronegativity of chlorine.
Sol. Average of bond energies of H2 and Cl2 is the calculated bond energy of HCl = 10458 2 + = = 81 kcal mol–1
Experimental bond energy of HCl = 100 kcal mol–1
D = Bond (resonance) stabilisation energy = 100 - 81 = 19 kcal mol–1
X1– X2 = 0.208 = 0.208 = 0.208 × 4.358 = 0.90
Since Pauling’s electronegativity of hydrogen is 2.1, that of chlorine = 2.1 + 0.9 = 3.0.
Try yourself:
Is it correct to say that pauling electronegativity of carbon is 2.5 in every compound?
Ans: No
3.7.5 Valency
The combining capacity of the elements is called valency.
The combining capacity of elements is often compared to that of hydrogen, chlorine, or oxygen. Valency is defined as the number of hydrogen atoms or the number of chlorine atoms or twice the number of oxygen atoms with which one atom of the element combines. Valencies of elements in some compounds are listed in Table 3.26.
Table 3.26 Valencies of some elements
Valency of all elements in a group is generally same.
The periodicity of valency of typical elements is provided in Table 3.27 in the form of hydrides, and oxides
1. Valency is zero for any element in its uncombined state. Valency of zero group elements under laboratory conditions is zero. The electrons present in the outermost shell of a representative element generally denotes its valency. These electrons are called valency electrons or valence electrons.
2. Generally, valency of elements with fewer number of electrons in the valence shell is the number of valence electrons. In case the number of valency electrons is four or more, the general valency of the elements is eight minus the number of valence electrons.
3. The m aximu m valency exhibited by an ele ment is its group number in roman notation. The highest valency ever known is eight. Valency of osmium in osmium tetroxide or valency of xenon in xenon tetroxide is eight.
Table 3.27 The periodicity of valency of typical elements
Period Compound
II
III
IV
Formula of the Compound
Hydride LiH BeH2 B2H6 CH4 NH3 H2O HF Ne
Oxide Li2O BeO B2O3 CO2 N2O5 - [OF2]
Fluoride LiF BeF2 BF 3 CF4 [NF3] [OF2]
Hydride NaH MgH2 (AlH3)n SiH4 PH 3 SH2 HCl Ar
Oxide Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7
Fluoride NaF MgF2 AlF 3 SiF4 PF 5 SF6 [ClF3], IF7
Hydride KH CaH2 Ga2H6 GeH4 AsH3 SeH2 HBr
Oxide K2O CaO Ga2O3 GeO2 As4O10 SeO2 Oxides of Br Kr
Fluoride KF CaF2 GaF3 GeF4 AsF5 SeF6 [BrF5]
4. The valency with respect to hydrogen increases from 1 to 4 and then decreases to 1 from IA group to VIIA group in a period.
The periodicity of valency can be seen from the successive valencies shown in hydrides or oxides or fluorides in each period.
Oxidation Number
The charge present or which appears to be present on an elementary atom in a given species is known as its oxidation state. Oxidation state may be +ve (or) – ve or zero or fractional.
Example: Alkali metals lose one electron and exhibit +1 and halogens gain one electron and exhibit – 1 oxidation state.
1. For Group IA elements oxidation number is + 1 and for Group IIA elements oxidation number is + 2
2. For p-block elements, oxidation number= Group number. (or) Group number -8.
Example: VA group elements can exhibit + 5 and – 3 states.
3. In p-block elements, down the group higher oxidation state becomes less stable due to inert pair effect.
The reluctance of ns electrons to participate in bond formation is known as inert pair effect.
Example : In IIIA group, +1 oxidation state of Tl is more stable than +3 oxidation state. Similarly, in IVA group, for Pb, +2 oxidation state is more stable than + 4 oxidation state, and in VA group, for Bi +3 oxidation state is more stable than +5 oxidation state.
4. d-block elements exhibit variable oxidation numbers, ranging form + 1 to + 8, due to their configuration. ns0-2, (n–1) d1 to 10
Example: Manganese exhibits + 2, + 3, + 4, + 6 and + 7 states.
5. The common oxidation number for d-block elements is + 2
6. In d-block elements, +1 state is shown by Cr, Cu, Ag, Au, Hg.
7. f-block elements exhibit + 3, + 2, + 4 states based on their configuration.
For lanthanides, common oxidation number is + 3.
8. The maximum oxidation number + 8 is shown by Ru, Os and Xe in their oxides (RuO4 ,OsO4, XeO4) Variation of oxidation number in transition elements is shown in Table 3.28
Table 3.28 Variation of oxidation states in transition elements
Q. Using the periodic table, predict the formula of compound formed between an element X of group 13 and another element Y of group 16.
Sol. The valency of X (group 13) = 3
The valency of Y (group 16) = 2
The compound has 2 atoms of X and 3 of Y.
Hence, the formula = X2Y3
Q. What are the valencies of K in K 2 O and S in H 2S? What should be the formula of compound of K and S in which the above valencies are reflected?
Sol. Two K atoms are in combined state with one oxygen atom. One K-atom combines with 1/2 atom of oxygen. Hence, valency of K is one. One S-atom combined with two H atoms. Hence, valency of S is two. Hence, the formula of the compound is K2S.
3.7.6
Metallic Nature
Electropositivity is the converse of electronegativity. It denotes metallic nature. It is described as the tendency of losing electron or electrons by an atom of the element. It is also described as the ability to form cation easily.
M → Mn+ + ne–
The stronger the tendency of the element to lose electrons, the more will be its electropostivity and more is the metallic nature. The electropositive nature of the element will be more, if the ionisation enthalpy is less. The electropositive nature of elements
can also be explained using the electrode potentials. Electropositive elements react with water or acids to liberate hydrogen. They react with non-metals usually to form electrovalent substances. Alkali metals have only one valence electron each and are highly electropositive. The evidence for strong electropositive nature of alkali metals is due to the following reasons:
1. Their ions do not undergo hydrolysis.
2. The only group elements whose bicarbonates exist in solid state (except in LiHCO3).
3. They have higher reduction ability.
4. Their oxides are highly soluble in water to give strong bases.
More than 80 elements of the long form of the periodic table are metals. Metallic elements are present in all blocks of the periodic table. Most electropositive metals are present in the s-block. Metals of p-block are less electropositive. All the d-block and f-block elements are metals.
Metallic nature increases down the group. However, a reverse trend is observed in case of transition elements. Metallic nature gradually decreases from left to right in a period.
There are certain elements which behave like metals as well as non-metals. They are called metalloids. Examples of metalloids are germanium in group-14, arsenic and antimony in group-15, selenium and tellurium in group-16.
3.7.7 Nature of Oxides
Based on the nature, oxides are classified into four types: basic oxides, acidic oxides, amphoteric oxides, and neutral oxides.
Generally, metallic oxides are basic. They neutralise acids like hydrochloric acid. They dissolve in water to give bases. They are called basic anhydrides, e.g., K 2O, MgO, Tl2O, etc.
Na2O + H2O → 2NaOH
CaO + H2O → Ca(OH)2
Generally, non-metallic oxides are acidic. They neutralise bases like sodium hydroxide. They dissolve in water to give acids. They are called acidic anhydrides, e.g., SO 2, P4O10, Cl2O7, etc.,
2223 COHOHCO +→
NOHOHNOHNO +→+
242 23
Oxides of metalloids are generally amphoteric. They react with both acids as well as bases e.g., GeO2, Sb4O6, TeO2, As2O3,etc.
Some of the metallic oxides are also amphoteric, e.g.,
ZnO, Al2O3, SnO2, etc.,
ZnO2HClZnClHO +→+
22
ZnO2NaOHNaZnOHO +→+
AlO6HCl2AlCl3HO +→+
AlO2NaOH2NaAlOHO +→+
Some of the non-metallic oxides are neutral. They do not react with acids as well as with bases e.g., CO, N2O, NO, etc.
Basic nature of the oxides increases generally with an increase in the electropositivity of metal forming oxide. Oxides of all elements of group1 are basic and of group 17 are acidic.
Down the group, basic nature of oxides increases and acidic nature decreases, as shown in Table 3.29.
Table 3.29 Nature of trioxides of group-15 Element
N non-metal N2O3 acidic
P non-metal P4O6 acidic
As metalloid As4O6 weakly acidic
Sb metalloid Sb4O6 amphoteric
Bi metal Bi2O3 basic
Across a period, basic nature of oxides gradually decreases and acidic nature gradually increases. The nature of the oxides across third period can be observed in Table 3.30.
If the same element is forming many oxides, acidic nature increases with an increase in the number of oxygen atoms form ing oxide.
Most basic oxide is caesium oxide. Caesium hydroxide is the strongest base.Most acidic oxide is chlorine heptoxide. Perchloric acid is the strongest acid.
Periodic Trends and Chemical Reactivity
All chemical properties are a manifestation of the electronic configuration of elements. The periodic trends in the properties of elements are diagramatically given in Fig.3.12
Fig.3.12. Periodic trends of elements
The ato mic radii generally decrease in a period from left to right. As a consequence, the ionisation enthalpies increase and electrons gain enthalpies become more negative.
Table 3.30 Nature of oxides of elements of third period
Na2O MgO Al2O3 SiO2 P4O10 SO3 Cl2O7
Strong base Basic Amphoteric Weak acid
Moderately acidic Strong acid Strong acid
NaOH Mg(OH)2 Al(OH)3 H2SiO3 H3PO3 H2SO4 HClO4
Strong base Basic Amphoteric Weak acid
Since ionisation potentials are less, alkali metals are very reactive. Similarly, halogens are also very reactive due to high electron affinity. Thus, high chemical activity is witnessed at the two extremes and the lowest in the centre of the periodic table.
Maximum chemical reactivity at the extreme left is exhibited by the formation of cation. This is referred to electropositivity and the elements act as good reductants.
Maximum chemical reactivity at the extreme right (not noble gases) is exhibited by the formation of anion. This is referred to non-metallic nature and the elements act as good oxidants.
Q. Is hydrogen electropositive?
Sol. Hydrogen has electropositivity. It is evidenced by the formation of proton.
H → e – + H+
However, hydrogen is not a metal. It is a common non-metal.
Q. In aqueous solutions lithium is the best reductant. Why?
Sol. Lithium cation is small and its hydration ability is high. The stronger reduction ability of lithium is also reflected in least standard potential of Li+/Li.
3.7.8 Anamolous properties of second period elements
Elements of second period exhibit anomalous behaviour. Lithium (group-1), beryllium (group-2), and boron to fluorine (groups 13 to 17) differ in many respects from the other members of their respective groups due
Moderately acidic Strong acid Strongest acid
to smal l size, high electro negativity, large charge/ radius ratio and absence of d orbitals.
The maximum covalency of a second period element is four. Furthermore, the first member of p-block elements displays greater ability to form multiple bonds.
Example:
1) CC,CC,NN,NN =≡=≡
2) CO,CN,CN,NO ==≡=
a) Lithium (IA) and beryllium (IIA), unlike o ther elements of same group form compounds with covalent character. b) Boron forms [BF4]– whereas other elements of this group can expand valence shell to accomodate more than four electron pairs, like [AlF6]3–.
Chemical similarities among elements are observed in the periodic table, going down a group. The member elements of a group together is called a chemical family and the relationships are referred to as vertical relationships. However, there are few diagonal relationships too.
Diagonal Relationships: Lighter elements of second period resemble in their properties with those of the third period elements, placed diagonally to each other. Diagonal relationship of elements is shown in Table 3.31
This similarity is called diagonal relationship. LiBeBC NaMgAlSi
Table 3.31 Diagonal relationship of elements
The ph enomenon of diagonal relationship does not appear after group 4 and also below third period of the periodic table. The diagonal relationship is due to similar electronegativity of the respective elements.
The elements that show diagonal relationship possess similar polarising power. Polarising power is the power of cation attracting the charge cloud of anion. Polarising power is given as the ratio of ionic charge to the square of ionic radius.
Polarising power = () Chargeofion 2 Ionicradius
On moving across a period, the charge on the ions increases and the size decreases, causing the polarising power to increase. On moving down a group, the size increases and the polarising power decreases. On moving diagonally, these two effects partially cancel each other, so that there is no marked change in properties.
The type of bond formed, the strength of the bond and properties of the compounds of diagonally related elements are often similar. However, valency is different for diagonally related pair of elements.
The polarising power of Be 2+ and Al 3+ is similar. The properties of beryllium and aluminium are also similar. Both elements dissolve in caustic soda solution. Oxides of both elements are amphoteric. The carbides of both beryllium and aluminium produce methane gas on hydrolysis.
Be2C + 4H2O → 2Be(OH)2 + CH4
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
Diagonal similarities are most important among lighter electropositive elements, but the line seperating the metals from the nonmetals also run diagonally.
Q. Compare the oxidation ability of sulphur and chlorine.
Sol. 2 CleCl
S2eS +→ +→
Chlorine is better oxidant than sulphur. Electron gain enthalpy is more for chlorine. Chlorine accepts electron easily and becomes stable chloride.
Q. Lithium is monovalent. Magnesium is divalent. But Li and Mg are diagonally related pair of elements. Why?
Sol. Lithium and magnesium have certain similarities in their properties. Hence, they are called a diagonally related pair of elements.
The reasons are:
(a) Li and Mg have similar electronegativity.
(b) Li+ and Mg2+ have similar polarising power.
TEST YOURSELF
1. The correct order of van der Waals radius of F, Cl and Br is
(1) F > Br > Cl (2) Br > Cl > F
(3) F > Cl > Br (4) Br > F > Cl
2. The correct arrangement of O, P, and N in order of increasing radii is
(1) O < N < P (2) P < O < N
(3) O < P < N (4) N < O < P
3. Covalent bond length of chlorine molecule is 1.98 Å. Covalent radius of chlorine is (1) 1.98 Åx (2) 1.7 Å
(3) 2.05 Å (4) 0.99 Å
4. The covalent and van der Waals radii of chlorine, respectively, are
(1) 1.80 Å and 0.99 Å (2) 0.99 Å and 1.80 Å
(3) 1.80 Å and 1.80 Å
(4) 0.99 Å and 0.99 Å
5. In the isoelectronic species, the ionic radii (A°) of N3–, O2–, F– are, respectively, given by
6. Very slight decrease in atomic radius occurs in a transition series when compared with that in a representative series. This is due to (1) shielding effect (2) penetrating effect (3) Compton effect (4) inert pair effect
7. If the radius of Fe++ is 0.76 A°, the radius of Fe+++ is
(1) 0.64 A° (2) 0.76 A°
(3) 0.88 A° (4) 1.08 A°
8. Among elements with the following electronic configurations, the one with the largest radius is
(1) 1s2 2s2 2p6 3s2
(2) 1s2 2s2 2p6 3s2 3p1
(3) 1s2 2s2 2p6 3s2 3p3
(4) 1s2 2s2 2p6 3s2 3p5
9. In a sodium atom, the screening is due to (1) 3s2 3p6 (2) 1s2 2s2 2p6 (3) 2s1 (4) 1s2 2s2
10. Which of the following orders is correct for the size of Fe3+, Fe, and Fe2+ ?
(1) Fe < Fe2+ < Fe3+ (2) Fe2+ < Fe3+ < Fe
(3) Fe < Fe3+ < Fe2+ (4) Fe3+ < Fe2+ < Fe
11. Similarity in the radius of Zr and Hf is explained on the basis of (1) lanthanide contraction (2) inert pair effect
(3) same outer shell configuration
(4) anomalous configuration
12. Metallic radius of Ca is 200 pm. Covalent radius of Ca is
(1) 200 pm (2) 230 pm
(3) 280 pm (4) 174 pm
13. The lanthanide contraction relates to (1) oxidation states (2) magnetic state (3) atomic radii (4) valence electrons
14. Which of the following ions has smaller size? (1) Cl– (2) S2– (3) Na+ (4) Ca2+
15. Which of the following series of elements has most nearly the same atomic radius? (1) Mg, Ca, Sr, Ba (2) Ca, Ge, As, Se (3) B, C, N, O (4) Cr, Mn, Fe, Co
16. Atomic radii of fluorine and neon in angstrom units are, respectively, (1) 0.72, 1.62 (2) 0.72, 0.72 (3) 1.2, 1.2 (4) 1.62, 0.72
17. The correct order of atomic radius of Li, Be and B is (1) B > Be > Li (2) B > Li > Be (3) Li > B > Be (4) Li > Be > B
18. Correct order of atomic radii is (1) N < C < P < S (2) C < N < S < P (3) C < N < P < S (4) N < C < S < P
19. How many ionisation energies can carbon have? Electronic configuration of carbon in ground state: 1s2 2s2 2px1 2py1 in excited state : 1s2 2s1 2px1 2py1 2pz1. (1) 1 (2) 2 (3) 4 (4) 6
20. First ionisation potential of magnesium is greater than that of aluminium because (1) aluminium atom is very large when compared to magnesium (2) aluminium has a stable electronic configuration (3) magnesium has a stable electronic configuration (4) The electron affinity of Magnesium is positive (energy is absorbed)
21. Electrons with the highest penetrating power are (1) p-electrons (2) s-electrons (3) d-electrons (4) f-electrons
22. The species with largest ionisation potential (1) Li+ (2) Mg+ (3) Al+ (4) Ne
23. Second ionisation energy is higher than first ionisation energy for an element. This is because (1) nuclear charge is high in cation (2) size of cation is higher than neutural atom (3) effective nuclear charge is more for cation (4) bond energy changes with charge
24. In modern periodic table, the groups table that possesses the highest and lowest ionisation energies, respectively, are (1) IA, VIIA (2) zero, IA (3) IA, IIA (4) VIIA, IA
25. In the lithium atom, screening effect of valence shell electron is caused by (1) electrons of K and L shell (2) electrons of K shell (3) two electrons of 1st and one of 2nd shell (4) electrons of L-shell
26. First four ionisation energy values of an element are 191, 578, 872, and 5972 kcals. The number of valence electrons in the element is (1) 4 (2) 3 (3) 1 (4) 2
27. Ionisation potential values of Li, Be and B, respectively, in kJ mol –1 are (1) 801, 899, 520 (2) 520, 801, 899 (3) 899, 801, 520 (4) 520, 899, 801
28. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn), and iron (Fe) are, respectively, 23, 24, 25, and 26. Which one of these may be expected to have the highest second ionisation enthalpy? (1) Cr (2) Mn (3) Fe (4) V
29. Second ionisation potential of oxygen is (1) equal to that of fluorine (2) less than that of fluorine (3) greater than that of fluorine (4) half of that of fluorine
30. Among the elements, A, B, C, and D have atomic numbers 9, 10, 11, and 12, respectively. The correct order of ionisation energies is (1) A > B > C > D (2) B > A > D > C (3) B > A > C > D (4) D > C > B > A
31. Successive ionisation potentials of an element M are 8.3, 25.1, 37.9, 259.3, and 340.1 eV. The formula of its bromide is (1) MBr 5 (2) MBr 4 (3) MBr 3 (4) MBr2
32. In a period, from left to right, (A) nuclear charge increases (B) effective nuclear charge increases (C) atomic size decreases (D) ionisation potential increases
Correct among the above statements are (1) A, B (2) B, C (3) A, C, D (4) A, B, C, D
33. If the ionisation potential (IP) of Na is 5.48 eV, the IP of K will be (1) 4.34 eV (2) 5.68 eV (3) 10.88 eV (4) 5.48 eV
34. Amongst the following elements (whose electronic configurations are given below), the one having the highest first ionisation energy is:
(1) [Ne]3s23p1 (2) [Ne]3s23p3
(3) [Ne]3s23p2 (4) [Ar]3s104s24p2
35. The ionisation potential of nitrogen is more than that of oxygen because of (1) the greater attraction of the electrons by the nucleus
(2) the extra stability of the half filled porbitals
(3) the smaller size of nitrogen
(4) more penetration effect
36. The element with highest electron affinity is
(1) fluorine (2) cesium
(3) helium (4) chlorine
37. Ionisation of energy of F– is 320 kJ mol–1. The electron gain enthalpy of fluorine would be
(1) – 320 kJ mol–1 (2) –160 kJ mol–1
(3) + 320 kJ mol–1 (4) 160 kJ mol–1
38. Which of the following is an endothermic process?
(1) First electron affinity of chlorine
(2) Second electron affinity of oxygen
(3) Formation of NaCl from gaseous ions
(4) Hydration of MgCl2
39. In a period from left to right, electron affinity (1) increases (2) decreases (3) remains constant (4) first increases and then decreases
40. Configuration that shows the highest energy released when an electron is added to the atom is
41. Which set of elements shows positive electron gain enthalpy?
(1) He, N, O (2) Ne, N, Cl
(3) O, Cl, F (4) N, He, Ne
42. The correct order of electron affinity of the elements of oxygen family in the periodic table is
(1) O > S > Se (2) S > O > Se
(3) S > Se > O (4) Se > O > S
43. Energy is released in the process of (1) Na(g) → Na+(g) + e–
(2) O–(g) + e → O–2(g)
(3) N–2(g) + e– → N–3(g) (4) O(g) + e → O–(g)
44. The formation of the oxide ion requires first an exothermic and then an endothermic step as shown below
O(g) + e– = O–(g); = – 142 kJ mol–1
O–(g) + e– = O2–(g) ; = 844 kJ mol–1
This is because
(1) O– ion will tend to resist the addition of another electron
(2) oxygen has high electron affinity
(3) oxygen is more electronegative (4) O– ion has comparatively larger size than oxygen atom
45. Screening effect influences (A) atomic radius
(B) ionisation enthalpy
(C) electron gain enthalpy
(1) A, B (2) B, C
(3) A, C (4) A, B, C
46. Pauling’s electronegativity is based on (1) electron affinity
(2) ionisation potential
(3) both IP and EA
(4) bond energies
47. The electronegativity value of chlorine and bromine are, respectively 3 and 2.8. Formula of a binary compound is best represented as (1) BrCl (2) ClBr3 (3) ClBr (4) ClBr5
48. Reference element for Pauling’s electronegativity is (1) H (2) C (3) Cl (4) He
49. What is the correct order of electronegativity?
(1) M+1 < M+2 < M+3 < M+4
(2) M+1 > M+2 > M+3 > M+4
(3) M+1 < M+2 > M+3 < M+4
(4) M+4 < M+2 < M+3 < M+1
50. In a period, electronegativity is highest for (1) chalcogen (2) halogen
(3) inert gas (4) alkali metal
51. The values that are useful in writing chemical formulae and in calculation of oxidation states are (1) ionisation potential
(2) electron affinity
(3) electronegativity
(4) metallic character
52. The electronegativity of elements X and Y on Pauling's scale is equal. Then, elements X and Y are, respectively, (1) C and S (2) Be and I (3) Al and I (4) S and Be
53. The electronegativities of C, N, Si, and P are in the order of
(1) P < Si < C < N (2) Si < P < C < N
(3) P < Si < N < C (4) Si < P < N < C
54. Pa ir of elements with equal values of electronegativiy is (1) Be, Al (2) Mg, Al (3) Mg, Ca (4) F, Ne
55. The difference in bond energy between the experimental and the calculated values of YH is 1.96 kcal mol–1. The electronegativity of Y is (electronegativity of H is 2.1)
(1) 1.90 (2) 1.81
(3) 1.78 (4) 1.75
56. All the following elements show both positive and negative oxidation states, except (1) N (2) H (3) O (4) F
57. An element with electronic arrangement as 2, 8, 18, 1 will exhibit the following oxidation states
(1) + 2 and + 4 (2) + 1 and + 2
(3) + 2 only (4) + 1 only
58. Oxidation state and covalency of Al in [AlCl(H2O)5]2+ are
(1) +2, 6 (2) +3, 6
(3) +2, 4 (4) +3, 4
59. If X(OH) > Y(OH) (acidic strength) and ‘X’ and ‘Y’ both belong to same group, then select the correct statement.
(1) Electron gain enthalpy of ‘Y’ must be greater than ‘X.’
(2) Atomic size of ‘X’ must be greater than ‘Y.’
(3) O–H bond will be easily broken in case of hydroxide of ‘X’.
(4) O–H bond will be easily broken in case of hydroxide of ‘Y.
60. Maximum oxidation state (+8) is exhibited by
(1) Co and Ni (2) Ru and Os
(3) Cl and I (4) Te and I
61. An element has electronic configuration 1s 22s 22p 63s 1 in its +2 oxidation state. The formula of its sulphide is (1) M3S2 (2) MS2 (3) MS3 (4) M2S3
62. Basic nature of the oxides of a period from left to right (1) increases (2) decreases (3) remains constant (4) first increases and then decreases
63. Oxide that is most acidic is (1) Cl2O7 (2) SO3 (3) P4O10 (4) N2O5
64. Generally, the nature of non-metal oxides is
(1) basic (2) acidic (3) amphoteric (4) neutral
65. Most acidic oxide in the periodic table is formed by an element in (1) 2nd period, Group VI A (2) 4th period, Group VII A (3) 3rd Period, VI A (4) 3rd period, VII A
66. Which of the following sets of oxides are amphoteric?
(1) CO2, SiO2, P4O10, N2O5
(2) K2O, BaO, CrO, FeO
(3) N2O3, CO, NO, N2O
(4) Al2O3, ZnO, SnO2, Sb4O6
67. Elements A, B, and C belong to the same period in the long form of the periodic table.
The nature of the oxides of A, B, and C is amphoteric basic, and acidic, respectively. The correct order of the atomic numbers of these elements is
(1) B > A > C (2) C > B > A
(3) C > A > B (4) A > B > C
68. Which of the following properties increases across a period?
(1) Reducing property
(2) Size of atom
(3) Acidic nature of oxides
(4) Metallic property
69. When an atom of an electronegative element becomes anion, which of the following occurs?
(1) It acts as a reducing agent.
(2) It loses electrons.
(3) Its ionic radius becomes larger.
(4) Effective nuclear charge increases.
70. Among Al2O3, SiO2, P2O3, and SO2, the correct order of acid strength is
(1) Al2O3 < SiO2 < SO2 < P2O3
(2) SiO2 < SO2 < Al2O3 < P2O3
(3) SO2 < P2O3 < SiO2 < Al2O3
(4) Al2O3 < SiO2 < P2O3 < SO2
71. The correct increasing order of metallic nature of Si, Be, Mg, Na, P is
(1) P < Si < Be < Mg < Na
(2) Si < P < Be < Mg < Na
(3) Si < P < Mg < Be < Na
(4) Na < Mg < Be < Si < P
72. The diagonal relationship phenomenon is not observed after
(1) I A Group
(2) II A Group
(3) III A Group
(4) IV A Group
73. Which of the following is not correct in the case of Be and Al?
(1) Both are rendered passive by conc. HNO3.
(2) Carbides of both give methane on hydrolysis.
(3) Both give hydroxides that are basic.
(4) Both give covalent chlorides.
74. The correct order of polarisability of ion is
(1) Cl– > Br– > I– > F–
(2) F– > I– > Br– > Cl–
(3) I– > Br– > Cl– > F–
(4) F– > Cl– > Br– > I–
75. The chemistry of lithium is very similar to that of magnesium, even though they are placed in different groups. Its reason is that (1) both are found together in nature (2) both have nearly the same size
(3) both have similar electronic configuration
(4) the ratio of charge and size is nearly same
76. Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in
(1) forming covalent halides
(2) forming polymeric hydrides
(3) exhibiting maximum covalency in compounds
(4) exhibiting amphoteric nature in their oxides
CHAPTER REVIEW
Genesis Of Periodic Classification
■ A triad is a group of tthree similar elements arranged in increasing order of their atomic mass.
■ In Newland's classification, every eight element is found similar to its first element, when they are arranged in increasing order of atomic masses.
■ In Lother Meyer curve is obtained when atomic volumes of elements are plotted against atomic masses.
Mendeleef’s Periodic Table
■ The first periodic table was constructed by Mendeleev.
■ Mendeleev’s periodic law is stated as the physical and chemical properties are, periodic functions of their atomic weights.
■ The elements with low atomic weights were found to be widely distributed in nature are referred as typical elements that are present in three short periods of Mendeleev’s table
■ The gaps in Mendeleevf’s table were for missing elements, called ‘Eka’ elem ents.
■ Eka-boron is now known as scandium, Ekasilicon as germanium, and Eka -aluminum as gallium.
■ Atomic weight of an element is the product of its equivalent weight and valency. Based on this, atomic weights of Be, In, and Au are corrected.
■ The increasing order of atomic weights of certain pairs of elements were not followed. Such pairs are called anomalous pairs.
■ Cobalt –nickel, argon – potassium, tellurium–iodine and thorium–protactinium are examples of anomalous pairs.
Modern Periodic Law
■ Modern periodic law states that the physical and chemical properties of elements are periodic functions of their electronic configurations.
Long Form of Periodic Table
■ Long form of the periodic table is called Bohr’s table. Here, the elements are arranged in the increasing order of atomic numbers.
■ There are 18 vertical columns in the long form, called groups, and 7 horizontal rows called periods.
■ First period is the shortest period. It has only 2 elements. Second and third periods are short periods with eight elements each. Fourth and fifth periods are normal periods with eighteen elements each.
■ Sixth period is the longest period with 32 elements and seventh period is incomplete.
■ The periodic number indicates the valence shell and the group number in Roman letters denotes the number of electrons in the outer most shell.
Nomenclature of Elements
■ IUPAC developed a system of given names that can be deriveed from atomic number
■ Element with atomic number, 101 is named as unnilumium.
■ For 0,1,2,3 digits of atomic number nil, un, bi, tri etc., names were proposed.
Classification of Elements
■ s-block elements have ns1-2 configuration. They include alkali and alkaline earth elements (groups IA and IIA).
■ p-block elements have ns 2 np 1-6 configuration, except He. The elements present are IIIA to VIIA groups and zero group.
■ Many gases, non-metals, and metalloids are present in the p-block.
■ The general configuration of d-block elements is (n-1)d1-10 n s1 or 2 .
■ There are four series of d-block elements. They are 3d-, 4d-, 5d- and 6d- series.
■ The general configuration of f-block elements is (n-2) f1-14 (n-1)d0 or1 n s2.
■ There are two series of f-block elements. 4fseries are called lanthanides and 5f-series are called actinides.
■ Noble gases belong to zero group. They have the outer s and p orbitals completely filled (n s2 n p6) He is 1s2
■ Noble gases are chemically inactive, and this is attributed to octet configuration. He exhibits duplet configuration, i.e., 1s 2
■ Representative elements, or main group elements, have outer shells partially filled.
■ All ‘s’ and ‘p’ block elements except ‘zero’ group are representative elements. They are chemically active.
■ Transition elements are all metals. They are hard with high melting, boiling points, and densities. They are all solids, except Hg.
■ Transition metals have characteristic properties: variable oxidation states, coloured salts or solution, paramagnetism, catalytic activity forming alloys and complexes.
■ Elements of IIB group, Zn, Cd, and Hg are not considered as transition elements as they do not contain unpaired electrons.
■ Elements in which three outer shells are partially filled are called inner transition elements. They are f-block elements.
■ Common oxidation state of transition elements is +2 and of inner transition elements is +3.
■ The cause for periodicity of properties is similar valence shell configuration.
■ The properties which are not periodic are atomic weight, atomic number, specific heat, refractive index, etc.,
Periodic Properties and trends
■ Atomic radius depends upon the size of bonded atoms, nature of bond, bond order, and the oxidation state.
■ Covalent radius is one-half of the inter nuclear distance between atoms of a homodiatomic molecule. It is used for non-metals.
■ Metallic radius or crystal radius is used for metallic elements. It is one-half of the internuclear distance between adjacent atoms of a metal.
■ van der Waals radius is used for molecular substances in the solid state. It is one-half of the internuclear distance between atoms facing each other in a djacent molecules.
■ Atomic radius increases down the group because of the presence of new shells.
■ In any period, in general, atomic radius is least for halogen and highest for noble gas element.
■ Radius is highest for the element ‘Cs’ and least for the element ‘H’.
■ In transition elements, as the atomic number increases, the atomic radius decreases slightly.
■ The decrease in radius is insignificant because of screening effect by inner ‘d’ orbitals.
■ Due to lanthanide contraction, the crystal structure and other properties of lanthanides are similar and are difficult to separate.
■ The radius of cation is less than the parent atom and that of the anion is greater.
■ Among isoelectronic groups, radius decreases with increase in the nuclear charge.
■ Energy required to separate the outermost electron of an isolated metal atom is called ionisation potential or ionisation energy.
■ Ionisation potentials are determined by discharge tube experiments.
■ The units of ionisation energy are eV/atom or kJ/mol. 1 eV atom–1 = 96.45 kJ mol–1
■ Ionisation potential values are dependent upon atomic radius, nuclear charge, screening effect, penetration ability of orbitals and electronic configuration.
■ Elements having completely filled or halffilled configurations are more stable and need more energy for ionisation.
■ In a group, ionisation energy decreases, as the size and screening effect increases.
■ In a period, ionisation energy increases as the nuclear charge increases and the size decreases.
■ In any period, alkali metal has least and noble gas has highest ionisation energies.
■ The element with highest first ionisation energy is helium and least is caesium.
■ The peaks of ionisation energy curves are occupied by noble gases.
■ The energy released when an electron is added to isolated atom is called electron affinity or electron gain enthalpy.
■ Uni-negative ion is formed on adding an electron.
■ Due to dispersed shape of f-orbitals and poor shielding effect, the atomic radius of lanthanides steadily decreases. This is called lanthanide contraction.
■ The ion prevents entry of further electron due to repulsive forces. Hence, second electron affinity is positive.
■ Halogens have high electron affinities. Electron affinity of noble gases is zero.
■ Electron affinity is highest for chlorine. It is represented as positive for alkaline earth metals.
■ In general, electron affinities of third period elements are more than that of second period elements.
■ Electron affinities of fluorine are less than chlorine due to small size and stronger inter electronic repulsions.
■ In a period, electron affinity gradually increases. In a group it generally decreases.
■ The tendency to attract bonded electrons by an atom in the bond is called electronegativity. It has no units.
■ Pauling’s scale of electronegativities is based on bond energies.
■ Bond energy of AB molecule is taken as the average of bond energies of A 2 and B2 molecules.
() AABB AB EE E 2 + ∆=− , where D is the difference in the bond e nergies
■ XA –XB = 0.1017 ∆ , where D is taken in kJ mol–1
■ Taking the electronegativity of hydrogen as 2.1, the values of other elements are determined.
■ Electronegativity of fluorine is highest. Its value is 4.
■ Electronegativity values of O, N and Cl are respectively, 3.5, 3 and 3. For noble gases, the value is taken as zero.
■ Electronegativity decreases in a group as the atomic size increases. It increases in a period as the nuclear charge increases.
■ If the difference in electronegativity of bonded atoms is 1.7 or more, the compound formed is generally ionic.
■ Electronegativity values are useful in writing chemical formulae, calculation of oxidation states, a nd determining bond polarity.
■ Valency is the combining capacity of an element.
■ Number of hydrogen or chlorine atoms or twice the number of oxygen atoms with which one atom of the element combines is taken as valency.
■ Elements in a group show generally same valency.
■ Across a period, the valence increases unit by unit. Transition metals show variable valency.
■ Maximum valency of an element is its group number.
■ Minimum valency is zero. Highest valency of 8 is exhibited by Os or Xe.
■ The tendency of an element to lose an electron is called electropositivity or metallic nature.
■ Electropositivity increase down the group, as the size increases, and decreases across a period as the size decreases.
■ Element with low electronegativity is a metal and with high electronegativity is a non-metal.
■ Electropositive elements react with water and acids to liberate hydrogen.
■ Ions of strongly electropositive alkali metals do not undergo hydrolysis.
■ Metallic nature increases and non-metallic nature decreases down the group. Metallic nature decreases across a period.
■ Best metal available is ‘Cs’ and best nonmetal is ‘F’.
■ Metallic oxides are generally basic and form alkaline solutions in water.
■ Non-metallic oxides are generally acidic and form acidic solutions in water.
■ Oxides of metalloids are normally amphoteric.
■ An element of a group in the second period has similar properties with the second element of the next group in the third period. This is called diagonal relationship.
■ The three diagonally related pairs of the elements are Li – Mg, Be – Al and B – Si.
■ The reasons for the diagonal relationship are similarity in the electronegativity and similar polarising powe r.
■ Elements that exhibit diagonal relationship have some similar properties.
■ Oxides of Be and Al are amphoteric. Carbides of both Be and Al produce CH 4 on hydrolysis.
Additional Information
Factors Effecting Electronegativity:
■ Ionisation enthalpy and electron affinity: As ionisation energy and electron affinity increases, electronegativity also increases.
■ s-Character: As s-Character increases, electro-negativity increases.
Order of electronegativity: sp > sp 2 > sp3
■ Charge on the ion: Electronegativity order: Cation > parent atom > anion.
Hanny and Smith Relationship
Percentage of ionic character = 16 (XA – XB ) + 3.5(XA – XB)2 where XA and XB are the electronegativities of two atoms A and B.
Effective Nuclear Charge
■ In a period, one electron is added and the shielding effect of the outermost orbit is 0.35. Therefore, with the increase in nuclear charge by one, the decrease in charge is 0.35 due to screening effect. Hence, there is an increase of 0.65 in the effective nuclear charge in a period.
■ The effective nuclear charge normally remains almost constant in a group.
■ In transition elements, the incoming electron goes into (n - 1) d orbital. This results in the actual increase in nuclear charge, only 0.15 (1-0.85 = 0.15), on going from left to right in a period.
■ In a group of transitional elements, the effective nuclear charge remains almost constant from top to bottom.
Exercises
NEET DRILL
Level-1
Genesis of Periodic Classification
1. Which of the following sets of elements follows Newland’s Octave rule
(1) Be, Mg, Ca (2) F, Cl, Br
(3) Na, K, Rb (4) B, Al, Ga
2. Which of the following is a transition triad?
(1) Li, Na, K (2) Cl, Br, I
(3) Fe, Co, Ni (4) Cr, Mn, Fe
Mendeleev Periodic Table
3. The atomic weights of “Be” and “In” were corrected by Mendeleev using the formula
(1) a(Zb)γ=−
(2) nh mvr 2 = π
(3) Atomic weight = Equivalent weight x valency
(4) Equivalent weight = Atomic weight x valency
4. Eka-silicon is now known as
(1) Scandium (2) Gallium
(3) Germanium (4) Boron
5. The places that were left blank in the periodic table by Mendeleev were later occupied by the elements:
(1) Aluminum and silicon
(2) Gallium and germanium
(3) Arsenic and antimony
(4) Molybdenum and tungsten
Modern Periodic Law
6. As per the modern periodic law, the physical and chemical properties of elements are the periodic functions of their
(1) atomic volume
(2) electronic configuration
(3) atomic weight
(4) atomic size
7. In the periodic table, the elements are arranged in the periods following the (1) Hund’s rule of maximum multiplicity
(2) Pauli’s exclusion principle
(3) Aufbau principle
(4) Both (1) and (2)
Long Form of Periodic Table
8. The long form of the periodic table consists of
(1) 8 horizontal and 7 vertical series
(2) 7 horizontal and 18 vertical series
(3) 7 horizontal and 7 vertical series
(4) 8 horizontal and 8 vertical series
9. The position of the element with z = 106 in the periodic table is
(1) d- block (2) s-block
(3) f- block (4) p-block
10. Atomic number of nitrogen is 7. The atomic number of the third member in the same family is ______.
(1) 23 (2) 15
(3) 33 (4) 51
11. Atomic numbers of actinides are ______.
(1) 57 to 71 (2) 80 to 103
(3) 58 to 71 (4) 90 to 103
12. Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table?
(1) 107 (2) 118
(3) 126 (4) 102
13. Set of elements with the following atomic numbers belong to the same group
(1) 9, 16, 35, 3 (2) 12, 20, 4, 38
(3) 11, 19, 27, 5 (4) 24, 47, 42, 55
14. The element that belong to 3rd period and IVA group of periodic table is (1) Silicon (2) Carbon (3) Germanium (4) Tin
15. An element with atomic number 17 belongs to group 17.What is the atomic number of the element belonging to same group and present in fifth period?
(1) 25 (2) 33 (3) 35 (4) 53
16. The sub- shells filled one by one for 4th period elements are
(1) 3d, 4s and 4p (2) 4s,4p and 4d
(3) 4s,3d and 4p (4) 3d,4p and 4s
Nomenclature of Elements
17. As per IUPAC rule, the atomic number for the element Un-bi-unium is_____.
(1) 120 (2) 121 (3) 112 (4) 122
18. IUPAC official symbol of the element with the symbol ‘Unu’ is ______.
(1) No (2) Md
(3) Lr (4) Fm
19. IUPAC official symbol of the element with the symbol ‘Uuu’ is (1) Ds (2) Cn (3) Rg (4) Nh
Classification of Elements
20. Which group and period does the element belong if the electronic configuration of an element in its -2 oxidation state is 1s22s22p63s23p6?
(1) Period 3, group 16
(2) Period 3, group 17
(3) Period 4, group 16
(4) Period 4, group 17
21. The 79th electron of an element ‘X’ with an atomic number 79 enters into the orbital
(1) s - Orbital (2) p – Orbital
(3) d- Orbital (4) f – Orbital
22. What is the position of the element in the periodic table satisfying the electronic configuration (n-1)d1ns2 for n=4 ?
(1) 3rd period and 3rd group
(2) 4th period and 4th group
(3) 3rd period and 2nd group
(4) 4th period and 3rd group
23. The statement that is not correct for periodic classification of elements is _______.
(1) The properties of elements are periodic function of their electronic configuration.
(2) Non-metallic elements are less in number than metallic elements.
(3) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
(4) The first ionisation enthalpies of elements generally increase with increase in atomic number as we go along a period.
24. An element has 18 electrons in the outer most shell. The element is
(1) transition metal
(2) rare earth metal
(3) alkaline earth metal
(4) alkali metal
25. Which of following is not correctly matched?
(1) d-block element: electronic configuration is n s0-2(n-1) d1-10
(2) p-block element: electronic configuration is ns1−2np1−6
(3) s-block element: electronic configuration is ns1−2
(4) Cerium(Ce) is the first member in the f-block in modern periodic table.
26. The outer most electronic configuration of the most electronegative element is
(1) ns2np3 (2) n s2np6(n-1)d2
(3) ns2np5 (4) ns2np6
27. In the sixth period, the orbitals that are filled are _______.
(1) 6s, 6p, 6d, 6f (2) 6s, 5d, 5f, 6p
(3) 6s, 5f, 6d, 6p (4) 6s, 4f, 5d, 6p
28. Two elements’ X ’ and ‘Y ’ have the following configuration
X = 1s22s22p63s23p64s2, Y = 1s22s22 p63s23p5 the compound formed by the combination of X ‘ and ‘Y‘ will be
(1) XY2 (2) X 5Y2
(3) X2Y5 (4) XY 5
Periodic Properties and Trends
Atomic Radius
29. The correct increasing order of the ionic radii is
35. The correct order of effective nuclear charge Zeff is
(1) B < C < N < O < F
(2) B = C = N = O = F
(3) B > C > N > O > F
(4) none of these
36. Screening by inner electrons will be more effective in
(1) Mg (2) K (3) Sr (4) Cs
Ionisation Enthalpy
37. The first ionisation potential of Na, Mg and Si respectively are 496,737 and 786 kJ/mole. The first ionisation potential of ‘Al’ in kJ/ mole is
(1) 577 (2) 487
(3) 856 (4) 768
38. The first ionisation enthalpy of the element C, N, P, Si are in the first order of
(1) C < N < Si < P (2) N < Si < C < P
(3) Si < P < C < N (4) P < Si < N < C
39. Consider the following changes:
A(g)→A(g) + + e– : E1 and A(g)→A(g) +2 + e– : E2
The energy required to pull out the two electrons are E 1 and E 2 respectively. The correct relationship between two energies would be
(1) E1 < E2 (2) E1 = E2
(3) E1 > E2 (4) E1 ≤ E2
40. Which of the order for ionisation energy is correct?
(1) Be < B < C < N < O
(2) B < Be < C < O < N
(3) Be < B < C < N > O
(4) B < Be < N < C < O
41. The five successive ionisation energies of an element ‘X’ are 800, 1427, 2658, 25024 and 32824 KJ mole−1 respectively. The valency of ‘X’ is______.
(1) 1 (2) 2 (3) 3 (4) 4
42. First, second and third ionisation energies are 737, 1045 & 7733 KJ/mol respectively. The element can be _______.
(1) Na (2) B
(3) Al (4) Mg
43. Elements X, Y and Z have atomic numbers 19, 37 and 55 respectively. Which of the following statements is true about them?
(1) Their ionisation potential would increase with increasing atomic number
(2) Y would have an ionisation potential between those of X and Z.
(3) Z would have the highest ionisation potential.
(4) Y would have the highest ionisation potential.
44. The element with highest I.P1 in the modern periodic table is_______.
(1) Cl (2) He (3) N (4) O
45. The group of elements with the highest second ionisation energy is______.
(1) II A (2) VIII A
(3) VII A (4) I A
Electron affinity
46. The correct order of electron gain enthalpy is
(1) O > S > Se > Te (2) Te > Se > S > O
(3) S > O > Se > Te (4) S > Se > Te > O
47. Which of the following is the correct order of electron affinity?
(1) I > Br > F > Cl (2) F > Cl > Br > I
(3) Br > I > F > Cl (4) Cl > F > Br > I
48. Which of the following has the highest electron gain enthalpy?
49. The correct order of electron affinity is ____.
(1) Be < B < C < N
(2) Be < N < B < C
(3) Be < N < B < C
(4) N < C < B < Be
50. In which of the following process, energy is liberated?
(1) Cl → Cl+ + e– (2) HCl →H ++ Cl–
(3) Cl + e– → Cl– (4) O– + e– → O2–
51. The element with least electron affinity is __.
(1) S (2) O
(3) Se (4) Te
Electro-Negativity
52. The correct order of electro negativity for given elements is :
(1) Br > C > At > P (2) P > Br > C > At
(3) C > P > At > Br (4) Br > P > At > C
53. The electro negativity of the following elements increases in the order
(1) S < P < N < O
(2) P < S < N < O
(3) N < O < P < S
(4) N < P < S < O
Metallic Nature
54. Which list includes elements in order of increasing metallic character?
(1) Si, P, S (2) As, P, N
(3) Al, Ge, Sb (4) Br, Se, As
55. The correct decreasing order for metallic character is
(1) Na > Mg > Be > Si > P
(2) P > Si > Be > Mg > Na
(3) Si > P > Be > Na > Mg
(4) Be > Na > Mg > Si > P
Valency
56. Maximum valency with respect to hydrogen shown by a third period element is
(1) 3 (2) 4
(3) 5 (4) 7
Nature of Oxides
57. Among a) Na 2 O; b) MgO; c) Al 2 O 3 ; d) P2O5 and e) Cl2O7, the most basic, most acidic and amphoteric oxide can be (1) a, b, c (2) b, e, c (3) a, e, c (4) e, c, a
58. Among Al2O3, SiO, P2O, and SO2 , the correct order of acid strength is _____.
(1) SO2 < P2O3 < SiO2 < Al2O3
(2) SiO2 < SO2 < Al2O3 < O3
(3) Al2O3 < SiO2 < SO2 < P2O3
(4) Al2O3 < SiO3 < P2O3 < SO2
59. In which pair of oxides both are acidic in nature?
(1) CaO, SiO2 (2) B2O3, SiO2 (3) B2O3, CaO (4) N2O, BaO
Diagonal Relation
60. The diagonal relationship phenomenon is not observed after
(1) I A Group (2) II A Group (3) III A Group (4) IV A Group
61. Diagonal relationship is present between the lighter elements of periods
63. Beryllium and aluminium exhibit many properties that are similar. But, the two elements differ in (1) forming covalent halides (2) forming polymeric hydrides (3) exhibiting maximum covalency in compounds (4) exhibiting amphoteric nature in their oxides
64. The chemistry of lithium is very similar to that of magnesium even though they are
placed in different groups. The reason is (1) both are found together in nature (2) both have nearly the same size (3) both have similar electronic configuration (4) the ratio of charge and size is nearly same
Level 2
Genesis of periodic classification
1. X,Y and Z are three elements forming Dobereiner’s triads. The increasing order of atomic masses of these elements is Y< X < Z. If the atomic masses of Y and Z are respectively 7 and 39, the atomic mass of X is _____.
(1) 23 (2) 80
(3) 35.5 (4) 46
2. The elements P, Q and R are one of the Dobereiner’s triad of elements in the increasing order of their atomic masses. If the atomic masses of P, Q and R are x, y and z respectively, then
(1) x + y + z = 3y (2) x + z = 3y
(3) x = 3y (4) z = 3y
3. In the Lothar Meyer graph; A, B, C, D and E elements are:
(1) halogen
(2) alkaline earth metals
(3) alkali metals
(4) transition metals
Mendeleev periodic table
4. Eka-aluminium and Eka-silicon are known as :
(1) gallium and germanium
(2) aluminium and silicon
(3) iron and sulphur
(4) proton and silicon
5. The following are some statements about Mendeleeff’s periodic table
i) It is based on increasing order of atomic numbers
ii) Mendeleeff corrected the atomic weight of some elements like Be, In etc
iii) (Ar; K), (Co; Ni), (Te; (1) i,ii, and iii are correct
(2) ii & iii are correct (3) iii is correct (4) i & iii are correct
Modern periodic law
6. The frequency of the characteristic X ray of Kα line of metal target ‘M’ is 2500 cm–1 and the graph between v Vs ‘z’ is as follows, then atomic number of M is
(1) 49 (2) 50 (3) 51 (4) 25
7. Frequency of X – ray emitted by an element is 100 sec –1. If constants in the moseley’s equation a = b = 1 the element will be ______ (1) Na (2) K (3) Mg (4) Ca
Long form of Periodic Table
8. Which of the following is incorrect?
Group Period
(1) Z= 38 IIA 5th
(2) Z= 58 IIIB 6th
(3) Z= 25 VIIB 4th
(4) Z= 22 VIB 4th
9. The last element of the p-block in 6th period is represented by the outermost electronic configuration.
22. Element with atomic number 38, belongs to (1) II A group and 5th period (2) II A group and 2nd period (3) V A group and 2nd period (4) III A group and 5th period
23. Lanthanum element with z = 57 belongs to (1) s-block (2) p-block (3) d-block (4) f-block
Periodic Properties and Trends
Atomic Radius
24. When the atoms Li, Be, B and Na are arranged in order of increasing atomic radius, what is the correct order?
(1) B, Be, Li, Na (2) Li, Be, B, Na (3) Be, Li, B, Na (4) Be, B, Li, Na
25. Atomic radius depends upon i) the number of bonds formed by the atom ii) nature of the bonding iii) oxidation state of the atom (1) i and ii (2) ii and iii (3) i and iii (4) i, ii, iii
26. The pair that has similar atomic radii is____. (1) Sc and Ni (2) Ti and Hf (3) Mo and W (4) Mn are Tc
Ionisation Enthalpy
27. Ionisation energy values of an atom are 495,767, 1250 and 4540 kJ mole-1 the formula of its sulphate is (1) MSO4 (2) M2SO4
(3) M2(SO4)3 (4) M(SO4)2
28. Which of the following order is not correct?
(1) IE(I) of Be > IE(I) of B but IE (II) of Be < IE(II) of B
(2) IE(I) of Be < IE(I) of B but IE(II) of Be < IE(II) of B
(3) IE(II) of O > IE(II) of N (4) IE(I) of Mg > IE(I) of AI
29. Screening effect influences
A) atomic radius
B) ionisation enthalpy
C) electron gain enthalpy
(1) A, B only (2) B, C only
(3) A, C only (4) A, B, C
Electron Affinity
30. The correct order of electron gain enthalpies of Cl, F, Te and Po is _______.
(1) F < Cl < Te < Po (2) Po < Te < F < Cl
(3) Te < Po < Cl < F (4) Cl < F < Te < Po
31. Which of the following is incorrect?
(1) Cesium is the most electro positive element while F is most electronegative element.
(2) Cl is the highest –ve electron gain enthalpy of all the elements.
(3) Electron gain enthalpy of N as well as that of noble gases is positive.
(4) In any period, the atomic radius of the noble gas is lowest.
32. The process requiring absorption of energy is_____.
(1) N → N (2) F → F–
(3) Cl → Cl– (4) H → H–
33. Electronic configuration of four elements A, B, C and D are given below:
(A) 1s22s22p6 (B) 1s22s22p4
(C) 1s22s22p63s1 (D) 1s22s22p5
Which of the following is the correct order of increasing tendency to gain electron?
(1) A < C < B < D (2) A < B < C < D
(3) D < B < C < A (4) D < A < B < C
34. Fluorine has the highest electro-negativity among the ns2np5 group on the Pauling scale, but the electron affinity of fluorine is less than that of chlorine because
(1) the atomic number of fluorine is less than that of chlorine
(2) fluorine, being the first member of the family, behaves in an unusual manner.
(3) fluorine can accommodate an electron better than fluorine by utilizing its vacant 3d-orbital
(4) small size, high electron density and an increased electron repulsion makes addition of an electron to fluorine less favorable than that in the case of chlorine in isolated state
35. Which of the following elements have electron affinity greater than ‘F’?
(1) O (2) S (3) Se (4) Cl
Electro-Negativity
36. The correct option with respect to the Pauling electro-negativity values of the elements is
(1) Ga < Ge (2) Si < Al
(3) P > S (4) Te > Se
37. A, B, and C are hydroxyl compounds of the elements X, Y, and Z respectively. X, Y, and Z are in the same period of the periodic table. A gives an aqueous solution of p H less than seven. B reacts with both strong acids and strong alkalies. C gives an aqueous solution that is strongly alkaline. Which of the following statements is/are true?
I: The three elements are metals.
II: The electro-negativities decreases from X to Y to Z.
III: The atomic radius decreases in the order X, Y and Z.
IV: X, Y and Z could be phosphorus, aluminium and sodium respectively.
(1) I, II, III only correct.
(2) I, III only correct.
(3) II, IV only correct.
(4) II, III, IV only correct.
38. Which of the following statements are not correct?
1) The electron gain enthalpy of F is more negative than that of Cl.
2) Ionisation enthalpy decreases in a group of periodic table.
3) The electro-negativity of an atom depends upon the atoms bonded to it.
4) Al2O3 and NO are examples of amphoteric oxides.
(1) 1, 3,and 4
(2) 1, 2 and 3
(3) 3 only
(4) 1, 2, 3 and 4
39. % Ionic character in the covalent bond A−B is________.
[E.N of XA = 2, XB = 3]
(1) 12.5% (2) 30%
(3) 19.5% (4) data is insufficient
Periodic Trends in Chemical Properties
40. In which of the following arrangements the order is not according to the property indicated against it?
(1) Al3+< Mg2+ < Na+ < F Increasing ionic size
(2) B < C < O < N Increasing first ionisation energy
(3) I < Br < F < Cl Increasing electronegativity
(4) Li < Ca < Al < Si Valence with respect to hydrogen
41. Which of the following statement is incorrect?
(1) H + is the smallest size cation in the periodic table.
(2) van der Waals radius of chlorine is more than covalent radius.
(3) Ionic mobility of hydrated Li+ is greater than that of hydrated Na + .
(4) He has the highest ionisation enthalpy in the periodic table.
42. Consider the following points :
(a) Cs is the strongest reducing agent in IA group elements (In aqueous solution)
(b) Be(OH)2 is amphoteric
(c) The density of potassium is less than sodium
FURTHER EXPLORATION
1. The electron affinity of Fluorine is 3.45 eV. How much energy in kcal is released when 2 g of atomic Fluorine is completely converted to Fluoride ion in a gaseous state ? (1 eV = 23.06 kcal /mol) (1) 4.8 Kcal (2) 2.8 Kcal
(d) In alkali metals Li, Na, K, and Rb, lithium has the maximum value of MP.
Correct statement are
(1) (a) & (b) are correct
(2) (a) & (b) & (c) are correct
(3) (b) & (c) are correct
(4) (b) & (c) & (d) are correct
Nature of Oxides
43. The correct order of acidic strength is
(1) K2O > CaO > MgO
(2) CO2 > N2O5 > SO3
(3) Na2O > MgO >Al2O3
(4) Cl2O7 > SO2 >P4O10
44. Identify the correct match
1) O < C< S < Se – Atomic size
2) Na < Al < Mg <Si – 1st IP
3) MgO < SrO < Cs 2 O < K 2 O –Basic character
4) P4O10 > SO3> Cl2O7 – Acidic character
(1) 1, 2, 3 (2) 2, 3, 4 (3) All (4) 1, 2
Diagonal Relation
45. Amphoteric-oxide combinations are in (1) ZnO, K2O, SO3 (2) ZnO,P2O5,Cl2O7 (3) SnO2,Al2O3,ZnO (4) PbO2,SnO2,SO3
46. Which of the following is not correct in the case of Be and Al ?
(1) Both are rendered passive by conc.HNO3
(2) Carbides of both give methane on hydrolysis
(3) Both give hydroxides which are basic
(4) Both give covalent chlorides
(3) 8.84 Kcal (4) 12.8 Kcal
2. According to slaters rule order of effective nuclear charge for last electron in case of Li, Na and K.
(1) Li > Na > K (2) K > Na > Li
(3) Na > Li > K (4) Li < Na = K
3. Elements P, Q, R, and S belong to the same group The oxide of P is acidic, the oxide of Q and R are amphoteric while the oxide of S is basic. Which of the following elements is the least electro-positive?
(1) R (2) Q (3) P (4) S
4. Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in
(1) Forming covalent halides
(2) Forming polymeric hydrides
(3) Exhibiting maximum covalency in compounds
(4) Exhibiting amphoteric nature in their oxides
Matching Type Questions
1.
Column-I Column-II
(A) Dobereiner (I) Law of Octaves (B) Newlands (II) Cylindrical table of elements
(C) Chancourtois (III) 1st periodic law
(D) Mendeleev (IV) Law of triads
(A) (B) (C) (D)
(1) IV I II III
(2) I IV II III
(3) IV I III II
(4) I IV III II
2. Match the atomic numbers of the elements given in Column I with the periods given in Column II and mark the appropriate choice.
Column-I (Atomic number) Column-II (Period)
(A) 31 (I) 5
(B) 50 (II) 3
(C) 56 (III) 4
(D) 14 (IV) 6
(A) (B) (C) (D)
(1) I II III IV
(2) II I III IV
(3) III IV I II
(4) III I IV II
3. Match column I (element) with column II (characteristic property)
Which of the following is correct option? (2020–I)
(A) (B) (C) (D)
(1) II I IV III
(2) III IV I II
(3) IV III I II
(4) IV III II I
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II)
(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 : 5th period of the long form periodic table is the longest period.
S–II : The longest period of the periodic table includes 32 elements.
2. S-I : In Cl2 molecule the covalent radius is double the atomic radius of chlorine.
S-II : Radius of anionic species is always greater than their parent atomic radius.
3. S -I : Metallic or electro-positive character of elements increases as the value of ionisation potential decreases.
S-II : In a group moving from top to bottom, metallic or electro-positive character increases.
BRAIN TEASERS
1. On the basis of given part of periodic table incorrect statement is (Z = 39)
A D E B C
(1) A is an alkaline earth metal.
(2) Atomic number of B is 103 which belongs to III B group
S-II : Oxygen uni-positive ion has a halffilled electronic configuration.
Assertion and Reason Type Questions
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
4. S-I : The first IP. of nitrogen is greater than oxygen, while the reverse is true for the second IP. values.
1. (A) : ‘He’ and ‘Be’ both have the same outer electronic configuration, like ns2 type.
(R) : Both are bhemically inert.
2. (A) : Mg2+ and Al 3+ are iso-electronic but ionic radius of Al 3+ is less than that of Mg2+.
(R) : The effective nuclear charge on the outershell electrons in Al 3+ is more than that in Mg2+
(3) Atomic number, group no. and period number of D are 72, IV B and 6th respectively
(4) ‘C’ is transuranic element
2. The first and second ionization enthalpies of a metal are 496 and 4560 kJ mol -1 respectively. How many moles of HCl and H2SO4 respectively, will be needed to react completely with 1 mole of the metal hydroxide
(1) 1 and 1
(2) 2 and 0.5
(3) 1 and 2
(4) 1 and 0.5
3. Consider the elements B, C, N, F and Si,the correct order of their non-metallic character is
(1) B > C > Si > N > F
(2) Si > C > B > N > F
(3)F > N > C > B > Si
FLASHBACK (Previous NEET Questions)
1. The incorrect statement among the following is (2021)
(1) Actinoids are highly reactive metals, especially when they are finely divided.
(2) Actinoid contraction is greater for element to element than lanthanoid contraction.
(3) Most of the trivalent lanthanoid ions are colourless in the solid state.
(4) Lanthanoids are good conductors of heat and electricity.
2. Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of (2021)
(1) having similar chemical properties
(2) belonging to same group
(3) diagonal relationship
(4) lanthanoid contraction
3. The IUPAC name of an element with atomic number 119 is (2022)
(1) unnilennium
(2) unununnium
(3) ununoctium
(4) ununennium
CHAPTER TEST
1. Three elements X, Y, Z are following Dobereiner’s triad rule. If the atomic weight of X and Y are 10 and 26 respectively, then
(4) F > N > C > Si > B
4. Zeff for last electron of 4S sub shell electron of Zn atom (Z = 30) is (1) 25.65 (2) 8.85 (3) 4.35 (4) 12.15
4. Identify the incorrect match (2020–I)
Name IUPAC official name
(A) Unnilunium (I) Mendelevium
(B) Unniltrium (II) Lawrencium
(C) Unnilhexium (III) Seaborgium
(D) Unununnium (IV) Darmstadtium
(1) (b), (ii)
(2) (c), (iii)
(3) (d), (iv)
(4) (a), (i)
5. Match the element in Column I with that in Column II. (2020–II)
Column-I Column-II
(A) Copper (I) Non-metal
(B) Fluorine (II) Transition metal
(C) Silicon (III) Lanthanoid
(D) Cerium (IV) Metalloid
Identify the correct match
(A) (B) (C) (D)
(1) I II III IV
(2) II IV I III
(3) II I IV III
(4) IV III I II
atomic weight of Z will be________.
(1) 34 (2) 40
(3) 42 (4) 19
2. According to Moseley, a straight line graph is obtained on plotting.
(1) the frequencies of characteristic line graph.
(2) the square of the frequencies of characteristic X-rays of elements against their atomic numbers.
(3) the square root of the frequencies of characteristic X-rays of elements against their atomic numbers.
(4) the reciprocal of the frequencies of characteristic X-rays of elements against their atomic numbers.
3. Atomic weight of second element is less than the first element in the following pair
(1) P, S (2) Fe, Co
(3) Be, B (4) Te, I
4. Identify correct statements from the following
A) Atomic weight of tellurium < iodine.
B) Melting point of Eka-aluminium is low.
C) Li, Na, K is a Dobereiner triad.
D) Eka-silicon is gallium.
E) forms an oxide of the formula, E 2O3
(1) A, B, C, D (2) B and C only
(3) A only (4) A and C only
5. The triad not present in Group VIII of Mendeleev’s table
(1) Li, Na, K (2) Fe, Co, Ni
(3) Ru, Rh, Pd (4) Os, Ir, Pt
6. The IUPAC symbol for the element with atomic number 199 would be_________.
(1) unh (2) uue
(3) uun (4) une
7. The IUPAC nomenclature of an element with electronic configuration [Rn]5f146d17s2 is :
8. The number of elements present in 2nd, 3rd , 4th and 5th periods of modern periodic table respectively are
9.
(1) 2, 8, 8 and 18 (2) 8, 8, 18 and 32 (3) 8, 8, 18 and 18 (4) 8, 18, 18 and 32
Element Group to which element belongs to
(A) N (P) VI A group
(B) F (Q) VA group
(C) O (R) VII A group
(D) C (S) IV A group
(A) (B) (C) (D)
(1) Q R S P
(2) Q R P S
(3) R Q P S
(4) P Q R S
10. What is the position of the element in the periodic table satisfying the electronic configuration?
(n–1)d1ns2 for n = 4?
(1) 3rd period and 3rd group
(2) 4th period and 4th group
(3) 3rd period and 2nd group
(4) 4th period and 3rd group
11. Element with atomic number 116 has recently been discovered. Which of the following electronic configuration will it possess and to which group will it belong to?
(1) [Rn]5f146d97s27p5−17th group
(2) [Rn]5f146d107s27p4−16th group
(3) [Rn]5f146d87s27p6−18th group
(4) [Rn]5f146d107s27p3−15th group
12. The electronic configuration of the element that is just above the element with atomic number of 43 in the same group is:
(1) 1s22s22p63s23p63d54s2
(2) 1s22s22p63s23p63d54s34p6
(3) 1s22s22p63s23p63d6 4s2
(4) 1s22s22p63s23p63d74s2
13. Elements a, b, c, d, and e have the following electronic configurations.
(a) 1s2, 2s2, 2p1
(b) 1s2, 2s2, 2p6, 3s23p1
(c) 1s2, 2s2, 2p6, 3s23p2
(d) 1s2, 2s2, 2p6, 3s23p5
(e) 1s2, 2s2, 2p6, 3s23p6
Which among these will belong to the same group in the periodic table?
(1) a and c (2) a and b
(3) a and d (4) d and e
14. Match the following lists and select the correct answer
List-I List-II
(A) 1s2, 2s22p6, 3s23p6,4s1 (I) d-block element
(B) 1s2, 2s22p6,3s23p6 (II) Halogen
(C) 1s2, 2s22p6, 3s23p63d6,4s2 (III) Alkali metal
(D) 1s2, 2s22p5 (IV) Noble gas
(A) (B) (C) (D)
(1) I II III IV
(2) III IV I II
(3) I III II IV
(4) II IV III I
15 Match List-I with List-II.
List -I (Block) List-II (General electronic configuration)
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) Q R S P
(2) R Q P S
(3) Q R P S
(4) S Q R P
16. Following are some statements about modern periodic table. Pick the correct ones
i) It consists of s, p, d and f blocks.
ii) The energy levels filling order in 6th period is 6s, 4f, 5d and 6p.
iii) IIIA group contains maximum number of elements.
(1) i & ii (2) only i
(3) ii & iii (4) all
17. Which one of the following is correct order of the size?
(1) I > I > I+ (2) I > I+ > I–
(3) I+ > I > I (4) I– > I > I+
18. In which of the following sets, elements have nearly same atomic radii?
(1) Li, Be, B (2) Mg, Ca, Sr
(3) Fe, Co, Ni (4) O, S, Se
19. If an element ‘x’ is assumed to have the type of radii then their order is
(1) crystal radius > van der Waals radius >covalent radius
(2) van der Waals radius > crystal radius>covalent radius
(3) covalent radius> crystal radius> Van der Waals radius
(4) van der Waals radius > Covalent radius > crystal radius
20. Given below are two statements
Statement–I: In Cl2 molecule the covalent radius is double of the atomic radius of chlorine.
Statement–II: Radius of anionic species is always greater than their parent atomic radius.
In light of the above 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.
21. Give below are two statements
One is labelled as Assertion (A) and the other is labelled as Reason (R) .
Assertion(A) : The ionic radii of O 2- and Mg2+ are same
Reason(R): Both the species have same nuclear charge
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.
22. The lanthanoid contraction is responsible for the fact that
(1) Zr and Y have about the same radius
(2) Zr and Nb have similar oxidation state
(3) Zr and Hf have about the same radius
(4) Zr and Zn have same oxidation state
23. Consider the following ionisation reaction: I.E(KJ/mol) I.E(KJ/mol)
A(g) → A+(g)+e–,A1 B(g) → B+(g)+e−,B1
B+(g) → B+2(g)+e ,B2 C(g) → C+(g)+e ,C1
C+(g)→C+2(g)+e ,C2 C+2(g)→C+3(g)+e ,C3
If uni positive ion of A, di positive ion of B and tri positive ion of C have zero electron. Then incorrect order of corresponding IE is
(1) C3 > B2 > A1 (2) B1 > A1 > C1
(3) C3 > C2 > B2 (4) B2 > C3 > A1
24. In which of the following arrangement the order is incorrect according to the property indicated against it ?
(1) Al+3 < Mg2+< Na+ <F–- increasing ionic size
(2) B < C < N < O – increasing first ionisation enthalpy
(3) I < Br < F < Cl -(increasing electron gain enthalpy (with negative sign)
(4) Li < Na < K < Rb – increasing metallic radius
25. The correct order of electron gain enthalpy of the given elements is
(1) S > Se > O (2) Se > S > O
(3) O > S > Se (4) S > O > Se
26. Ionisation energy of F– is 320 KJ mol–1. The electron gain enthalpy of fluorine will be
(1) –320 KJ mol–1 (2) –160 KJ mol–1
(3) +320 KJ mol–1 (4) +160 KJ mol–1
27. This graph of electron gain enthalpy (magnitude) vs atomic number is of which group of periodic table?
Consider : P,Q,R,S are respective elements down the group
(1) Gp : 15 (2) Gp : 16
(3) Gp : 17 (4) Gp : 13
28. Which one of the following arrangements of the incorrect representation of the property indicated with it?
(1) Br < C < F : Electro-negativity
(2) F < Br < C : Electron-affinity
(3) F2 < Br2 < C 2: Bond energy
(4) Br2 < C 2 < F2: Oxidising strength
29. The elements ‘X’, ‘Y’, and ‘Z’ form oxides that are acidic, basic and amphoteric, respectively. The correct order of their electro negativity is
(1) X > Y > Z (2) Z > Y > X
(3) X > Z > Y (4) Y > X > Z
30. Calculate the % ionic character for molecule AB when E.N difference is 2.0.
(1) 46% (2) 36%
(3) 30% (4) 50%
31. In the periodic table, the maximum chemical reactivity is at the extreme left (alkali metals) and extreme right (halogens). Which properties of these two groups are responsible for this?
(1) Least ionisation enthalpy on the left and highest negative electron gain enthalpy on the right.
(2) Non-metallic character on the left and metallic character on the right.
(3) High atomic radii on the right and small atomic radii on the left.
(4) Highest electro-negativity on the right.
32. Given below are two statements.
Statement– I: Li and Mg show similar chemical properties.
Statement–II: Li + and Mg 2+ have nearly same Polarising power.
In light of the above 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.
33. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R :
Assertion (A): Both Be and Al exhibit similar properties in their compounds.
Reason (R) : Due to almost the same Polarising power of cations, some pairs of elements like Be and Al are diagonally related in their properties of compounds.
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.
34. Given below are two statements.
Statement-I : ‘Tl’ in it’s ‘+1’ oxidation state is more stable than that of its ‘+3’ oxidation state.
In light of the above 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.
35. Which of the following oxide is amphoteric?
(1) CrO
(2) Cr2O3
(3) CrO3
(4) CrO5
36. Some oxides are shown in List-I and their nature is shown in List-II
List-I
List-II
(A) MgO (I) Amphoteric
(B) BeO (II) Acidic
(C) P2O5 (III) Neutral
(D) CO (IV) Basic
The correct match is
(A) (B) (C) (D)
(1) I II III IV
(2) IV I II III
(3) IV I III II
(4) II III IV I
37. Which of the following order presents the correct sequence of the increasing basic nature of the given oxides?
(1) Al2O3 < MgO < Na2O < K2O
(2) MgO < K2O <Al2O3 < Na2O
(3) Na2O < K2O < MgO < Al2O3
(4) K2O < Na2O < Al2O3 < MgO
38. Given below are two statements: One is labelled as Assertion A and other is labelled as Reason R :
Assertion (A): The decreasing order of acidic character of CO 2 , N2O5, SiO2 and SO3 is CO2 > N2O5 > CO2 > SiO2
Reason (R) : As the electro-negativity difference between the central atom and oxygen decreases, the acidic character of the oxide increases.
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.
39. An element A of group I shows similarity to an element B belonging to group II. If A has maximum hydration enthalpy in group I then B is_______.
(1) Mg (2) Be
(3) Ca (4) Sr
40. A sudden jump between the values of second and third ionisation energies of an element is associated with configuration________.
(1) 1s22s22p63s1 (2) 1s22s22p63s23p1
(3) 1s22s22p63s23p2 (4) 1s22s22p63s2
41. Given below are two statements
Statement–I: Beryllium and Al are placed in different periods and groups, but they show similar properties.
Statement–II: On moving diagonally, i.e., from Be to Al these two effects partly cancel each other and so there is no marked change in properties.
In light of the above 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.
42. Given below are two statements
Statement–I: For transition elements, on moving from left to right in a transition series, ionisation energy increases.
Statement-II: As the atomic number increases, the effective nuclear charge also increases.
In light of the above 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.
43. The Ionisation potential and electron affinity of fluorine are17.42 and 3.45eV respectively. Calculate the electro-negativity of fluorine in Paulings scale
(1) 3.726 (2) 4.726
(3) 2.726 (4) 1.726
44. The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2 g of chlorine is completely converted to Cl – ion in a gaseous state? (1 eV = 23.06 kcal molñ1)
(1) 48 kcal (2) 2.8 kcal
(3) 8.4 kcal (4) 4.8 kcal
45. The correct order of ionic radius is
(1) Ti4+ < Mn7+ (2) 35Cl– > 37Cl–
(3) K+ > Cl– (4) P3+ > P5+
46. Which is the correct order of ionic sizes?
(At. no. : Ce = 58, Sn = 50, Yb =70 and Lu = 71)
(1) Ce > Sn > Yb > Lu
(2) Sn > Yb > Ce > Lu
(3) Sn > Ce > Yb > Lu
(4) Lu > Yb > Sn > Ce
47. Amongst Cd, Hg, Ag, Au the I.P value order will be
(1) Ag < Cd < Au < Hg
(2) Ag < Au < Cd < Hg
(3) Ag < Hg < Au < Cd
(4) Ag = Cd < Au < Hg
48. Which is correct in the following?
(1) Radius of Cl atom is 0.99 Å, while that of Na+ ion is 1.54 Å.
(2) Radius of Cl atom is 0.99 Å while that of Na atom is 1.54 Å.
(3) The radius of Cl atom is 0.95 Å while that of Cl– ion is 0.81 Å.
(4) Radius of Na atom is 0.95 Å, while that of Na+ ion is 1.54 Å.
49. Which of the following processes do not involve absorption of energy?
(A) S(g)+e–→ S–(g)
(B) O–(g)+e– → O2– (s)
(C) Cl(g)+e– → Cl–(g)
(D) O (g) +e–→ O–(g)
(1) (A), (C), (D)
(2) (A), (B), (D)
(3) (B), (C), (D)
(4) (A), (B), (C), (D)
50. Mark the correct statement out of the following
(A) Helium has the highest first ionisation enthalpy in the periodic table.
(B) Fluorine has less negative electron gain enthalpy than chlorine.
(C) In any period, the atomic radius of the noble gas is the highest .
(D) Hg and Br are liquids at room temperature.
(1) (A), (B) only
(2) (A), (C) only
(3) (A), (B), (C) only
(4) (A), (B), (C), and (D) only
ANSWER
(31)4 (32) 1 (33)
Further Exploration (1) 2 (2) 1 (3) 1 (4) 2
Matching Type Questions (1) 1 (2) 4 (3) 1 (4) 1
Statement Type Questions (1) 4 (2) 4 (3) 1 (4) 1
Brain teasers (1) 2 (2) 4 (3) 3
Chapter Test
CHAPTER 4
CHEMICAL BONDING AND MOLECULAR STRUCTURE
Chapter Outline
4.1 Electronic Theory of Valency, Ionic Bonding and Lattice Energy
4.2 Kossel Lewis Approach to Chemical Bonding
4.3 Covalent Bond
4.4 Coordinate Covalent Bond
4.5 Valence Bond Theory
4.6 Hybridisation
4.7 VSEPR Theory
4.8 Resonance
4.9 Molecular Orbital Theory
4.10 Hydrogen Bonding
4.11 Bond Parameters
4.12 Bond Polarity and Dipole Moment
Atoms combine with atoms of similar type (or) dissimilar type to form molecules or compounds in order to get stability. Thus, a variety of substances are formed. The nature of attractive forces between the atoms depend on the combined atoms. In this chapter different theories pertaining to these chemical bonds are discussed.
4.1 ELECTRONIC THEORY OF VALENCY, IONIC BONDING AND LATTICE ENERGY
Atoms of the same element or different elements combine to form molecules. An
atom or a group of atoms having independent existence in nature is defined as a molecule.
The attractive force that holds two or more atoms or ions together in a molecule or ion is called a chemical bond. In the formation of chemical bond, atoms follow the behaviour of material systems. Any system that tends to pass from a higher energy state into a lower energy state becomes more stable. Energy of a system of two hydrogen atoms as a function of internuclear separation is given in Fig.4.1.
atoms''
Fig. 4.1 Energy of system of two ‘H’ atoms
It may be noted that bond is said to be formed at the lowest energy of the curve. At distances shorter than the equilibrium distance denoting lowest energy, repulsive forces operate predominantly.
When a pair of atoms combine to form a bond, a certain amount of energy is released. The chemical species so formed is relatively much more stable than the constituent atoms. The same amount of energy is to be supplied to break the particular bond.
4.1.1 Formation and Cleavage of Chemical Bonds
Consider two hydrogen atoms A and B approaching each other, having nuclei N A
and N B , and electrons present in them are represented by eA and eB.
Attractive Forces Arise Between:
(i) nucleus of one atom and its own electron i.e., NA – eA and NB– eB.
(ii) nucleus of one atom and electron of other atom i.e., NA– eB, NB– eA.
Repulsive Forces Arise Between:
(i) electrons of two atoms like e A – eB, (ii) nuclei of two atoms N A – NB
Attractive forces tend to bring the two atoms close to each other, whereas repulsive forces tend to push them apart.
Experimentally, it has been found that the magnitude of new attractive force is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm. Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of isolated hydrogen atoms. The energy so released is called as bond enthalpy. ()()() 1 gg2g HHH435.8kJmol +→+
4.2 KOSSEL-LEWIS APPROACH TO CHEMICAL BONDING
The electronic theory of valency was proposed by Kossel and Lewis on the basis of Bohr’s atomic structure. This is also called chemical bond theory or modern theory of valency. Kossel explained the formation of electrovalent
bond. Lewis explained the formation of covalent bond. The important postulates of the chemical bond theory are given here.
1. The electrons present in the outermost shell of an atom are called valence electrons. The electrons present in the inner shells of the atom are called core electrons.
The nucleus, along with core electrons form the positively charged kernel.
Example: Sodium atom has one valency electron and ten core electrons.
2. The valence electrons are responsible for the chemical activity of the atom. The core electrons generally do not participate in the chemical reactions.
3. All elements in zero group have stable electronic configuration.
4. Except helium, atoms of all noble gases have eight electrons (octet) in their outermost shells. It was suggested that the inertness of elements is due to octet structure. Even though helium has two electrons, it is highly stable and chemically inert as its only orbit is filled (1s 2)
5. All elements other than zero group elements are chemically reactive because they have less than eight electrons in their outermost energy levels.
6. Generally, every atom tends to acquire eight electrons in its outermost energy level. The principle of acquiring eight electrons in the outermost shell of an atom to attain stability is called octet rule.
Lewis introduced simple notations to represent valence electrons in an atom. The valence electrons, in this system, are represented by dots (one for each electron). These notations are called Lewis symbols.
Example:
Li, Be, B , C , N , O , F , Ne
Octet configuration is acquired in two types: By the transfer of electrons (ionic bond) or by mutual sharing of electrons (covalent bond). The valencey due to transfer of electrons is called electrovalency. The valencey due to sharing of electrons is called covalency.
If the difference in electronegativities of the bonded atoms is high, an ionic bond is formed, and if it is less, a covalent bond is formed.
Metallic bond is formed between atoms of low electronegativity (metals)
Strength of the metallic bond depends on the number of valence electrons involved, and it is inversely proportional to size of the metal atom.
Stronger the metallic bond, more is the enthalpy of atomization and higher is the melting point.
Variation of the nature of the bond with the difference in electronegativity of the bonded atoms is given in Table 4.1
4.2.1 Ionic Bond
1. T he electrostatic attraction between the oppositely charged ions formed by the transfer of one or more electrons from the outermost shell of a metal atom to the outermost shell of a nonmetal atom is called electrovalent or ionic bond.
2. The ionic bond is formed by the transfer of electrons between an atom of low
ionisation energy and an atom of high electron affinity.
3. The difference in the values of electronegativity should be greater than 1.7 for this bond formation. The electrovalent bond is not possible between similar atoms. This type of bonding requires two elements of opposite nature, one should have a tendency to lose electron or electrons (electropositive in nature) and the other should have the tendency to accept electron or electrons (electronegative in nature).
4. Atoms of elements with low ionisation energies tend to form cations, while those with high electron affinities tend to form anions. In general, elements most likely to form cations in ionic compounds are the alkali metals and alkaline earth metals. The elements most likely to form anions are halogens (VIIA group) and chalcogens (VIA group).
Illustrative Examples
1. Consider the formation of ionic compound, sodium chloride, from sodium and chlorine. The electronic configuration of sodium and that of chlorine are:
Na (Z = 11) is 1s2 2s2 2p6 3s1
Cl (Z = 17) is 1s2 2s2 2p6 3s2 3p5 When sodium and chlorine atoms come in contact with each other, the valence electron (3s1) of sodium is transferred to chlorine atom. This transfer leads to the formation of ions.
Na(g) → Na+(g) + e–Cl(g) + e– → Cl–(g)
Table 4.1 The electronegativities of the elements A and B and nature of bond A–B formed
Electronic configuration of Na + is 1s2 2s2 2p6.
Electronic configuration of Cl– is 1s2 2s2 2p6 3s2 3p6
The oppositely charged ions are held together by the coulombic attractive force. It is termed as ionic bond.
Na+ + Cl– → Na+Cl– (or) NaCl
2. Consider the formation of ionic compound, magnesium fluoride from Mg and F 2.
The magnesium (having 1s 2 2s 2 2p 6 3s 2 configuration) loses two electrons to get the configuration of neon (having 1s 22s 22p 6 configuration). To gain these two electrons, two fluorine atoms are required. Each fluorine can accept one electron to acquire the nearest inert gas, neon configuration. This means that every magnesium atom caters to the needs of two fluorine atoms. Thus, magnesium atom changes to Mg +2 and two fluorine atoms change into fluoride ions. Both the fluoride ions are bound to Mg+2 ion by ionic linkages. ; ; 2 MgMg2eFeF +−−−→++→ or 22 22 Mg2FMgFMg(F) +−+− +→
Ionic bonds are easily formed between alkali metals and halogens. The most easily formed ionic compound is caesium fluoride, CsF. This is because of the fact that the formation involves high electropositive metal, caesium and the highly electronegative nonmetal, fluorine. Formation of some other ionic compounds are given in Table 4.2.
In the formation of ionic compounds, the number of electrons an atom of the element loses or gains is known as its electrovalency or ionic valency of that element.
Example:
1) The electrovalency of magnesium in MgF2 is two and that of fluorine is one.
2) The electrovalency of sodium in Na 2O is one and that of oxygen is two.
Factors Favourable for the Formation of Ionic Bond
Only when cation and anion are formed, they will have ionic bond. The following are the factors which favour their formation.
Formation of Cation
Generally, the metal a toms lose electron or electrons to form positively charged ions called cations. The factors favourable for the formation of cation are given below.
Low Ionisation Energy:
An atom with low ionisation energy loses electron or electrons to form cation readily and can form ionic compound easily. Ionisation energy of potassium is 495.57 kJ mol –1, while that of sodium is 519.82 kJ mol–1. Therefore, K+ ion is more readily formed than Na + ion. The order of ease of formation of cation is : Na+ < K+ < Rb+ < Cs+.
Low Charge on the Ion
The cation must have low positive charge. The ions of lower positive charges are easily produced than those of higher charges. An atom which has to lose more electrons to attain stable configuration does not readily form cation or ionic bond. Ease of formation of cations is in the order:
Al3+ < Mg2+ < Na+.
Magnitude of cationic charge
ionicnature
Large Atomic Size
The metal atom must have large atomic radius. As the size of the atom increases, the force of attraction of the nucleus on valence electrons decreases. A cation is easily formed from large atom. Cs is the largest atom among the alkali
Table. 4.2 Formation of some other ionic compounds
Compound Method
Lewis dot method
method
Lewis dot method
method
Configuration before the transfer of electron (atom) (Two lines only)
Configuration after the transfer of electron (ion)
(Helium Configuration)
(Neon configuration)
(Neon configuration)
(Argon configuration)
Lewis dot method
method
Lewis dot method
(Neon configuration)
(Neon configuration)
(Neon configuration)
(Neon configuration)
metals. The order of atomic size of alkali metals is: Li < Na < K < Rb < Cs.
Cations with Inert Gas Configuration
The cation formed must have ns2 np 6 outer electron configuration. The cations which have inert gas configuration of eight electrons in the valence shell readily form ionic compounds. Though the formation of Mg + is easy, ionic compounds of Mg2+ are only known. Eg. Ca2+ (2, 8, 8) has inter gas configuration while Zn2+ has pseudo inert gas configuration. Therefore Ca2+ is formed more readily than Zn 2+ .
The ions with 18 electrons in their outermost energy level are said to possess “nickel group configuration” or pseudo inert gas configuration (ns 2np6nd10).
Formation of Anion
Generally, non-metal atoms gain electron or electrons to form negatively charged ions called anions. The factors favourable for the formation of anion are given below.
High Electron Affinity
Electron affinity is a measure of the tendency of an atom to accept electrons. Therefore, the atoms with high values of electronegativity and electron affinity form anions readily, and hence, ionic compounds. The order of ease of formation of anions is: Cl – > Br– > I–.
Low Charge on the Anion
The anion must have low magnitude of negative charge. As the negative charge on the anion increases, its stability decreases. Anions with low negative charges, like F–, are more readily formed than anions with high negative charges, like O 2– and N 3– . The compounds formed by anions with low negative charge are relatively more ionic in nature. The order of ease of formation of anions is:
F – > O2– > N3–.
Small Atomic Size of the Non–Metal
The atomic radius of the non metal should be small. Small atoms can hold the electrons gained by them strongly and form anions easily. Chloride is formed more easily than bromide or iodide. The order of ease of formation is:
FCBrI. >>> l
Anions with Inert Gas Configuration
The anions must have ns2np6 outer electron configuration. The anions which have eight electrons in the outer most shell readily form ionic compounds. Though O– is easy to form, ionic compounds of O2– are only known.
The difference between the electronegativities of the bonding atoms can decide the bond nature. When the difference is large, an ionic bond is formed. If it is less, a covalent bond is formed. Fig.4.2. shows the variation of the nature of the bond, i.e. ionic nature of the bond, with the difference in electronegativities of the bonded atoms.
1 2 3 4
Difference in electronegativity
Fig.4.2. Variation of Ionic Nature of the Bond Versus Difference in Electronegativities.
Ionic Nature
Generally, the bond formed between a metal and a non-metal is ionic. However, covalent bonds are also known to form between a metal and a non-metal.
During the formation of an ionic bond, when two oppositely charged ions of unequal
size closely approach each other, the ion smaller in size attracts outermost electrons of the other ion and repels its nuclear charge. The net result is distortion or deformation or polarisation of the larger ion.
The distortion is generally done by the cation as its size is smaller than anion. The electron cloud of anion no longer remains symmetrical but is elongated towards the cation. The ability of a cation to polarise the nearby anion is called its polarising power or polarisation ability.
The tendency of an anion to get distorted or deformed or polarised by the cation is said to be polarisability. As a result of polarisation, sharing of electrons takes place between two ions to some extent and the bond shows some covalent character.
Cation with higher charge and smaller size has high polarising power, which gives more covalent character to the compound.
i.e., Polarising power of cation charge size α
Polarising power of the cations decreases down the group and increases across a period.
Anion with higher charge and larger size is easily polarisable (polarisability), which gives more covalent character to the compound.
i.e., Polarisability of anion a magnitude of the charge and size.
Polarisability of anions by a given cation increases down the group and decreases across a period.
The magnitude of the polarisation or the increased covalent character is favoured by a number of factors. These factors are given as Fajan’s rules, and they are useful in predicting the nature of the bond formed between two atoms.
Fajan’s Rules
1. Increase in size of the cation increases the ionic nature of the bond.
Size: Li+ < Na+ < K+ < Rb+ < Cs+
Ionic bond of nature: LiCl < NaCl < KCl < RbCl < CsCl.
2. Increase in size of the anion favours the covalent bond.
Size: F– < Cl– < Br– < I–
Ionic bond nature: CaF 2 < CaCl2 < CaBr2 < CaI2
Among the halides of calcium, CaF 2 is more ionic and CaI2 is more covalent.
3. Highly charged cation and anion form covalent bonds.
NaCl is more ionic compared to the ionic nature of AlCl3. In NaCl, both cation and anion have low charges, while in AlCl 3, the cation Al3+ is of a higher charge than Na+, while the anion is the same in both compounds. According to Fajan’s rule, AlCl3 is more covalent in nature.
4. Cations with inert gas configuration form ionic compounds, while cations with pseudo inert gas configuration favour the formation of covalent compounds.
NaCl is ionic because Na + ion in NaCl has inert gas configuration. CuCl is more covalent because Cu+ ion has pseudo inert gas configuration.
Most favourable conditions for ionic bond are: (a) large cation, (b) small anion and (c) small magnitudes of charge on both the ions.
Among the halides of lithium, LiI is least ionic. AgCl is less ionic than NaCl. LiF is ionic, while BeF2 is more covalent.
The order of increasing covalent character is: NaCl < MgCl2 < AlCl3 < SiCl4, because the charge on the cation increases.
Energy Changes in ionic Bond Formation
Mere formation of oppositely charged ions does not lead to an ionic compound. In an ideal ionic compound, there is a considerable
4: Chemical Bonding and Molecular Structure
electro-static force operating between the two oppositely charged ions (cations and anions). When these ions approach each other in required numbers, they will arrange themselves in a regular pattern so that a close packed arrangement is acquired. While attaining such a structure, energy is liberated and the system gets stabilized.
Crystal lattice energy is another important factor that influences ionic bond formation. Let us study the formation of Na+Cl–, which shows that crystal lattice energy is a contributing factor to stabilize ionic crystals.
Sodium chloride is formed by the union of Na and Cl2. To form an ionic solid, the energy changes are summarized as follows:
Na ionizes to Na+ ions by absorbing energy.
(s) (g)1
gg
(1)
Chlorine accepts the electron exothermically.
between sodium metal and chlorine gas cannot take place on its own (since it is endothermic). But it is well known that sodium and chlorine combine very vigorously to form sodium chloride.
The sum of the two, 147.1 kJ mol -1 , is more than compensated for by the enthalpy of lattice formation of NaCl(s) (–788 kJ mol–1). Therefore, the energy released in the processes is more than the energy absorbed. Thus, a qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state.
Lattice Energy
Lattice is a systematic three dimensional arrangement of oppositely charged ions. This arrangement leads to the stability of the ionic substances. The lattice energy of an ionic crystal is the amount of energy released when one mole of crystal is formed from the oppositely charged gaseous ions. This is denoted by “U”. It can also be defined as the amount of energy needed to dissociate one mole of ionic crystal into isolated constituent gaseous ions.Lattice energies of ionic crystals cannot be measured directly, but experimental values are obtained from thermodynamic data using the Born-Haber cycle.
(2)
By combining (1) and (2), the energy changes involved in the formation of a mole of gaseous Na+Cl– is obtained. i.e., + (g)(g) (g)
NaClNaCl ; +→
524 H[HH] ∆=∆+∆ = + 147.1 kJ mol-1 . . (3)
It can be readily concluded from equation (3) that energy is absorbed during the reaction to form Na+Cl–(g). This means that the reaction
The magnitude of lattice energy is described by coulomb’s law, and it is equal to constant K times the charges on the ions, Z + and Z –, divided by the distance between their centers, ‘d’.
ZZ UK d +− −=×
T he value of constant K depends on the geometrical arrangement of ions in the specified compounds, and it is different for different substances.
Within a series of compounds that have the same anion but different cations, lattice energy increases as cation size decreases. So, lattice energies follow the order: LiF > NaF > KF. Similarly, within a series of compounds that have the same cation but different anions, lattice energy increases as anion size decreases. So, lattice energies follow the order:
LiF > LiCl > LiBr > LiI.
The compounds of ions with higher charges have large lattice energies than compounds of ions with lower charges. So, lattice energy order is:
NaF < MgO < AlN
Properties of Ionic Compounds
Crystalline nature: All ionic compounds are crystalline solids. The solid state is due to close packing of oppositely charged ions in the crystal lattice. As a result no isolated, discrete molecule exists in the crystal lattice.
NaCl crystal lattice is described as face centred cubic lattice (FCC), as shown in Fig. 4.3
between oppositely charged ions, which keep them in their allotted positions. The brittleness of the crystals is due to movement of layer of crystal on the other layer by application of external force.
Melting and boiling points:
The melting and boiling points of ionic compounds are high. M.P. of NaCl is 803°C. The high values are due to strong electrostatic attraction between the oppositely charged ions in a crystal. Large energy is required to overcome the interionic forces of attraction in a crystal lattice. Melting point of ionic solid is sharp and fixed. The order of melting points in sodium halides is: NaF > NaCl > NaBr > NaI
Solubility
The ionic compounds are polar because of the presence of oppositely charged ions. They are soluble in polar solvents, like water, liquid NH 3 and liquid HF. They are insoluble in non-polar solvents, like benzene, ether and CCl4. For instance, when NaCl is dissolved in water, the constituent ions, Na + and Cl–, get solvated with water. Water has high dielectric constant. It weakens the strong electrostatic attraction between the oppositely charged ions in a crystal. The ions are separated in an aqueous solution. Each ion is surrounded by water molecules in aqueous solution (Fig.4.4.) of ionic compound. This process is called hydration. The energy liberated during the process is called hydration energy. In case of soluble ionic compounds, the hydration energy is greater than the lattice energy.
One unit cell of NaCl is associated with 27 ions, in which 14 ions are of one type and 13 ions are of other type. However, the stoichiometric ratio of constituent ions in sodium chloride crystal is 1:1.
Hardness and Brittleness:
The ionic compounds are hard in nature. The hardness is due to strong forces of attraction
NaCl dissolves in water due to the formation of Na + (aq) and Cl –(aq) . Salts like CaF2, BaSO4, AgCl, PbSO4 and Ca3(PO4)2 are insoluble in water because of their high lattice energies.
Electrical Conductivity
The ionic solids do not conduct electricity. The reason is that the ions, on account of
Fig.4.3. Unit cell of NaCl
4: Chemical Bonding and Molecular Structure
electrostatic forces of attraction, remain intact, occupying fixed positions in the crystal lattice. The ions thus do not move when electric current is applied. However, ionic compounds conduct electricity either in fused state or in aqueous solution because the ions can freely move. The ionic compounds are known as electrolytes.
Ionic Reactions
The electrovalent compounds furnish ions in solution. The chemical reactions which involve ions are called ionic reactions. Such reactions are instantaneous and they have very high reaction rates.
Space Isomerism
The ionic compounds do not exhibit space isomerism or stereo isomerism as the ionic bonds are non-directional.
Q. Lattice energy in sodium chloride is x kJ.
Assuming the same interionic distance, what will be the lattice energy of magnesium sulphide?
Sol. The charge magnitudes of ions in NaCl are 1 and 1, respectively. The product of q1 and q2 = 1 × 1 = 1
The charge magnitudes of ions in magnesium sulphide are 2 and 2 respectively.
The product of q1 and q2 = 2 × 2 = 4
Hence, lattice energy of MgS = 4 x kJ.
TEST YOURSELF
1 Ease of formation of anion is favoured by (1) lower value of ionisation energy (2) lower value of electronegativity (3) lower value of electron affinity (4) higher value of electron affinity
2. Which of the following is easily formed?
(1) Calcium chloride
(2) Calcium bromide
(3) Potassium chloride
(4) Potassium bromide
3. Which of the following is a favourable condition for the formation of ionic bond?
(1) Small cation with small charge
(2) Small anion with large charge
(3) Large difference in the electronegativity
(4) Small cation with high charge
4. Number of electrons transferred from one Al atom during bond formation in Aluminium fluoride is
(1) 1 (2) 2 (3) 3 (4) 4
5. Which of the following is least ionic?
(1) CaF2 (2) CaBr2
(3) CaCl2 (4) CaI2
6. The strongest ionic bond is present in (1) LiF (2) NaF (3) RbF (4) CsF
7. The compound with highest melting point is
(1) NaF (2) NaCl (3) NaBr (4) NaI
8. CuCl has more covalent character than NaCl because
(1) Na+ has more polarizing power than Cu+
(2) Cu+ has more polarizing power than Na+
(3) Cl– has pseudo inert gas electron configuration
(4) Na + has pseudo inert gas electron configuration
9. The electronic structure of four elements a,b,c and d are
(a) 1s2 (b) 1s2, 2s2, 2p2
(c) 1s2, 2s2, 2p5 (d) 1s2, 2s2, 2p6.
The tendency to form electrovalent bond is greatest in
(1) a (2) b (3) c (4) d
10. If Na+ ion is larger than Mg2+ ion and S2– ion is larger than Cl– ion, which of the following will be least soluble in water?
(1) NaCl (2) Na2S (3) MgCl2 (4) MgS
11. Lattice energy of NaCl is ‘ X ’. If the ionic size of A+2 is equal to that of Na +, and B–2 is equal to Cl–, then lattice energy associated with the crystal AB is
(1) X (2) 2 X
(3) 4 X (4) 8 X
12. For which of the following sets are all the compounds ionic?
(1) NaF, BF3, MgF2
(2) NaBr, MgBr2, MgO
(3) Al2O3, MgO, SO3
(4) NCl3, BeCl2, AlCl3
13. Among the following compounds, the one with greatest ionic character is (1) NaCl (2) KCl
(3) CsCl (4) RbCl
Answer Key
(1) 4 (2) 3 (3) 3 (4) 3
(5) 4 (6) 1 (7) 1 (8) 2
(9) 3 (10) 4 (11) 3 (12) 2
(13) 3
4.3 COVALENT BOND
Irving Langmuir modified the Lewis postulates and introduced the term, colvalent bond. Sometimes Atoms do not gain or lose electrons due to lower difference of their electron grativity (less than 1.7). In such cases the formation of covalent bond is possible is possible between the atoms.
The force of attraction which binds atoms by mutual sharing of electrons (which are contributed by the participating atoms) is called covalent bond.
Examples:
1. In chlorine molecule, their is one Cl-Cl single covalent bond. In its formation, each Cl-atom contributes one electron for sharing, the resultant pair of electrons is
shared between them, thus each Cl-atom gets octet after sharing.
Cl Cl x x x x x x x
2. Similarly in H2, H-F, single covalent bonds are formed in each case.
Polar
and Non-Polar covalent bonds:
1. In between two same non-metal atoms, equal sharing of electron pairs takes place, and the bond formed is a nonpolar covalent bond. Example: H 2 ,Cl 2 , etc.
2. In between two different non-metal atoms, unequal sharing of electron pairs takes place, and the bond formed is a polar covalent bond. Example: HCl, ICl, etc.
3. Some molecules like H 2 O 2 and N 2 H 4 possess both polar and non-polar covalent bonds.
4.3.1 Lewis Dot Model
Lewis dot model of bonding is a very simple model of explaining the structures of covalent molecules. Each valence electron of the atom involved in the bond formation is denoted by a ‘dot’. Atoms share electrons in order to acquire ‘octet’ configuration. Hydrogen atom, however, acquires ‘duplet’ configuration. The ‘octet’ or ‘duplet’ can be illustrated by drawing a circle around each atom.
Multiple Bonds
Multiple bonds formation
■ When a pair of electrons is shared between two atoms, the resultant bond is called single covalent bond.
■ When two pairs of electrons are shared between two atoms, the resultant bond is
double bond when three electron pairs are shared between the atoms, the resultant bond formed is called triple bond. These bonds are multiple bonds.
■ Atomic oxygen has six valence electrons. It has tendency to gain the configuration of Neon. Therefore in O2-molecule each atom contributes two electrons for sharing. Thus sharing of two pairs of electrons gives rise a double bond in O2, O=O.
Oxygen(O2)
Ozone(O3)
Nitrogen (N2)
Carbon dioxide (CO2)
Carbon monoxide (CO)
Boron trichloride (BCL3)
iv. In N 2 molecule, N ≡ N, a triple bond is formed as a resultant of sharing of three pairs of electrons.
Octet Rules:
Lewis formulated the octet rule. Octet rule states that an atom must possess eight electrons in its outermost energy level for its stability. The octet rule also applies to molecules with more than one electron pair shared. Some other examples of Lewis dot structures are listed in Table. 4.3
The octet rule though useful, is not universal. It applies mainly to the second period elements.
Exceptions to the Octet Rule
Incomplete Octet
In some compounds, the number of electrons surrounding the central atom in a stable molecule is fewer than eight. Examples are BeCl2, BF3, AlCl3 etc.
The expanded Octet
There are more than eight valence electrons around the central atom in a number of compounds. This is termed as expanded octet. In addition to 3s and 3p orbitals, elements in
Methane (CH4)
Ethylene (C2H4)
Ammonia(NH3)
Ammonium cation (NH4+)
Carbonate anion (CO32–)
Nitrite anion (NO2–)
Nitric acid (HNO3)
Phosphorus Pentachloride (PCl5)
Sulphur hexafluoride(SF6)
Nitrogen trifluoride (NF3)
Table. 4.3 Some examples of Lewis dot formulae
the third period also have 3d orbitals that can be used in bonding. Examples are SCl 4, PCl5, SF6, IF7, H2SO4, etc.
Odd Electron Molecules
Some molecules contain an odd number of electrons. Examples are, NO, NO 2 , ClO 2 , etc. Octet configuration has even number of electrons. The octet rule clearly can never be satisfied for molecules with odd number of electrons.
Cations
Some of the transition metal ions are known in which the octet is not complete. Examples are Fe2+, Fe3+, Co3+, Ni2+, Cu2+ etc.
Guidelines to Write the Lewis Dot Formula
■ The total number of electrons available for writing the structures is obtained by adding the valence electrons of the combining atoms.
Example: CH 4 molecule has 8 valence electrons (4 electrons from C and 4 electrons from four hydrogens).
■ For anions, to the total number of electrons from step (a), the number of electrons equal to the overall negative charge is to be added.
Example: In CO32– , the available number of electrons = 4 from C + 18 (3 × 6) from oxygens + 2 charges = 24
■ For cations, the number of electrons equal to the overall positive charge is to be subtracted from the total number of electrons from step (a).
Example: For NH4 + ion, the number of electrons available = (5 from Nitrogen + 4 from four Hydrogens –1 charge) = 8.
■ Distribute the total number of electrons as bond pairs between the atoms. For this, the skeletal structure of the compound is written.
■ In general the least electronegative atom is made the central atom of the molecule or ion.
Example: In NF3, nitrogen is the central atom and fluorines are bonded atoms.
■ Single bonds are constructed first with bond pairs. Then, the remaining electron pairs are utilized either for multiple bonds or for the non-bonded pairs, i.e., lone pairs. In each case, each of the bonded atoms (except hydrogen) gets an octet of electrons.
Let us consider methyl nitrite, CH3ONO.
Step 1: Total number of valency electrons = 24, the number of electron pairs = 12
Step 2: H | HCONO | H
Step 3: 6 electron pairs are used as bond pairs. Now, the remaining six electron pairs are to be placed as lone pairs on the atoms in the descending order of electronegativity.
| HCO | H O N O
Nitrogen does not satisfy the octet rule.
Setp 4: Convert one lone pair on terminal oxygen as p bond.
| HC | H O N O
Now, all the atoms in the molecule are satisfying the octet rule. Thus, it is the correct structure.
4.3.2 Formal Charge
The charge on a molecule is zero. In case of polyatomic ion, the net charge is possessed
4: Chemical Bonding and Molecular Structure
by the ion as a whole and not by a particular atom. It is, however, feasible to assign a formal charge on each atom.
Formal charge is a factor based on a pure covalent bond formed by the sharing of electron pairs equally by neighbouring atoms. Formal charge of an atom in a polyatomic molecule or ion is defined as the difference between number of valence electrons of the atom in free state and number of electrons assigned to the atom in the Lewis structure.
Formal charge is denoted by Qf and is given by Qf = NA – NM = NA – (NLp + 1/2 NBp)
NA is the number of valence electrons in the free atom,
N M is the number of valence electrons belonging to the atom in the molecule, NLp is number of electrons in lone pairs and NBp is number of electrons in bond pairs. The counting of number of electrons is based on the assumption that the atom in the molecule owns one electron of each shared bonded pair and both the electrons of unshared lone pair.
Illustrative Examples
1. PH 3 molecule: The Lewis dot structure is P H H H or P H H H
Formal charge of P:
=−=−+ fAMALP BP Q[NN][N(N1/2N)]
= {5 – 2 – 1/2(6)} = (5 – 5) = 0
Formal charge of H :
=−=−− fAMALP BP Q[NN][N(N1/2N)]
= {1 – 0 – 1/2(2)} = (1 – 1) = 0
2. N2O molecule: The Lewis structure is
(1) (1) (2) (2)
Formal charge of first N atom is Q f = 5 –4 – 1/2(4) = –1.
Formal charge of second N atom is Qf = 5 – 0 – 1/2(8) = +1
Formal charge of O atom is Qf = 6 – 4 – 1/2(4) = 0
(or)
Formal charge on O(1)
Formal charge on O(2)
Formal charge on O(3)
Note:
Formal charge of central oxygen atom is +1 only and terminal oxygen atoms may be 0 or -1
Importance of Formal Charge
The formal charges do not represent real ionic charges. It is possible to select a structure of lowest energy for the given molecule from the various Lewis structures of the molecule with the help of formal charges. The lowest energy structure is the one with the smallest formal charges on the atoms.
Covalency
The number of electrons contributed by an atom of the element for sharing with other atoms or number of electron pairs shared so as to achieve the nearest noble gas configuration is known as covalency of the element. It can also be defined as the number of covalent bonds formed by the atom of the element with other atoms in the formation of molecule.
The covalency of oxygen in H 2O is 2. The covalency of carbon in CH4 is 4. The covalency of non-metallic elements except hydrogen is
equal to (8-G), where G is the group (Roman) number of the element. Common co-valence of IVA, VA, VIA and VIIA group elements are 4, 3, 2 and 1. Generally, the covalency of an element is equal to the total number of unpaired electrons in s and p orbitals of the valency shell. Common co-valence of N(1s22s22p1x 2p1y 2p1z), O(1s22s22p2x 2p1y 2p1z) and F(1s22s22p2x 2p2y 2p1z) are, respectively 3, 2 and 1.
The elements, which have vacant ‘d’ orbitals in their valency shell, exhibit variable covalency by increasing the number of unpaired electrons under excited conditions. Such a shifting is not possible in case of H, N, O and F because of the absence of ‘d’ orbitals in their valence shell.
4. Phosphorus in ground state has the configuration [Ne] 3s 2 3p3.
P(Z=15) [Ne] 3p 3d
3s
It has three unpaired electrons. Hence, the covalency is 3. Phosphorus in the excited state has the configuration [Ne] 3s 13p33d1.
3s [Ne] 3p 3d
It has five unpaired electrons. Hence, the covalency is 5. Thus, phosphorus exhibits covalency 3 and 5.
Sulphur exhibits covalency 2, 4, and 6 in the ground, first excited, and second excited states, respectively. Chlorine exhibits covalency 1, 3, 5 and 7 in the ground, first excited, second excited and third excited states, respectively.
Q. Will the formal charges of an atom in a molecule remain the same ? Why or why not?
Sol. No. the formal charge on an atom may not be same. It changes with the structural environment of the atom in the molecule. In resonance structures, the electronic environment changes. Hence, the formal charges also may change.
Example:
N2O has the resonance structures and
The formal charges in structure I were of N (1), N(2) are –1 and +1. In structure (II) the formal charges of N(1), N(2) and O are 0, +1 and – 1, respectively.
Q. In which excited state sulphur forms six covalent bonds in SO3
Sol. Electron configuration of S = 1s2 2s2 2p6 3s2 3p4
In 1st excited state it has four unpaired electrons (1s2 2s2 2p6 3s2 3p3 3d1). In 2nd excited state its configuration is 1s2 2s2 2p6 3s1 3p3 3d2. In this excited state it forms six bonds.
TEST YOURSELF
1. In molecule of COCl2, between carbon and oxygen there are four dots in Lew’s dot formula. What does it indicate?
(1) four non-bonding electrons
(2) a double bond ond between carbon and oxygen
(3) between carbon and oxygen the bond formed is odd electron bond
(4) valency shown by carbon is 4
2. Which of the following has odd electron?
(1) SO2 (2) ClO3 (3) N2O4 (4) H2O
3. Which of the following molecules obeys octet rule?
(1) BCl3 (2) BeCl2 (3) NCl3 (4) SF4
4. Which is the total number of valence electrons involved in bonding by all atoms in the compound, H2CO3?
(1) 16 (2) 12 (3) 24 (4) 10
5. In drawing Lewis dot formula of CN– ion, how many number of dots is expected to appear?
(1) 10 (2) 11 (3) 9 (4) 13
6. What are the formal charges of carbon in 2 3 CO and CN– ions respectively
(1) Zero and +1 (2) Zero and –1 (3) +1 and –1 (4) +1 and –4
7. The bond between two identical non-metal atoms has a pair of electrons (1) unequally shared between the two (2) transferred fully from one atom to another (3) with identical spin (4) equally shared between them
8. Covalent compounds are generally soluble in (1) polar solvents (2) non-polar solvents (3) concentrated acids (4) all solvents
9. Which of the following boils at higher temperature?
(1) CCl4 (2) CO2
(3) C6H12O6 (4) KCl
10. CCl4 is insoluble in water because (1) H2O is non-polar (2) CCl4 is non-polar (3) they form intermolecular H-bonding (4) they form intramolecular H–bonding
11. Which of the following is very much volatile? (1) Diamond (2) Sodium chloride (3) Calcium (4) Dry ice
12. The following are some statements about the characteristics of covalent compounds
a) The combination of a metal and nonmetal must give a covalent compound.
b) All covalent substances are bad conductors of electricity.
c) All covalent substances are gases at room temperature.
The distinction between ionic and covalent bonds is presented in Table. 4.4.
Table 4.4 Differences between ionic and covalent bonds
Ionic bond Covalent bond
1. Ionic bond is formed by the transfer of one or more electrons.
2. It is formed between metal and non-metal.
3. It is also called electrovalent bond and is due to electro-valency.
4. Ionic bond consists of electrostatic force of attraction between the oppositely charged ions.
5. Ionic bond is non rigid and non directional.
6. Ionic bond is polar in nature.
1. Covalent bond is formed by sharing of electrons.
2. It is formed between nonmetals.
3. It is called electron pair bond and is due to covalency.
4. Covalent bond consists of shared pair or pairs of electrons which are attracted by both the nuclei.
5. Covalent bond is rigid and directional except for H2
6. Covalent bond may be polar or non polar.
The characteristic properties of covalent substances are presented below.
Physical State
Under normal conditions of temperature and pressure, covalent substances exist as gases or volatile liquids. For example, H2, O2, N2, CH4 are gases but H2O, CHCl3, C6H6 are liquids. Very weak. Van der Waals forces exist between discrete molecules. Some covalent substances exist as soft solids, if their molecular masses are high. Sulphur, iodine, sucrose, etc., are soft solids.
Melting and Boiling Points
In covalent substances, weak Van der Waals’ forces of attraction operate between the molecules. A small amount of energy is sufficient to overcome the attractive forces. Hence, they have low melting and boiling points.
Crystal Structure
The lattices of the covalent substances are composed of neutral individual molecules with no strong electrostatic force between them. Thus covalent molecules are not closely
Physical state
Melting and boiling points.
Solubility
Conductivity
Nature of reactions
Isomerism
packed.
However, diamond, silicon carbide, silica, aluminium nitride, etc., are very hard and they possess high melting points, because of their net work structures.
Non-Brittle Nature
Covalent substances with high molecular weights are solids. They are soft and nonbrittle as they do not contain any ions in their crystal lattice.
Solubility
Generally, the covalent substances are insoluble in polar solvents like water, but soluble in nonpolar solvents like benzene, hexane, alcohol, ether, etc. Some of the covalent compounds, like alcohols and amines, dissolve in water due to hydrogen bonding. Covalent solids that have giant molecules are insoluble in water.
Isomerism
Covalent molecules exhibit isomerism because they possess rigid and directional bonds.
Differences in the properties of ionic and covalent substances are presented in Table. 4.5.
Ionic compounds are crystalline solids at room temperature.
They possess high melting and high boiling points
Freely soluble in polar solvents like water and liquid ammonia
Bad conductors of electricity in solid state, but good conductors in molten state and in aqueous solution
They undergo ionic reactions. The ionic reactions are fast and instantaneous. Ionic compounds do not exhibit isomerism, as electrovalent bond is non directional.
Covalent substances are usually gases, low boiling liquids and soft solids under ordinary conditions.
Possess low melting and low boiling points with the exception of gaint molecules
Generally, insoluble in water but soluble in nonpolar solvents
Bad conductors of electricity, with a few exceptions, like graphite
they undergo molecular reactions. The reactions are slow and the rates are low. Covalent substances exhibit isomerism, as the covalent bonds are directional.
Table 4.5 Differences between ionic and covalent substances
Electrical Conductivity
Covalent substances do not conduct electricity either in the molten state or in aqueous solutions because there are neither free electrons nor ions present in them. Therefore, they are non electrolytes. Graphite can conduct electricity.
Rates of Reactions
Covalent substances undergo molecular reactions. During these reactions, bond breaking and bond making take place. These reactions are generally slow and rarely proceed to completion. Hence, catalysts are used to speed up these reactions.
4.4 COORDINATE COVALENT BOND
1. Coordinate bond is a special type of covalent linkage, which is formed by mutual sharing of pair of electrons between the two atoms but the pair is contributed by onoly one of the two atoms.
2. The Lewis concept was extended by Sidgwick to explain the mechanism of the formation of coordinate bond. The atom which contributes the electron pair for sharing is called electron pair donor and the atom that accepts the electron pair for sharing is referred to electron pair acceptor. The coordinate bond is shown by an arrow, → pointing from donor atom to acceptor atom.
An atom to act as a donor must have at least one lone pair in its valence shell.
Example:
one lone pair two lone pairs one lone pair
An atom to act as an acceptor must have an empty orbital to accommodate the accepted lone pair of electrons.
Example:
empty orbital empty orbital empty orbital
3. Compounds containing unshared pair of electrons readily form dative bonds. For example, when ammonia is mixed with gaseous hydrogen chloride, white clouds of ammonium chloride are formed. The reaction involves the formation of a dative bond between N atom in NH3 containing lone pair and H+ ion from HCl.
A similar reaction occurs, and hydronium ion is formed when hydrogen chloride dissolves in water.
4. Once a dative bond is formed, it is indistinguishable from a covalent bond. The only difference between the two is mode of formation.
5. A dative bond is formed between ammonia and boron trifluoride because N in NH3 has lone pair of electrons while boron in BF 3 has only six electrons in the valence
shell. sp3 hybrid orbital of nitrogen having a lone pair overlaps the vacant p-orbital of boron.
Some other examples
1. CO C O
2. HNO3 H O N=O O
[Cu(NH3)4]2+
4. [Ag(CN)2]– CN–→ Ag+ ← CN–
5. I3 I–I –I
6. O3 O O O
7. SO2 S O O (according to octet rule)
8. SO3 s O O O (according to octet rule)
Properties of Coordinate Compounds
The properties of coordinate compounds are intermediate between the properties of ionic and covalent compounds.
1. Coordinate compounds exist as gases, liquids and solids under ordinary conditions.
2. The melting and boiling points of coordinate compounds are higher than
those of covalent and lower than those of ionic substances.
3. They are less soluble in polar solvents like water, but readily soluble in non polar solvents like benzene and carbon tetrachloride.
4. They are as stable as covalent compounds.
5. Like covalent compounds, coordinate compounds are poor conductors of electricity. They undergo molecular reactions, which are slow.
6. The dative bond is a strong bond with directional nature.
TEST YOURSELF
1. In the molecule, HNO3, which atom acts as electron pair donor?
(1) oxygen (2) Nitrogen
(3) Hydrogen (4) NO atom
2. In NH3BF3 addition compound formation, Boron is electron pair acceptor because (1) Boron has five valence electrons (2) Boron is a non-metal
(3) Boron has vacant orbital and electron deficient
(4) NH3 can donate electron pair
3. Which one of the following has no coordinate bond?
(1) Ni(CO)4 (2) 4 NH +
(3) CCl4 (4) CO
4. Which one of the following is not the property of coordination compounds?
(1) They have directional bond
(2) They are soluble in polar solvents
(3) They have polar bonds
(4) They are not good conductors of electricity
Answer Key (1) 2 (2) 0 (3) 3 (4) 2
4.5 VALENCE BOND THEORY
The va lence bond theory was proposed by Heitler and London to explain the formation of a covalent bond by the application of Schrodinger wave equation. The theory was extended by Pauling and Slater to explain the shapes of the molecules and the direction of the bonds in the molecules.
4.5.1 Postulates
■ A covalent bond is formed by the overlap of atomic orbitals of valence shell of the two atoms.
■ Only half-filled atomic orbitals can enter into overlapping process.
■ As a result of overlap, the density of shared electron pair is concentrated mostly on the internuclear axis, an imaginary line joining the nuclei of the bonded atoms.
■ The resulting bond acquires a pair of electrons with opposite spins.
■ The greater the extent of overlap, the stronger is the covalent bond formed.
■ The bonds are formed mostly in the direction in which the electron clouds are concentrated.
■ The electrons in bonding orbitals are attracted by both the nuclei of the bonded atoms.
■ As a result, the electron cloud is concentric and symmetric in the space between the nuclei so that the repulsions between the nuclei are reduced and the stability is attained by the molecules.
■ The amount of energy released per mole of gaseous molecules during overlapping process is called bond energy. This energy stabilises the system. Here, the molecule formed has less energy and consequently, it is more stable than the isolated atoms.
Formation of H2 Molecule
Each hydrogen atom provides ‘1s’ orbital with unpaired electron for bonding. The orbital overlap present in hydrogen molecule is s-s overlap. The molecule is shown as H..H or H–H.
Formation of Cl2 molecule
s–s Overlap
Each chlorine atom provides ‘3p’ orbital each having an unpaired electron for bonding. The orbital overlap present in chlorine molecule is p-p overlap. The molecule is shown as Cl Cl or Cl–Cl.
Formation of HCl Molecule
Hydrogen atom provides ‘1s’ orbital, while chlorine atom provides ‘3p’ orbital each having an unpaired electron for bonding. The orbital overlap present in hydrogen chloride molecule is s-p overlap.
1s 3p s-p overlap
I n a given energy level, the overlapping efficiency is in the order: s-s < s-p < p-p.
The overlap between atoms of the same element is called homoatomic overlap. The molecule formed is called homoatomic or homonuclear molecule. eg. H 2, F 2, Cl 2, Br 2, I 2 , O 2 , N 2 etc. The overlap between atoms of different elements is called heteroatomic overlap.
The molecule formed is called heteroatomic or heteronuclear molecule. Eg. HF, HCl, ICl, etc. Homoatomic molecules are non-polar but hetrodiatomic molecules are polar.
Based on the difference in the overlapping, covalent bonds are of two types sigma (s) bond and pi ( p ) bond.
Sigma Bond
A covalent bond formed between two atoms by the end-to-end or axial or linear or headon overlap of half-filled orbitals along their axes is called sigma bond. A sigma bond is the first bond formed between bonded atoms. Sigma bond is strong because the extent of overlapping between electron clouds is maximum. The electron cloud of s bond is symmetrical about the line joining the nuclei of the two atoms. S–S, S-P and P-P coaxial overlaping give s bonds.
Pi Bond
A covalent bond formed by the sidewise or lateral or parallel or conjugate overlap of the half-filled orbitals is called pi bond. The axes of the orbitals involved in overlapping should be parallel to one another and perpendicular to the internuclear axis.
The bonding orbital consists of two charged clouds above and below the plane of the participating atoms. Each additional bond
Sigma (s ) bond
1. Sigma bond is formed by the overlap of orbitals along their axes. It includes s-s, s-p and p-p overlapping.
formed between bonded atoms is a pi bond. These pi bonds are formed by p or d orbitals, but not by s-orbitals. A pi bond is weak bond when compared with bond, because the extent of overlapping is relatively small.
Free rotation about the bond is possible but free rotation is not possible around p bond. s bond can be formed independently but p bond exists along with s bond only. Hence, the shape of the molecule is decided by s bond but not by p bond. Substances with only s bonds are less reactive but substances with p bonds are more reactive. Hybrid orbitals and s orbitals can form s bond but not p bond.
A single bond consists of only one s bond. It is formed by sharing one pair of electrons.
A double bond consists of one s bond and one p bond. It is formed by sharing two electron pairs.
A triple bond consists of one s bond and two p bonds. It is formed by sharing three electron pairs. Double bond and triple bond are collectively known as multiple bonds. Without forming a s bond between two atoms, a pi bond can never be formed. The difference between a s bond and a p bond are given in Table 4.6.
Pi (p ) bond
1. Pi bond is formed by the side-wise or lateral overlap of orbitals.
2. s bond is relatively stronger bond. 2. p bond is relatively weaker bond.
3. Electron cloud is symmetrical about the inter nuclear axis. 3. Electron cloud is unsymmetrical, but present in two lobes are in equidistant to the axis.
4. There can be free rotation of the atoms around the bond. 4. Free rotation is not possible around the bond.
5. Substances with bonds are less reactive. 5. Substances with p bonds are more reactive.
6. The shape of the molecule is described by bonds. 6. p bonds do not affect the shape of the molecule.
7. s electrons are referred to as localised electrons. 7. p electrons are referred to as mobile
8. s bond has independent existence. 8. p bond exists along with the bond.
Table 4.6 Differences between s bond and p bond
Localised Orbital
An orbital with bonded electron cloud between the nuclei of the bonded atoms is called localised orbital. The electron pair is called bond pair or localised electron pair. The electron pair that is present in the valence shell and not used for bonding is known as non bonded electron pair or unshared pair or lone pair.
Formation of Oxygen Molecule
The electronic configuration of oxygen atom, O (Z=8), is 1s2 2s2 2px2 2py1 2pz1. Oxygen atom has two half filled ‘p’ orbitals in the valence shell, which are perpendicular to each other. When the two oxygen atoms are brought nearer to each other, the half filled 2py orbitals of two oxygen atoms overlap along the same axis to form a strong s bond. The half-filled 2p z orbitals on both the oxygen atoms are parallel to one another. They undergo lateral overlap to form a weak p bond, as shown in Fig. 4.4
Fig.4.4. A Double Bond in O2 Molecule
Formation of Nitrogen Molecule
The electronic configuration of nitrogen atom, N (Z=7) is 1s 2 2s 2 2p x 1 2p y 1 2p z 1 . Nitrogen atom has three half filled ‘p’ orbitals, which are perpendicular to each other. When two nitrogen atoms are brought nearer to one another, the half filled 2p x orbitals of two nitrogen atoms overlap along the same axis to form a strong s bond. The remaining half filled 2p y and 2p z orbitals on the nitrogen atoms undergo two lateral overlappings along the y axis and z axis to form two p bonds as shown in Fig. 4.5
atom
Formation of H2O Molecule
Water molecule has two O–H bonds. The two bonds are bonds. They are the result of overlap 2p y and 2p z orbitals of oxygen containing unpaired electrons separately with the 1s orbitals of two hydrogen atoms. The angle between these s s-p bonds should be 90° (because 2py and 2pz orbitals are perpendicular to each other and they are involved in overlapping). But the experimental results show that it has 104°30΄ This angle could be explained by VSEPR theory.
Formation of CH4 Molecule
To explain the tetravalence of carbon, excited state electronic configuration (1s22s12px1 2py1 2p z1) is to be used. The energy required for excitation comes from the energy released during the C–H bond formation. One C–H bond is s-s and the remaining three C–H bonds are s s-p .
According the VBT, one bond is non directional and the angle between the three directional bonds is 90°. But actually, in CH4, all the four C–H bonds are identical and bond angle is 109°28΄. Later it was explained by hybridisation and VSEPR theory.
Directional Property of Covalent Bond
Hydrogen is the simplest molecule with a nondirectional 1s–1s overlap. All other covalent bonds in simple as well as in polyatomic molecules involve directional properties.
Fig.4.5 A triple bond in N2 molecule
Depending on the characteristics of atomic orbitals, the overlap involved in the bond formation may be zero overlap, positive overlap or negative overlap. Zero overlap is that which does not lead to the covalent bond formation. No bond is resulted between a p x orbital and a py orbital. Similarly, bond is not formed between an s orbital and a p x orbital in the direction of y-axis or z-axis. Positive or negative overlaps lead to the bond formation, as shown in Table 4.6. The differences between a s bond and a bond are given in Table 4.7.
Table 4.7 Positive or negative overlapping
Drawbacks
Valence bond theory explains the strength of covalent bond and its directional characteristics. However, it fails to explain the identical nature of two valencies of Be, three valencies of B and four valencies of carbon. The theory also fails to explain the shape and bond angles of even simple molecules like CH4, NH3, H2O, etc.
Q. What kind of a bond is formed when the orbitals of two atoms A and B undergo (i) s-s overlap (ii) s-p overlap?
Q. Can px overlap a py orbital? State the reason?
Sol. No. They cannot overlap because their orientations and symmetries are not same.
Try yourself
Q. On the basis of valence bond theory predict the bond angle in NH3(experimental bond angle is 107°) Answer:
TEST YOURSELF
1. Most important concept of valence bond theory is (1) overlap of atomic orbitals results in the bond
(2) sharing of odd number of electrons for bonding
(3) sharing of electrons follow the octet rule (4) transfer of electrons follow the octet rule
2. The type of overlap present in the bonds of hydrogen sulphide molecule is (1) s–p (2) s–s
(3) p–p (4) sp3–s
3. Iodine monochloride molecule is formed by the overalap of (1) s-s orbitals (2) s-p orbitals
(3) p-p orbitals end to end (4) p-p orbitals sideways
4. Choose the molecule which has only one pi bond
(1) CH2 = CH – CH = CH2
(2) CH2 = CHCOOH
(3) CH3CH = CH2
(4) CH ≡ CH
5. Double bonds are present in (1) CO2 (2) BeCl2
Sol. In both (i) and (ii), only s -bonds are formed. Because s-orbitals are nondirectional in nature, always overlap along the internuclear axis only. So s-orbitals form s-bonds only. The s-bonds in (i) and (ii) are denoted as and , respectively.
(3) HgCl2 (4) MgO
6. According to VB theory, the bonds in methane are formed due to the overlapping (1) 1 s s–s, 3 s s–p (2) 1 s s–p, 3 s s–s
(3) 2 s s–s, 2 s s–p (4) σ 3 sp 4
7. The bond between chlorine and bromine in BrCl is
(1) ionic
(2) non-polar
(3) polar with negative end on Br
(4) polar with negative end on Cl
8. Some statements about valence bond theory are given below
a) The strength of bond depends upon extent of overlapping.
b) The theory explains the directional nature of covalent bond.
c) According to this theory oxygen molecule is paramagnetic in nature.
Choose the correct option.
(1) all are correct
(2) only a and c are correct
(3) only a and b are correct
(4) all are wrong
9. In the electronic structure of acetic acid, there are
■ I n order to explain the characteristic geometrical shapes of polyatomic molecules
(such as H 2 O, NH 3 , CH 4 etc.,) pauling introduced the concept of Hybridization.
■ The phenomenon of intermixing of orbitals having nearly the same energy to form the same number of orbitals of equivalent energies and shape is called hybridization.
■ The equivalent orbitals formed are called hybrid orbitals.
4.6.1 Rules of Hybridisation
■ The orbitals of one and the same atom only participate in the hybridisation. Orbitals of different atoms do not hybridise.
■ The energies of the atomic orbitals involving in hybridisation must be nearly equal.
■ Half-filled, completely-filled and empty atomic orbitals can participate in hybridisation.
■ If an orbital is present in the outermost orbit, it is not a must for it to involve in hybridisation.
■ The number of hybrid orbitals formed is equal to the number of atomic orbitals participating in hybridisation.
■ The hybrid orbitals are symmetrically arranged around the nucleus of the atom so that they have maximum stability.
■ The angle between any two adjacent hybrid orbitals is same, and it is generally the bond angle of the molecule.
■ The distribution of electrons in hybrid orbitals takes place as per Pauli’s exclusion principle and Hund’s rule of maximum multiplicity.
■ Hybrid orbitals form only s bonds. They do not form p bonds, in their participation in bonding.
4.6.2 Types of Hybridisation
Various types of hybridisation: They are sp, sp2, sp3, dsp2, sp3d, sp3d2 and sp3d3. hybridisation and shapes of some examples are given in Table 4.8.
Table 4.8 Common hybridisations and corresponding shapes
S. No. Pure orbitals involved in hybridisation Resulting hybrid orbitals (their no.)
Shape of the molecules as a result
Bond angles in them. Examples
1. one s; one p sp (2) Linear 1800; BeCl2; C2H2
2. one s; two p sp2 (3) Trigonal planar 1200; BCl3; C2H4
3. one s; three p sp3 (4) Tetrahedral 109028’; CH4 ; [NiCl4]2–, MnO4–(NH3 and H2O bond angle decreases)
4. one s; two p one d dsp2 (4) Square planar 900 and 1800; [Cu(NH3)4]2+ ; [PdBr4]2–, [Ni (CN)4]2–,
5. one s; three p one d sp3d (or) dsp3 (5) Trigonal bipyramidal
6. one s; three p; two d sp 3 d 2 (or) d2sp3 (6) Square pyramidal (or) Octahedral
7. one s; three p; three d sp3d3 Pentagonal bipyramidal
sp
Hybridisation
The intermixing of one ‘s’ orbital and one ‘p’ orbital to give two equivalent hybrid orbitals is known ‘sp’ hybridisation or Diagonal hybridisation. The two ‘sp’ hybrid orbitals have linear shape with an angle 180°, as shown in Fig.4.6 . Each sp hybrid orbital has 50% ‘s’ character and 50% ‘p’ character.
Formation of BeCl2
The electronic configuration of beryllium atom in its excited state is 1s 2 2s1 2p1
Be(Z=4) is 2px2py2pz
1s 2s
In the excited beryllium atom, ‘2s’ and ‘2px’ orbitals intermix to give two equivalent ‘sp’ hybrid orbitals. The electronic configuration of Cl is 1s2 2s2 2p6 3s2 3px2 3py2 3pz1. It has one half filled ‘p’ orbital. The half filled 3pz orbitals of two chlorine atoms overlap with ‘sp’ hybrid orbitals of beryllium atom in their axes to form two s sp-p - bonds.
BeCl 2 molecule so formed has linear shape. The bond angle in BeCl 2 is 180°.
Formation of Acetylene Molecule
There are two carbon atoms and two hydrogen atoms in the molecule of acetylene. The electronic configuration of carbon in its excited state is 1s22s12px12py1 2pz1.
In the excited carbon atom, one ‘2s’ orbital and one 2p orbital intermix to give two equivalent ‘sp’ hybrid orbitals. They have linear shape. In each carbon, there remains two unhybridised ‘p’ orbitals. They are perpendicular to each other and to the ‘sp’ hybrid orbitals.
In the formation of acetylene molecule, one ‘sp’ orbital of one carbon overlaps with one ‘sp’ orbital of other carbon in the same axis to form bond. The remaining sp orbital on each carbon overlaps with ‘1s’ orbital of hydrogen atom to give bond. There are 2py and 2p z orbitals on both the carbon atoms which are parallel to one another respectively. They overlap laterally to form two p bonds respectively. The combination of one bond and two bonds between carbon atoms is known as triple bond as shown in Fig. 4.7.
Each sp2 hybrid orbital has 33.33% ‘s’ character and 66.66% ‘p’ character. The set of sp2 hybrid orbitals have trigonal planar structure. The angle between any two sp2 hybrid orbitals is 120°.
Formation of BCl3
The central atom in BCl 3 is boron. The electronic configuration of boron atom in its excited state is 1s2 2s1 2p2
B(Z=5) is
Fig.4.7. Acetylene Molecule
The acetylene molecule so formed has a linear structure. There are 3 s and 2 p bonds in the acetylene molecule. The bond angle in C 2H2 HCH is 180°. The C–C bond length is 1.20A°.
sp2 Hybridisation
The intermixing of one ‘s’ orbital and two ‘p’ orbitals to form three equivalent sp 2 hybrid orbitals is known as sp 2 hybridisation. The three sp2 hybrid orbitals are directed towards the corners of an equilateral triangle with an angle of 120° as show n in Fig. 4.8
's' orbital +
sp2 sp2 sp2 1200
Two 'p' orbitals sp2 hybrid orbitals
Fig.4.8. sp2 Hybridisation
In the excited boron atom ‘2s’ orbital and two ‘2p’ orbitals intermix to give three equivalent sp2 hybrid orbitals. In the formation of BCl 3 molecule, three sp2 hybrid orbitals of boron overlap with half filled 3pz orbitals of three chlorine atoms in their axes to give three bonds. BCl3 molecule so formed has trigonal planar structure. The bond angle in BCl3 is 120° formation of BCl 3
Formation of Ethylene Molecule
There are two carbon atoms and four hydrogen atoms in ethylene molecule. The electronic configuration of carbon in its excited state is 22111 xyz 1s2s2p2p2p .
In the excited carbon, one 2s and two 2p orbitals intermix to give three equivalent sp 2 hybrid orbitals. In each carbon, there is unhybridised 2p z orbital, which is perpendicular to the plane of sp 2 orbitals. In the formation of ethylene molecule one sp 2 hybrid orbital of one carbon overlaps with
one sp2 hybrid orbital of other carbon in the same axis to give 22 spsp σ bond.
The remaining two sp 2 hybrid orbitals on each carbon overlap with ‘1s’ orbitals of two hydrogen atoms in their axes to give two 2 sps σ bonds. The two 2p z orbitals on both the carbon atoms, which are parallel to one another, overlap laterally to give a bond. The combination of one s and one p bond between the two carbon atoms is called double bond, as shown in Fig. 4.9
'S' orbital + px, py and Pz, orbitals sp3 hybrid orbitals sp3 sp3 sp3 1090
Fig.4.10 sp3 hybridisation
Formation of Methane Molecule
The central atom in CH 4 is carbon. The electronic configuration of carbon atom in its excited state is 1s 22s12p3 1s 2s
C(Z=6) is 2px2py2pz
Fig.4.9 Ethylene molecule
Th us there are 5 s and 1 p bonds in the molecule of ethylene. C 2 H 4 so formed has planar structure. The bond angle in ethylene HCH is 120°(expected). The C–C bond length is 1.34A°.
The C–H bond is sp 2–s sigma with bond length 108 pm. The H– C–H bond angle is 117.6°, while the H–C–C angle is 121°.
sp3 Hybridisation
The intermixing of one ‘s’ orbital and three ‘p’ orbitals to give four equivalent sp 3 hybrid orbitals is known as sp 3 hybridisation. The four sp3 hybrid orbitals are directed towards the four corners of a regular tetrahedron. The angle between any two sp3 hybrid orbitals is 109° 28΄, as shown in Fig. 4.10.
I n the excited carbon atom, one ‘2s’ orbital and three ‘2p’ orbitals intermix to give four equivalent ‘sp 3’ hybrid orbitals. In the formation of methane molecule, four ‘sp 3 ’ hybrid orbitals of carbon overlap with ‘1s’ orbital of four hydrogens in their axes to give four sps 3 σ bonds. Methane molecule so formed has tetrahedral shape. The bond angle in CH4 is 109°28΄.
Note:
Total number of H–C–H bond angles in CH4 are six.
Formation of Ethane Molecule
There are two carbon atoms and six hydrogen atoms in the molecule of ethane. Each carbon atom in the molecule undergoes sp3 hybridisation. In the excited carbon atom, one ‘2s’ orbital and three ‘2p’ orbitals intermix to give four equivalent sp 3 hybrid orbitals. They have tetrahedral shape. In the formation of ethane molecule, one sp 3 orbital of one carbon overlaps with one sp 3 orbital of other
carbon in the same axis to give spsp 33 σ bond. The remaining three sp 3 hybrid orbitals on each carbon overlap with 1s orbitals of three hydrogen atoms in their axes to form three sps 3 σ bonds. Thus, there are seven bonds in ethane molecule, as shown in Fig. 4.11
Fig.4.11. Ethane Molecule
The molecule so formed has a shape of two tetrahedron fused at the corners. The bond angle in ethane, HCH and HCC is 109°.281. The C–C bond length is 1.54 A°.
Structure of ammonia molecule
The central atom in NH3 molecule is nitrogen. The electronic configuration of nitrogen is 1s 2 2s 2 2p x 1 2p y 1 2p z 1 . In the nitrogen atom 2s, 2px, 2py and 2pz orbitals intermix to give four equivalent sp 3 hybrid orbitals. These are directed towards the four corners of a regular tetrahedron. Among the four sp 3 hybrid orbitals, one is occupied by a lone pair of electrons and the remaining three are half filled. The half filled sp3 hybrid orbitals overlap with ‘1s’ orbitals of three hydrogen atoms to form three sps 3 σ bonds. The expected shape of NH3 is tetrahedral, but NH3 has pyramidal shape with three hydrogen atoms occupying the base, and lone pair forms the apex of the pyramid, as shown in Fig. 4.12. The observed bond angle is HNH 107°. sp3 sp3-s sp3-s ssp3-s s H N N H H H H H lone pair lone pair s 107°
Fig.4.12. Ammonia Molecule
Structure of Water Molecule
The central atom in water molecule is oxygen. The electronic configuration of oxygen is 1s 2 2s 2 2p x 2 2p y 1 2p z 1 . In oxygen atom, 2s, 2p x , 2py and 2pz intermix to give four equivalent ‘sp3’ hybrid orbitals. Among these sp 3 hybrid orbitals, two are occupied by lone pairs of electrons and two are half filled. The two half filled ‘sp3’ orbitals of oxygen overlap with 1s orbitals of two hydrogen atoms in their axe s to form two sps 3 σ bonds. Now oxygen atom is surrounded by two lone pairs and two bond pairs of electrons. The molecule is expected to have a tetrahedral shape. As a result of the presence of two lone pairs, the water molecule assumes ‘V’ or angular or bent shape, as shown in Fig. 4.13 sp3 sp3 sp3 sp3 lone pairs lone pairs S S H O O H 104031 H
Fig.4.13. Water Molecule
The bond angle in water molecule HOH is 104.5°.
Hybridisation Involving d-Orbitals
The atoms of the elements present in 3 rd period contain ‘d’ orbitals in addition to s and p orbitals. The energy of the 3d orbitals is comparable to the energy of 3s and 3p orbitals. The energy of 3d orbitals is also comparable to those of 4s and 4p orbitals. As a consequence, the hybridisation involving either 3s, 3p and 3d or 3d, 4s and 4p is possible.
dsp2 Hybridisation
The orbitals involved in dsp2 hybridisation are xy 22 d, s and two p orbitals. The four orbitals intermix to give four equivalent dsp 2 hybrid orbitals. These hybrid orbitals have square planar geometry. The angle between any two dsp2 orbitals is 900
Example: [Ni(CN)4]–2, [Cu(NH3)4]2+. etc
sp3d Hybridisation
The intermixing of one s, three p and one d orbitals to form five equivalent sp 3d hybrid orbitals is called sp 3d hybridisation. These sp3d hybrid orbitals are oriented in trigonal bipyramidal manner.
Gaseous PCl5 is covalent. The central atom in PCl5 is phosphorus. The electronic configuration of phosphorus in its first excited state is P (Z=15) : 1s2 2s2 2p6 3s1 3px1 3py1 3pz1.Then phosphorus undergoes sp 3d hybridisation. Each of the sp3d hybrid orbital overlaps with 3p z orbital of chlorine, forming five spdp 3 σ bonds. Thus, PCl 5 molecule has trigonal bipyramidal geometry in Fig. 4.14
This is because axial P–Cl bonds experience greater repulsion from equatorial P-Cl bonds. PCl5 is highly reactive and in the solid state exists as [PCl4]+ and [PCl6]– ions.
sp3d2 Hybridisation
The intermixing of one ‘s’ three ‘p’ and two ‘d’ orbitals to form six equivalent hybrid orbitals is known as sp3d2 hybridisation. These hybrid orbitals have lobes directed towards the corners of a regular octahedron.
The central atom in SF6 molecule is sulphur. The electronic configuration of sulphur in its ground state is [Ne] 3s2 3px2 3py1 3pz1. In the 2nd excited state of sulphur atom, one ‘3s’ and one ‘3px’ electrons are promoted to empty 2 z 3d and orbitals, respectively. In the 2 nd excited state, sulphur atom undergoes sp 3d2 hybridisation. Each of the sp3d2 hybrid orbital overlaps with half- filled 2pz orbital of fluorine atom to form six spdp 32 σ bonds. SF 6 molecule so formed has octahedral shape as shown below. This structure is completely regular with bond angles of 90° in Fig. 4.15
It may be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. Three bonds are in one plane at an angle of 120° to one another, as in a triangular planar arrangement. These bonds are termed as equatorial bonds. Of the remaining two bonds, one lies above and the other below the equatorial plane making an angle of 90° with the plane. These bonds are called axial bonds. It may be remembered that this geometry is not symmetrical.
The axial bonds (2.19A0) have been found to be larger than equatorial bonds(2.04A0).
Fig.4.15. Sulphur Hexaflouride
sp3d3 Hybridisation
The intermixing of one ‘s’, three ‘p’ and three ‘d’ orbitals to form seven equivalent sp 3 d 3 hybrid orbitals is called sp3d3 hybridisation. Hybrid orbitals are orientated towards the corners of a pentagonal bipyramid and angles between them are 72° and 90°. Eg.IF 7. In this bond angle between equitorial F–I–F bonds is 72°, bond angle between equitorial and axial bond is 90°.
Fig.4.14. Phosphorous Penta Chloride
Q. Discuss the hybridisation of carbon atoms in allene, C3H4, and show the orbital overlaps.
Sol. Allene is : CH2 = C = CH2
I II III
Carbon atoms I and III are sp2 hybridised, while carbon II is sp hybridised. The two unhybridised orbitals of carbon II overlap sidewise with unhybridised p orbitals of carbon I and III to form bonds.
Q. Calculate the ratio of number of pure and hybrid orbitals used for bonding in an acetylene molecule.
Sol. Acetylene has two ‘C’ atoms and two ‘H’ atoms. Carbon undergoes sp hybridisation.
Number of hybrid orbitals = 2 × 2 = 4
Number of pure orbitals = 4 (from C) + 2 (from H) = 6
Ratio of number of pure and hybrid orbitals = 4:6 = 2:3
TEST YOURSELF
1. The % of ‘p’ character in hybrid orbital of the central atom of water molecule is (1) 25% (2) 50% (3) 75% (4) 33.3%
2. Which hybridisation is found in NH 3 and H2O?
(1) sp3 (2) dsp2 (3) sp (4) sp2
3. Octahedral shape is due to the hybridisation (1) sp3d (2) sp3d2 (3) sp3 (4) sp
4. The type of hybrid orbitals used by the central atom in perchlorate ion is (1) sp3 (2) sp2 (3) sp (4) dsp2
5. The hybrid orbitals have a bond angle of 109.50. The percentage of p-character in the hybrid orbital is nearly (1) 25% (2) 33% (3) 50% (4) 75%
6. In BCl3 molecule, the Cl-B-Cl bond angle is (1) 90° (2) 120° (3) 109°.28’ (4) 180°
7. The hybridisation of oxygen in OF2 molecule is
(1) sp (2) sp2 (3) sp3 (4) dsp2
8. In hydrazine (N2H(4), the hybridisation of nitrogen is (1) sp (2) sp2 (3) sp3 (4) dsp2
9. Increasing order of size of hybrid orbitals is (1) sp, sp2, sp3 (2) sp3, sp2, sp (3) sp2, sp3, sp (4) sp2, sp, sp3
10. A square planar complex is formed by hybridization of which atomic orbitals. (1) s, px, py, 2 z d (2) s, px, py, 22 xy d (3) s, px, py, dyz (4) s, px, pz, d xy
11. The type of overlaping not observed in the formation of ethylene molecule is (1) 22 spsp σ (2) 2 spp σ (3) 2 sps σ (4) p p - p p
12. The d-orbitals involved in hybridisation of central atom in are (1) 222 zxy d,d (2) 22 xy,yz,zx xy d,ddd (3) 2 xy,yz,zx z d,ddd (4) xy,yz,zx ddd
4.7 VSEPR THEORY [VALENCE SHELL ELECTRON PAIR REPULSION THEORY]
Lewis concept is not able to explain the shapes of molecules.
This VSEPR theory provides a procedure to predict the shapes of different polyatomic molecules.
This theory was proposed by sidgwick and
powell. this was further developed by Nyholm and Gillespic.
A central atom of a covalent molecule has electron pairs in its valence shell. (bond pairs and some times lone pairs). These electrons pairs are oriented in space in such a way that the repulsions among them is minimized. Thus the molecular assumes a particular shape.
When central atoms of molecules have no lone pairs and surrounded by similar atoms, the shapes predicted are as follows Table 4.9.
Table 4.9 Shapes based on VSEPR model
7 Pentagonal bipyramid 720 and 900
4.7.1 Postulates of VSEPR Theory
■ The shape of the molecule is determined by repulsions between all electron pairs present in the valence shell of the central atom.
■ A lone pair of electrons takes up more space around the central atom than a bond pair because the lone pair (LP) is attracted to one nucleus, while the bond pair (BP) is shared by two nuclei.
■ It follows that the repulsion between two lone pairs is greater than repulsion between a lone pair and a bond pair, which in turn is greater than the repulsion between two bond pairs.
LP–LP > BP–LP > BP–BP
■ The presence of lone pairs on the central atom causes slight distortion of the bond angles from the ideal shape. If the angle
between a lone pair and a bond pair is increased, it follows that the actual bond angle between the atoms must be decreased.
■ The magnitude of repulsions between bond pairs of electrons depends on the electronegativity difference between the central atom and the bonded atom.
■ Double bonds cause more repulsion than single bonds, and triple bonds cause more repulsion than double bonds. Multiple bonds between two atoms count the same as single bond because it does not matter how many electrons are present in each direction.
In studying molecular geometry it is convenient to divide different molecules into two categories:
i) Molecules in which the central atom has no lone pairs
ii) Molecules in which the central atom has one or more lone pairs
4.7.2 Molecules with no Lone Pairs on Central Atom
For simplicity, consider molecules that contain atoms of only two elements, A and B, of which A is the central atom. The molecules have the general formula AB x, where x is an integer 2,3,4,5,6 and so on.
AB2 (Beryllium Chloride)
BeCl2 has two bond pairs. Because the bond pairs repel each other, they must be at opposite ends of a straight line in order for them to be as far apart as possible. Thus the ClBeCl angle in BeCl2 is 180° and the molecule is linear.
Cl Cl Be 1800
AB3(Boron Trifluoride)
BF 3 contains three covalent bonds or bond pairs. In the most stable arrangement, the three B–F bonds point to the corners of an
4: Chemical Bonding and Molecular Structure
equivalent triangle with boron atom in the centre of the triangle. Thus, each of the three FBF angles in BF3 is 120° and all four atoms lie in the same plane. The geometry of BF 3 is trigonal planar.
AB4 (Methane)
Since there are four bonding pairs, the geometry of CH4 is tetrahedral. This is a three dimensional shape with each angle 109°28΄ .
Iso-electronic species usually have the same structure. This may be extended to species with the same number of valence electrons. Thus, BF4, and NH4+ are all tetrahedral. CO32–, NO3– and SO3 are all trigonal planar and CO2, N3–, BO2– and NO2 + are all linear.
4.7.3 Molecules with Lone Pairs on the Central Atom
We designate molecules with lone pairs as AB x E y . Here, x and y are integers, which indicate the number of surrounding atoms and number of lone pairs on the central atom, respectively.
AB5 (Phosphorus Pentachloride)
PCl 5 has five bond pairs. The only way to minimise the repulsive forces among the five bonding pairs is to arrange the P–Cl bonds in the form of a trigonal bipyramid. The two Cl atoms in the axial position show slightly different chemical reactivity than the three in equatorial positions.
I.AB 2 E (Sulphur dioxide): The sulphur dioxide molecule can be viewed as consisting of three electron pairs on the central sulphur atom since in our simplified scheme, we treat double bonds as if they were single bonds. Of these, two are bond pairs and one is lone pair. The overall arrangement of three electron pairs is trigonal planar. But because one of the electron pairs is a lone pair, the SO2 molecule has a bent shape. Due to lone pair and bonding pair repulsion, the two sulphur- oxygen bonds are pushed together slightly and OSO angle is less than 120°. By experiment, OSO bond angle is found to be 119.5°.
AB3E (Ammonia)
AB6 (Sulphur Hexafluoride)
SF 6 has six bond pairs. The octahedral arrangement of the six bonding pairs provides the most stable form for the SF 6 molecule in which each of the FSF angles is 90° and 180°.
The ammonia molecule contains three bond pairs and one lone pair. The overall shape of four electron pairs is tetrahedral. But in NH 3 one of the electron pairs is a lone pair, so the geometry of NH 3 is trigonal pyramidal. Because the lone pair repels the bond pairs more strongly, the three N–H bonding pairs are pushed closer together, thus the HNH angle in ammonia 107°is smaller than, the ideal tetrahedral angle of 109.5°.
F F F F F S F
Table 4.10 Structures of some molecules and ions with lone pairs on central atom
A water molecule contains two bonding pairs and two lone pairs. The overall arrangement of the four electron pairs in water is tetrahedral, water has two lone pairs on the central ‘O’ atom. These lone pairs tend to be as far from each other as possible. Consequently, the two O–H bond pairs are pushed towards each other, so that we can predict an even greater deviation form the tetrahedral angle than in NH 3. This prediction is correct and HOH bond angle is 104.5°.
Structures of some molecules with lone pairs on the central atom are given in Table 4.10.
Q. SnCl2 is angular, but HgCl2 is linear. Why? Write their VSEPR notations.
Sol. SnCl2 has two bond pairs and one lone pair. Hence, it is angular. HgCl2 has 2 bond pairs and no lone pair. Hence, it is linear. SnCl2 is denoted as AB2E. HgCl2 is denoted as AB2.
Q. In SF4 molecule, lone pair of electrons occupies equatorial position but not axial position Why?
Sol.
In (a) the lp is present at axial position so there are three lp – bp repulsions at 90°. In (b) the lp is present at equatorial position, and there are only two lp – bp repulsions at 900. Hence arrangement (b) is more stable i.e lp occupies equatorial position.
TEST YOURSELF
1. Maximum number of planar atoms in SF6 molecule is (1) 5 (2) 4 (3) 6 (4) 7
2. The angle between two covalent bonds is minimum in (1) H2O (2) CO2 (3) NH3 (4) CH4
3. Octahedral molecule among the following is (1) SO3 (2) CHCl3 (3) SF6 (4) PCl5
4. In PCl5, bond angle in plane is (1) 90° (2) 120° (3) 180° (4) 109°.28΄
5. The shape of water molecule and bond angle in water molecule according to Hybridisation concept, respectively, are (1) Angular, 90° (2) Angular, 104°30΄ (3) Tetrahedral, 109°28΄ (4) Tetrahedral, 104°30΄
6. The geometry of H3O+ ion is (1) planar (2) triangular (3) pyramidal (4) tetrahedral
7. The shape of AB3E type molecule is (1) pyramidal (2) tetrahedral (3) angular (4) linear
8. Which one of the following is a correct set? (1) H2O,sp3 hybridisation, angular (2) H2O, sp2 hybridisation, linear (3) NH4+, dsp2 hybridisation, square planar (4) CH4, dsp2 hybridisation, tetrahedral
9. In which of the following molecules, sigma bonds formed by the overlap of sp3d and p-orbitals are absent
(1) PCl5 (2) ClF3 (3) SbCl5 (4) HClO4
10. Which of the following is not tetrahedral?
(1) BF 4 – (2) NH4+ (3) CO32– (4) SO42–
11. Which of the following has square pyramidal structure ?
(1) BrF 5 (2) ClF3 (3) IF 7 (4) ClF
12. Structure of ICl2 – is
(1) trigonal (2) octahedral (3) square planar (4) linear
13. The lone pairs in ClF3 and SCl4 are present respectively, at (1) axial, axial (2) axial, equatorial (3) equatorial, axial (4) equatorial, equatorial
14. Which has triangular planar shape? (1) CH3+ (2) CIO2–(3) H3O+ (4) CIO3–
It is often observed that a single Lewis structure of a molecule cannot satisfactorily explain all the properties of the molecule. In such cases, the concept of resonance is applied. If two or more alternate Lewis structures can be written for a molecule, the actual structure is said to be a resonance hybrid of all these alternate structures.
For example, the ozone molecule, O 3 can be equally represented by the structures I and II, as shown in Fig. 4.16.
Fig.4.16. Resonance in Ozone
In b oth structures we have O–O single bond and O=O double bond. Normal O–O and O=O bond lengths are 148 pm and 121 pm, respectively. The experimentally determined oxygen-oxygen bond lengths in ozone molecule are same and is 128 pm. Thus the oxygen-oxygen bonds in ozone molecule are intermediate between a double bond and a single bond. Obviously, this cannot be represented by either of the two Lewis structures.
Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, position of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid, which describes the molecule accurately. The phenomenon of writing two (or) more structures for a given molecule is called Resonance. For O3, the structures I and II are canonical forms structure - III is its resonance hybrid.
Other examples of resonance structures are those for the carbonate ion Fig. 4.17
According to experimental evidence, all carbon to oxygen bonds are equivalent. Therefore, the properties of the carbonate ion are best described by considering its resonance structures together, not separately.
The concept of resonance applies equally well to organic systems. A well known example is the benzene molecule. If benzene actually existed as one of its resonance forms, there would be two different bond lengths between adjacent carbon atoms, one characteristic of the single bond and the other of the double bond. In fact, the distance between adjacent carbon atoms in benzene are all 139 pm, which is between the length of C–C single bond (154 pm) and that of a C=C double bond (134 pm).
A single way drawing the structure of benzene molecule or compound containing benzene ring is to show only the skeleton and not the carbon and hydrogen atoms.
In general, it may be stated that, resonance stabilises the molecule as the energy of the resonance hybrid is less than the energy of any single canonical structure. It can also be stated that resonance averages the bond characteristics as a whole.
Resonance Energy
Resonance energy is the difference between the actual bond energy of the molecule and that of the most stable one of the canonical forms. The difference is denoted by E and is called resonance energy. Higher the resonance energy, greater is the stability of the molecule or ion.
Fig.4.17. Resonance in CO3–2
Suppose, E 1 , E 2 and E 3 are the energies of the three structures for a molecule. E 3 is the lowest energy and most stable canonical form. If E0 is the experimentally determined energy of the molecule, the resonance energy, as shown in Fig. 4.18, is E = E3 –E 0
Fig.4.18. Resonance Energy
It may be noted that the canonical forms have no real existence. The molecule does not exist for a certain fraction of time in one canonical form and for other fraction of time in another canonical form.
The real molecule cannot be depicted by a single Lewis structure.
4.8.1 Bond Order
T he number of bonds formed between two atoms in a molecule is known as bond order. For example, bond order in H 2 is one, in O2 is two and in N 2 it is three and in CO, it is also three.
H H H Bond order =1 or H, –
Single shared electron pair in H 2.
O O , O O = Bond order =2 or
Two shared electron pairs in O 2.
N N N N , Bond order =3 or ≡
Three shared electron pairs in N 2.
C C O O , Bond order =3 or ≡
Three shared electron pairs in CO.
It may be noted that iso-electronic species have identical bond orders. N2, CO, and NO+ are iso-electronic, each of these has 14 electrons. The bond order in N2, CO or in NO+ is three.
The bond order of a molecule having resonance structures is giving by bond order = x/y, Here, ‘x’ is the total number of bonds between specified atoms and ‘y’ is the number of canonical structures.
As the bond order increases, bond energy increases and bond length decreases. Bond orders in some molecules and ions are listed in Table 4.11 . Bond orders are now calculated by applying most familier theory of bonding, called molecular orbital theory.
Table 4.11 Bond orders in some molecules and ions
Q. Explain the structure of CO 2 molecule.
Sol. The experimentally determined carbon, oxygen bond length in CO2 is 115 pm. The lengths of a normal carbon oxygen double bond (C=O) and carbon to oxygen triple bond (C≡O) are 121 pm and 110 pm, respectively. The carbon-oxygen bond lengths in CO2 (115 pm) lie between the values for C = O and . The structure of CO 2 is best described as a hybrid of the canonical or resonance forms I, II and III.
4.9 MOLECULAR ORBITAL THEORY
The molecular orbital theory (MO theory) was developed by Hund and Mulliken.
4.9.1 Postulates
■ Just as electrons of an atom are present in various atomic orbitals, electrons of a molecule are present in various molecular orbitals.
Q. What is the nitrogen-oxygen bond order in NO3– ion?
Sol. NO3– The structure of NO3 – ion is:
It has four bonds and three canonical structures. Bond order = 1.33.
TEST YOURSELF
1. How many resonance structures are possible for NO3– ion.
(1) five (2) three
(3) four (4) six
2. According to the concept of resonance, among the ions OCl–, ClO2–, ClO3– and ClO4–, the most stable ions is
(1) OCl– (2) ClO2–
(3) ClO3– (4) ClO4–
3. In the resonance structures of CO 3 –2 not observed.
(1) Every structure has the same number of lone pairs.
(2) In one structure central atom is oxygen.
(3) In all the structures one C=O exists per structure.
(4) Every structure has –2 units of charge.
4. C–O bond order in CO2 molecule is (1) 3 (2) 4 (3) 2 (4) 1
Answer Key
(1) 2 (2) 4 (3) 2 (4) 3
■ Molecular orbitals are formed by the combination of atomic orbitals of comparable energies and proper symmetry.
■ While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital, it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus an atomic orbital is monocentric, while a molecular orbital is polycentric.
■ The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital, while the other is called anti-bonding molecular orbital.
■ The bonding molecular orbital has lower energy and hence, greater stability than the corresponding anti-bonding molecular orbital.
■ Just as the electron probability distribution around the nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecule is given by molecular orbital.
■ The molecular orbitals like atomic orbitals are filled in accordance with the Aufbau principle, obeying Pauli’s exclusion principle and Hund’s rule of maximum multiplicity.
■ Just as the letters s, p and d are used to denote the atomic orbitals, the greek letters, and are used to denote the molecular orbitals.
4.9.2 Formation of Molecular Orbitals
Molecular orbitals are formed by the combination of atomic orbitals of the bonded atoms. We have learnt that according to wave mechanics, the atomic orbitals can be expressed by wave functions. The wave functions represent the amplitude of the electron waves. These are obtained from the solutions of Schrodinger wave equation. Similarly we can write Schrodinger wave equation for the whole molecule. However, it is very difficult to solve the wave equation for molecules. To overcome this problem, scientists have used an approximate method known as linear combination of atomic orbitals abbreviated as LCAO method.
Consider the application of LCAO method to the homonuclear diatomic hydrogen molecule. The two hydrogen atoms in hydrogen molecule, H2 may be labelled as A and B. Each hydrogen atom has one electron in 1s orbital. These atomic orbitals may be represented by the wave functions Ψ A and ΨB. Mathematically the function of molecular orbitals is described by the linear combination of atomic orbitals that can take place by addition and by subtraction of wave functions of individual atomic orbitals.
Therefore two molecular orbitals s and s* are formed as shown in Fig. 4.19. s = Y A + Y B; s * = Y A – Y B
The molecular orbital, formed by the addition of atomic orbitals is called the bonding molecular orbital and s* by subtraction
of atomic orbitals is called anti-bonding molecular orbital.
Qualitatively, the formation of molecular orbitals can be understood in terms of the constructive or destructive interference of the electron waves of the combining atoms.
In the formation of bonding molecular orbital, the two electron waves of the bonding atoms reinforce each other due to constructive interference. While in the formation of antibonding molecular orbital, the electron waves cancel each other due to destructive interference.
The energy of the anti-bonding orbital is raised above the energy of the parent atomic orbitals that have combined. The energy of the bonding orbital has been lowered compared to that of the parent atomic orbitals. The total energy of the two molecular orbitals, however, remains the same as that of two original atomic orbitals.
Main differences between atomic orbitals and molecular orbitals are given in Table 4.12.
1. They belong to one specific atom only They belong to all the atoms in a molecule
2. They are the internal characteristic of an atom They result when atomic orbitals of similar energies combine
3. They have simple shapes of geometries They have complex shapes
4. The atomic orbitals are named as s, p, d, f, .... etc. The molecular orbitals are named as s,p,d , . . . etc.
5. The stabilities of these orbitals are less than bonding and more than the anti-bonding orbitals
The stabilities of these orbitals are either more or less than the atomic orbitals.
Fig.4.19 Order of Energies of Bonding and Anti-bonding Orbital
Table 4.12 Main differences between atomic and molecular orbitals
S.No.
The order of energies of the orbitals is Bonding < Non–bonding < Anti–bonding.
Conditions for the combinations
■ The combining atomic orbitals must have the same or nearly the same energy.
■ The combining atomic orbitals must have the same symmetry about the molecular axis. Z axis is taken as molecular axis by convention.
■ The combining atomic orbitals must overlap to the maximum extent. Higher the extent of overlap, greater will be the electron density between the nuclei of a molecular orbital.
When s-orbitals of two atoms or when p-orbitals of two atoms approach along z-axis, sigma molecular orbitals are formed as shown in Fig. 4.20
When atomic orbitals overlap in side wise fashion the molecular orbitals formed have a nodal plane
and lobes of the orbitals are on either side of the intermolecular axis. The bond formed with no cylindrical symmetry is known as p–bond. p–bonds due to p-p overlaps are shown in Fig. 4.21
Main difference between sigma and pi molecular orbitals are given in Table. 4.13.
The sequence of increasing order of energy of various molecular orbitals for O 2 and F2 is
Electronic Configuration and Molecules
Behaviour
Stability of Molecules
Let Nb be the number of electrons occupying the bonding orbitals and Na be the number of electrons occupying the anti-bonding orbitals.
Fig.4.20 Sigma bonding and anti-bonding molecular orbits
Table 4.13 Differences between s and p Molecular Orbitals.
1. Formed by the end on overlap along the inter nuclear axis. Formed by the side wise overlap, perpendicular to the internuclear axis.
2. Overlapped region is very large. Overlapped region is small.
3. Rotation about the internuclear axis is symmetrical. Rotation about the internuclear axis is unsymmetrical.
4. Strong bonds are favoured. Weak bonds are favoured.
| CHAPTER 4: Chemical Bonding and Molecular Structure
Fig.4.21. Pi bonding and anti-bonding molecular orbits
1) If Nb is greater than Na, the molecule is stable.
2) If Nb is less than N a or Nb = Na,the molecule is unstable.
Bond Order
The bond order is defined as one half of the difference between the number of electrons present in bonding and in anti bonding orbitals.
Bond order ba 1 NN 2 =−
A positive bond order means a stable molecule while a negative bond order means an unstable molecule. The integral values of bond order 1,2 or 3 correspond to single, double or triple bonds respectively.
Bond Length
The bond order between two atoms is taken as an approximate measure of the bond length. As the bond order increases, the bond length decreases.
Magnetic Behaviour
If all the molecular orbitals in a molecule are occupied by two electrons each, the substance is said to be diamagnetic. However, if one or more molecular orbitals are singly occupied by electrons, it is said to be paramagnetic.
Bonding in Some Homonuclear Diatomic Molecules
In this section we shall discuss bonding in some homonuclear diatomic molecules as follows.
Hydrogen Molecules (H2)
It is formed by the combination of two hydrogen atoms. Each hydrogen atom has one electron in 1s orbital. Therefore, in all there are two electrons in hydrogen molecule which are present in molecular orbital. So, electronic configuration of hydrogen molecules is H 2 : ( s 1s2)
The bond order of H 2 molecule can be calculated as given below:
Bond order NNba 20 1 22 ===
This means that the two hydrogen atoms are bonded together by a single covalent bond. The bond dissociation energy of hydrogen molecule has been found to be 438 kJ mol –1 and bond length equal to 74 pm. Since no unpaired electron is present in hydrogen molecule, therefore, it is diamagnetic.
Helium Molecule (He2)
The electronic configuration of helium atoms is 1s 2. Each helium atom contains 2 electrons. therefore in He 2 molecule, there would be 4 electrons. These electrons will be accommodated in s 1s and s* 1 s molecular orbitals, leading to electronic configuratio n:
2*2 2 He:(1s)(1s) σσ
Bond order of He2 is 1/2 (2–2) = 0
He 2 molecule is therefore unstable and does not exist.
Similarly, it can be shown that Be2 molecule 2*22*2 (1s)(1s)(2s)(2s) σσσσ also does not exist.
Lithium Molecule (Li2)
The electronic configuration of lithium is 1s 2, 2s 1. There are six electrons in Li 2. The electronic configuration of Li 2 molecule, therefore, is 2*22 2 Li:(1s)(1s)(2s) σσσ
The above configuration is also written as KK(2s)2 σ , where KK represents the closed K shell structure 2*2 (1s)(1s) σσ
From the electronic configuration of Li 2 molecule, it is clear that there are four electrons present in bonding molecular orbitals and two electrons present in anti-bonding molecular orbitals. Its bond order, therefore, is 1/2(4–2) = 1. It means that Li2 molecule is known to exist in the vapour phase.
Carbon Molecule (C2)
The electronic configuration of carbon is 1s2 2s22p2. There are twelve electrons in C2. The electronic configuration of C 2 molecule, therefore, is 2*22*222 2y C:(1s)(1s)(2s)(2s)(2p2P) σσσσπ=π x or 2*222 xy KK(2s)(2s)(2p2p) σσπ=π
The bond order of C2 is 1/2(8–4) = 2 and C 2 should be diamagnetic. Diamagnetic C 2 molecules have indeed been detected in vapour phase. It is important to note that double bond in C2 consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals.
Nitrogen Molecule (N2)
The electronic configuration of nitrogen atom is 1s 2 2s 2 2p 3 . Each nitrogen atom has 7 electrons. Hence, in N2 molecule, there are 14 electrons. The electronic configuration of N2 molecule, therefore, is
or 2*2 2 222 xyz KK(2s)(2s) N: (2p2p),(2p) σσ
From the electron configuration of N 2 molecule, it is clear that ten electrons are present in bonding molecular orbitals and four electrons are present in anti bonding molecular orbitals. Its bond order, therefore, is three.
Bond order ba 11 [NN][104]3 22 =−=−=
So, in nitrogen molecule, atoms are held by a triple bond. N2 is diamagnetic because of no unpaired electron.
Oxygen Molecule (O2)
The electronic configuration of oxygen is KK 22222 zxy (2s)(*2s)(2p)(2p)(2p) σσσπ≡π 11 xy (*2p)(*2p) π≡π
Bond order 106 2 =
The electrons in p *2px and p *2py orbitals remain unpaired. In oxygen molecule, the atoms are held by a double bond. Moreover, the molecule contains two unpaired electrons. It should be paramagnetic. This is an experimental observation. The bond dissociation energy of oxygen is 495 kJ mol–1 The bond length in O2 is 121 pm.
The electronic configuration of O 2 + is KK ( s 2s) 2 ( s* 2s) 2 ( s 2p z) 2 ( p 2p x) 2 ( p 2p z) 2 ( p* 2px)1
Bond order 83 2.5 2 ==
The bond strength of O2+ is more than that of O2 and bond length of O2 + is less than that of O2. O2+ is paramagnetic.
The electronic configuration of O 2 – is KK ( s 2s) 2 ( s* 2s) 2 ( s 2p z) 2 ( p 2p x) 2 ( p 2p y) 2 ( p 2px)2 ( p* 2px)2 ( p* 2py)1
Bond order 853 1.5 22 ===
The bond order of O2 – is less than that of O2. The bond in O2 – is weaker than in O2 and the bond length of O 2 – is larger than that of O2. O2– is paramagnetic.
The electronic configuration of O22– is KK (s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p2px)2 ( p* 2px)2 ( p* 2py)2
Bond order 86 1 2 ==
Since the bond order of peroxide, O22– ion is less than that of O2, the peroxide bond will be weaker than in O2. The bond length of O22– is larger than that of O 2
Order of bond energies is: O2+ > O2 > O2– > O22–Order of bond lengths is: O22– > O2– > O2 > O2+
Configuration and magnetic properties of some diatomic molecules and their ions are in given Table 4.14.
Electronic arrangement in molecules are in given Table 4.15.
Table 4.14. Configuration and Magnetic Properties of Diatomic Molecules and Ions Molecule
Q. By the use of molecular orbital consideration, account for the fact that oxygen is paramagnetic.
Sol. Electronic configuration of molecular orbitals of O2 is
( s 1s)2 ( s* 1s)2 ( s 2s)2 ( s 2pz)2 ( p 2px)2 ( p 2py)2 ( p* 2px)1 ( p* 2pz)1
The last two electrons are in p*2px and p*2py anti bonding orbitals so as to maximize pi electrons in accordance with Hund’s rule. These two unpaired electrons confer paramagnetism in oxygen.
Q. Use molecular orbital theory to explain why Be2 molecule does not exist.
Sol. The atomic no.(z) of Be is 4. This means that 8 electrons are to be filled in the M.O. of Be 2
The configuration is : s 1s2 s* 1s2 s 2s2 s*2 s2;
As bond order of Be2 is zero, the molecule of Be2 does not exist.
TEST YOURSELF
1. The para magnetic nature of oxygen is best explained by (1) V.B. theory (2) Hybridisation (3) M.O. theory (4) VSEPR theory
3. Number of anti-bonding electrons in O 2 molecule are (1) 10 (2) 6 (3) 4 (4) 2
4. In which pair, the stronger bond is found in the first species?
a) O,O22 b) N,N22+ c) NO,NO+− (1) a only (2) b only (3) a and c only (4) b and c only
5. Bond energy is maximum in (1) F2 (2) N2 (3) O2 (4) Br2
6. Bond order is 2.5 in a) O2+ b) N2+ c) NO (1) a, b only (2) b, c only (3) a, c only (4) a, b and c
7. Which of the following molecular orbital has the lowest energy for O 2 molecule?
(1) s 2pz (2) 2px σ
(3) 2pxπ (4) 2px * σ
8 The molecular orbital shown in the diagram can be described as (1) s (2) s *
(3) p (4) p *
9. In O2 molecule, the correct order of energy of molecular orbitals is (1) yz 2p2pπ>π (2) yz 2p2p = ππ (3) 2s2s * σ<σ (4) 2s2px **σ>σ
10. Maximum number of electrons that can be present in any molecular orbital is (1) 3 (2) 6
(3) 8 (4) 2
11. The molecular electronic configuration of B2 is
(1) 2*211 xy KK(2s)(2s)(2p)(2p) σσππ
(2) 2*22 KK(2s)(2s)(2p)x σσπ
(3) 2*22 KK(2s)(2s)(2p) σσπ
(4) 2*211 KK(2s)(2s)(2p)(2p) σσσπ
12. When N2 goes to N2+, the N–N bond distance ........ and when O2 goes to O2+, the O–O bond distance ........:
(1) increases, decreases (2) decreases, increases
(3) increases, increases
(4) decreases, decreases
13 Which combination of atomic orbitals is not allowed according to MO theory?
(1) px – px (2) px – py
(3) py – py (4) pz – pz
14. A bonding molecular orbital is produced by (1) Destructive interference of wave functions
(2) Constructive interference of wave functions
(3) Pairing of electrons with opposite spins
(4) Combination of +ve and –ve wave functions
15. If Z-axis is the molecular axis, then pimolecular orbitals are formed by the overlap of
(1) s + pz (2) px + py
(3) pz + pz (4) px + px
16. Which one of the following is correct regarding s molecular orbital?
(1) The rotation along the inter nuclear axis is symmetric
(2) It is formed by the partial overlap of atomic orbitals at right angle to internuclear axis
(3) It is very weak bond
(4) It has less overlapping region
17. p 2px differs from p 2py molecular orbital in which of the following?
(1) Number of nodal planes
(2) Energy
(3) Symmetry
(4) Shape
18. In the formation of a homo-diatomic neutral molecule, if ‘N’ atomic orbitals combine, then the total number of bonding molecular orbital formed is (1) 2N (2) N (3) N/2 (4) N /4
The idea of hydrogen bond was introduced by Latimer and Rodebush to explain the nature of association in liquid state of substances, like water, ammonia, hydrogen fluoride, formic acid, etc.
When hydrogen is bonded to highly electronegative atom, such as F or O or N by a covalent bond, the bond electron pair is attracted towards electronegative atom so strongly that a dipole results. When a number of such molecules are brought nearer to each other, the positive end of one molecule and negative end of the other molecule will attract each other and a weak electrostatic force will develop. Thus, these molecules associate together to form a cluster of molecules.
XH...XH....XH....XH δ−δ+δ−δ+δ−δ−δ−δ+
The electrostatic force of attraction between a partially positively charged hydrogen atom of a molecule and a highly electronegative atom of the same molecule or of a different molecule is known as hydrogen bond.
For the existence of hydrogen bond, the atom which is bonded to hydrogen atom must be highly electronegative and small sized (F, O or N).
The magnitude of hydrogen bond depends on physical state of compound, it is maximum in solid state minimum in gaseous state.
CHAPTER 4: Chemical Bonding and Molecular Structure
The energy of hydrogen bond ranges between 12-50 kJ mol –1 . The energy of hydrogen bond is less than that of covalent bond.
H–N...H < H–O.....H < H–F....H
8.4 kJ mol–1 29.4 kJ mol–1 42 kJ mol–1
The bond length of hydrogen bond is larger than that of covalent bond. The O–H covalent bond length in water is approximately 1Aº, while O....H hydrogen bond length is 1.76Aº.
Note:
Deuterium also forms Hydrogen bonds.
4.10.1 Types of Hydrogen Bonding
Hydrogen bonding is of two types — Intermolecular and Intramolecular hydrogen bonding
Intermolecular Hydrogen Bonding
A hydrogen bond formed between partially positive hydrogen atom and electronegative atom of different molecules of the same or different substances is called intermolecular hydrogen bond.
Intermolecular hydrogen bonding increases the boiling point and solubility of the compound in water. The increase in boiling point is due to association of several molecules of the compound. Intermolecular hydrogen bonding is also responsible for increase in the viscosity. Greater the extent of hydrogen bonding, more is the viscosity.
1. Hydrofluoric acid
2. Water
3. Ammonia
4. Ammonia in water
5. Formicacid
6. Parafluorophenol
7. Metanitrophenol
8. Para hydroxy benzaldehyde
Intramolecular Hydrogen Bonding
A hydrogen bond formed between partially positive hydrogen atom and electronegative atom present in the same molecule of the substance is called intramolecular hydrogen bonding. It leads to chelation. Due to intramolecular hydrogen bonds, no molecular association occurs. Boiling points will be normal and volatility will be high for the substances with intramolecular hydrogen bonding.
1. Salicylaldehyde
(o–hydroxybenzaldehyde)
2. Orthonitrophenol
3. Orthofluorophenol
4. Orthonitroaniline
5. Salicylic acid
Consequences of Hydrogen Bonding
The substances which exhibit hydrogen bonding have abnormal properties.
■ The physical state may change. For example, water is expected to be a gas at room temperature, but it is a liquid.
■ The melting and boiling points of water, ammonia, hydrogen fluoride etc. are abnormally higher than expected.
■ Solubility of lower alcohols like CH3OH, C 2 H 5 OH in water is due to hydrogen bonding.
■ Para hydroxybenzaldehyde exhibits intermolecular hydrogen bonding, and it is soluble in water. It cannot be purified by steam distillation and has high boiling point. Ortho hydroxybenzaldehyde exhibits intra molecular hydrogen bonding. It is less soluble in water and has low boiling point. It is volatile in steam and hence, it can be purified by steam distillation.
■ Sulphuric acid is denser and viscous on account of molecular association due to intermolecular hydrogen bonding.
■ Water has higher boiling point than hydrogen fluoride, through stronger hydrogen bonds, exist in HF. This is because larger number of Hydrogen bonds are in water as compared to those in HF. In the formation of water vapour all its hydrogen bonds are broken but HF exists as a polymer, (HF)n is vapour state without breakage of all its Hydrogen bonds.
■ Ammonia has higher boiling point than hydrogen chloride, even though nitrogen and chlorine have nearly the same values of electronegativity. Ammonia forms hydrogen bonds, HCl does not form hydrogen bonds due to larger size of Clatom.
■ HF2 (or KHF2) exists but HCl 2 (or KHCl2) does not exist because there is hydrogen bonding in HF but not in HCl.
It may be concluded that the electronegativity, number of hydrogen bonds and the size of the atom are the important factors that influence not only the formation of hydrogen bonds but also the properties.
Hydrogen bonding is not observed in the hydrides of group 14 elements. Ammonia, water and hydrogen fluoride (hydrides
4: Chemical Bonding and Molecular Structure
of group 15, 16 and 17 elements) have hydrogen bonding. The influence of hydrogen bonding on the boiling points of these hydrides is shown graphically in Fig. 4.22
Boiling points ( 0
Fig.4.22. Boiling Points of Hydrides
Q. Why is salicylaldehyde less soluble in water?
Sol. Salicylaldehyde has chelate ring formed due to intramolecular hydrogen bonding.
Salicylaldehyde O C H H O
Solute-solute attractions are stronger. Hence, solubility in water is less.
Q. Why is orthofluorophenol more volatile than its meta and para isomers?
Sol. Intramolecular hydrogen bonding is present in orthofluorophenol, which does not lead to association of molecules.
Hence, it is volatile.
Intermolecular hydrogen bonding is present in meta and para isomers, which leads to molecular association.
TEST YOURSELF
1. Intermolecular hydrogen bonding is absent in (1) H2O (2) NH3
(3) C2H5OH (4) CH4
2. Which among the following has the highest volatility?
(1) H2O (2) H2S
(3) H2Se (4) H2Te
3. Among the following, which has the highest boiling point ?
9. Boiling point is highest for (1) hydrogen fluoride (2) water (3) ammonia (4) methane
10. Hydrogen bonds are present even in vapour state of (1) H2O (2) HF (3) p-hydroxy benzaldehyde (4) C2H5OH
11. KF combines with HF to form KHF2. The compound contains the species
(1) K+, F– and H+ (2) K+, F– and HF (3) K+ and [HF2]–1 (4) [KHF]+ and F–
Answer Key
(1) 4 (2) 2 (3) 1 (4) 3
(5) 2 (6) 1 (7) 4 (8) 1 (9) 2 (10) 2 (11) 3
4.11 BOND PARAMETERS
Chemical bonds are identified with certain measurable properties. They are also follows
4.11.1 Bond Length
The atoms of a covalent molecule are not stationary but always vibrate about their equilibrium positions. This is due to the fact that the attractive as well as repulsive forces are acting between the atoms. Therefore, the average distance between the nuclei of the two bonded atoms in a covalent molecule is taken as the bond length.
The equilibrium distance between the nuclei of the two covalently bonded atoms is defined as bond length or bond distance.
Bond lengths are measured spectroscopically or by X-ray and electron diffraction techniques. Bond length is expressed in Angstrom units, A° or picometers, (pm). In a covalent molecule, f or a
b ond between two atoms, the bond length is taken as the sum of their covalent radii, as shown in Fig.4.23 . Bond lengths of some molecules are listed in Table 4.15.
Bond length A–B = ra + rb where r a and rb are the covalent radii of the atoms A and B respectively.
Important Features
1. The magnitude of the bond length between the same atoms in different molecules is the same. O–H bond length in H 2 O or H2O2 or C2H5OH is equal to 96 pm. If the bond type between specific atoms changes, the bond length also changes.
2. Bond length increases with an increase in the size of the bonded atoms.
Fig.4.23. The Bond Length in a Covalent Molecule AB
Table 4.16 Bond lengths in some molecules
F–F < Cl–Cl < Br–Br < I–I
H–F < H–Cl < H–Br < H–I
3. The bond length of a homonuclear diatomic molecule is twice the covalent radius.
Covalent radius of H atom is 37 pm. H–H bond length is 74 pm.
4. A double bond is shorter than a single bond. A triple bond is shorter than a double bond. This is because of an increase in the number of bonds. Single bond > double bond > triple bond.
The more the number of bonds present between bonded atoms, the stronger is the overlapping and the shorter is the bond length.
C–C bond length in ethane is 154 pm, C=C bond in ethylene is 134 pm and C≡ C bond in acetylene is 120 pm. Average bond lengths for some single, double and triple bonded molecules are given in Table 4.17
Table 4.17 Average bond lengths for some molecules
Bond
5. The bond length decreases with the increase in ‘s’ character or decrease in ‘p’ character in case of hybrid orbitals.
6. The bond length of sp 3 C–H is 110 pm (25%, s character), sp 2 C–H is 109 pm (33% s character) and sp C–H is 106 pm (50% s character).
7. Bond length of a polar bond is smaller than the theoretical bond length (calculated by the addition of covalent radii).
4.11.2 Bond Enthalpy
Diatomic Molecule
The amount of energy required to break one mole of diatomic molecules in the gaseous state into their free constituent atoms is called bond enthalpy. It is also called bond energy or bond strength.
Bond energy is generally expressed at room temperature. The unit of bond energy is kJ mol–1. It is generally observed that shorter the bond length, greater is the bond strength or energy of the bond.
HH2H;H436kJmol −→∆=
()() 1 gg
(H–H bond length, 74 pm)
()() 1 gg ClCl2Cl;H242kJmol −→∆=
(Cl–Cl bond length, 198 pm).
B ond dissociation enthalpy ( ∆ H) is always considered to have positive values because energy must be supplied to break a bond. Conversely, the amount of energy released on forming a bond is always ∆ H with negative value. Bond energies of some covalent bonds are listed in Table 4.18.
Table 4.18
Bond energies
Polyatomic Molecule
In a poly atomic molecule, bond enthalpy is different from bond dissociation enthalpy. In the molecule, the bond energy is the average of the various bond dissociation energies of similar bonds, of it.
Example:
In CH4, the bond dissociation energies of the four C–H bonds are also follows:
()()()4g 3gg CHCHH;→+∆H
CHCHH;→+∆H = 349 kJ mol–1
()()()3g 2gg
CHCHH;→+∆H = 451 kJ mol–1
()()()2g gg
CHCH;→+∆H = 347 kJ mol–1
()()() ggg
CHC4H;→+∆ H ≅ 1663 kJ mol–1
()()() 4ggg
Hence, average C–H bond energy is 1663/4 = 416 kJ mol –1. The energy required for heterolytic cleavage of bond is higher than that required for homolytic cleavage.
In case of H2O molecule, the enthalpy needed to break the two O – H bonds is not the same because of changed chemical environment.
2(g)(g)(g)a1 HOHOHH kJ mol 1 ;502→+∆=
(g)(g)(g)a1 OHHOH kJ mol 1 ;427→+∆=
Important Aspects
1. As the bond energy increases, the stability of the bond also increases. The chemical reactivity of the substance decreases.
N 2 is very stable gas in the atmosphere because the bond energy is high (944 kJ mol–1).
2. s bonds are stronger than p bonds.
3. As the number of bonds between same atoms increases, the overall bond strength also increases.
Bond Energy Order
kJ mol–1 610.7 kJ mol–1 827.6 kJ mol–1
4. The bond energies of similar type of bonds gradually decrease down the group.
HF > HCl > HBr > HI
5. As the number of lone pairs of electrons on bonded atoms increases, the bond energy decreases.
(341 kJ) (163 kJ) (146 kJ)
6. The heat of reaction can be estimated from the values of bond energy.
∆ H = Heat of reaction = total bond energy of reactants – total bond energy of products.
7. Among the halogens, chlorine has highest bond energy. Cl2 > Br2 > F2 > I2.
4.11.3 Bond Angle
The average angle between two adjacent atoms bonded to the central atom in a molecule is called bond angle.
Bond angle is expressed in degrees. The bond angle can be determined by X-ray diffraction and spectroscopic methods.
Larger the bond angle higher is the stability of the molecule. The bond angle gives some idea regarding the distribution of orbitals around the central atom in a molecule or complex ion. Hence bond angle helps to determine the shape of the molecule or ion.
Important Factors
1. The bond angle depends upon the geometry of the molecule.
2. It is observed that, as ‘s’ character in hybrid orbital increases, the bond angle also increases.
3. The bond angle is affected by the presence of lone pairs of electrons on the central atom.
The bond angle in NH 3 is 107°, inspite of the fact that ‘N’ atom undergoes sp 3 hybridisation.
4. If the electronegativity of the central atom decreases, the bond angle also decreases. In case the central atom remains the same, bond angle increases with the decrease in
the electronegativity of the surrounding atoms.
Bond angles in OF 2 and Cl 2 O are respectively 103º and 111°.
Q. If the C=O bond length is 121 pm, what is the distance between the nuclei of oxygen atoms in carbondioxide molecule?
Sol. Carbon dioxide is a linear molecule with two C=O bonds opposite to each other.
O = C = O
121pm 121pm
The distance between the nuclei of oxygen atoms in CO2 molecule is 121 + 121 = 242pm.
Q. The As–Cl bond distance in AsCl3 is 2.20A°. Estimate the single bond covalent radius of Arsenic. (Covalent radius of Cl is 0.99A°).
Sol. Internuclear distance - radius of chlorine atom = radius of arsenic atom
2.20 – 0.99 = 1.21 A° Covalent radius of As = 1.21A°.
TEST YOURSELF
1. Bond energy is least in (1) HF (2) HCl (3) HBr (4) HI
2. Bond length of H2 is 0.074 nm, bond length of Cl2 is 1.98A°. Bond length of HCl is (1) 2.72 A° (2) 136 pm (3) 1.027 nm (4) 0.136 A°
3. The O-H bond length in H 2O is xA°. The O-H bond length in H2O2 is (1) < x A° (2) x A° (3) > x A° (4) 2x A°
4. Bond energy is highest in the molecules of (1) F2 (2) Br2 (3) I2 (4) Cl2
5. Bond energy of C C bond is highest in (1) H3C –CH3 (2) H2C = CH2 (3) CH ≡ CH (4) C2H5Cl
6. The bond dissociation energy of the molecules A 2, B 2, C 2 are 498, 158, 945 kJ/ mole respectively. So, the correct decreasing order of their bond orders is (1) A2, B2, C2 (2) C2, B2, A2 (3) C2, A2, B2 (4) B2, C2, A2
7. The order of bond length of ( O – O) in O2, O3, H2O2 is (1) O2 > H2O2 > O3 (2) O3 > H2O2 > O2 (3) H2O2 > O3 > O2 (4) O2 > H2O2 > O3
8. The correct order of increasing bond angles is
(1) PF 3 < PCl3 < PBr 3 < PI 3
(2) PF 3 < PBr 3 < PCl3 < PI 3
(3) PI 3 < PBr 3 < PCl3 < PF 3
(4) PF 3 > PCl3 < PBr 3 < PI 3
9. The correct order of increasing C-O bond length of CO, 2 CO3 and CO2 is
(1) 2 CO3 < CO2 < CO
(2) CO2 < CO< 2 CO3
(3) CO < 2 CO3 < CO2
(4) CO < CO2 < 2 CO3
10. Which of the following has least bond energy?
(1) F2 (2) H2
(3) N2 (4) O2
Answer Key
(1) 4 (2) 2 (3) 2 (4) 4
(5) 3 (6) 3 (7) 3 (8) 1
(9) 4 (10) 1
4.12 BOND POLARITY AND DIPOLE MOMENT
Ionic and covalent bonds represent only two extremes of continuous spectrum of possibilities. Between these two extremes, we have a large number of bonds in which the bonding electrons are shared unequally
between two atoms but are not completely transferred. Such bonds are said to be polar covalent bonds. Polar covalent bonds lie between the two extremes of the bonding.
Compare the three substances NaCl, HCl and Cl2. The bond in NaCl is largely ionic, between Na + and Cl –. NaCl bond is only about 80% ionic, and the electron transferred from Na to Cl still spends some of its time near sodium.
Na+ Cl– is an ionic bond.
The bond in HCl molecule is polar covalent. The chlorin e atom attracts the bonding electron pair more strongly than hydrogen does, resulting in an unsymmetrical distribution of electrons. Chlorine thus has a partial negative charge and hydrogen has partial positive charge.
Table 4.19 Bond polarity and ionic nature
Difference in electro-negativity
Percent ionic character
Nature of the bond
0 0 Pure covalent
Polar Molecule
A molecule which has oppositely charged poles is called polar molecule or dipole.
Bond polarity is described in terms of ionic character in Table 4.19
The ionic character increases with the increasing difference in the electro-negativity (EN) between bonded atoms.
The following equation was proposed by Hanny and Smith for calculating the percentage of ionic character in A–B bond, on the basis of values of electronegativity of the atoms A and B.
Percentage of ionic character =
XA = electronegativity of the atom A
HCl
δ+δ− is a polar covalent bond.
The bond in Cl 2 molecule is non-polar covalent, with the bonding electrons centered between the two identical chlorine atoms and attracted equally to both atoms.
Cl–Cl is a non-polar covalent bond.
A covalent bond, in which the electrons are shared unequally and the bonded atoms acquire partial charges is called polar covalent bond. Bond polarity is due to difference in electro-negativity.
XB = electronegativity of the atom B
Dipole Moment
The magnitude of polarity of a molecule is expressed in terms of dipole moment. The product of the magnitude of the charge on any one of the poles in a dipole and the distance between the poles is defined as dipole moment.
The dipole moment (m ) is mathematically expressed as m=d× d , where is charge on one of the poles and d is distance between the poles.
The electric charge is in the order of 10 –10 esu and the distance is in the order of 10–8 cm.
10 –18 esu cm is known as Debye unit. 1 Debye = 3.336 × 10 –30 C m. The SI unit of dipole moment is coulomb meter, (C-m)
Every bond between two hetero atoms has a definite dipole moment called bond moment. Dipole moment is a vector. It has both magnitude and direction.
Dipole moment is a vector quantity and it is depicted by a small arrow with tail on the positive centre and head pointing towards the negative centre.
Dipole moment of HX as in Table 4.20.
Table 4.20 Dipole moments of hydrogen halides
May be represented as H X
The shift in electron density is symbolised by crossed arrow ( ) above the Lewis structure to indicate the direction of the electron shift.
In case of polyatomic molecules, the dipole moment not only depends upon the individual dipole moments of bonds known as bond dipoles but also on the spatial arrangement of various bonds in the molecule. In such a case, the dipole moment of a molecule is the vector sum of the dipole moments of various bonds.
Homoatomic molecules and symmetrical molecules having normal shapes, such as linear, trigonal, tetrahedral possess zero dipole moment.
Molecules having distorted shapes, like angular, pyramidal, see-saw etc. show dipole moments.
For example: H2O, SO2, H2S, NH3, SF4
In H 2 O molecule, which has a bent structure, the two O–H bonds are oriented at an angle of 104.5º. Net dipole moment of 6.1710–30 C m (1D = 3.33564 10–30 C m) is the resultant of the dipole moments of two O – H bonds.
(a) (b)
Bond dipole Resultant dipole moment
Net Dipole moment, m = 1.85 D
= 1.85 × 3.33564 × 10 –30 Cm
= 6.17 × 10–30 Cm
The dipole moment in case of BeF2 is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect of each other.
F F + Be
BeF2 molecule
In tetraatomic molecule, for example in BF3, the dipole moment is zero although the B – F bonds are oriented at an angle of 120° to one another, the three bond moments give a net sum of zero, as the resultant of any two is equal and opposite to the third.
B F F F + ( ( (A) (B)
BF3 Molecule; Representation of (a) Bond Dipoles and (b) Total Dipole moment
Let us study an interesting case of NH3 and NF 3 molecule. Both the molecules have pyramidal shape with a lone pair of electrons on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of NH3 (4.90 × 10–30 C m) is
greater than that of NF 3 (0.8 × 10 –30 C m). This is because, in case of NH 3, the orbital dipole due to lone pair is in the same direction as the result ant dip ole moment of the N-H bonds, whereas in NF, the orbital dipole is in the direction opposite to the resultant dipole moment of the three N–F bonds.
The orbital dipole, because of lone pair, decreases the effect of the resultant N-F bond moments, which results in the low dipole moment of NF3.
Resultant dipoles of NH 3 and NF 3 are represented diagramatically.
Resultant dipole moment in NH3=4.9 × 10–30 Cm
Resultant dipole moment in NF3=0.80×10–30Cm
Calculation of Resultant Bond Moments
Le t AB and AC have two polar bonds of a molecule inclined at an angle q . Let m 1 and m 2 have their dipole moments, respectively. The resultant dipole moment, may be obtained by vectorial method.
C B q
R12122cos µ=µ+µ+µµθ
When q = 0, the resultant is maximum. (cos 0 = 1)
R12 µ=µ+µ
When 180º, the resultant is minimum. (cos 180 = – 1) R12 µ=µ−µ
Dipole moments of some covalent molecules are listed in Table 4.21.
Table 4.21 Dipole moments of some molecules
1.85 D
0.3 D
Pyramidal 1.43 D
Pyramidal 0.23 D
PH 3 Pyramidal 0.57 D
CH4 Tetrahedral 0
CHCl3 Tetrahedral 1.04 D
Applications of Dipole Moment
■ T he percentage of ionic character of a covalent bond can be calculated from the ratio of the observed dipole moment to the dipole moment calculated on the basis of complete electron transfer.
Percentage ionic character × Observeddipolemoment 100 Dipolemomentforcompleteioniccharacter
■ The bond angles can be calculated and inference can be drawn about the symmetry of the molecule. If q is the angle of AB 2 molecule,
2(bondmoment)cos 2 θ µ=××
■ To distinguish cis- and trans- forms of geometrical isomers:
For example, the following two molecules have the same molecular formula C2H2Cl2, the same type and number of bonds, but different molecular structures.
1,2-dichloroethene 1,2-dichloroethene
Since cis- dichloroethane is a polar molecule, it has =1.89D, but trans dichloroethene is not polar, = 0 . They can readily be distinguished by a dipole moment measurement.
■ To determine orientation in benzene ring. Dipolemoment is useful to ascertain the orien-tation of substituents. In general, dipole moment of disubstituted benzenes follows the order: ortho > meta > para
(non-linear angular shape)
vii) Dipole moment of PCl 3F2 is 0
Q. The dipole moments of SO2 and CO2 are 5.37 × 10–30 Cm and zero, respectively. What can be said about the shapes of the two molecules?
Sol. Oxygen is considerably more electronegative than either sulphur or carbon. The sulphuroxygen and the carbon-oxygen bond should be polar with a net negative charge residing on the oxygen.
SO2 is angular, as the S = O bond moments do not cancel.
CO2 is linear. Though C = O bonds are polar, the bond moments cancel each other.
Q. The dipole moment of HBr is 2.60 × 10–30 C m and the interatomic spacing is 1.41A°. What is the percent ionic character of HBr?
Sol. The dipole moment of 100% assumed ionic molecule, HBr at the given internuclear distance = (1.60 × 10–19C) (1.41 × 10–10m) = 2.26 × 10–29 C m
Order of Dipole Moment: i) NH3 > NCl3 > NF3
ii) CH3Cl > CH2Cl2 > CHCl3 > CCl4 iii) CH3Cl
The actual dipole moment is less. The percent ionic character =
2.2610Cm × ×= ×
2.6010Cm 10011.5%
TEST YOURSELF
1. One Debye (D) equals to (1) 1 × 10–4 esu.cm
(2) 1 × 10–18 esu.cm
(3) 1 × 10–10 esu.cm (4) 1 × 10–16 esu.cm
2. Carbon tetrachloride has no dipole moment because of (1) its regular tetrahedral structure (2) its planar structure (3) similar sizes of carbon and chlorine atoms (4) similar electron affinities of carbon and chlorine
3. Molecule with dipole moment among the following is (1) SF6 (2) SF4 (3) CCl4 (4) CH4
4. The molecule having zero dipole moment is (1) CHCl3 (2) CH2Cl2 (3) CCl4 (4) CH3Cl
5. The dipole moment of CO2 is zero, because its bond angle is (1) 120° (2) 180° (3) 130° (4) 90°
6. S.I. unit for dipole moment is (1) e.s.u-cm (2) Coulomb-cm (3) coulomb-metre (4) e.s.u - metre
7. Molecule with zero dipole moment is (1) BCl3 (2) BeCl2 (3) CCl4 (4) All of these
CHAPTER REVIEW
Electronic Theory of Valency, Ionic Bonding, and Lattice Energy
■ The attractive force that holds two constituent atoms or oppositely charged ions together is called bond or chemical bond.
■ Every atom tries to acquire eight electrons in the outermost orbit or to have the configuration of zero group elements in the valance shell.
8. Molecule with dipole moment among the following is (1) SF6 (2) PCl5 (3) CCl4 (4) BF 3
9. In which of the following pairs, both molecules possess dipole moment? (1) CO2, SO2 (2) BCl3, PCl3 (3) H2O, SO2 (4) CO2, CS2
10. The dipole moment of HX is 1.2 D. If the % ionic character of the bond is 25%, then its bond length is
■ Ionic bond is formed by transfer of electrons from one atom to another atom.
■ Electrovalency is the number of electrons lost or gained by a species during ionic bond formation.
■ The electrostatic attraction or coulombic attraction between oppositely charged ions is called the ionic bond or the electrovalent bond.
■ When the difference in electronegativity between two atoms is 1.7 or more, generally, an ionic bond is formed between them.
■ Conditions favourable for the formation of an ionic bond are: large cation, small anion, small magnitude charge.
■ Ions involved in ionic bond posses the nearest inert gas configuration.
■ Pseudo inert gas configuration is also called nickel group configuration.
■ The smallest portion of the space lattice, which by repetition of itself in all the three directions generates the entire space lattice, is known as the unit cell.
■ Chlorides, bromides and iodides of Na, K or Rb are in Fcc lattice.
■ Oxides and sulphides of Mg, Ca, Sr or Ba are also in Fcc lattice.
■ CsCl, CsBr and CsI are in BCC lattice.
■ The shape of the crystal lattice and the coordination number depend upon the ratio of the radii of cation and anion.
■ The amount of energy released when one mole of an ionic crystalline substance is formed from gaseous ions that are infinite distance apart is called the crystal lattice energy.
■ Crystal lattice energy can be calculated indirectly using the Born–Haber cycle.
■ Lattice energy is the sum of the attractive and the repulsive forces in the crystal.
■ With an increase in the charge on the ion, the crystal lattice energy increases.
■ With an increase in the radius of the ion, the crystal lattice energy decreases.
■ Ionic substances are all crystalline, hard, but they are brittle
■ Reactions involving ionic substances are spontaneous and are usually fast.
■ Ionic substances are polar. They are soluble in polar solvents like water but insoluble in non-polar solvents like benzene.
Covalent Bond
■ The bond formed by the sharing of electrons is called a covalent bond.
■ The number of electrons of an atom of the element contributes to the formation of covalent compound and this is known as covalency.
■ Lewis dot moledel was introduced first to explain the formation of covalent bond.
■ During the bond formation, atoms acquires octet for attaining stability.
■ Molecules like BeF2, PCl5 and SF6 do not obey the octet rule.
■ BeF2, BCl3 etc. are short of the octet. Hence they are sometimes described as electron deficient molecules.
■ Formal charge may be regarded as the charge that an atom in a molecule would have if all the atoms had the same electronegativity.
■ Formal charge may or may not approximate the real ionic charge.
Coordinate Covalent Bond
■ When the shared electron pair is contributed by only one atom but shared by two atoms the bond is called a co-ordinate covalent bond.
■ Coordinate covalent bond is also called semipolar bond or dative bond.
■ The essential condition for dative bond formation is that the donor should have at least one lone pair of electrons and the acceptor should have a vacant orbital.
Valence Bond Theory
■ In the valence bond theory, the strength of the bond depends upon the extent of overlapping
■ A bond formed by the linear overlapping or axial overlap of the atomic orbitals is known as a s -bond.
■ A p -bond is formed by lateral or parallel overlapping of two atomic orbitals.
■ A p -bond is weaker than a bond.
■ The double bond in the oxygen molecule consists of a s- bond ( s p–p) and a p bond ( p p–p).
■ The triple bond in the nitrogen molecule consists of a s bond and two p bonds.
■ According to valence bond theory, in the formation of CH4, molecule there are two types of overlappings.
■ The expected bond angle in CH4 molecule is 900. but actual bond angle is 109 0.281.
■ An ionic bond is non-directional. A covalent bond is a directional bond and acts only in one direction in space.
■ The mixing of orbitals of nearly equal energy of an atom to form an equal number of equivalent (identical) orbitals is known as hybridisation
Hybridisation
■ The orbitals formed in the intermixing are known as hybrid orbitals.
■ In BeCl2, the beryllium atom undergoes sp hybridisation.
■ In BF 3 the boron atom undergoes sp 2 hybridisation. BCl 3 is a plane triangular molecule with a bond angle of 120°. BeF 2 involves sp-p overlap, where as BF 3 involves sp 2 -p overlap. Greater the S-character, stronger is the bond. Therefore BeF2 is more stable than BF 3.
■ In PCl 5 the phosphorus atom undergoes sp 3 d hybridisation. Trigonal bipyramid structure is not a regular structure. Hence, PCl 5 molecule is unstable because of the non-uniformity of the bond angles.
■ During the formation of SF6 molecule, the 3s, 3px, 3py, 3pz, 22 3 xy d and 2 3 z d orbitals are involved in hybridisation. SF6 molecule has the shape of a regular octahedron.
VSEPR Theory
■ VSEPR theory explains the structural aspects based on electron pair repulsions.
■ The repulsion between the lone pairs and bond pairs is in the order. Lone pair - Lone pair > Lone pair - bond pair > Bond pair - Bond pair.
■ In NH3 molecule, nitrogen atom undergoes sp3 hybridisation. The bond angle is 1070. This is due to greater repulsion between bond pair - lone pair than bond pair - bond pair.
■ In H2O molecule, oxygen atom undergoes sp3 hybridisation. The bond angle in water is 104°31’. This is less than the tetrahedral angle because of greater repulsion between the two lone pairs.
■ In molecules like HgCl 2 and ions like [Ag(CN)2] , [Ag(NH3)2]+ etc., the central atom contains two bond pairs in its valence shell. Hence, these have a linear shape and the central atom undergoes sphydridisation.
■ In 3 I , the central iodine atom contains 2 bond pairs and 3 lone pairs. It undergoes sp3d hybridisation. Since the lone pairs occupy the equatorial positions, it has a linear shape.
■ In molecules like SF6 and ions like [SiF6]2, [PtCl 6 ] 2– , [SnCl 6 ] 2– , [Co(NH 3 ) 6 ] 3+ , the central atom contains 6 bond pairs. They have an octahedral shape.
Resonance
■ Writing two or more structural formulae for a given molecule is called Resonance.
■ Resonance structures differ only in the position of lone pairs and pi bond electrons
■ More the possible resonating structures, more is the stability of molecule.
Molecular Orbital Theory
■ Molecular orbitals with energy lower than atomic orbitals are known as bonding molecular orbitals, while those with higher energy are known as anti bonding molecular orbitals.
■ Orbitals which are not involved in linear combination are called non bonding orbitals.
■ The order of energies of molecular orbitals: bonding < non-bonding < anti-bonding molecular orbitals.
■ The sequence of energy levels of molecular orbitals for oxygen and other heavier elements is:
s 1s < s 1s < s 2s < s 2s < s 2p x < p 2p y = p 2pz < p 2py = p 2pz < s 2px
■ Oxygen is paramagnetic due to the presence of two unpaired electrons.
■ The sequence valid for lighter elements like B,C and N is
s1s < s1s < s2s < s2s < p2py = p2pz < s2px < p 2py = p 2py < s 2px.
■ Bond order is given as:
Bonding electrons - Antibonding electrons 2
■ Bond order in F2 is 1, O2 is 2 and in N2 is 3.
Hydrogen Bonding
■ The el ectrostatic attraction between covalently bonded hydrogen atom with a highly electronegative atom, like fluorine, oxygen or nitrogen, is called hydrogen bond.
■ Hydrogen bonding is two types: intramolecular and intermolecular.
■ A water molecule can form four hydrogen bonds with four other water molecules.
■ A HF molecule can form only two hydrogen bonds with two other HF molecules.
■ The hydrogen bonds between water molecules, NH 3 molecules and HF molecules are called inter-molecular hydrogen bonds.
■ Hydrogen bond is longer than that of a covalent bond.
■ The bond energy is less for hydrogen bond than covalent bond. It is in the order of 10 to 50 kJ mol–1 .
■ o-nitrophenol is steam volatile and has a lower boiling point than p-itrophenol because there are intra-molecular hydrogen bonds in o-nitrophenol and inter-molecular hydrogen bonds in p-nitrophenol.
■ Compounds having intra-molecular hydrogen bonds (o-nitrophenol) are less soluble in water than those in which there are intermolecular hydrogen bonds (p-nitrophenol).
■ Ring formation or chelation through intra molecular H-bonding results in greater volatility, lower boiling point and lower solubility in water.
■ Intermolecular hydrogen bonding has a striking effect on physical properties like melting points, boiling points, enthalpies of vapourisation and sublimation.
■ Though alcohols, carboxylic acids, amines and carbohydrates are covalent, they are water soluble due to inter mole cular hydrogen bonding.
Bond Parameters
■ The dista nce between the nuclei of the atoms n a molecules is known as bond length.
■ Bond length is equal to the sum of the covalent radii of the two atoms that are bonded in homo atomic molecule.
■ As the number of bonds between two atoms increases, the bond length decreases.
■ C-C length is 1.54A0, C = C length is 1.33A0 and length is 1.20 A0
■ The energy liberated when a bond is formed between two gaseous atoms is called bond energy. This is expressed in KJ/mole.
■ The energy required to break the bond and form separate gaseous atoms is called bond dissociation energy. It is expressed in KJ/mole.
■ In the case of diatomic molecules bond energy and bond dissociation energy have an equal magnitude.
■ In a polyatomic molecule containing similar bonds, the bond energy of each bond is taken as the average of the bond dissociation energies of similar bonds.
■ The bond dissociation energies of the four bonds in CH4 are not equal.
■ The average of the bond dissociation energies of the four C-H bonds is taken as the bond energy of each C-H bond.
■ The energy required for heterolytic cleavage is more than that of homolytic cleavage.
■ The strength of a bond depends upon the extent of overlapping of the orbitals.
■ In a p -bond, the extent of overlapping of the orbitals is less than that in a s bond. Hence, a p bond is weaker than a s bond.
■ A double bond is stronger than a single bond and a triple bond is stronger than a double bond.
■ Molecules containing an odd number of electrons are called odd electron molecules. NO, NO 2, ClO 2 etc., are example of odd electron molecules.
■ The strength of the bonds is in the order : ionic bond > covalent bond > metallic bond > hydrogen bond > dipole-dipole attraction > van der Waals’ force.
Bond Polarity and Dipole Moment
■ When there is a difference in electronegativity between the two atoms in a molecule, the shared electron pair will be nearer to the more electronegative atom.
■ With unequal sharing of electron pair of the bond is called a polar covalent bond and the molecule is known as a dipole.
■ Dipole moment D µ = Charge x distance between the two poles (bond length).
■ SI unit of dipole moment is coulomb -metre. 1 Debye = 3.33 × 10–30 coul. m.
■ The present ionic character of a bond increases with an increase in the difference of electronegativities of the two atoms in the molecule
■ A polar covalent bond is stronger than a pure covalent bond.
■ The dipole moment of a polyatomic molecule is the vectorial sum of the dipole moments of the individual bonds in the molecule.
■ Though a molecule contains polar bonds, the net dipole moment of the molecule is zero, if it is a symmetrical molecule.
■ B-Cl bond is a polar bond. But BCl 3 molecule is non-polar because it is a symmetrical molecule (plane triangle)
■ CO2, CH4 and CCl4 are non-planar because they are symmetrical.
■ H2O, NH3, NCl3 , NF3 etc. are polar because they are unsymmetrical molecules.
■ The dipole moment of NCl3 is greater than that of NF 3 because the lone pair on the nitrogen atom compensates for the sum of the dipole moments due to N-F bond which act in the opposite direction.
■ A knowledge of the dipole moment of a molecules helps in knowing the shape of the molecule.
Exercises
NEET DRILL
Level 1
Electronic Theory of Valency ,Ionic bonding and Lattice Energy.
1. Which molecule do not have octet?
(1) Hydrogen (2) Oxygen
(3) Chlorine (4) Nitrogen
2. Which of the following molecule, has octet on its central atom ?
(1) CH4 (2) BF 3 (3) PF 5 (4) ICl3
3. If the values of Madelung constants of the following compounds are equal, then their lattice energy values decreases in the order:
(1) NaF > KCl > CaO > Al2O3
(2) Al2O3 > CaO > KCl > NaF
(3) Al2O3 > CaO > NaF > KCl
(4) KCl > NaF > CaO > Al2O3
4. The strongest ionic bond is present in (1) LiF (2) NaF (3) RbF (4) CsF
5. Which of the following has highest lattice energy?
(1) AlF 3 (2) NaF
(3) MgF2 (4) NaCl
6. The electronegativity of H and Cl are 2.1 and 3.0 respectively. The correct statement about the nature of HCl is/are (1) 17% ionic (2) 83% ionic (3) 50% ionic (4) 100% ionic
7. Lattice energy of NaCl is ' X '. If the ionic size of A+2 is equal to that of Na+ and B-2 is equal to Cl-, then lattice energy associated with the crystal AB is
(1) X (2) 2X* (3) 4X (4) 8X
8. Coordination number of cation is 6 in (1) NaCl (2) CsCl (3) ZnO (4) KCl
9. Which combination will give strongest ionic bond?
(1) Na+and Cl– (2) Mg2+ and Cl–(3) Na+and O2– (4) Mg2+ and O2–
10) Electrovalent bond is mainly favoured between alkali metals and (1) halogens (2) Chalcogens
(3) metalloids (4) Boron family
Kossel Lewis Approach to Chemical Bonding
11. Two elements 'X' and 'Y' have the following configuration
X = 1s2 2s2 2p6 3s2 3p6 4s2
Y = 1s2 2s2 2p6 3s2 3p5
The compound formed by the combination of 'X' and 'Y' will be ______.
(1) XY2 (2) X 5Y2 (3) X2Y5 (4) XY 5
12. Given electronic configurations of four elements E 1, E2, E3 and E4 are respectively 1s2, 1s2 2s2 2p2, 1s2 2s2 2p5 , and 1s2 2s2 2p6 . The element that is capable of forming ionic as well as covalent bonds is (1) E1 (2) E2 (3) E 3 (4) E 4
13. The molecule that deviates from octet rule is
(1) CCl4 (2) BF 3 (3) MgO (4) NCl3
14. In molecule, the formal charges of oxgen atoms 1, 2, 3 are respectively (1) –1, 0, +1 (2) 0, –1, +1 (3) 0, +1, –1 (4) +1, 0, –1
Covalent Bond
15. Polarisation is the distortion of the shape of an anion by an adjacently placed cation. Which of the following statements is correct? (1) Maximum polarisation is brought about by a cation of high charge
(2) Minimum polarisation is brought about by a cation of low radius
(3) A large cation is likely to bring about a large degree of polarisation
(4) A small anion is likely to undergo a large degree of polarisation
16. The following are some statements about the characteristics of covalent compounds
(i) The combination of a metal and nonmetal must give a covalent compound.
(ii) All covalent substances are bad conductors of electricity.
(iii) All covalent substances are gases at room temperature.
(1) i, ii, ii are correct
(2) only i and ii are correct
(3) only ii and iii are correct
(4) i, ii, iii are incorrect
17. In the formation of SF6 molecule, the sulphur atom is in
(1) First excited state
(2) Second excited state
(3) Third excited state
(4) Fourth excited state
18. Which contains both polar and non-polar covalent bonds?
(1) NH4Cl (2) HCN
(3) H2O2 (4) CH4
Co-ordinate Covalent bond
19. Dative bond is present in (1) CH4 (2) O3
(3) NaCl (4) C6H12O6
20. PH3and BF3 form an adduct readily because they form
(1) Co-ordinate covalent bond
(2)A covalent bond
(3) An ionic bond
(4) A hydrogen bond
21. Dative bond is not present in (1) Ozone
(2)Nitrous oxide
(3) Sulphur tetraoxide
(4) Borazole
Valence Bond Theory
22. Which of the following is called negative overlap? (1) (2) (3)
(4)
Hybridisation
23. Correct statements regarding hybridisation are A) orbitals of different atoms mix with each other to form hybrid orbitals.
B)vacant orbitals cannot involve in hybridisation.
C) Each hybrid orbital can accommodate a maximum of two electrons with opposite spin.
D) Hybrid orbitals cannot form sigma bonds.
(1) A, B, C and D (2) C only
(3) C and D only (4) A and C only
24. In which of the following sets the central atom is involved in SP3 hybridisation?
(1) BF 4–, ClO4– (2) BF3, ClO3–
(3) XeO4, XeO3 (4) SO3, XeO3
25. Sp3d2 hybridisation is not displayed by?
(1) SF6 (2) PF 5 (3) IF 5 (4) BrF 5
26. Hybridisation of I in ICl2+is
(1) Sp (2) Sp2 (3) Sp3 (4) dSp2
VSEPR theory
27. A central atom in a molecule has two lone pairs of electrons and form three single bonds. The shape of the molecule is :
(1) T – shaped
(2) see – saw
(3) Trigonal pyramidal
(4) planar triangular
28. Identify the pair in which the geometry of the species is T-shape and square-pyramidal, respectively
(1) ClF3 and IO4– (2) ICl2–and ICl5
(3) XeOF2 and XeOF4 (4) IO4– and IO2F2–
29. Shape of BrF 5 molecule is
(1) pentagonal bipyramidal
(2) trigonal pyramidal
(3) square pyramidal
(4) trigonal bipyramadal
30. Which is not correctly matched?
(1) XeO3– Trigonal bipyramidal
(2) ClF3 – bent T – shape
(3) XeOF4 – square pyramidal
(4) XeF2 – Linear shape
31. In which of the following molecules, all the atoms lie in one plane?
(1) CH4 (2) BF 3 (3) PF 5 (4) NH3
Resonance
32. Resonance in carbonate ion (CO 3 2– )
Which of the following is true?
(1) All these structures are in dynamic equilibrium with each other
(2) Each structure exists for equal amount of time
(3) CO32– has a single structure i.e., resonance hybrid of the above three structures
(4) It is possible to identify each structure individually by some physical or chemical method
Molecular Orbital Theory
33. The correct order of bond orders of C 2 2–, N22– and O22– is, respectively
(1) C22– < N22– < O22– (2) O22– < N22– < C22–
(3) C22– < O22– < N22– (4) N22– < C22– < O22–
34. In which of the following transformation, the bond order has increased and the magnetic behaviour has changed.
(1) C2+ → C2 (2) NO+ → NO
(3) O2 → O2+ (4) N2 → N2+
35. Some statements about valence bond theory are given below
i) The strength of bond depends upon extent of overlapping.
ii) The theory explains the directional nature of covalent bond.
iii) According to this theory oxygen molecule is paramagnetic in nature.
(1) all are correct
(2) only i and iii are correct
(3) only i and ii are correct
(4) all are wrong
Hydrogen Bonding
36. All the following molecules involve in intra molecular hydrogen bonding except (1) o–nitro phenol
(2) Salicylic acid
(3) p–nitro aniline
(4) o–hydroxy benzaldehyde
37. The intramolecular hydrogen bonding is present in
(1) Phenol
(2) Benzoic acid
(3) Para–nitrophenol
(4) 2–hydroxybenzoic acid
38. Intramolecular hydrogen bond is present in (1) orthohydroxy benzaldehyde (2) parahydroxy benzaldehyde
(3) ethyl alcohol
(4) hydrogen fluoride
Bond Parameters
39. Which among the following species has unequal bond lengths?
(1) [XeF5]– (2) [BF4]–
(3) SiF4 (4) NF3
40. What is the correct order of bond angles in the following
i) OF2 ii) Cl2O iii) H2O
(1) OF2 < H2O < Cl2O
(2) OF2 < Cl2O < H2O
(3) H2O < OF2 < Cl2O
(4) Cl2O < OF2 < H2O
41. Bond enthalpy is maximum for (1) Br2 (2) F2 (3) Cl2 (4) I2
42. For which of the following molecule the resultant dipole moment μ ≠ 0?
Level 2
Electronic Theory of Valency ,Ionic bonding and Lattice Energy.
1. A, B and C are atoms of elements with atomic number Z, Z + 1 and Z + 2 respectively. If 'B' has octet configuration, the bond formed between 'A' and C predominantly is (1) Covalent bond (2) Ionic bond
(3) Dative bond (4) Hydrogen bond
2. The order of relative ease of formation of various ions is
(1) F– > O–2 > N–3 (2) N–3 > O–2 > F–
(3) O–2 > N–3 > F– (4) F– > N–2 > O–2
3. The electro-negativities of two elements are 1.0 and 3.5. Bond formed between them would be (1)Electrovalent (2)Polar covalent (3)Pure covalent (4) Metallic
4. Maximum ionic character is present in (1)BeCl2 (2)NaCl (3)KF (4) MgO
(1) Only (i) (2) (i) and (ii)
(3) Only (iii) (4) (iii) and (iv)
Bond polarity and Dipole Moment
43. Which of the following molecule has the highest value of dipole moment?
32. Correct statement regarding molecules SF4, CF4 and XeF4 are : (1) 2, 0 and 1 lone pairs of central atom respectively (2) 1, 0 and 1 lone pairs of central atom respectively (3) 0, 0 and 2 lone pairs of central atom respectively (4) 1, 0 and 2 lone pairs of central atom respectively
33. Shape of BrF5 molecule is (1) Pentagonal (2) Trigonal pyramidal (3) Square pyramidal (4)Trigonal bipyramidal
34. Total lone pair of electrons present on the central atom of I3 - is (1) 6 (2) 9 (3) 4 (4) 3
CHAPTER 4: Chemical Bonding and Molecular Structure
35. The shape of AB3E type molecule is (1) pyramidal (2) tetrahedral (3) angular (4) linear
Resonance
36. A molecule is described by three Lewis structures having energies E 1 , E 2 & E 3 respectively . The energies of these structures follow the order E1 > E2 > E3. If resonance hybrid energy of the molecule is E 0 , the resonance energy is
(1) E 0 – E1
(2) E 0 – E2
(3) E 0 – E 3
(4) E 0 – (E1 + E2 + E3)
37. The resonance hybrid of nitrate ion is
(1) O N O O -1/2 -1/2 -1/2
(2) O N O O -2/3 -2/3 -2/3
(3) O N O O -1/3
(4) O N O
Molecular orbital theory
38. If E is the energy of atomic orbitals involved to form molecular orbitals. E b & E a are the energies of bonding and antibonding molecular orbitals formed respectively then
(1) Eb > E a > E
(2) E a > E > Eb
(3) E > E a > Eb
(4) E a > Eb > E
39. The correct molecular orbital diagram for F2 molecule in the ground state is (1)
40. Among the following molecules or ions C2–2, N2–2, O2–2, O2 which one is diamagnetic and has the shortest bond length?
(1) O2 (2) N2–2
(3) O2–2 (4) C2–2
Hydrogen Bonding
41. Which of the following can form hydrogen bonds even in vapour
(1) HF (2) HCl (3) NH3 (4) H2O
42. Intramolecular H-bonding is present in (1) meta nitrophenol (2) salicylaldehyde (3) hydrogen chloride (4) benzophenone
43. Which of the following molecules / ions can exist?
(1) H4S (2) OF4 (3) KHF2 (4) BeF5-3
44. Ortho-Nitrophenol is less soluble in water than p - and m - Nitrophenols because (1) o-Nitrophenol is more volatile in steam than those of m - and p -isomers
(2) o-Nitrophenol shows Intramolecular H - bonding
(3) o-Nitrophenol shows Intermolecular H-bonding
(4) Melting point of o- Nitrophenol is lower than those of m- and p -isomers.
Bond parameters
45. The correct order of bond angle in the given molecules is
(1) PH 3 < NH3 < H2O < NH4+
(2) PH 3 < H2O < NH3 < NH4+
(3) NH3 < PH 3 < H2O < NH4+
(4) NH4+ < NH3 < H2O < PH3
FURTHER EXPLORATION
1. Consider the following statements
(A) NF 3 molecule has a trigonal planar structure.
(B) Bond length of N2 is shorter than O2
(C) Isoelectronic molecules or ions have identical bond order
(D) Dipole moment of H2S is higher than that of water molecule.
46. Which among the following species has unequal bond lengths?
(1) XeF4 (2) SiF4
(3) SF4 (4) BF 4 -1
47. Of the species, NO, NO+, NO2+ and NO- the one with minimum bond strength is (1) NO (2) NO2+
(3) NO- (4) NO+
48. The bond length of triple bond in carbon monoxide is 1.20 Å and in carbon dioxide it is 1.34 Å. Then the C = O bond length in CO32- will be ___ Å
(1) 0.95A0 (2) 1.25A0 (3) 1.36A0 (4) 1.34A0
Bond polarity and Dipole Moment
49. Which of the following molecule is polar molecule?
(1) N2 (2) CH4 (3) CHCl3 (4) CCl4
50. Which one of the following possess highest melting point ?
(1) Chlorobenzene
(2) O–dichlorobenzene
(3) m–dichlorobenzene
(4) p–dichlorobenzene
51. The molecule with zero dipole moment among the following is (1) NH3 (2) H2O (3) SO2 (4) CCl4
52. A diatomic molecule has a dipole moment of 1.2D. If the bond length is 10 -8cm, what fraction of charge does exist on each atom?
(1) 0.1 (2) 0.2 (3) 0.25 (4) 0.3
(1) (A) and (D) are correct
(2) (B) and (C) are correct
(3) (A) and (B) are correct
(4) (C) and (D) are correct
2. The bond length of HCl bond is 2.29 x 10-10m. The percentage ionic character of HCl, if measured dipole moment 6.226x 10-30cm, is (1) 8% (2) 20% (3) 17% (4) 50%
3. A polar molecule AB have dipole moment 3.2 D (Debye) while the bond length is 1.6
A. Find the percentage ionic character in the molecule.
(1) 31% (2) 41.6%
(3) 39.6% (4) 20%
MATCHING TYPE QUESTIONS
1. List-1 contains increasing order different compounds against the properties in List-II
(IV) RbI < NaBr < MgF2 < CaO (S) Electrical conductivity in fused state
(T) Solubility in water
Choose the correct answer from the options given below:
(I) (II) (III) (Iv)
(1) S R Q P
(2) Q T S R
(3) P Q T S
(4) R P T Q
2. Match List-I with List-II
List - I species
List - II Geometry/Shape
(A) H3O+ (I) Tetrahedral
(B) Acetylide anion (II) Linear
(C) NH4+ (III) Pyramidal
(D) ClO2– (IV) Bent
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) III IV I II
(2) III I II IV
4. I+3 and I-3 have same (1) Geometry
(2) Number of lone pair (s)
(3) Bond angle
(4) None of these
(3) III IV II I
(4) III II I IV
3. Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II and select the most appropriate option
List - I (Shape)
List - II (Molecules)
(A) T-shaped (I) XeF4
(B) Trigonal planar (II) SF4
(C) Square planar (III) ClF3
(D) See-saw (IV) BF 3
(A) (B) (C) (D)
(1) I II III IV
(2) III IV I II
(3) III IV II I
(4) IV III I II
4. Match List I with List II :Choose the correct answer from the options given below:
List - I (molecule) List - II (hybridisation; shape)
(A) XeO3 (I) sp3d ; linear
(B) XeF2 (II) sp3; pyramidal
(C) XeOF4 (III) sp3d3; distorted octahedral
(D) XeF6 (IV) sp3d2; square pyramidal
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) II I IV III
(2) II IV III I
(3) IV II III I
(4) IV II I III
5. Match list I with list II and select the correct answer using the codes given below.
List - I List - II
(P) CS2 (1) Bent (Q) SO2 (2) Linear
(R) BF 3 (3) Trigonal planar
(S) NH3 (4) Trigonal bipyramidal (5) Trigonal pyramidal
Choose the correct answer from the options given below:
(P) (Q) (R) (S)
(1) 2 1 3 5
(2) 1 2 3 5
(3) 2 1 5 4
(4) 1 2 5 4
6. List-I contains molecules or ions and List-II contains shapes or lone pairs
List - I List - II
(I) XeF4, ClF3 (P) Same shape (II) NH3, H3O+ (Q) Different shape
(III) XeF2, BeCl2 (R) Same number of lone pair of electrons on central atom
(IV) SF4, SiF4 (S) Different number of lone pair of electrons on central atom (T) Isoelectronic species
Choose the correct answer from the options given below: (I) (II) (III) (IV)
(1) P, Q Q,R,T P,S P,S
(2) R,S P,T R,S P,Q
(3) Q,R P,R,T P,S Q,S
(4) P,R,S P,Q,T P,R R,T
7. Match List-I with List-II
List - I List - II
(A) XeF4 (I) See-saw
(B) SF4 (II) Square planar
(C) NH4+ (III) Bent T – shaped
(D) BrF 3 (IV) Tetrahedral
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) II I IV III
(2) IV III II I
(3) II I III IV
(4) IV I II III
8. Match List-I with List-II
List - I List - II
(P) IF 7 (1) planar and polar (Q) SO2 (2) Non- planar and polar (R) SF4 (3) planar and non - polar (S) CS2 (4) non planar and nonpolar
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) 4 1 2 3
(2) 4 2 1 3
(3) 3 1 2 4
(4) 3 1 4 2
9. Match the molecule/ion given in List-I with the type of hybridisation of its central atom given in List-II
List - I List - II
(I) SF6 (P) sp (II) I 3 – (Q) sp3d
(III) NO2+ (R) sp2
(IV) XeO4 (S) sp3d2 (T) sp3
Choose the correct answer from the options given below:
(I) (II) (III) (IV)
(1) P Q R S
(2) S Q P T
(3) T R P S
(4) S P Q T
10. Match the molecule in the List-I with the type of orbital overlapping in the List-II
List - I List - II
(I) CH4 (P) sp – p
(II) C2H6 (Q) sp2 – sp2
(III) C2H4 (R) sp – sp (IV) C2H2 (S) sp3 – s
STATEMENT TYPE QUESTIONS
Each question has two statements: statement I (S-I) and statement II (S-II)
(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: NaF is more ionic than Na 2O
S-II : formation of O -2 from O -1 is exothermic
2. S-I : SO2 and H2O both possess V – shaped structure
S-II : The bond angle of SO2 is less than that of H2O .
3. S-I : Carbon sub oxide is bent molecule
S-II : Each carbon atom is in sp –hybrid state
4. S-I : Hyper conjugation is a permanent effect.
S-II : Hyper conjugation in ethyl cation involves the overlapping of NN O
== bond with empty 2p orbital of other carbon.
5. S-I : In C2 vapours, the bond order is two
(T) sp3 – sp3
Choose the correct answer from the options given below:
(I) (II) (III) (IV)
(1) S T Q R
(2) Q R S T
(3) S P T R
(4) T P R S
with one sigma bond and one pi bond.
S-II : Formal charge on terminal nitrogen atom of NN O
== is ''+1''.
6. S-I : Bond energy of N2 – < N2+
S-II : Anti bonding electrons are more in N2– than in N2 +
7. S-I : Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.
S-II : The crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecules.
8. S-I : CO2 has no dipolemoment where as SO2 and H2O have dipolemoment.
S-II : SnCl 2 is ionic, where as SnCl 4 is covalent which of the following is correct
9. S-I : The actual structure of the benzene molecule is said to be a resonance hybrid.
S-II : Alternative structures are referred to as resonance structures or canonical forms
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). Mark the correct answer as
(1) (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
1. (A): Formal charge on all nitrogen atoms in N3– (azide ion) is same in all different possible lewis structures
(R) : Formal charge on an atom is independent of lewis structure
2. (A : POF3 exist but NOF3 does not exist
(R) : ‘P’ cannot form five bonds by expanding its octet while ‘N’ can expand its octet to form five bonds.
3. (A) : Water is a good solvent for ionic compounds but poor one for covalent compounds
(R) : Hydration energy of ions releases sufficient energy to overcome lattice energy and break hydrogen bonds in water, while covalent bonded compounds interact so weakly that even van der Waal’s forces between molecules of covalent compounds cannot be broken
4. (A) : Zero orbital overlap is an out of phase overlap.
(R) : It results due to different orientation/ direction of approach of orbitals.
5. (A) : The shape of N ( SiH 3 ) 3 is trigonal planar.
(R) : In N(SiH3)3 the hybridisation of ' N ' is sp2
6. (A) : [ I ( CN ) 2 ] – is planar whereas [I(CH3)2 ]– is not " planar
(R) : Iodine atom is sp hybridized in both the species
7. (A) : Diamagnetic C 2 molecule has been detected in vapour phase in which double bond consists of both Pi bonds
(R) : In the above C2 molecule four electrons are present in two Pi molecular orbitals
8. After understanding the assertion and reason, choose the correct option.
(A) : In the bonding molecular orbital (MO) of H 2, electron density is increased between the nuclei.
(R) : The bonding MO is ψ A + ψ B, which show destructive interference of the combining electron waves.
9. (A) : Bond order in a molecule cannot be negative,or fractional value
(R) : Bond order depends only on the number of electrons in the bonding but not in anti-bonding orbitals.
10. (A) : Among the two O – H bonds in H2O molecule, the energy required to break the first O – H bond and the other O – H bond is different.
(R) : This is because the electronic environment around oxygen is different after breakage of one O – H bond.
11. (A) : The dipole moment helps to predict whether molecule is polar or non–polar.
(R) : The dipole moment helps to predict the geometry of molecules.
12. (A) : The electronic structure of azide ion (N3–) is N –= N + = N –
(R) : is not a resonating structure of azide ion, because the position of atoms cannot be changed.
BRAIN TEASERS
1. Consider the lewis dot structure that satisfies octet configuration for both ''C'' and ''O'' atoms in CO and identify correct statement among the following (A) Formal charge on oxygen atom is – 2 (B) Formal charge on carbon atom is +2 (C) One lone pair electron is present on each carbon and oxygen atom (D) Three bond pair electrons are present between carbon and oxygen (1) A,B,C and D (2) A and B (3) C and D (4) A,B and C
2 The type of bonds present in Ni(CO) 4 are (1) electrovalent and covalent (2) electrovalent and coordinate (3) electrovalent, covalent and coordinate (4) covalent and coordinate
3. Lattice energy of NaCl is ' X '. If the ionic size of A+2 is equal to that of Na+ and B–2 is equal to Cl–, then lattice energy associated with the crystal AB is (1) X (2) 2X (3) 4X (4) 8X
4. Which of the following is incorrect match
Hybridisation Geometry
Orbital use
1 sp3d Trigonal bipyramidal s + p x + p y + pz+ d z 2
2 sp3d3 Pentagonal bipyramidal
FLASH BACK (Previous NEET Questions)
1. The correct sequence of bond enthalpy of 'C – X' bond is
(1) CH3 – Cl > CH3 – F > CH3 – Br > CH3 – I
(2) CH – F < CH3 – Cl < CH3 – Br < CH3– I
(3) CH3 – F > CH3 – Cl > CH3 – Br > CH3 – I
(4) CH3 – F < CH3 – Cl > CH3 – Br > CH3 – I
2. Which of the following molecules is non –polar in nature?
(1) NO2 (2) POCl3
(3) CH2O (4) SbCl5
3 sp3d2 distorted octahedral
4 sp3 Tetrahedral (1) 1 (2) 2 (3) 3 (4) 4
5. Hybridisation of 'P' in solid PCl 5 (1) sp3d (2) sp3d2, sp3 (3) sp3d2, sp2 (4) sp2
6. The following graph is given between total energy and distance between the two nuclei for species H2+, H2, He2+ and He2 which of the following statement is correct?
(1) He2+ is more stable than H2 +
(2) Bond dissociation energy of H2+ is more than bond dissociation energy of He 2 +
(3) Since, bond orders of He2 + and H2+ are equal hence both will have equal bond dissociation energy
(4) Bond length of H2 + is less than bond length of H2
3. Match List-I with List-II
List - I List - II
(a) PCl5 (i) Square pyramidal
(b) SF6 (ii) Trigonal planar
(c) BrF 5 (iii) Octahedral
(d) BF 3 (iv) Trigonal bipyramidal
Choose the correct answer from the options given below:
(a) (b) (c) (d)
(1) iv iii ii i
(2) iv iii i ii
(3) ii iii iv i
(4) iii i iv ii
4. BF 3 is planar and electron deficient compound. Hybridisation and number of electrons around the central atom, respectively are
(1) sp2 and 8 (2) sp3 and 4
(3) sp3 and 6 (4) sp2 and 6
5. Amongst the following which one will have
CHAPTER TEST
1. Which of the following species is neither hypervalent nor hypovalent?
(1) CLO4– (2) BF 3
(3) SO42– (4) CO32–
2. In ammonium ions the covalency of nitrogen is
(1) 3 (2) 4 (3) 2 (4) 5
3. Which of the following contains unshared electron on central atom ?
(1) NO2 (2) CO2 (3) NO2– (4) CN–
4. Which of the following is the most likely Lewis structure of nitrosyl chloride, NOCl
(1) (2)
(3)
(4)
5. The formal charge on nitrogen atoms 1 and 2 respectively
maximum 'lone pair – lone pair' electron repulsions?
(1) IF 5 (2) SF4 (3) XeF2 (4) ClF3
6. Which amongst the following is incorrect statement?
(1) C2 molecule has four electrons in its two degenerate p molecular orbitals
(2) H2+ ion has one electron.
(3) O2+ ion is diamagnetic
(4) The bond orders of O2+, O2, O2–and O22are 2.5, 2, 1.5 and 1 , respectively.
(1) –1, –1 (2) –1, +1
(3) +1, –1 (4) +1, +1
6. Covalent compounds are generally soluble in (1) polar solvents
(2) non–polar solvents
(3) concentrated acids (4) all solvents
7. Dative bond is absent in (1) BF 4 – (2) NH4+ (3) O3 (4) CO3–2
8. Which compounds contain both ionic and covalent bonds?
(P) BaSO4 (Q) Ca(NO3 )2 (R) NH4Cl
(1) P only (2) P and R only
(3) Q and R only (4) P, Q and R
9. Oxidation numbers of A , B , C are +2, +5 and –2 respectively. Possible formula of compound is.
(1) A2(BC2)2 (2) A3(BC4)2
(3) A3(BC2)2 (4) A3(B2C)2
10. Correct order of Lattice enthalpy is
(1) NaCl > MgCl2 > AlCl3
(2) MgCl2 > NaCl > AlCl3
(3) AlCl3 > MgCl2 > NaCl
(4) AlCl3 > NaCl > MgCl2
11. The lattice energies of KF, KCl, KBr and KI
follow the order:
(1) KF > KCl > KBr > KI
(2) KI > KBr > KCl > KI
(3) KF > KCl > KI > KBr
(4) KI > KBr > KF > KCl
12. Which of the following is least ionic?
(1) CaF2 (2) CaBr2
(3) CaCl2 (4) CaI2
13. Which of following incorrect about Ionic compounds?
(1) It has high Melting and Boiling points. (2) Ionic compounds are poor conductors in solid state.
(3) It exhibits Isomerism.
(4) Most of the Ionic compounds soluble in polar solvents like H 2O.
14. The compound with highest melting point is
(1) NaF (2) NaCl
(3) NaBr (4) NaI
15. Compound having lowest melting point.
(1) BeCl2 (2) MgCl2
(3) CaCl2 (4) SrCl2
16. Given below are two statements.
Statement-I : large number of ionic solids are more soluble in water than in common organic solvents
Statement-II : dielectric constant of water is more when compared with common organic solvents
In light of the above statements, choose the correct answer from the options given below.
(1) Both statement I and II are correct
(2) Both statement I and II are incorrect
(3) Statement I is correct and statement II is incorrect
(4) Statement I is incorrect and statement II is correct
17. The correct order of decreasing polarisability of ion is
(1) Cl –> Br– > I –> F –
(2) F –> I –> Br– > Cl –
(3) I –> Br– > Cl –> F –
(4) F –> Cl– > Br –> I –
18. If Na + ion is larger than Mg 2+ ion and S 2–ion is larger than Cl – ion, which of the following will be least soluble in water?
(1) NaCl (2) Na2S
(3) MgCl2 (4) MgS
19. Number of bonding electron pairs and number of lone pairs of electrons in xenon difluoride, sulphur tetrafluoride, and iodine pentafluoride respectively are
(1) (3, 2)(4, 2)(5, 2)
(2) (3, 1)(4, 1)(5, 2)
(3) (3, 1)(4, 2)(5, 1)
(4) (3, 2)(4, 1)(5, 1)
20. Which of the following structures are correctly represented according to VSEPR theory?
(1) A, B and C only (2) B and C only
(3) A and C only (4) C only
21. In which of the following species the maximum number of atoms lie in the same plane?
(1) XeF2O2 (2) PCl5
(3) AsH4+ (4) XeF4
22. For central atom 'A', let 'E' be the lone pair.
Column - I Column - II
(A) AB2E (P) T–shape
(B) AB 3E2 (Q) Square pyramid (Distorted)
(C) AB 5 E (R) Bent
(D) AB2E2 (S) See–saw (T) Non–zero dipole moment
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) R, T S S T
(2) R S, T S, T P, T
(3) R, T P, T Q, T R, T
(4) P, T Q, T Q, T R
23. Which of the following is zero over lapping of orbitals; if z – axis is inter nuclear axis
(1) s – pz (2) pz – pz
(3) px – px (4) s – px
24. The species that cannot exist is
(1) SiF62- (2) BF63- (3) SF6 (4) AIF63-
25. According to valence bond theory, the bonds in methane are formed due to the overlapping
(1) 1σs – s, 3σs – p (2) 1σs – p, 3σs – s
(3) 2
26. Match the Column – I with Column – II and mark the appropriate choice.
Column - I Column - II
(A) C2H2 (i) sp3d2 Hybridisation
(B) SF6 (ii) sp3d3 Hybridisation
(C) SO2 (iii) sp Hybridisation
(D) IF 7 (iv) sp2 Hybridisation
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) i iii ii iv
(2) iii i iv ii
(3) ii iii i iv
(4) iv i iii ii
27. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion(A) : Xe in XeF2 is d2sp3 hybridised Reason(R) : XeF2 molecule follows octet rule
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.
28. A square planar complex is formed by hydridization of which of the following atomic orbitals?
(1) s, px, py, d z2
(2) s, px, py, d x2 – y2
(3) s, px, py, d xy
(4) s, py, pxdzx
29. The ratio of the number of sp, sp2 & sp3 hybrid orbitals in the compound
CH3 – CH = C = CH – C º C – CH3 is___
(1) 1 : 1 : 1 (2) 2 : 2 : 1
(3) 3 : 2 : 1 (4) 3 : 3 : 4
30. In which of the following molecules all bonds are not equal ?
(1) ClF3 (2) SF4
(3) PCl5 (4) All
31. Specify the coordination geometry around and hybridisation of N and B atoms on the 1: 1 complex of BF3 and NH3 :
(1) N : tetrahedral, sp3, B: tetrahedral, sp3
(2) N : pyramidal, B: pyramidal, sp 3
(3) N : pyramidal, sp3, B : planar, sp2
(4) N : pyramidal, sp2, B : tetrahedral, sp3
32. Given below are two statements. One is labelled
Assertion (A) and the other is labelled Reason (R).
Assertion (A): Double bond in C 2 molecule (vapour phase) consists of both Pi bonds
Reason (R): Four electrons are present in two Pi molecular orbitals.
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.
33. Match the substance in the List-I with the type of bond given in List-II
List - I List - II
(I) Nacl (P) Non–polar covalent only
(II) PCl3 (Q) Ionic only
(III) NH4Cl (R) covalent, ionic and dative
(IV) KHF2 (S) Polar covalent only
(T) Ionic, covalent, H–bond
Choose the correct answer from the options given below:
(I) (II) (III) (IV)
(1) P Q T S
(2) Q P R T
(3) Q S R T
(4) T S R P
34. Which of the following statement is wrong
(1) In the formation of s s * (anti bonding) M.O, one extra nodal plane that is not present in s – atomic orbital will be form combined
(2) In the formation s p M.O, no extra nodal plane is formed than present in p – atomic orbitals combined
(3) In the formation of s p(anti bonding M.O),
one extra nodal plane is formed than present in p – atomic orbitals combined
(4) In the formation of p M.O, one extra nodal plane is formed present than in pure p – atomic orbitals combined
35. Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists.
List - I List - II
(1) p–dπ antibonding
(2) d – d s antibonding
(3) p – dπ bonding
(4) d – d s bonding
Choose the correct answer from the options given below:
36. Given below are two statements.
Statement–I: According to MOT, the process N2(g) → N2+(g) + e – is more endothermic than the process O2 (g) → O2 + + e –
Statement–II: Because electron is removed from bonding molecular orbital in
N 2 where as electron is removed from antibonding molecular orbital in O2.
In light of the above 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 correct but statement II is incorrect.
(4) Statement I incorrect but statement II is correct.
37. Decreasing order of stability of O 2 , O 2, O2+and O22– is
(1) O22– > O2– > O2 > O2+
(2) O2 > O2+ > O22– > O2–
(3) O2– > O22– > O2+ > O2
(4) O2+ > O2 > O2– > O22–
38. According to molecular orbital theory which of the lists ranks the nitrogen species in terms of increasing bond order?
(1) N2–2 < N2– < N2 (2) N2 < N2–2 < N2–
(3) N2– < N2–2 < N2 (4) N2– < N2 < N2–2
39. In which of the following process, the value of magnetic moment does not change?
(1) (i) and (ii) (2) (iii) and (v) (3) (iii) and(iv) (4) (v) and (ii)
40. In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?
41. The bond order and magnetic behavior of O2–2 ion are, respectively: (1) 1.5 and paramagnetic (2) 1.5 and diamagnetic (3) 2 and diamagnetic (4) 1 and diamagnetic
42. The correct order of N – O bond length in NO, NO2–And NO3– will be (1) NO > NO2 > NO3–
(2) NO > NO3– > NO2–
(3) NO3– > NO2– > NO
(4) NO2– > NO3– > NO
43. When the molecules N2, N2O and N2O4 are arranged in order of decreasing N – N bond length, which order is correct
45. The correct order of increasing CO bond length of CO, CO3 2– and CO2
(1) CO32– < CO2 < CO
(2) CO2 < CO32– < CO
(3) CO < CO32– < CO2
(4) CO < CO2 < CO32–
46. Which of the following order is correct against the given property?
(1) H2O > NH3 > NF3 : dipole moment
(2) O22– > O2 > O2+: bond order
(3) AlF 3 > MgF2 > NaF : ionic character
(4) BF 3 < BCl3 < BBr 3 : bond angle
47. The electronegativity difference between N and F is greater than then between N and H yet the dipole moment of NH 3 (1.5D) is larger than that of NF 3 (0.2D). This is because
(1) in NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite directions
(2) in NH 3 as well as NF 3 , the atomic dipole and bond dipole are in opposite directions
(3) in NH3 the atomic dipole and bond dipole are in the opposite directions whereas in NF3, these are in the same direction
(4) in NH 3 as well as in NF 3, the atomic dipole and bond dipole are in the same direction
48. Given below are two statements.
Statement-I: The oxygen-oxygen bond length in Ozone molecule is intermediate between a single and double bond.
Statement-II: Resonance averages the bond characteristic as a whole.
In light of the above 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.
ANSWER KEY
NEET Drill
– 1
Level – 2
2 (2) 1 (3) 1 (4) 3 (5) 2
Further Exploration
2 (2) 3 (3) 2 (4) 1
Matching Type Questions
(3) Statement I correct but statement II is incorrect.
(4) Statement I incorrect but statement II is correct.
49. How many resonance structures are possible for perchlorate ion, [ClO 4–]?
(1) 2 (2) 3 (3) 4 (4) 1
50. Identify the species having one p – bond and maximum number of canonical forms from the following:
