RANKERS CHEM ADV

CHEMISTRY Chemical Bonding

ADVANCED XI

CHEMICAL BONDING

BUILDING UPON PRIOR KNOWLEDGE

The following concepts are important for understanding the fundamentals of chemical bonding and should be reviewed first

1. Electronic configuration of elements : specially Non-metals

2. Orbital shapes : with wave Function diagram

3. Knowledge of EN

4. Lone pairs

5. Bond length

6. Valency Expansion.

7. Periodicity of atomic radius.

• Chemical Bond

• Ionic Bond

• Polar and Non Polar Covalent Bond

• VSEPR

• Valence Bond Theory

• Hybridisation

• Molecular Orbital Theory (MOT)

• Hydrogen Bond

1. ELECTRONIC THEORY OF VALENCY

➢ The theory of valency explains chemical combination in terms of electrons.

➢ The theory was developed independently by W. Kossel and G.N. Lewis (1916) and extended by Irving Langmuir (in 1919).

➢ The chemical behaviour of an atom is determined to a large extent by the number and arrangement of electrons in the outer orbitals of the atom.

➢ Only outer electrons are involved in chemical combination and so these are called the valence electrons.

2. COMPLETED ELECTRON OCTET OR DUPLET

➢ Group 0 of the periodic table contains the noble gases. With the exception of helium which has a 1s2 electron arrangement others have ns2 np6 configuration in the outer orbitals.

➢ Since the atoms of the noble gases were not known to form chemical bonds, it was argued that the presence of 8 electrons (an electron octet) in the valence shell makes the atom stable. Therefore all other atoms must undergo bonding by gaining or losing or sharing electrons so as to acquire the electronic configuration of the nearest inert gas. The presence of 8 electrons gives the name octet rule to this concept. In the case of the first few elements such as hydrogen, lithium and beryllium the atoms combine in such a way as to attain the stable structure of helium with 2 electrons (duplet) in its only one valence shell. There are, however, many exceptions to the octet rule. Also compounds of noble gases, especially xenon, have been synthesized. The various types of chemical bonds are discussed below.

2.1 Ionic Bond (or Electrovalent Linkage)

➢ An ionic bond is formed by the complete transfer of electron(s) from one atom to another. Atoms of metals generally lose electrons and those of non-metals gain electrons.

2.1.1 Formation of Sodium Chloride, NaCI

➢ A sodium atom (Z = 11) transfers its valence electron to a chlorine atom (Z = 17). The sodium atom by losing an electron acquires the electronic configuration of neon (1s22s22p6) and becomes sodium ion Na+ carrying a unit positive charge. The chlorine atom by gaining an electron acquires the stable configuration of argon (1s22s22p63s23p6) and becomes a chloride ion, Cl , with a unit negative charge.

➢ The transfer of electron results in the formation of the ionic bond.

Na + .Cl: → Na+ + :Cl: .. .. .. ..

[Ne]3s1 [Ne]3s23p5 [Ne] [Ar]

➢ Here we have used Lewis dot symbols in which the symbol of an element is surrounded by dots (or crosses) to represent electrons in the outermost (valence) shell. Formulae of compounds using Lewis symbols are called Lewis formulae.

➢ When atoms form a bond by electron transfer, the number of electrons lost and gained must be equal, because the resulting ionic compound is neutral.

➢ The number of electrons lost or gained by an atom in the formation of an ionic bond is its valency. Thus, Na and Cl have a valency of 1.

➢ Loss of electron is called oxidation; thus Na is oxidised to Na+. The gain of electron is reduction; thus Cl is reduced to Cl . Formation of an ionic bond from elements is an oxidation - reduction or redox reaction. Generally the metal is oxidised and the non-metal is reduced.

➢ Na+ and Ne are isoelectronic: since they contain the same number of electrons. Similarly Cl and Ar are isoelectronic.

➢ Because Na+ and Cl carry opposite charges, electrostatic forces of attraction hold them together Sodium chloride may be represented as Na+CI

2.1.2 Formation of Magnesium Oxide, MgO

Mg: + O: → Mg2+ + :O: .. .. 2

[Ne] 3s2 [He]2s22p4 [Ne] [Ne]

2.1.3 Conditions for the Formation of Ionic Bond

➢ The difference between the electronegativity of two combining atoms must be greater than two.

(i) Low Ionization Energy of the Metal:

Low ionization energy means that the metal atom requires only a small amount of energy to release its valence electron. For example, sodium, which has a low I.E. readily gives up its loosely held electron and forms Na+ ion. Metals of s-block have low ionization energies and so readily form the corresponding cations.

Ionization energy of an element with a single electron in its valence shell is less than that with two electrons. In going across a period of the periodic table from left to right, I.E. increases and the formation of the cation is less likely. On-going down a group, the outermost electron gets further away from the nucleus, and hence is more easily removed i.e., I.E. decreases; the formation of the cation becomes more likely.

(ii) High Electron Affinity of the Non-Metal:

An atom with a high electron affinity releases a lot of energy when it takes up an electron and forms an anion. For example, chlorine which has a high electron affinity, readily takes up an electron from the Na atom and forms Cl ion. Non-metals of groups VI A and VII A have high electron affinity and can form ionic bonds.

In going across a period from left to right, electron affinity (energy released) increases and so the formationofthe negative ion becomesmore likely. On-going downa group, electron affinity decreases and so the formation of anion becomes less likely.

(iii) High Lattice Energy of the Crystal:

In the formation of sodium chloride crystal, the Na+ ion attracts the Cl ion to form an ion-pair Na+Cl . Since the electrostatic force of attraction is present in all directions, this ion-pair will attract other ionpairs and build up into a crystal lattice. A crystal lattice is three dimensional basic patterns of points, in which each point corresponds to a unit of the crystal, say an ion (atom or molecule). As the lattice builds up, energy is released. The energy released when sufficient number of cations and anions come together to form 1 mole of the compound is called the lattice energy of the compound. Therefore, an ionic compound is formed when the energy released exceeds the energy absorbed.

Know Yourself

The formation of O2– ion is accompanied by considerable absorption of energy [U] O(g) + 2e →O2 ; H = + 844 kJ Still MgO is readily formed Explain.

2.1.4 General Characteristics of Ionic Compounds

➢ Generally ionic compounds are hard solids. As single ions of a metal are not associated in the solid with single ions of a non-metal, separate units of ionic compounds do not exist. It is, therefore, wrong to talk of a molecule of an ionic compound. The formula only indicates the ratio of number of ions and the crystal consists of a very large number of oppositely charged ions. Thus in NaCl crystal each Na+ ion is surrounded by 6Cl ions and vice versa (in an octahedralarrangement).TheattractionbetweenNa+ and Cl ions is quite large.

➢ As a good deal of thermal energy is required to overcome the large electrostatic forces of attraction in an ionic crystal, ionic compounds have high melting and boiling points.

➢ Ionic compounds are commonly soluble in water and other polar solvents (which separate the ions). They are practically insoluble in organic solvents such as benzene, carbon tetrachloride, etc., as there is no attraction between ions and the molecules of the non-polar liquids.

➢ Ionic compounds are electrolytes. In the presence of an ionizing solvent such as water, the electrostatic forces between the ions are so greatly reduced that the ions get separated. (This is due to the electrostatic attraction between the ions and the polar molecules of the solvent.) The free ions in solution conduct electricity and on passing a current, the ionic compound undergoes chemical decomposition (called electrolysis). When an ionic compound is melted, the crystal lattice structure is broken and free ions are produced. It is the free movement of ions, which makes an ionic compound a conductor and to undergo electrolysis in the molten condition.

➢ When an ionic compound dissolves in water, the ions get solvated (in this case hydrated). The energy released is called solvation energy. This energy counters wholly or in part the high lattice energy of the ionic compound. Insoluble ionic compounds (eg., sulphates, phosphates and fluorides of Ca, Sr and Ba) have very high lattice energies and the solvation energy of the constituent ions is insufficient to counteract the high lattice energies and make them soluble.

➢ The chemical properties of an ionic compound are the properties of its constituent ions. Thus all chlorides give the characteristic reactions of the chloride ion (reactions with conc. H2SO4, AgNO3 solution, etc). All acids, which contain H+ ions give the same reactions (change blue litmus to red, effervesce with a carbonate, etc).

➢ Reactions between solutions of ionic compounds are almost instantaneous, because they are reactions between ions (and do not involve the breaking up of bonds as in covalent compounds, q.v.). For example, when silver nitrate solution is added to sodium chloride solution, silver chloride is immediately precipitated. The reaction may be represented thus:

➢ Na+ + Cl + Ag+ 3NO → AgCl + Na+ + 3NO .

Know Yourself

2. Ionic fluorides outnumber ionic iodides. Explain [A]

2.2 Covalent Bond

➢ A covalent bond is formed by the sharing of a pair of electrons between two atoms, each atom contributing one electron to the shared pair. The shared pair of electrons should have opposite spins and they are localized between the two nuclei concerned. A covalent bond is usually represented by a shortline(i.e.,adash)betweenthetwoatoms.Notethatthecovalent bondconsistsofapairofelectrons shared between two atoms, and occupying a combination of two stable orbitals, one of each atom; the shared electronsof each covalent bondarecounted foreach ofthetwoatomsconnectedbythecovalent bond.

➢ The difference between the electronegativities of the combining atoms is less than two.

2.2.1 Formation of the Hydrogen Molecule

➢ Each hydrogen atom requires 1 electron to become isoelectronic with helium, the nearest inert gas. The hydrogen atoms share their electrons thus:

H + H → H H or H H . . . .

➢ Once the covalent bond is formed, the two bonding electrons are attracted by the two nuclei (instead of one) and the bonded state is more stable than the non-bonded state. The resultant attraction is responsible for the strength of the covalent bond.

2.2.2 Formation of Hydrogen Fluoride

➢ The hydrogen atom has in its orbital 1 electron. It can achieve the helium configuration by forming a single covalent bond with another atom. Fluorine has 7 electrons in its outer, i.e., L shell. F can acquire the Neon configuration by forming a single covalent bond using its unpaired electron. This may be represented as follows. H + F: → H :F: or H F

➢ The single covalent bond holds the H and F atoms firmly together. Similarly we can explain the formation of HCl, HBr and HI.

2.2.3 Formation of Water, H2O

:O: + 2H → H .. :O: .. H or

H | :O .. H

➢ Similarly we can explain the formation of H2S (hydrogen sulphide), H2Se (hydrogen selenide) and H2Te (hydrogen telluride).

2.2.4 Formation of Ammonia, NH3 ...... . .N3HH:N:HorHNH .... | H H

+⎯⎯→−−

➢ The structures of phosphine (PH3), arsine (AsH3) and stibine (SbH3) are similar to that of ammonia.

2.2.5 Formation of Carbon Tetrachloride, CCI4

2.2.6 Formation of Methane, CH4

2.2.7 Formation of Ethane, C

➢ The number of electrons needed by an atom to acquire its octet (C

4, N

3, O

2, Cl

1) is equal to the number of covalent bonds commonly formed.

➢ When two pairs of electronsare shared betweentwo atoms, there is a double bond as in ethylene, C2H4

➢ When three pairs of electrons are shared between two atoms, there is a triple bond as in acetylene, C2H2

H C C H or H C  C H . . . . . . . .

➢ Generally all atoms involved in covalent bonding have completed octets (except hydrogen, which has a duplet of electrons). Sometimes an atom forms more than 4 covalent bonds.

➢ An example is phosphorus pentachloride, PCI5 In this molecule phosphorus atom is surrounded by 5 chlorine atoms, with each of which it forms a covalent bond. In this compound, the phosphorus atom seems to use 5 of the nine orbitals of the M shell (rather than only 4 of the most stable orbitals). It seems likely that of the nine or more orbitals in the M, N and O shells, four are especially stable, but one or more others may be occasionally utilized.

Another example is SF6

2.2.8 General Characteristics of Covalent Compounds

➢ In a purely covalent compound, the electrons in the bond are shared equally between the atoms linked by the bond; the resultant particles formed are not electrically charged. So, separate molecules of the covalent compounds exist. Covalent compounds may therefore be expected to be gases or low boiling liquids or soft, low melting solids at ordinary temperature. In the solid state, they may be amorphous or present as molecular crystals, the molecules being held together by weak van der Waals’ forces of attraction.

➢ Since the molecules are held together by weak van der Waals’ forces, covalent compounds (except those consisting of giant molecules) have low melting and boiling points; very little thermal energy is needed to overcome these weak intermolecular forces.

➢ They are non-electrolytes, i.e., they do not contain ions.Even in giant molecules such as diamond there are no free electrons. So, they are very poor conductors of electricity.

➢ They are generally soluble in organic (non-polar) solvents such as benzene or carbon tetrachloride but are insoluble in water or other ionizing solvents. (The solubility of covalent compounds is very much dependent onthe size of themolecules; giant molecules are practically insoluble in nearly all solvents.)

➢ Reactions between covalent compounds are slow and often incomplete and reversible. This is so because the reaction involves breaking and making of bonds i.e., energy considerations are involved for reactants, activated complexes and products.

➢ A covalent bond is a space-directed bond and it may exhibit isomerism.

2.2.9 Polar Covalent Bonds Electronegativity

➢ The shared pair of electrons may be shared equally between two atoms; then the covalent bond is said to be non-polar. Equal sharing occurs between identical atoms, as in H H or Cl Cl (i.e., in homonuclear molecules) or between identical atoms with identical neighbours as in H3C CH3. When the two bonded atoms are dissimilar (i.e., in heteronuclear molecules) the sharing is unequal. For example, a chlorine atom has a greater electron attracting power than a hydrogen atom; so in H Cl, the shared pair of electrons are drawn more towards chlorine and away from hydrogen. The result is separation of charges within the molecule, the chlorine end acquiring a slight negative charge and the hydrogen end a slight but equal positive charge:

➢ H + Cl −

. Such covalent bonds are said to be polar (i.e., bonds formed by sharing a pair of electrons between two atoms but displaced towards the nucleus of one of the bonded atoms).

➢ Thenettendencyofabondedatom inacovalentmoleculetoattract thesharedpairofelectronstowards itself is known as electronegativity. Thus, F is highly electronegative, but F , which has already an extra electron, is not. Electronegativities

➢ To assess the tendency of an atom of a given element to attract electrons towards itself in a covalent bond, relative electronegativity values are used. Above table gives the relative electronegativity values of atoms calculated by Pauling (adopting arbitrarily the value of 4 units for the electronegativity of F).

Electronegativity values increase across a period and decrease down a group.

Smaller atoms have greater electronegativity than larger ones and so they attract electrons more towards them than larger ones. Alkali metals have low electronegativities and halogens high electronegativities.

Atoms with nearly filled shells of electrons (e.g., halogens) have greater electronegativity than those with sparsely occupied shells.

Elements with low electronegativity values such as Cs (0.8) and Rb (0.8) tend to form positive ions, i.e.,thesearemetals.Elementswithhighelectronegativity values suchas F(4.0) andO(3.5)tend to form negative ions, i.e., these are non-metals.

Electronegativity value may be used to make rough predictions of the type of bonding to be found in a compound. The larger the difference between electronegativity values of two combining atoms, the morepolarthecovalentbond.Ifthedifferenceisgreaterthan2,thegreaterthechanceforionicbonding (i.e., the chance of covalent bond assuming 100% ionic character). From this point of view ionic bond may be considered to be an extreme case of a polar bond (with total separation of charges).

If the difference between the electronegativities of the combining atoms is zero or small, the bond is essentially non-polar.

Let XA and XB represent the electronegativities of two atoms A and B. If XB XA = 1.7, the covalent bond A B is said to have 50% ionic character. On this basis, the % ionic character in some typical bonds is calculated (Table 3.2b). These calculations are very qualitative.

2.3 Lewis Structures of Molecules

➢ The formula of a molecule shows the number of atoms of each element but does not show the bonding arrangement of the atoms.

➢ To represent the bonding pattern in a molecule, the electron dot symbols of the elements are arranged such that the shared pairs and unshared pairs (called lone pairs) are shown and the octet rule (or duet for hydrogen) is satisfied. For example, a molecule of fluorine is shown as : F : F : .. ..

.. .. and a molecule of hydrogen fluoride is shown as H : F : or H F : .. ..

.. ..

.. .. or : F F : .. ..

➢ Arrangement of dot symbols used to represent molecules are called Lewis structures.

➢ Lewis structures do not convey any information regarding the shape of the molecule.

➢ Usually, the shared pairs of electrons are represented by lines between atoms and any unshared pairs are shown as dot pairs.

➢ Lewis structures are written by fitting the element dot symbols together to show shared electron pairs and to satisfy the octet rule. For example,

In water (H2O), one H . and two O .. . . : complete their duet and octet respectively as

In ammonia (NH3), three H . and one N .. . . . fit together and satisfy their duet and octet respectively as

In carbon tetrachloride (CCl4), four : Cl .. .. . and one C . . . . complete their octet as

➢ For the given molecules, we have adopted hit & trial method to fit the dot symbols together and satisfy the octet rule. But remember that hydrogen form one bond, oxygen forms two bonds, nitrogen three bonds and carbon forms four bonds.

➢ For simpler molecules, the hit & trial method works perfectly but for slightly complicated polyatomic species, this may give us more than one possible structure.

➢ Thus, a systematic approach is needed to design the Lewis structures of such polyatomic species. But before proceeding further, let us understand the limitation of this approach.

2.3.1 Limitations of Lewis Theory of Drawing Structure:

➢ This method would be applicable to only those molecules/species, which follow octet rule except hydrogen.

➢ There are three kinds of molecules/species, which do not follow octet rule.

• Molecules, which have contraction of octet. Such molecules are electron deficient. For example, BH3, BF3, BCl3, AlCl3, GaCl3 etc.

• Molecules, which have expansion of octet. Such species have more than eight electrons in their outermostshell.Thisispossibleinthosemolecules,whichhavevacant d-orbitals,thustheycanexpand their octet. For example, PCl5, SF6 etc.

• Molecules containing odd number of electrons (in total) cannot satisfy octet rule. Such species are called odd electron species and are paramagnetic in nature due to presence of unpaired electron. For example, NO, NO2 and ClO2.

2.3.2 Method of Drawing Lewis Structures:

➢ To draw the Lewis structures of polyatomic species, follow the given sequence.

First calculate n1

n1 = Sum of valence electron of all the atoms of the species  net charge on the species.

Foranegativelychargedspecies,electronsareaddedwhileforpositivelychargedspecies,the electrons are subtracted. For an uninegatively charged species, add 1 to the sum of valence electrons and for a dinegatively charged species, add 2 and so on.

Then calculate n2.

n2 = ( 8  number of atoms other than H) + (2  number of H atoms)

Subtract n1 from n2, which gives n3.

n3 = n2 n1 = number of electrons shared between atoms = number of bonding electrons.

3 21 n nn 22 = = number of shared (bonding) electron pairs = number of bonds.

Subtracting n3 from n1 gives n4.

n4 = n1 n3 = number of unshared electrons or non-bonding electrons.

4n 2 = 13nn 2 = number of unshared electron pairs = number of lone pairs.

Identify the central atom. Generally, the central atom is the one, which is least electronegative of all the atoms, when the other atoms do not contain hydrogen. When the other atoms are hydrogen only, then the central atom would be the more electronegative atom. However some exceptions are possible, for example Cl2O.

Now around the central atom, place the other atoms and distribute the required number of bonds (as calculated in step (iii)) & required number of lone pairs (as calculated in step (iv)), keeping in mind that every atom gets an octet of electrons except hydrogen.

Then calculate the formal charge on each atom of the species.

Formal charge on an atom = number of valence electrons of the atom number of bonds formed by that atom number of unshared electrons (2  lone pairs) of that atom.

When two adjacent atoms get opposite formal charges, then charges can be removed by replacing the covalent bond between the atoms by a dative (co-ordinate) bond. This bond will have the arrowhead pointing towards the atom with positive formal charge. It is not mandatory to show the dative bonds unless required to do so.

The given Lewis structure should account for the factual aspects of the molecule like resonance (delocalization), bond length, p d back bonding etc.

Sometimes, there are more than one acceptable Lewis structure for a given species. In such cases, we select the most plausible Lewis structure by using formal charges and the following guidelines:

Key Concept

• For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.

• Lewis structures with large formal charges (+2, +3 and/ or 2, 3 and so on) are less plausible than those with small formal charges

• Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms

Illustration-1

Determine Lewis structure of 3NO ion. Solution: (i)

3) + 1 = 24

 Number of bonds = 8 2 = 4

(iv)

4 = n1 n3 = 24 8 = 16

 Number of lone pairs = 16 2 = 8

(v) Nitrogen is the central atom (as it is less electronegative than O). Arranging three O atoms around it and distributing 4 bonds and 8 lone pairs as

(vi) Calculating formal change on each atom.

Formal charge on N = 5 4 0 = + 1

even accounts for resonance in 3NO ion.

Illustration-2

Determine Lewis structure of CN ion.

 Number of bonds = 6 2 = 3

= 10 6 = 4

 Number of lone pairs = 4 2 = 2

(v) Carbon is the central atom (C is less electronegative than N) and arrange N, number of bonds and number of lone pairs around it as

C N: :

(vi) Formal charge on C = 4 3 2 = 1, Formal charge on N = 5 3 2 = 0 Thus, final Lewis structure of CN would be

C N: :

Illustration-3

Draw Lewis structure for + 4NH ion.

Solution:

(i) n1 = 5 + (4  1) 1 = 8

(ii) n2 = (8  1) + (2  4) = 16

(iii) n3 = n2 n1 = 16 8 = 8

 Number of bonds = 8 2 = 4

(iv) n4 = n1 n3 = 8 8 = 0

 Number of lone pairs = 0

(v) Nitrogen being the central atom, distributing other atoms (H) around it, and 4 bonds with the 4 H atoms, the structure looks like

(vi) Formal charge on N = 5 4 0 = + 1

Formal charge on H(a)/H(b)/H(c)/H(d) = 1 1 0 = 0

Thus, final Lewis structure of + 4NH would be

Know Yourself

Draw Lewis structures for the following molecules/species [U]

2.4 Dipole Moment

➢ A dipole consists of a positive and an equal negative charge separated by a distance within a molecule.

➢ The degree of polarity of a bond is given by the dipole moment(), whichistheproduct ofeithercharge (e)andthedistance(d)betweenthem.

➢  = d  e where ‘e’ is of the order of magnitude of the electronic charge, i.e., about 10 10 esu and d is the distance between the atomic centres, i.e., about 10 8 cm.

➢ Dipole moments may be expected to have values around 10 10 x 10 8 = 10 18 esu-cm.

➢ It is a general practice to express dipole moments in Debye units (D), 1 D = 10 18 esu-cm.

Electric pole d

Electric pole

➢ If the charge is in SI units (Coulombs) and d in metre,  will be coulomb-metre (C  m) units.

1D = 3.336  10 30 C m.

➢ Any covalent bond which has a certain degree of polarity will have a corresponding dipole moment, though it does not follow that compounds containing such bonds will have dipole moments, for the polarity of the molecule as a whole is the vector sum of the individual bond moments. For example, CO2 has zero dipole moment, although the C = O bond is a polar bond. This shows that CO2 is a linear molecule, O = C = O, so that the dipole moments of the two C = O bonds cancel out.

The C → Cl bond has a definite polarity and a definite dipole moment but carbon tetrachloride has zero dipole moment because it is a tetrahedral molecule, and the resultant of the 4 C – Cl bond moments is zero.

On the contrary, CH3Cl, CH2Cl2 and CHCl3 have definite dipole moments.

2.4.1 Applications of Dipole Moment Measurements:

➢ Dipole moment is a measure of the electrical dissymmetry (polarity) in the molecule and so its measurement provides valuable information concerning the shapes of molecules. Conversely, when the symmetry of the molecules is known, dipole moment could be estimated fairly.

(A) Inorganic substances:

Monatomic molecules: such as He, Ne, etc., have zero dipole moment because they are symmetrical. Diatomic molecules: such as H2, Cl2 and N2 have no dipole moment; so these molecules are symmetrical.

Triatomic molecules: Some of these molecules possess zero dipole moment; so they have a symmetricallinearstructure,e.g.,CO2,CS2,HgCl2.Otherslikewater andsulphurdioxidehavedefinite dipole moments. They are said to have angular or bent structure or V-shaped structure.

Tetratomic molecules: Some molecules like BCI3 have zero dipole moment. They are said to possess a flat and symmetrical (triangular) structure; other examples are BF3, BBr3, 2 33 COandNO

..

PCl3,AsCl3,NH3,PH3,AsH3,H3O+ haveappreciabledipolemoments.Theypossesstrigonalpyramidal structures. P Cl Cl Cl

Illustration-4

Both CO2 and N2O are linear but dipole moment of CO2 in zero but for N2O it is non-zero, why?

Solution:

The answer lies in the structure of these molecules. CO2 is a symmetrical molecule while N2O is unsymmetrical. Thus for N2O, dipoles do not cancel each other, leaving the molecule with a resultant dipole, while the bond moment of CO2 cancel each other, so CO2 has no net dipole moment.

N  N → O ; O = C = O

Illustration-5

Compare the dipole moment of NH3 and NF3.

Solution:

Let’s draw the structure of both the compounds and then analyse their dipole directions.

The structure of both NH3 and NF3 are pyramidal with three bond pairs and one lone pair. In NH3, as N is more electronegative than hydrogen, so the resultant bond dipole is towards N, which means that both the lone pair and bond pair dipoles are acting in the same direction and are summed up. In case of NF3, the bond dipole (of N F bonds) is acting towards fluorine, (as fluorine is more electronegative than N) so in NF3 the lone pair and bond pair dipoles are acting in opposition, resulting in a decreased dipole moment. Thus, NH3 has higher dipole moment than NF3

(B) Organic Substances

Methane and CCl4 have zero dipole moment. So theypossess symmetrical tetrahedral structures with C atom at the centre of the tetrahedron.

Benzene has zero dipole moment. All the 6 C and 6 H atoms are assumed to be in the same plane (symmetrical hexagonal structure).

Measurement of dipole moments will enable us to detect cis-and trans-isomers of organic compounds. The trans isomer, which is symmetrical, has zero dipole moment while the cis isomer has a definite dipole moment.

cis-dibromoethylene ( = 1.4 D)

Dipole moment in aromatic ring system

trans-dibromoethylene ( = 0)

The dipole moments of the aromatic compounds present a very good illustration of dipole moment. We know that when substituted benzene is treated with reagent different products (namely ortho, meta and para products) areformed. The dipolemoments ofthese products are different sincetheorientation of the groups is different.

Let us take an example to clarify it. Let us take three isomers, o-nitrophenol,m-nitrophenol and pnitrophenol. We have also have three other isomers o-aminophenol, m-aminophenol and paminophenol. We want to arrange these isomers in the order of their dipole moments.

In those cases where X = Y, the para isomer becomes symmetrical and have zero dipole moment. In order to find their dipole moment, we need to know about the nature of the groups linked to the benzene ring. In nitro phenols, one group (OH) is electron pushing and the other (NO2) is electron withdrawing while in aminophenols, both the groups (OH and NH2) attached are electron pushing. So, depending on the nature of the groups attached, the isomers have different dipole moment. Then calculation of dipole moment follows as:

Case-1:

➢ When X and Y both are electron pushing or electron withdrawing:

➢ Let the bond dipole of C X bond is represented by 1 and that of C Y bond by 2. Now let us assume that the electron pushing groups have +ve bond moment and the electron withdrawing groups have ve bond moment. The net dipole moment is the resultant of two bond dipoles at different orientations.

22o meta12122cos120=++

➢ From the above expressions of 0, m and p, it is clear that when both X and Y are of the same nature i.e., both are electron withdrawing or both are electron pushing the para product has the least dipole moment and ortho product has the highest dipole moment. When X = Y, 1 = 2, thus p would be zero.

Case-2:

➢ When X is electron pushing and Y is electron withdrawing or vice versa.

➢ Let the bond moment of C X dipole is 1 and that of C Y dipole is 2.

22o o1212()2()cos60=+−+−

2 o1212 ()3 =+−

()2()cos120

➢ Looking at the expressions of O, m and p, it is clear that the para isomer has the highest dipole moment and ortho has the least.

Know Yourself

1. Benzene forms 3 different dinitrobenzenes. Their dipole moments are (a) 3.90 (b) 0 and (c) 6.00 D respectively. Classify them as ortho, meta and para [U]

2. p dichlorobenzene is non polar while p dihydroxybenzene is polar. Explain [U]

2.4.2 Dipole Moment and Percentage Ionic Character:

➢ The measured dipole moment of a substance may be used to calculate the percentage ionic character of a covalent bond in simple molecules.

1 unit charge = Magnitude of electronic charge = 4.8  10 10 e.s.u.

1 D = 1  10 18 e.s.u cm.

 % ionic character = Observeddipolemoment 100 Theoreticaldipolemoment 

➢ Theoretical dipole moment is confined to when we assume that the bond is 100 % ionic and it is broken into ions while observed dipole moment is with respect to fractional charges on the atoms of the bond.

2.5 Transition From Ionic to Covalent Bond – Fajans’ Rule

➢ Just as a covalent bond may have partial ionic character, an ionic bond may also show a certain degree of covalent character.

➢ When two oppositely charged ions approach each other closely, the cation would attract the electrons in the outer shell of the anion and simultaneously repel its nucleus.

➢ This produces distortion or polarization of the anion, which is accompanied by some sharing of electrons between the ions, i.e., the bond acquires a certain covalent character.

➢ The formation of a covalent bond between two ions may be illustrated with reference to formation of AgI.

2.5.1 Factors Influencing Ion Deformation or Increasing Covalent Character:

(i) Large charge on the ions:

The greater the charge on the cation, the more strongly will it attract the electrons of the anion. For example, Al3+ can distort Cl ion more than Na+ ion. So aluminium chloride is a covalent compound whereas NaCl, AlF3, AgF are ionic.

(ii) Small cation and large anion:

For a small cation, the electrostatic force with which its nucleus will attract the anion will be large. Moreover a large anion cannot hold the electrons in its outermost shell, especially when they are attracted by a neighbouring cation. Hence there will be increased covalence with a small cation and a large anion, as in AgI

(iii) Cation with a non-inert gas type of electronic configuration:

A cation with a 18 electron outermost shell such as Ag+ ([Kr] 4d10) polarizes anions more strongly than a cation with a 8 electron arrangement as in K+. The ‘d’ electrons in Ag+ do not screen the nuclear charge aseffectivelyas the‘s’ and‘p’electronshell inK+. Thus AgIismorecovalent than KI, although Ag+ and K+ ions are nearly of the same size. Cuprous and mercurous salts are covalent. The above statements regarding the factors, which influence covalent character, are called Fajans’ rules. It can thus be seen easily that there is nothing like a purely ionic compound or a purely covalent compound.

2.6 Co-Ordinate Covalent Bond or Dative Bond

➢ We have seen that in the formation of a covalent bond between two atoms, each atom contributes one electron to the shared pair. Sometimes both the electrons of the shared pair may come from one of the atoms. The covalent bond thus formed is a co-ordinate bond or dative bond

(i) Formation of Ammonium ion

The ammonia molecule has a lone pair of electrons i.e., an unshared pair. The hydrogen ion H+, has an empty s orbital. The lone pair comes to be shared between the nitrogen and hydrogen atoms:

Nitrogen atom is called the donor and H+, the acceptor. The arrow-head in N → H shows that Natom is electron donor and H–atom is electron acceptor. NH3 is a neutral molecule. H+ carries a unit

positive charge; so 4NH+ ion carries a unit positive charge. Once the 4NH+ ion is formed, all the N H bonds become identical.

(ii) Hydronium ion, H3O+

(iii) Aluminium chloride, Al2Cl

(iv) Nitromethane, CH

2.6.1 General Characteristics of Coordinate Covalent Compounds

➢ As is to be expected the properties of coordinate covalent compounds are mostly similar to the properties of covalent compounds.

➢ The nuclei in coordinate covalent compounds (such as in + 4NH ) are held firmly by shared electrons and so do not form ions in water.

➢ Their covalent nature makes them sparingly soluble in water and more soluble in organic solvents.

➢ The coordinate bond is also rigid and directional, just like covalent bonds.

3. RESONANCE

➢ Carbon dioxide may be represented by Lewis dot formula as :O::C::O: or O = C = O …(i)

➢ The bond length of C = O is 1.22 Å, but the actual measured value is 1.15 Å. Further CO2 is quite stable and does not show the characteristic reactions of the carbonyl group, as shown by aldehydes and ketones.Without shifting, the relative positions of atoms of CO2 canberepresented bytwo more Lewis formulae:

O C O O C O

(ii) (iii)

➢ In (ii) and (iii), the two bonds between C and O are different, one being a triple bond and the other a single bond. Both the C O bonds in CO2 are identical. It is now obvious that none of these structures actually represents CO2. To explain this difficulty the concept of resonance was introduced, according to which CO2 cannot be accurately depicted by any Lewis formula. The actual structure of CO2 is a resonance hybrid of the three structures:

O = C = O  → O C → O  → O  C O

➢ These different structures are called the canonical or contributing structures. The actual structure of CO2 is different from the canonical structures and although it is closely related to them, the actual structure cannot be represented on paper using the accepted symbols. All the molecules of CO2 have the same structure. Usually, a double headed arrow  → is used between the canonical structures.

3.1 Conditions for Resonance

➢ Resonance can occur when the canonical structures have the constituent atoms in the same relative positions; have the same number of unpaired electrons but differ in the distribution of electrons around the constituent atoms; molecules or ions are planar.

3.2 Resonance Energy

➢ The resonance hybrid is a more stable structure than any of the contributing structures. This means that resonance hybrid has less energy than any of the contributing structures. The difference in energy between the actual observed energy of the resonance hybrid and the most stable of the contributing structures is called resonance energy.

➢ For CO2, structure (i) has less energy than structure (ii).

3.3 Other Examples of Resonance

➢ Phosphate ion, 3 4PO

Bond order = Total No.ofbondsformedbetweentwoatomsinallstructures

Total No.ofresonatingstructures

= 1211 4 +++ = 5/4

➢ Benzene, C6H6. It is a resonance hybrid of the following structures (hexagonal, planar).

C–C bond length is 1.54 Å; C=C bond length is 1.34 Å. In benzene, all the C–C bonds are identical in length, 1.39 Å, i.e., intermediate between those of single and double bonds.

Note that shortening of bond length and therefore increased stability is an indication of the existence of resonance.

Resonance energy of benzene is –152 kJ/mol.

4. VBT, OVERLAPPING OF ORBITALS

4.1 Modern Concept of Covalent Bond (VBT):

➢ Introduced by Heitler and london.

➢ This theory is bassed on the knowledge of atomic orbitals electronic configuration of electrons, the overlap criterion of atomic orbitals, the hybridization of atomic orbitals and the principles of variation and superposition.

➢ Consider two hydrogen atoms A and B approaching each other having nuclei HA and HB and electrons present in them are represented by EA and EB.

➢ When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms approach each other, new attractive and repulsive forces begin to operate.

➢ Attractive forces arise between:

nucleus of one atom and its own electron that is HA – EA and HB – EB

nucleus of one atom and electron of other atom i.e., HA – EB. HB– EA.

➢ Similarly repulsive forces arise between :

electrons of two atoms like EA – EB

nuclei of two atoms HA – HB

➢ 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. Ultimately a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. 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, which is corresponding to minima in the curve depicted in Fig. Conversely. 435.8 kJ of energy is required to dissociate one mole of H2 molecule.

4.2 Orbital Overlap Concept

➢ In the formation of H2 Molecule, the minimum energy state when two H atoms are so near that their atomic orbitals undergoes partial interpenetration, this partial merging is called overlapping.

➢ According to orbital overlap concept, the formation of a covalent bond between two atoms results by pairing of electrons present in the valence shell having opposite spins.

➢ The extent of overlap decides the strength of a covalent bond.

➢ Greater the overlap, stronger is the bond formed between two atoms.

4.3 Directional Properties of Bonds

➢ The valence bond theory explains the formation and directional properties of bonds in polyatomic molecules like CH4, NH3 and H2O, etc. in terms of overlap and hybridisation of atomic orbitals.

4.4 Overlapping of Atomic Orbitals

➢ When two atoms come close to each other there is overlapping of atomic orbitals. This overlap may be positive, negative or zero depending upon the properties of overlapping of atomic orbitals.

➢ The various arrangements of s and p orbitals resulting in positive, negative and zero overlap are depicted in the figure given below.

➢ The criterion of overlap, as the main factor for the formation of covalent bonds applies uniformly to the homonuclear/heteronuclear diatomic molecules and polyatomic molecules.

Positive overlap

Negative overlap

Zero overlap

4.5 Types of Overlapping and Nature of Covalent Bonds

➢ The covalent bond may be classified into two types depending upon the types of overlapping : sigma() bond, and pi () bond

➢ Sigma () bond:

This type of covalent bond is formed by the end to end (head-on) overlap of bonding orbitals along the internuclear axis. This is called as head on overlap or axial overlap.

This can be formed by any one of the following types of combinations of atomic orbitals.

s-s overlapping:

In this case, there is overlap of two half-filled s-orbitals along the internuclear axis as shown below:

s-p overlapping:

This type of overlap occurs between half-filled s-orbitals of one atom and half-filled p-orbitals of another atom.

p-p overlapping: This type of overlap takes place between half-filled p-orbitals of the two approaching atoms.

➢ pi() bond:

In the formation of  bond, the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to sidewise overlapping consists of two saucer type charged clouds above and below the plane of the participating atoms.

4.6 Strength of Sigma and Pi Bonds:

➢ Basically the strength of a bond depends upon the extent of overlapping- In case of sigma bond, the overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent.

➢ It is important to note that pi bond between two atoms is formed in addition to a sigma bond. It is always present in the molecules containing multiple bond (double or triple bonds).

5. MOLECULAR GEOMETRY AND VSEPR THEORY

➢ Molecular geometry is the three-dimensional arrangement of atoms in a molecule.

➢ A molecule’s geometry affects its physical and chemical properties such as meltingpoint, boiling point and the types of reactions it undergoes.

➢ In general, bond length and bond angles are determined by experiments. However, there is a simple procedure to predict the overall geometry of a molecule or ion with considerable accuracy, if we know the number of electrons surrounding a central atom in its Lewis structure.

➢ The basis of this approach is the assumption that electron pairs in the valence shell (outermost electron occupied shell of an atom) of an atom repel one another.

➢ In a polyatomic species, the repulsion between electrons in different bonding pairs causes them to remain as far as possible.

➢ Thus, the geometry assumed by the species ultimately minimizes the repulsion. This approach is called valence shell electron pair repulsion (VSEPR) theory because it accounts for the geometric

arrangements of electron pairs around a central atom in terms of the electrostatic repulsion between electron pairs.

➢ Molecules in this theory are divided into two categories, depending on whether the central atom has lone pair of electrons or not.

5.1 Molecules In Which The Central Atom Has No Lone Pairs

➢ For simplicity, we will consider molecules that contain only two types of atoms, A and B, of which A is the central atom. These molecules have the general formula ABx, where x is an integer 2, 3 …..(if x = 1, the molecule will be diatomic, which is linear by definition).

(a) Molecules having general formula AB2

BeCl2 isrepresentingthegeneralformulaAB2.TheLewisstructureofberylliumchlorideinthegaseous state is : Cl Be Cl: .. .. .. 180°

Since the bonding 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, ClBeCl bond angle is predicted to be 180° and the molecule is linear.

(b) Molecules having general formula AB3

ThegeneralformulaAB3 isrepresentedbythemoleculeBF3.BF3 hasthreebondingpairs,whichpoints to the corners of an equilateral triangle with boron at the center of the triangle.

This geometry of BF3 is referred as trigonal planar with FBF bond angle to be 120°. In this structure, all four atoms lie in the same plane,

(c) Molecules having general formula AB4

Methane (CH4) represents the best example of this class of molecules. The Lewis structure of CH4 is

The four bonding pairs in CH4 can be arranged to form a tetrahedron, so as to minimize the repulsion between them. A tetrahedron has four faces, all of which are equilateral triangles. In a tetrahedral molecule, the central atom (carbon) is located at the center of the tetrahedron and the other four atoms (H) are at the corners. The HCH bond angles are all 109°28’.

(d) Molecules with general formula AB5

The general formula AB5 is represented by the molecule PCl5. The Lewis structure of PCl5 (in gas phase) is

The only way to minimize the repulsive forces among the five bonding pairs is to arrange the P Cl bonds in the form of a trigonal bipyramid. Joining two tetrahedrons along a common triangular base can generate a trigonal bipyramid.

The central atom (P) is at the center of the common triangular with the surrounding atoms positioned at the five corners of the trigonal bipyramid.

The atoms that are above and below the triangular plane are said to occupy axial positions and those, which are in the triangular plane, are said to occupy equatorial positions.

The angle between any two equatorial bonds is 120°, that between an axial bond and an equatorial bond is 90° and that between two axial bonds is 180°.

(e) Molecules having general formula AB6

The molecule SF6 exhibits the general formula AB6. The Lewis structure of SF6 is

The most stable arrangement of the six S F bonding pairs is in the shape of an octahedron.

An octahedron has eight faces and can be generated by joining two square pyramids on a common base.

The central atom (S) is at the center of the square base and the surrounding atoms (F) are at the six corners.

All bond angles are 90° except the one made by the bonds between the central atom and the pairs of atoms that are diametrically opposite to each other, which is 180°. Since, all the bonds are equivalent in an octahedral molecule, the terms axial and equatorial are not used here.

5.2 Molecules in Which the Central Atom Has One or More Lone Pairs

➢ In such molecules, there are three types of repulsive interactions between bonding pairs, between lone pairs and between a bonding pair and a lone pair.

➢ In general, according to VSEPR theory, the repulsive forces decrease in the following order: lone pair lone pair repulsion > lone pair bond pair repulsion > bond pair bond pair repulsion.

➢ Bond pair electrons are held by the attractive forces exerted by the nuclei of the two bonded atoms. These electrons have less “spatial distribution” than lone pairs i.e., they take up less space than lone pair electrons, which are associated with only one nuclei (or one atom). Because lone pair electrons

in a molecule occupy more space, they experience greater repulsion from neighbouring lone pairs and bonding pairs.

➢ To keep track of total number of bonding pairs and lone pairs, we designate molecules with lone pairs as ABxEy, where A is the central atom, B is the surrounding atoms and E is a lone pair on A. Both x and y are integers, x = 2, 3….. and y = 1, 2 ….. Thus, x and y denote the number of surrounding atoms and number of lone pairs on the central atom, respectively.

(a) Molecules with general formula AB2E

(b)

.. ..

VSEPR theory treats double bond and triple bonds as though they were single bonds. Thus, SO2 molecule can be viewed as having three electron pairs on the central atoms, of which, two are bonding pairs and one is a lone pair.

:O .. O: .. .. . . S 

This shape is referred as bent or angular.

The shape is determined only by the bonding pairs and not by lone pairs. Since lone pair repels the bonding pairs more strongly, the SO bonds are pushed together slightly and the OSO angle is less than 120°.

Molecules having general formula AB3E

The general formula AB3E is exhibited by the molecule NH3

Ammonia has overall four electron pairs, of which three are bonding pairs and one is lone pair. The overall arrangement of four electron pairs is tetrahedral but since one of the electron pairs is a lone pair, so the shape of NH3 is trigonal pyramidal.

. . N H H H

107.3°

Example of such a molecule is H2O. A water molecule has 2 bonding pairs and two lone pairs H O H . ..

..

The overall arrangement of the four electron pairs in water is tetrahedral. However, unlike NH3, H2O has 2 lone pairs on the central O atom. These lone pairs tend to be as far from each other as possible.

Consequently, the two OH bonding pairs are pushed toward each other and H2O shows even greater deviation from tetrahedral angle than in NH3

The shape of H2O is referred as bent or angular with HOH bond angle of 104.5°.

(d) Molecules having general formula AB4E

Example to this class of molecule is SF4. The Lewis structure of SF4 is

The S atom in SF4 has 5 electron pairs, which can be arranged as trigonal bipyramidal. In SF4, since one of the electron pair is a lone pair, so the molecule can have any one of the following geometries:

In (a), the lone pair occupies an equatorial position and in (b), it occupies an axial position. Repulsion between the electrons pairs in bonds only 90° apart are greater than repulsion between electron pairs in bonds 120° apart. Each axial bond has three electron pairs 90° away while each equatorial bond has only two electron pairs 90° away. Thus axial bonds (electron pairs) experience greater repulsion than the equatorial bonds.

Thus, atoms at the equatorial positions are closer to the central atom than atoms at the axial positions i.e. equatorial bond lengths are smaller than axial bond lengths.

Thus, when the central atom also has lone pairs along with the bonding pairs, it will occupy a position where the repulsions are less, so lone pairs in trigonal bipyramidal are more comfortable at equatorial positions. Thus, (a) is the appropriate structure of SF4. It is referred as see saw shaped or irregular tetrahedron.

(e) Molecules with general formula AB3E2

Example of this type is ClF3. The Lewis structure of ClF3 is

The Cl atom in ClF3 has 5 electron pairs, of which 2 are lone pairs and 3 are bonding pairs. The molecule can have any of the following three geometries:

In structure (a), there are 6 lone pair-bond pair repulsions at 90° and one lone pair-lone pair repulsion at 180°.

In structure (b), 1 lone pair-lone pair repulsion is at 90° and there are 3 lone pair bond pair repulsions at 90°, 2 at 120° and 1 at 180°.

While in structure (c), there are 4 lone pair bond pair repulsions at 90°, 2 at 120° and one lone pair lone pair repulsion at 120°.

The structure (b) is out rightly ruled out since the lone pair lone pair repulsion is of highest magnitude. Among structures (a) and (c), each structure has 4 lone pair bond pair repulsions at 90°. Apart from these repulsions, (a) has 1 lone pair lone pairrepulsion at 180° and 2 lone pair bond pair repulsions at 90° while (c) has 1 lone pair lone pair repulsion at 120° & 2 lone pair bond pair repulsions also at 120°. So, the structure (c) has overall lesser repulsions than (a). Thus, (c) is the appropriate structure of ClF3. It is called T shaped structure.

5.3 Prediciting Geometry of Species Using VSEPR Theory

➢ With the help of VSEPR theory, we can predict the geometry of various species in a systematic way. The scheme makes use of the following steps:

Identify the central atom and count the number of valence electrons on the central atom. Add to this, the number of other atoms (which form single bonds only). Here, oxygen atoms are not added as they form two bonds.

If the species is an anion, add negative charges and if it is a cation, subtract positive charges. This gives us a number, which we refer as N.

Divide N by 2 and we get the sum of bonding and non-bonding electron pairs.

N 2 = Number of other atoms + number of lone pairs.

withthevaluegiven in table, correspondingtothe givennumber of lone pairs.

Key Concept

XeF6 does not have octahedral structure. It’s structure is capped octahedron

Let us see the usefulness of the VSEPR theory to predict the geometry of few molecules/ions.

(i) BeCl2 Molecule:

The central atom is Be and it has two other Cl atoms.

N22 2 22 + ==

Since, the number of other atoms are 2, so the number of lone pairs are zero. Thus, shape of BeCl2 is linear.

Cl Be Cl

(ii) BF3 Molecule:

In BF3, central atom is boron and it has three other atoms.

N33 3 22 + ==

Since, the number of other atoms is three, so the number of lone pairs is zero. Therefore, shape is trigonal planar.

(iii) –2NO ion:

The central atom in 2NO is N and it has two other atoms.

N51 3 22 + ==

Since, the number of other atoms is 2, so the number of lone pairs would be 1. Thus, shape of 2NO ion is angular or bent.

(iv) 2 4BeF ion:

In 2 4BeF , the central atom is Be and it has four other F atoms.

N242 4 22 ++ ==

The number of lone pairs are zero, as the number of other atoms are 4. Therefore, shape of 2 4BeF is tetrahedral.

(v) NH3 Molecule:

In NH3, the central atom is N and it has 3 other H atoms.

N53 4 22 + ==

Since, the number of other atoms is 3, so the number of lone pairs would be 1. Therefore, shape of NH3 molecule is trigonal pyramidal.

(vi) H2S Molecule:

The central atom is S and there are 2 other H atoms in H2S molecule.

N62 4 22 + ==

Since, the number of other atoms is 2, so the number of lone pairs would be 2. Thus, the shape of H2S is angular or bent.

(vii) ClO ion:

The central atom is Cl and it has one other atom. 

N71 22 + = = 4

The number of lone pairs would be 3 as the other atom is only one. Thus, the shape of ClO is linear.

(viii) PCl5 Molecule:

The central atom in PCl5 is P and it has 5 other Cl atoms.

N55 5 22 + ==

The number of lone pairs would be zero, as the number of other atoms is 5. Thus, the shape of PCl5 is trigonal bipyramidal.

(ix) 4 I +F ion:

The central atom in 4FI + ion is I and it has 4 other F atoms. 

N741 22 +− = = 5

Since, the number of other atoms is 4, so the number of lone pairs would be 1. Thus, the shape of + 4FI ion is see saw or irregular tetrahedron.

(x) BrF3 Molecule:

In BrF3 molecule, the central atom is Br and it has 3 other F atoms.

N73 22 + = = 5

Since, the number of other atoms is 3, so the number of lone pairs is 2. Therefore, the shape of BrF3 is T shaped.

(xi) –3Br ion:

The central atom is a Br atom and it has 2 other Br atoms as surrounding atoms.



N721 22 ++ = = 5

Since, the number of other atoms is 2, so the number of lone pairs is 3. Thus, the shape of Br3

ion is linear.

(xii) SF6 Molecule:

The central atom in SF6 is S and it has 6 other F atoms as surrounding atoms.

N66 6 22 + ==

Since, the number of other atoms is 6, so the number of lone pairs are zero. Thus, the shape of SF6 molecule is octahedron or square bipyramidal.

(xiii) BrF5 Molecule:

In BrF5, the central atom is Br and 5 F atoms are the surrounding atoms.

N75 6 22 + ==

Since, the number of other atoms is 5, so the number of lone pairs would be 1. Therefore, the shape of BrF5 is square pyramidal.

(xiv) 4ICl ion:

In 4ICl ion, the central atom is I and 4 Cl atoms are the surrounding atoms.

N741 6 22 ++ ==

The number of lone pairs would be 2 as the number of other atoms is 4. So, the shape of 4ICl ion is square planar.

(xv) IF7 Molecule:

In IF7 molecule, the central atom is I and 7 F atoms are the surrounding atoms. N77 7 22 +

Since, the number of other atoms is 7, so the number of lone pairs would be zero. Thus, the shape of IF7 molecule is pentagonal bipyramidal.

Know Yourself

Illustration-6

Write the geometry of XeF4 and OSF4 using VSEPR theory and clearly indicate the position of lone pair of electrons and hybridization of the central atom.

Solution:

XeF4: N 2 = + 84 2 = 6

There are two lone pairs. Structure is octahedral and shape is square planar.

OSF4: N 2 = + 64 2 = 5

Structure is irregular trigonal bipyramidal with less electronegative element occupying equitorial position. There is no lone pair

➢

In addition to all this, VSEPR theory can also used to determine the geometry of other covalently bonded molecules and their bond angles. In order to predict these, following generalizations would be helpful.

Lone pair causes greater repulsions than a lone electron. For example,

Repulsions exerted by bond pair’s decrease as the electronegativity of the bonded atom increases. For example,

OH2 (104.5°) > OF2 (103.2°)

NH3 (107.3°) > NF3 (102°)

PI3 (102°) > PBr3 (101.5°) > PCl3 (100°)

AsI3 (101°) > AsBr3 (100.5°) > AsCl3 (98.4°)

Repulsion between bonded electron pairs in filled shells is greater than those between electron pairs in incompleted shells.

OH2 (104.5°) >> SH2 (92°) > SeH2 (91°) > TeH2 (89.5°)

NH3 (107.3°) >> PH3 (93.3°) > AsH3 (91.8°) > SbH3 (91.3°)

N(CH3)3, 109° > P(CH3)3, 102.5° > As(CH3)3, 96°

When an atom with a filled valence shell & one or more lone pairs is bonded to an atom with an incomplete valence shell, or a valence shell that can become complete by electron shifts, there is a tendency for the lone pairs to be partially transferred from the filled to the unfilled shell.

PH3 (93.3°) < PF3 (104°)

AsH3 (91.8°) < AsF3 (102°)

The size of a bonding electron pair decreases with increasing electronegativity of the ligand. Also, the two electron pairs of a double bond (or the three electron pairs of a triple bond) take up more room than does the one electron pair of a single bond.

6. SIGMA () AND PI () BONDS

➢ Valence bond theory explains that a covalent bond is formed by the overlapping of the electron clouds of the atomic orbitals of the constituent atoms. The greater the overlap, the stronger the bond.

6.1 Formation of Hydrogen Molecule

➢ Hydrogen (1s1) has only one electron in its 1s-orbital. When two hydrogen atoms come together, overlap of their s orbitals takes place (s s overlap), energy is released (bond energy) and a covalent bond called the  bond is formed.

➢ The electrons shared between the two atoms are to a large extent located in the region of space between the two nuclei. So the region of overlap is the region of high electron density.

➢ The electron density (or electron cloud) is distributed symmetrically about the bond axis, i.e., the line joining the nuclei. Such a bond formed by the axial overlapping of the orbitals is called a sigma () bond.

6.2 Formation of Hydrogen Fluoride Molecule

H : 1s1 F : 1s22s22p5

➢ The 1s-orbital of the hydrogen atom and one of the 2p-orbitals of the fluorine atom contain only one electroneach.

➢ The 1s orbital of the hydrogen atom and the partly filled p orbital of the fluorine atom overlap axially and form a  bond (s p overlap).

6.3 Formation of Chlorine Molecule

Cl : 1s22s22p63s23p5

➢ Oneofthe3p orbitalsofthechlorineatomcontainsonlyoneelectron.Thehalf filledp orbital ofa chlorine atom overlaps axially with the half-filled p orbital of the other chlorine atom and forms a  bond (p p overlap).

6.4 Formation of Nitrogen Molecule

N : 1s22s2 12px 1 y2p 1 z2p

➢ There are three unpaired electrons in the 2p orbitals of a nitrogen atom. When two nitrogen atoms combine the three 2p orbitals of one atom mutually overlap with those of the other atom and form three bonds.

➢ Suppose the orbitals approach along the x-axis the px orbitals overlap axially and form a  bond (p – p overlap).Thepy andpz orbitalsoftheNatomscannot overlapaxially and so make a lateral (side to side) overlap forming two Pi () bonds.

➢ Generally in any multiple bond between two atoms one bond is a  bond and the others  bonds.

➢ A double bond will consist of a  bond and a  bond and a triple bond will consist of a  bond and two  bonds. In a  bond formed between two p-orbitals, the upper lobe overlaps the upper lobe and the lower lobe overlaps the lower lobe. Together they constitute  bond. The  electron cloud will lie above and below the plane of the bond.

7. HYBRIDISATION

7.1 Sp3 Hybridisation

➢ Carbon atom has the electronic configuration 1s22s2 1 x2p 1 y2p . It has two half-filled orbitals.

➢ It should be expected to show a covalency of 2. In its millions of compounds, carbon shows tetracovalency. To explain this, the concept of hybridisation is introduced. Consider the formation of methane, CH4

➢ It may be supposed just for the sake of a picture that one of the electrons in the 2s-orbital is promoted to the vacant px orbital (excited state). This is possible because energy released during bond formation will compensate for this.

➢ Then the four orbitals, one s and three p-orbitals, get mixed up and form four new ‘hybrid’ orbitals, of equal energy which are called sp3 hybrid orbitals, as they are formed by the mixing up (or blending) of one s and three p-orbitals.

➢ So hybridisation is nothing but combination of a certain number of atomic orbitals of slightly different energies to form the same number of new (hybrid) orbitals of equal energy.

sp3 hybrid orbital

➢ Each sp3 hybrid orbital is like a p-orbital, but with 2 lobes of unequal size (In figures, the small lobe is usually omitted).

➢ Since the hybrid orbitals repel one another, they orient themselves with an angle of 10928 between them and point to the four corners of a regular tetrahedron.

➢ Each hybridised orbital overlaps the 1s-orbital of a hydrogen atom and forms a  bond.

➢ Each sp3 hybridised orbital has one fourth s character and three-fourths p character. Note that a hybrid atomic orbital from s and p-orbitals can form only  bonds. (4 C H  bonds.)

Formation of Ethane

➢ In this case there is sp3 sp3 overlap resulting in the formation of the C

C bond and sp3 s overlap forming C – H bonds. (1 C C  bond and 6 C H  bonds.)

Formation of Ethane

7.2 Sp2 Hybridisation

Formation of boron trifluoride/trichloride, BF3/BCI3

➢ Boron has the electronic configuration 1s22s2 1 x2p . One of the s electrons is promoted to a vacant py orbital (excited state).

➢ Then one s-orbital and two p-orbitals hybridise to form three sp2 hybrid orbitals of equivalent energy. This kind of hybridisation is called sp 2 (trigonal) hybridisation.

Formation of BCl3

➢ The three sp2 hybrid orbitals are co-planar and are at angles of 120 to each other.

➢ Each hybrid orbital overlaps with the vacant p-orbital of the chlorine atom and forms a  bond.

➢ The other halides or Boron have similar structures. An sp2 hybrid orbital has one-third s character and two-thirds p character. (3  bonds.)

Formation of Ethylene molecule

of carbon

➢ In the formation of ethylene, carbon atom undergoes sp2 hybridisation. Two of the sp2 orbitals of each atom form  bonds with 1s-orbitals of hydrogen atoms by axial overlapping.

➢ The sp2 sp2 overlap results in the formation of a C C  bond. The two carbon atoms and the four hydrogen atoms are all in the same plane and the bond angles are 120

Ethylene molecule

➢ At right angles to this plane there remains the 2pz, orbital of each carbon atom which overlap laterally to form a  bond between the two carbon atoms. The double bond between the two carbon atoms consists of a  bond and a  bond. (4 C H  bonds, 1 C C  bond and 1  bond)

7.3 Sp Hybridisation

➢ One s and one p-orbital combine to form two hybrid orbitals known as sp (or linear or diagonal) orbitals. They are of equal energy and are collinear. Each sp-orbital has one-half s character and onehalf p character.

(i) Formation of Beryllium Chloride

The sp hybrid orbitals of beryllium atom overlap with the vacant px-orbitals of two chlorine atoms and form two  bonds.

(ii) Formation of Acetylene

Hybridization of the one 2s and 2p carbon orbitals leads to the formation of two sp hybrid orbitals. sp–sp overlap between two carbon atoms form a  bond between them. The other sp orbital on each carbon atom forms a  bond with the 1s orbital of a hydrogen atom.

Acetylene molecule

Each of the carbon atom has two remaining p-orbitals which are mutually at right angles to each other. They laterally overlap and form two  bonds, sometimes pictured as a cylindrical sheath about the line joining the nuclei. One triple bond between 2 carbon atoms contains one C–C  bond and 2 bonds. (1 C – C  bond, 2 C – H  bonds and 2 bonds)

The carbon-carbon triple bond is made up of one strong  bond and two weaker  bonds; it has a total strength of 198 kcal/mole. It is stronger than a carbon-carbon double bond (163 kcal) or C–C single bond in ethane (88 kcal) and therefore is shorter than either.

7.4 Sp3d Hybridization

➢ In this type of hybridization, one ‘s’, three ‘p’ and one ‘d’ orbitals of the same shell mix to give five sp3d hybrid orbitals. These five sp3d hybrid orbitals orient themselves towards the corners of a trigonal bipyramidal.

Key Concept

Bent’s rule states that, “Atomic s character tends to concentrate in orbitals that are directed toward electropositive groups and atomic p character tends to concentrate in orbitals that are directed toward electronegative groups”.

Or

We can also describe bent rule as “More electronegative substituent's prefer hybrid orbitals having less s-character and more electropositive substituent's prefer hybrid orbitals having more s-character.”

Important to remember some facts which are given below.

s orbitals has low energy than p orbital.

Greater s character decrease the energy of bonding orbitals and hence they have shaped more like a s orbital and greater p character increase the energy of bonding orbitals and hence they have shaped more like a p orbitals.

s orbitals are closer to the nucleus, so it stabilize the lone pair. Due to more s character less repulsion and less hybridisation energy and less bond angle and reverse observation is true for more p character.

The most stable arrangement would be to utilize pure p orbitals for bonding and letting the lone pair into the pure s orbital.

s-Orbitals are more penetrating and electron density is less available for bonding. Thus, more electronegative atoms would be able to withdraw more electron density from p orbitals than from s orbitals.

Examples of Bent Rule and Energetics of Hybridization

sp3d Hybridisation in PCl3F2 : In sp3d Hybridisation one s, three p and one d-orbitals form 5 sp3d hybridised orbital. All these five hybrid orbitals are not of the same type so they can be divided into two nonequivalent sets which is discussed below.

The first set is known as equatorial set of orbitals. It is formed from one s, one px and one py orbitals.

The second set is known as axial set. It is formed from one pz and one d orbitals. It is experimentally observed that the more electronegative substituent ‘F’ occupies the axial position (as it has less s character) and less electronegative substituent's ‘Cl’ is equatorially .

Less bond angle in CH2F2 :In the case of sp3 hybridisation, CH4 or CCl4 formed tetrahedral geometry with bond angle 109.5° . But in the case of CH2F2, the F-C-F bond angle is less than the normal tetrahedral angle, this explained on the basis of Bent’s rule. F is more electronegative than H therefore F-C-F s-character is less than 25% while in H-C-H it is more than 25%. Due to less s-character bond angle is less in CH2F2

7.5 Sp3d2 Hybridization

➢ In this type of hybridization, one ‘s’, three ‘p’ and two ‘d’ orbitals of the same/different shell mix to give six sp3d2 hybrid orbitals. These six sp3d2 hybrid orbitals orient themselves towards the corners of an octahedron. This type of hybridization is exhibited by SF6, SCl6 etc.

7.6 Scheme for Determining Hybridization of the Central Atom of A Species

➢ Identify the central atom of the species.

➢ Write outermost electronic configuration of the central atom.

➢ Determine oxidation state of the central atom.

➢ Excite the electrons (if necessary) to the orbitals of higher energy in order to make the number of unpaired electrons equal to the oxidation state of the central atom.

➢ Now start putting orbitals into your bag, beginning from ‘s’ orbitals. The number of orbitals added to the bag must have orbitals with unpaired electrons equal to number of other atoms of the species.

➢ All the orbitals added to the bag (including s orbital, whether it has paired or unpaired electrons) are now summed. If there are one ‘s’ and one ‘p’ orbital in the bag, then the hybridization is sp. If there are one ‘s’ and ‘2p’ orbitals in the bag, it is sp2 hybridization, and so on.

➢ Each unpaired electron left (outside the pocket) will form a pi bond.

➢ Each orbital with paired electrons in the pocket will exist as lone pair on the central atom. The working of this scheme can be seen in the following illustrations.

Illustration-7

Find out hybridization of the central atom in 3ClO and draw its structure.

Solution:

(i) The central atom is Cl.

(ii) Outer most electron configuration of Cl = [Ne]3s23p5

3s 3p 3d

(iii) Oxidation state of Cl is +5.

(iv) Outer most electron configuration of Cl after excitation:

3s 3p 3d

(v) Now start adding orbitals into your pocket beginningfrom s, and thereafter 3p orbitals. We will stop adding orbitals to the pocket after adding all three ‘3p’ orbitals because then the orbitalswithunpairedelectronsinthepocketwouldbecomeequal to thenumberofotheratoms in 3ClO .

(vi) So, the hybridization of Cl in 3ClO is sp3

(vii) Shape of 3ClO would be tetrahedral with one lone pair (pyramidal). Each unpaired electron left (outside the pocket) will form a  bond and there will be one lone pair on Cl.

Illustration-8

Find out hybridization of the central atom in 2ClO and draw its structure.

Solution:

(i) The central atom is Cl.

(ii) Outer most electron configuration of Cl = [Ne]3s23p5

3s 3p 3d

(iii) Oxidation state of Cl is +3.

(iv) Outer most electron configuration of Cl after excitation:

3p 3d

(v) Now start adding orbitals into your pocket beginningfrom s, and thereafter 3p orbitals. We will stop adding orbitals to the pocket after adding all three ‘3p’ orbitals because then the orbitalswithunpairedelectronsinthepocketwouldbecomeequal to thenumberofotheratoms in 2ClO

(vi) So, the hybridization of Cl in 2ClO is sp3

(vii) Shape of 2ClO would be tetrahedral with two lone pairs (angular). Each unpaired electron left (outside the pocket) will form a  bond and there will be two lone pair on Cl.

Find out hybridization of the central atom in ClO and draw its structure.

Solution:

(i) The central atom is Cl.

(ii) Outer most electron configuration of Cl = [Ne]3s23p5

3s 3p 3d

(iii) Oxidation state of Cl is +1.

(iv) There is no need for excitation of electrons because the number of unpaired electrons is already equal to oxidation state of Cl.

(v) Now start adding orbitals into your pocket beginningfrom s, and thereafter 3p orbitals. We will stop adding orbitals to the pocket after adding all three ‘3p’ orbitals because then the orbitalswithunpairedelectronsinthepocketwouldbecomeequal to thenumberofotheratoms in ClO

(vi) So, the hybridization of Cl in ClO is sp3

(vii) Shape of ClO wouldbetetrahedral withthreelone pairs (linear). Chlorineatom will havethree lone pairs.

Illustratioin-9

Illustration-10

Find out hybridization of the central atom in SO2 and draw its structure.

Solution:

(i) The central atom is S.

(ii) Outer most electron configuration of S = [Ne]3s23p4

3s 3p 3d

(iii) Oxidation state of S is +4.

(iv) Outer most electron configuration of S after excitation:

3s 3p 3d

(v) Now start adding orbitals into your pocket beginningfrom s, and thereafter 3p orbitals. We will stop adding orbitals to the pocket after adding two ‘3p’ orbitals because then the orbitalswithunpairedelectronsinthepocketwouldbecomeequal tothenumberofotheratoms in SO2.

(vi) So, the hybridization of S in SO2 is sp2 .

(vii) Shape of SO2 would be trigonal planar with one lone pair (angular). Each unpaired electron left (outside the pocket) will form a  bond and there will be one lone pair on S.

Illustration-11

Find out hybridization of the central atom in XeOF2 and draw its structure.

Solution:

(i) The central atom is Xe.

(ii) Outer most electron configuration of Xe is 5s25p6 5s 5p 5d

(iii) Oxidation state of Xe is +4.

(iv) Outer most electron configuration of Xe after excitation: 5s 5p 5d

(v) Now start adding orbitals into your pocket beginning from s, and thereafter 5p orbitals and then d orbitals. We will stop adding orbitals to the pocket after all three ‘5p’ orbitals and one ‘5d’ orbital because then the orbitals with unpaired electrons in the pocket would become equal to the number of other atoms in XeOF2.

(vi) So, the hybridization of Xe in XeOF2 is sp3d.

(vii) Shape of XeOF2 would be trigonal bipyramidal with two lone pairs (T shaped). Each unpaired electron left (outside the pocket) will form a  bond and there will be two lone pairs on Xe.

Illustration-12

Find out hybridization of the central atom in XeOF4 and draw its structure.

Solution:

(i) The central atom is Xe.

(ii) Outer most electron configuration of Xe is 5s25p6 5s 5p 5d

(iii) Oxidation state of Xe is +6.

(iv) Outer most electron configuration of Xe after excitation: 5s 5p 5d

(v) Now start adding orbitals into your pocket beginning from s, and thereafter 5p orbitals and then d orbitals. We will stop adding orbitals to the pocket after all three ‘5p’ orbitals and two ‘5d’ orbital because then the orbitals with unpaired electrons in the pocket would become equal to the number of other atoms in XeOF4.

(vi) So, the hybridization of Xe in XeOF4 is sp3d2 .

(vii) Shape of XeOF4 would be octahedron with one lone pair (square pyramidal). Each unpaired electron left (outside the pocket) will form a  bond and there will be one lone pair on Xe.

PCl3F2

PCl2F3

PCl3F2 is non-polar molecule as all three Cl atoms are at equatorial position and both F atoms in axial position.

PCl2F3 is polar molecule as both Cl atoms and one F atom are at equatorial position and both F atoms in axial position.

Pentagonal planar ion with two nonbonding electron pairs above and below the plane of the pentagon.

Key Concept

DRAGO’S RULE : According to Drago’s rule, If central atom is in third row of or below in periodic table then lone pair will occupy in stereo-chemically inactive Sorbital and bonding will be through p-orbital and near 90° angles. If electronegativity of surrounding atom is less than or equal < 2.5.

8. BOND PARAMETERS

8.1 Bond Length:

➢ The internuclear distance between two covalently bonded atoms is known as bond length.

Factors Affecting the Bond Length:

1. Size of atoms: with the increase in the size of atoms, Bond length increases.

C–H (109pm) < H–Cl (127pm)

2. Hybridisation state of the bonded atoms: If s character increases, bond length decreases.

sp3 – sp3 > sp2 – sp2 > sp – sp

3. Multiple bonds: Bond length decreases with multiplicity of bonds.

C – C 154 pm C = C 134 pm C  C 120 pm

8.2 Bond Angle:

➢ It is defined as the angle between the orbitals containing bonding electronpairsaround thecentral atom in a molecule/complex ion.

Factors Affecting the Bond Angle:

1. Hybridization of the central atom: Compounds having different hybridization states will have different bond angles.

For Example: CCl4 : sp3 (109° 28’) BCl3 : sp2 (120°) BeCl2 : sp (180°)

2. Electronegativity of central atom: Bond angle is directly proportional to electronegativity.

For Example: NH3 > PH3

3. Multiple bonds: Double bond repulsion will be higher than in single bond, due to more electron density.

4. Size of the atoms attached to the central atom: when size of the terminal atoms increase, bond angle increases.

For Example: PF3 < PCl3

Bond Enthalpy:

The amount of energy required to break 1mole of the compound and separate the bonded atoms is known as bond enthalpy.

Factors affecting Bond enthalpy:

1. Sizeoftheatomsinvolvedinbonding:Largersizeleadstolesseroverlapping,whichinturndecreases the bond strength and hence the bond enthalpy.

2. Multiple bonds: As the bond order increases, the bond enthalpy increases.

3. Hybridization of the central atom: As the s character increases, bond enthalpy increases too.

4. Electronegativity: As the difference of electronegativity between bonded atom increases, bond enthalpy increases.

5. Lone pairs: As the repulsion between lone pairs of bonded atoms increases, bond enthalpy decreases.

9. HYDROGEN BOND

➢ A hydrogen atom normally forms a single bond. In some compounds, however, the hydrogen atom may be located between two atoms acting as a bridge between them.

➢ Hydrogen atom is now involved in two bonds, one a normal covalent bond, the other a hydrogen bond.

➢ A hydrogen bond is always formed between two small, strongly electronegative atoms such as fluorine, oxygen and nitrogen.

9.1 Intermolecular Hydrogen Bonding – Molecular Association

(i) Hydrogen Fluoride:

From molecular measurements, it is known that hydrogen fluoride is associated (i.e., many molecules are joined together). HF is a polar molecule, with the fluorine atom acquiring a slight negative charge and the hydrogen atom acquiring an equal positive charge. The electrostatic attraction between the oppositely charged ends results in hydrogen bonding as shown below.

H – F ... H – F … H – F …

Many H – F units are held together, as (HF)n, by hydrogen bonding. The covalent H – F bond is much shorter than the F … H hydrogen bond; so a hydrogen bond is much weaker than a covalent bond. Fluorine, with the highest electronegativity forms the strongest hydrogen bond. The nature of the hydrogen bond is considerably electrostatic.

(ii) Water:

The high boiling point compared to that of hydrogen sulphide is due to molecular association through hydrogen bonding.

The crystal structure of ice shows a tetrahedral arrangement of water molecules. Each oxygen atom is surrounded tetrahedrally by 4 others. Hydrogen bonds link pairs of oxygen atoms together as shown in Figure. The arrangement of water molecules in ice is a very open structure and this explains the low density of ice. When ice melts, the structure breaks down and the molecules pack more closely together so that water has a higher density; this packing goes to a maximum up to a temperature of 4C.

(iii) Ammonia:

Ammonia is also associated through hydrogen bonding; hence it has higher boiling point than PH3 or AsH3.

(iv) Alcohols and phenols:

Lower alcohols and phenolsare associated due to intermolecular hydrogen bonding. Methanol, ethanol and phenol have relatively much higher boiling points than methane or chloromethane, ethane or chloroethane, benzene or chlorobenzene respectively.

(v) Carboxylic acids:

Some carboxylic acids exist as dimers e.g., the dimer of acetic acid is represented as

In aqueous solution molecules of a carboxylic acid link up with water molecules through hydrogen bonding rather than form dimers.

9.2

Intramolecular Hydrogen Bonding

➢ Sometimes hydrogen bonding may take place within a molecule; this is known as intramolecular (or internal) hydrogen bonding. It may lead to the linkage of two groups to form a ring; such an effect is known as chelation, in the case of complex compounds.

(i) Nitrophenols

Because of the proximity of

OH and

NO2 groups in o-nitrophenol there is intramolecular hydrogen bonding which prevents intermolecular hydrogen bonding between two or more molecules. Since molecular association cannot take place, the boiling point of o-nitrophenol is lower than that of the other two. Because of the distance between –OH and –NO2 groups in m– and p–nitrophenols there is no possibility of intramolecular hydrogen bonding. Intermolecular hydrogen bonding may take place to a certainextent which causes somedegree of molecular association; this is responsible for the higher boiling points of the two nitrophenols.

Further the formation of intramolecular hydrogen bonding in o-nitrophenol prevents it from entering into intermolecular hydrogen bonding with water and this explains its reduced solubility.

(ii) Other molecules in which intramolecular hydrogen bonding is present are o-hydroxybenzaldehyde, o-chlorophenol and o-hydroxybenzoic acid.

Illustration-13

H2O is a liquid at ordinary temperature while H2S is a gas although both O and S belong to the same group of the periodic table.

Solution:

H2O is capable of forming intermolecular hydrogen bond. This is possible due to high electronegativity and small size of oxygen. Due to intermolecular H bonding, molecular association takesplace.Asaresult theeffectivemolecular weight increasesandhencetheboilingpoint increases, so H2O exist in liquid phase. But in H2S, no hydrogen bonding is possible due to large size and less electronegativity of S. So it’s boiling point is that of an isolated H2S molecule and therefore it is a gas with low boiling point.

Illustration-14

The salt KHF2 is known but KHCl2 is not known. Explain.

Solution:

The formation of KHF2 involves reaction of 2HF with KOH. Similar is the case with KHCl2. So the main factor is the formation of 2HF or 2HCl ion.

H F + F → F H F

H Cl + Cl → Cl H……Cl (not possible)

Due to higher electronegativity and small size of fluorine, it is capable of forming H bond resulting in the formation of 2HF and thereby KHF2 exists. But with chlorine, there is no possibility of H bonding, so there is no possibility of existence of KHCl2

o hydroxy benzaldehdye is more volatile than p hydroxy benzaldehyde.

Solution:

More volatiliity means compound has lower boiling point. p hydroxy benzaldehyde remains associated through intermolecular hydrogen bonding. But in o hydroxy benzaldehyde, intramolecular H bonding takes places, as a result of which there is no association. So p hydroxy benzaldehyde, which remains as an associated species has got higher boiling and so less volatile while o hydroxy benzaldehyde is highly volatile.

10. MOLECULAR ORBITAL THEORY AND BOND ORDER

➢ The Valence Bond Theory (V.B. Theory) with the concepts of hybridisation and resonance is used to explain the structure and properties of several molecules, but there are limitations. For example, the V.B. theory in its original form is not able to explain the paramagnetic behaviour of O2 molecule.

➢ Hence the Molecular Orbital Theory (or M.O. Theory) was proposed by Hund and Mulliken.

➢ The following are the essential features of the M.O. Theory.

In the M.O. model, all the electrons are taken together and considered as moving in the field of all the nuclei. (In the V.B. model, only the bonding electrons are considered and they are taken to move in the field of the nuclei involved in bonding.)

The atomic orbitals are combined to form, what are called molecular orbitals and electrons are fed into these orbitals. (In the V.B. model, electrons are fed into the atomic orbitals, which are then supposed to overlap.)

The number of combining atomic orbitals is equal to the number of molecular orbitals formed. When two atomic orbitals combine, two M.O’s are formed, of which one has a lower energy, while the other has a higher energy. The former is known as the bonding orbital and the latter antibonding.

Mathematically, if 1 represents the wave function corresponding to orbital 1 and 2 for orbital 2, the total function is a linear combination of 1 and 2 i.e.,  = 1  2 (omitting the constants).

This is known as linear combination of atomic orbitals (L.C.A.O.). Of these, 1 + 2 corresponds to the bonding M.O., while 1 – 2 corresponds to antibonding M.O. i.e.,

b = 1 + 2 and a = 1

2. The electron density or probability of finding an electron is directly proportional to 2

For the bonding orbital, 2222 b121212 ()2 =+=++ , which is greater than 22 12+ i.e., the electron density between the two nuclei is concentrated when the bonding M.O. is formed, than when no such combination of orbitals is made.

electron density between the nuclei is withdrawn in an antibonding

In the bonding M.O., since the electron density between the two nuclei is large, it holds the two nuclei together; hence the name bonding orbital; in the antibonding M.O. the bonding of the nuclei is poor. There are different notations for representing bonding and anti-boding M.O’s obtained from A.O’s. We give a simple notation below.

We have assumed the two pz orbitals to overlap end to end, so that the M.O. formed is of the ‘‘ type (similar to the ‘‘ bond in V.B. theory); then the two px atomic orbitals, as also the two py orbitals will overlap laterally to give M.O’s of the  type.

When electrons are successively placed in the M.O’s, Aufbau principle, Hund’s rule and Pauli’s principle are followed, as in the case of the atomic orbitals

Aufbau Principle: M.O’s are occupied in the order of increasing energy. The following is the general arrangement of M.O’s in the order of increasing energy.

The above is only a general order and slight variations often occur due to interaction between s and p orbitals (s-p mixing). For example, sometimes, (2px) = (2py) < (2pz).

Hund’s Rule of Maximum Multiplicity:

The degenerate M.O’s are occupied singly, before any pairing could occur. The maximum capacity for each M.O. is 2 electrons.

Only atomic orbitals of equal or nearly equal energies combine to give the M.O’s. In the case of homonuclear diatomic molecules, energies of corresponding A.O’s of the two atoms are equal. So the above condition of combination of A.O’s assumes special significance in the case of heteronuclear diatomic molecules and it has to be used with caution.

Further,foreffective combination oroverlap,the A.O’sshouldhavethesamesymmetry.Thuswehave thes–s,pz –pz,py –py andpx –px overlapstogivebondingandantibondingorbitalsaspointedoutearlier. Regarding the s – p overlap, a 2s orbital may overlap with a 2pz orbital as shown in Figure (a) below, since both have axial symmetryaround the internuclearaxis.However,the2s–2px or2s–2py overlapmakes nocontributionto bonding, as shown in Figure (b) below, where the constructive overlap in one region isexactlycancelled by theeffect ofthedestructiveoverlap inthe other.

It has been already pointed out that electrons in the bonding M.O’s tend to pull the nuclei together and that the electrons in the antibonding M.O’s tend to separate them. Hence the combined influence of bonding and antibonding electrons may either stabilize or destabilize the molecule, depending on

the relative number of these two types of electrons. The stabilizing power is expressed in terms of “bond order”.

numberofelectronsnumberofelectrons 1 Bond Order= inthebondingM.O.intheantibondingM.O. 2

The greater the bond order, the greater the bond stability and the shorter the bond distance. The M.O’s are also named on symmetry grounds. For homonuclear diatomic molecules, the symbols ‘g’ (‘gerade’ meaning ‘even’) and ‘u’ (‘ungerade’ meaning ‘uneven’ or ‘odd’) are used. The symbol ‘g’ is used, if the orbital has a centre of symmetry. i.e., if along any straight line passing through the centre (This is called the centre of inversion), at equal distances from it, the electron densities are equal and the orbital signs are the same (i.e., the wave function has the same amplitude and sign at the two points which are opposite and equidistant from the centre). If the electron densities are equal, but the orbital signs are opposite at the two points mentioned above, the symbol ‘u’ is used. Inthe‘’typeoforbitals,thebondingorbitalsare‘g’andtheantibonding‘u’;inthe‘‘typeoforbitals, the bonding orbitals are ‘u’ and the antibonding ‘g’. Figure (a) and (b) below correspond to overlap of ‘s’ orbitals to form ‘’ type of M.O’s; Figure (c) and (d) correspond to the formation of ‘’ type of M.O’s from ‘p’-orbitals.

The equivalence of the different notations used to represent the M.O’s is shown for a few cases (the x, y, z subscripts in the case of p-orbitals are dropped).

For elements with Z 7

For elements with Z > 7

10.1 Homonuclear Diatomic Molecules

➢ We shall now consider the electronic configuration of a few homonuclear diatomic molecules.

H2: Electronic configuration of H atom: 1s1

 in H2 molecule there are 2 electrons. M.O. configuration of H2 is 1s2. There is no electron in antibonding M.O.  Bond order (B.O.) = 20 1 2 =

He2: Electronic configuration of He atom: 1s2

 in He2 molecule there are 4 electrons. The M.O. configuration for He is 1s2 , *1s2 .

 Bond order = 22 0 2 = . Therefore, He2 molecule is not stable.

Taking 2He+ , the structure is 1s2 , *1s1 .

 B.O. = 211 22 = . Therefore, + 2He is stabler than He2.

Li2: (Li atom: 1s2 2s1). Total number of electrons in Li2 = 6.

Molecular orbital configuration of Li

where KK corresponds to filled (1s) and *(1s) levels. The contribution of [KK] to bond order is zero. Therefore, we can ignore it and consider only the valence electrons.

B.O. = 20 1 2 =

 Li2 is stable and it is found to exist to some extent in lithium vapour.

Be2: (Be: 1s22s2)

Molecular orbital structure of Be2 is: [KK] 2s2 , *2s2. Ignoring [KK],

B.O. = 22 0 2 = . Therefore, Be2 is not stable.

B2: (B: 1s2 2s2 2p1)

M.O. structure of B2 is: [KK] 2s2 , *2s2 , 11 yz 2p,2p

Note that though (2p) orbital is usually more energetic than (2p), there is an inversion of the order here due to mixing (hybridization) of (2s) and (2p) orbitals. Also note that the degenerate M.O’s (2py) and ((2pz) have one electron in each according to Hund’s principle.

B.O. = 42 2 = 1. Therefore, B2 is stable.

Since there are two unpaired electrons in the molecule, B2 is paramagnetic. C2: (C: 1s2 2s2 2p2).

M.O. picture of C2 is: [KK] 2s2 , *2s2 , 22 yz 2p,2p

Here again (2p) orbital is less energetic than (2p).

B.O. = 62 2 = 2 and the molecule is stable.

Since there is no unpaired electron, C2 is diamagnetic. [If (p) orbital had been more energetic than (p), C2 would have been paramagnetic, which is contrary to experimental observation.]

N2: (N: 1s22s22p3).

Ignoring the subscripts x, y and z for the p-orbital and considering, only valence electrons, the representation is

B.O. = 82 2 = 3. The molecule is diamagnetic.

For 2N+ , B.O. = 72 2 = 2.5.

O2: [O: 1s22s2 2p4] M.O. structure of O

B.O. = 84 2 = 2

Due to the presence of unpaired electrons in the two antibonding orbitals (Hund’s rule), O2 is paramagnetic. The M.O. theory here is superior to the V.B. theory, which does not explain the paramagnetic behaviour of O2.

Let us now compare the bond strengths of O2, 2 222 O,O,O+−− . For O2, B.O. = 2.

For 2O+ : one electron from the antibonding M.O. has been removed.

 B.O. = 83 2 = 2.5

For 2O : one electron is added to the antibonding M.O.

 B.O. = 85 2 = 1.5

For 2 2O : B.O. = 86 2 = 1

 the bond is strongest in 2O+ and the bond length the least.

EXERCISE-1

Valency

1. Which of the species follows octet rule?

(a) IBr5 (b) N3– (c) SF4 (d) Pb+4

2. Which of the following compound exist?

(a) B2F6 (b) B2H6 (c) Al2F6 (d) B2Cl6

3. Which one of the following element will never be obeyed octet rule?

(a) Na (b) F (c) S (d) H

4. In which of the following molecule electron of 2s in carbon is not promoted to 2p sub shell?

(a) HCN (b) CO2 (c) CS2 (d) CO

5. Which is not an exception to octet rule?

(a) BF3 (b) SnCl4 (c) BeI2 (d) ClO2

VBT and Overlapping

6. In a triple bond, there is sharing of

(a) 3-electrons (b) 4-electrons (c) Several electrons (d) 6-electrons

7. The triple bond in CO = is made up of

(a) Three sigma bonds

(c) One sigma and two -bonds

8. Variable covalency is exhibited by

(a) P and S (b) N and O

9. A sigma bond is formed by the overlapping of

(a) s – s orbital only

(b) s and p orbitals only

(b) Three -bonds

(d) Two sigma and one -bond

(c) N and P (d) F and Cl

(c) s – s, s – p or p – p orbitals along internuclear axis

(d) p – p orbital along the sides

10. Which is not characteristic of a -bond?

(a) -bond is formed when a sigma bond already formed.

(b) -bond are formed from hybrid orbitals.

(c) -bond may be formed by the overlapping of p-orbitals

(d) -bond results from lateral overlap of atomic orbitals.

Resonance and Formal Charge

11. Higher is the bond order, greater is the

(a) Bond dissociation energy

(c) Bond length

(b) Covalent character

(d) Paramagnetism

12. In PO4 3–, the formal charge on each oxygen atom and the P – O bond order respectively are

(a) –0.75, 0.6 (b) –0.75, 1.0 (c) –0.75, 1.25 (d) –3, 1.25

13. In which of the following, lewis dot structure is written with incorrect formal charge?

(a) (b) (c) (d)

14. The species having no p-p bond but has bond order equal to that of O2, is (a) 3ClO (b)

PO

Hybridisation, Vsepr Theory and Geometry

15. In the protonation of H2O, change occurs in (a) Hybridisation state of oxygen

Shape of molecule (c) Hybridisation and shape both

16. The d-orbitals involved in sp3d hybridisation is

17. A sp3 hybrid orbital contains

18. Among the following species, identify the isostructural pairs

of these

19. When the hybridization state of carbon atom changes from sp3, sp2 and sp, the angle between the hybridized orbitals

(a) decrease considerably

increase progressively (c) decrease gradually

20. Which order of decreasing bond angle is correct?

all of these

21. By the hybridization of one s & one p orbitals, it will be obtained

(a) Two orbitals mutually at 90ºangle

(c) Two orbitals mutually at 120ºangle

(b) Two orbitals mutually at 180ºangle

(d) Two orbitals mutually at 150º angle

22. In compounds X, all the bond angles are exactly 109°28’, X is

(a) Chloromethane

(c) Iodoform

Dipole Moment

23. Which statement is correct?

(b) Carbon tetrachloride

(d) Chloroform

(a) All the compounds having polar bonds, have dipole moment

(b) SO2 is non-polar.

(c) H2O molecule is non-polar, having polar bonds

(d) PH3 is polar molecule having non-polar bonds.

24. BeF2 has zero dipole moment where as H2O has a dipole moment because

(a) Water is linear

(c) F is more electronegative than O

(b) H2O is bent

(d) Hydrogen bonding is present in H2O

25. Which of the following species are symmetrical?

(a) XeF4 (b) XeF6 (c) SO2 (d) 4NH+

Correct answer is

(a) a and b (b) b and c (c) c and d (d) a and d

26. Which of the following molecule have zero dipole moment?

(a) BF3 (b) CH2Cl2 (c) NF3 (d) SO2

27. The dipole moment of NH3 is

(a) Less than dipole moment of NCl3 (b) Higher than dipole moment of NCl3

(c) Equal to the dipole moment of NCl3. (d) None of these

28. Which of the following order of polar molecules is correct?

(a) HF > NH3 > PH3

(b) CH4 > NH3 > H2O

(c) CH3Cl < CH2Cl2 < CHCl3 (d) BF3 > BeF2 > F2

Coordinate Bond

29. Which of the following contains co-ordinate and covalent bonds?

(i) N2H5 + (ii) H3O+ (iii) HCl (iv) H2O

Correct answer is

(a) (i) & (iv) (b) (i) & (ii) (c) (iii) & (iv) (d) Only (i)

30. In co-ordinate bond, the acceptor atoms must essentially contain in its valency shell an orbital

(a) With paired electron

(b) With single electron

(c) With no electron (d) With three electron

31. Number of  and  bonds in C2(CN)4 are respectively (a) 9

32. Dative bond is present in (a)

33. Correct statement regarding this reaction BF3 + NH3 ⎯⎯→ [F3B  NH3] is

(a) Hybridization of N is changed. (b) Hybridization of B is changed.

(c) B – F bond length increases. (d) (b) & (c) both

34. In the neutralization process of NH3 and AlCl3, the compound formed will have the bonding

(a) Ionic (b) Covalent (c) Co-ordinate (d) Hydrogen

35. In the co-ordinate valence

(a) electrons are equally shared by the atoms

(b) electrons of one atom are shared by two atoms

(c) hydrogen bond is formed

(d) none of the above

36. The compound containing co-ordinate bond is

(a) H2SO4 (b) O3 (c) SO3 (d) All of these Back Bond

37. In BF3, the B-F bondlengthis 1.30 Å, when BF3 is allowed to be treatedwith Me3N, itformsan adduct, Me3N → BF3, The bond length of B – F in the adduct is

(a) Greater than 1.30 Å (b) Smaller than 1.30 Å

(c) Equal to 1.30 Å (d) None of these

38. Which of the following structure correctly represents the boron trifluoride molecule?

(a)

(b)

(c) (d)

39. Trisilylamine has a

(a) Planar geometry

(c) Pyramidal geometry

(b) Tetrahedral geometry

(d) None of these

40. The geometry with respect to the central atom of the following molecules are N(SiH3)3, Me3N, (SiH3)3P

(a) Planar, pyramidal, planar

(c) pyramidal, pyramidal, pyramidal

(b) planar, pyramidal, pyramidal

(d) Pyramidal, planar, pyramidal

41. In which of the following compounds, B – F bond length is shortest?

(a) BF4 –

(b) BF3  NH3

(c) BF3 (d) BF3  N(CH3)3

Ionic Bond

42. The compound which contains both ionic and covalent bonds is

(a) Al4C3 (b) HCl

(c) NH4Cl (d) KCl

43. Electrovalent compounds do not show stereoisomerism. The reason is

(a) Presence of ions

(c) Brittleness

(b) Strong electro static force of attraction

(d) Non-directional nature of ionic bond

44. Nature of bond formed when two elements react, depends on

(a) Ionisation potential

(c) Oxidation potential

(b) Electronegativity

(d) Electron-affinity

45. The electronic configuration of metal 'M' is 1s2 2s1 and of the non-metal 'X' is 1s2 2s2 2p3. When these two elements combines, the formula of the compound will be

(a) MX3 (b) M3X

Solubility

46. KCl easily dissolves in water because

(a) It is a salt of K.

(c) It hydrolysed with water.

(c) M2X3 (d) M3X2

(b) It reacts with water.

(d) Its ions are easily solvated.

47. Capacity of solvent to neutralise charge on ionic compound is called

(a) Solvation energy (b) Dielectric constant (c) Dipole moment (d) Solubility

48. The force responsible for dissolution of ionic compound in water is

(a) Dipole – dipole forces

(b) Ion – dipole force

(c) Ion – ion force (d) Hydrogen bond

49. The hydration of ionic compounds involves

(a) Evolution of heat

(c) Dissociation into ions

(b) Weakening of attractive forces

(d) All of these

50. A metal M readily forms its sulphate MSO4, which is water soluble. It forms an insoluble hydroxide M(OH)2 which is soluble in NaOH solution, then M is

(a) Mg (b) Ca (c) Be (d) Ba

51. The hydration energy of Mg+2 is greater than the hydration energy of

(a) Al+3

(b) Mg+3 (c) Na+ (d) Be+2

52. LiCl is soluble in organic solvent while NaCl is not because

(a) Lattice energy of NaCl is less than that of LiCl.

(b) Ionisation potential of Li is more than that of Na.

(c) Li+ has more hydration energy than Na+ ion.

(d) LiCl is more covalent compound than that NaCl.

53. Which of the compound is least soluble in water?

(a) AgF (b) AgCl (c) AgBr (d) AgI Melting Point and Boiling Point

54. Which of the following has the lowest melting point?

(a) SrF2 (b) BeF2 (c) BaF2 (d) MgF2

55. Which of the following substance will have highest b.p.?

(a) He (b) CsF (c) NH3 (d) CHCl3

56. As compared to covalent compounds, electrovalent compounds generally possess

(a) High m.p. and high b.p.

(c) Low m.p. and high b.p.

(b) Low m.p. and low b.p.

(d) high m.p. and low b.p.

57. In which of the following pair of compounds, their melting point are closest to each other

(a) LiCl, NaCl (b) RbCl, LiCl (c) CsCl, NaCl (d) LiCl, CsCl Lattice Energy

58. Which of the following is not a correct statement about an ionic compound?

(a) Higher the lattice energy, higher is melting point.

(b) Higher the dipole moment of solvent, more the solubility.

(c) Higher the lattice energy, more the solubility.

(d) More difference in electronegativity, more is the ionic nature.

59. Born Haber cycle is mainly used to determine

(a) Lattice energy

(c) Ionisation energy

(b) Electron affinity

(d) Electronegativity

60. Among the following which compounds will show the highest lattice energy?

(a) KF (b) NaF (c) CsF (d) RbF

61. The lattice energy of the lithium halides is in the following order

(a) LiF > LiCl > LiBr > LiI

(c) LiBr > LiCl > LiF > LiI

(b) LiCl > LiF > LiBr> LiI

(d) LiI > LiBr > LiCl > LiF

62. Which of the following compound has highest Lattice energy?

(a) AlF3

(b) Na2S

(c) Al2O3

(d) CaF2 Fajan’s Rule

63. The pair of elements, which on combination are most likely to form an ionic compound is

(a) Na and Ca (b) K and O (c) O and Cl (d) Al and I

64. Out of the following, the compound with maximum ionic nature are

(a) Metal oxide (b) Metal chloride (c) Metal phosphide (d) Metal sulphide

65. The most covalent halide is

66. AlCl3 is covalent, while AlF3 is ionic. This is justified by

(a) Molecular orbital theory (b) Valency bond theory

(c) Fajan rule (d)

67. Which pair in the following has maximum and minimum ionic character, respectively?

(a)

68. The most stable carbonate is

69. Which of the following does not give an oxide on heating?

70. Which metal hydroxide decomposes on heating? (a)

Which of the following carbonate will not decompose on heating?

72. The ion that is isoelectronic with CO and having same bond order is

73. Which of the following is paramagnetic?

74. Which has the bond order in fraction?

Bond order in

+ is

76. Which of the following molecules have unpaired electron?

Hydrogen Bond

77. The hydrogen bond is strongest in

78. H2O boils at higher temperature than H2S, because it is capable of forming

79. Intermolecular hydrogen bonds are not present in

80. Correct order of volatility is

JEE-Main Level Questions

Single Correct Answer Type Questions

1. PCl5 exists but NCl5 does not because

(a) nitrogen has no vacant d orbitals

(c) nitrogen atom is much small

(b) NCl5 in unstable

(d) nitrogen is highly inert

2. If a molecule MX3 has zero dipole moment, the sigma bonding orbitals used by M

(atomic number < 21) are

(a) pure p

sp hybridised

sp

hybridised

3. Of the three molecules XeF4, SF4, SiF4 one which have tetrahedral structures is

(a) all the three

(c) both SF4 and XeF4

4. The hydrogen halide having maximum dipole moment is

5. The state of hybridization of Xe in XeF4 is

6. Dipole moment is exhibited by (a) 1,4 Dichlorobenzene

(c) trans 1,2 dichloroethene

7. The pair of species having identical shape is

8. The species having octahedral shape is

9. Which of the following has fractional bond order?

10. According to Fajans’ rules, ionic bonds are formed when

(a) cations have low positive charge and large size.

(b) cations have low positive charge and small size.

only SiF4

only SF4 and XeF4

sp

1,2 Dichlorobenzene

(c) cations have high positive charge and large size, and anions have a small size.

(d) cations have a low positive charge and large size, and anions have a small size.

11. In an 2H+ ion

(a) one electron is bound to two protons. (b) two electrons are bound to two protons.

(c) three electrons are bound to two protons. (d) none of these happens.

12. AlCl3 is covalent while AlF3 is ionic. This can be justified on the basis of

(a) the valence-bond theory

(c) the molecular-orbital theory

(b) Fajan’s rules

(d) hydration energy

13. Among the following, the molecule with the highest dipole moment is

14. Orthonitrophenol is steam volatile but paranitrophenol is not because

(a) orthonitrophenol hasintramolecular hydrogen bondingwhileparanitrophenol has intermolecular hydrogen bonding.

(b) both ortho- and paranitrophenol have intramolecular hydrogen bonding.

(c) orthonitrophenol has intermolecular hydrogen bonding and paranitrophenol has intramolecular hydrogen bonding.

(d) Vander Waals forces are dominant in orthonitrophenol.

15. The shape of 4PCl+ , 4PCl and AsCl5 are respectively

(a) square planar, tetrahedral and see-saw

(b) tetrahedral, see-saw and trigonal bipyramidal

(c) tetrahedral, square planar and pentagonal bipyramidal

(d) trigonal bipyramidal, tetrahedral and square pyramidal

16. Which of the following pairs are isostructural?

17. The correct order of dipole moment is

18. Which of the molecule is T shaped?

19. Amongst LiCl, BeCl2, MgCl2 and RbCl the compounds with greatest and least ionic character, respectively are

20. In which molecule sulphur atom is not sp3 hybridised

21. Which of the following orders regarding the bond order is correct?

22. Which of the following species is electron deficient?

(a) BCl3

23. The species that does not contain peroxide ions are

24. The hybridization of carbon atoms in C C single bond of HC  C CH = CH2 is

25. The shape of XeOF4 is

(a) square pyramidal

(c) distorted octahedral

26. The geometry of XeO2F2 is

(a) plane triangular (b) see saw

(b) square antiprismatic

pentagonal bipyramidal

square planar (d) tetrahedral

27. The hybridisation of P in 3 4PO is same as in

(a) I in 4ClI (b) S in SO3 (c) N in 3NO (d) S in 2 4SO

28. The shape of CO2 molecule is similar to

(a) H2O

(i) CN

BeF2

SO

none of these 29. Which of the following have identical bond order

(a) (i) and (iii) (b) (ii) and (iv) (c) (i) and (iv) (d) (ii) and (iii) 30. Which of the following has highest ionic character?

(a) MgCl2 (b) CaCl2

BaCl2 (d) BeCl2 31. Which of the following pair has electron deficient compounds? (a)

Valency expresses generally (a) Total e– in an atom

(c) Oxidation number of an element

Which overlapping is involved in HCl molecule?

(a) s – s overlap

p – p overlap

Atomicity of an element

Combining capacity of an element

s

Which of the following bonds will have directional character?

(a) Ionic bond

(c) Covalent bond

d overlap (d) s – p overlap

Metallic bond

Both covalent & metallic

In which following set of compound/ion, has linear geometry?

There is no effect of lone pair on bond angle in

Which part(s) has same bond angles?

Correct option are

38. Which set of molecules is polar?

An example of a polar covalent molecule is

40. Number of co-ordinate bonds present in sulphuric acid are

41. The number of co-ordinate bonds present in SO3 molecule are

CHEMICAL BONDING_XI

42. Which of the following is not possible due to back bonding?

(a) State of hybridisation may change

(c) Bond angle always decreases

(b) Bond order increases

(d) Lewis acidic strength decreases

43. Among following molecule, N – Si bond length is shortest in

(a) N(SiH3)3

(c) NH2(SiH3)

(b) NH(SiH3)2

(d) All have equal N-Si bond length

44. An atom with atomic number 20 is most likely to combine chemically with the atom whose atomic number is

(a) 11 (b) 16 (c) 18 (d) 10

45. Least ionic bond is

(a) P – Cl

(b) S – Cl

46. Acetic acid exists as dimer in benzene due to

(a) Condensation reaction

(c) Presence of carboxyl group

47. Glycerol is more viscous than glycol, the reason is

(a) Higher molecular weight

(c) More extent of hydrogen bonding

48. The correct order of volatility is

(a) NH3< H2O

(c) CH3OH > CH3 – O – CH3

49. Maximum number of H–bonding is shown by

(a) H2O

50. Intramolecular H-bond

(a) Decreases volatility

(c) Increases viscosity

(b) H2Se

(c) I – Cl (d) Cl Cl

(b) Hydrogen bonding

(d) None of the above

(b) More covalent

(d) Complex structure

(b) p-nitro phenol < o-nitro phenol

(d) HF > HCl

(c) H2S (d) HF

(b) Increases melting point

(d) Increases vapour pressure

JEE Advanced Level Questions

Single Correct Answer Type Questions

1. What is the oxidation state of C in CO molecule?

(a) +2 (b) +3

(c) +4 (d) +1

2. An oxide of chlorine which is an odd electron molecule is ………… .

(a) ClO2

Cl2O6

3. Boron compounds behave as Lewis acids because of their

(a) Acidic nature

(c) Electron deficient character

Cl2O7 (d) Cl2O

(b) Covalent nature

(d) Ionising property

4. The strength of bonds by 2s – 2s, 2p – 2p and 2p – 2s overlapping has the order

(a) s – s > p – p > s – p

(c) p – p > s – p > s – s

(b) s – s > p – s > p – p

(d) p – p > s – s > p – s

5. Which of the following configuration shows second excitation state of iodine?

6. Which of the following compounds has a 3-centered –2 e–-bond?

7. Linear combination of two hybridized orbitals belonging to two atoms and each having one electron leads to a

(a) Sigma bond

(c) Co-ordinate covalent bond

(b) Double bond

(d) pi bond

8. Which trihalide ion is unknown due to the absence of vacant d-orbital?

Bond order and formal charge of perchlorate ion are

11. Hybridisation state of I in ICl2 + is

12. The type of hybrid orbitals used by chlorine atom in ClO

13. The AsF5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are

CHEMICAL BONDING_XI

14. Select the correct matching: List-I List-II

(A) XeF4 (i) Pyramidal

(B) XeF6 (ii) T-shape

(C) XeO3 (iii) Distorted octahedral

(D) XeOF2 (iv) Square planar

(a) (iv) (iii) (i) (ii)

(b) (i) (ii) (iii) (iv)

(c) (ii) (i) (iii) (iv)

(d) (iv) (i) (iii) (ii)

15. The xenon compound(s) that are iso-structural with 23IBr and BrO respectively, are

(a) Linear, XeF2 and pyramidal, XeO3

(c) Bent, XeF2 and planar, XeO

16. Which of the following molecule is having 2p

17. Which is the hybridisation on the central atom of SiO2? (a) sp

sp2

Bent, XeF2 and pyramidal, XeO3

Linear, XeF

and tetrahedral, XeO3

18. Among the following orbital/bonds, the angle is minimum between

(a) sp3 bonds

(c) H–O–H bond in water

px and py orbitals

sp bonds

19. In which of the following change, adjacent bond angle increases? (a) BeF2 + 2F– →

20.  F – As – F Bond angle in AsF3Cl2 molecules is

(a) 90° and 180° (b) 120°

21. Which of the following species are polar?

(i) C6H6 (ii) XeF

(v) SF6

Correct answer is

(a) (ii) and (iv) (b) (i), (ii) and (v)

22. The correct order of dipole moment is

(a) CH4 < NF3 < NH3 < H2O

(c) NH3 < NF3 < CH4 < H2O

23. Which species do not exists?

(a) AlF6 –3 (b) BF4 –

(i) and (v)

(iii) and (iv)

24. The pair of compounds which can form a co-ordinate bond is

(a) (C2H5)3B and (CH3)3N (b) HCl and HBr

(c) BF3 and NH3 (d) (a) & (c) both

25. Which of the following has no co-ordinate bond?

(a) PH3 (b) P2H6 2+

26. The correct order of B-F bond length follows the sequence

27. Back bonding is absent in

28. Formula of a metal oxide is MO, formula of its phosphate will be

29. Ionic conductances of hydrated M+ ions are in the order

(a) Li+(aq) > Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)

(b) Li+(aq) > Na+(aq) < K+(aq) < Rb+(aq) < Cs+(aq)

(c) Li+(aq) > Na+(aq) > K+(aq) > Rb+(aq) < Cs+(aq)

(d) Li+(aq) < Na+(aq) < K+(aq) < Rb+(aq) < Cs+(aq)

30. Which of the following halides has the highest melting point?

(a) NaCl (b) KCl

31. Pick out the wrong statement

(a) LiF has less solubility in water than LiI.

(b) Lattice energy of MgO is greater than Na2O.

(c) LiH is more stable than KH.

(d) KO2 is diamagnetic and colourless.

32. The correct expected order of decreasing lattice energy is (a) CaO > MgBr2 > CsI

33. Which bond is least covalent?

34. Which oxide is the most ionic?

35. Which of the following forms metal oxide on heating?

36. Increasing order of stability

of the given carbonates is

II < III < I

NaBr (d) NaF

37. In which of the following set, the value of bond order will be 2.5?

Which of the following group of molecules have 2½ bond order?

39. In a homonuclear molecule, which of the following set of orbitals are degenerate?

40. Which of the following LCAO represent formation of bonding molecular orbital?

One or More Correct Answer Type Questions

1. Which of the following molecules or ions is/are not linear?

2. Which of the following are non–polar but contain polar bonds?

3. The 90° angles between bond pair bond pair of electrons exist in

4. The shape of ClO4 – is

Square planar

5. Which of the following species is/are paramagnetic?

6. Geometry or shape of 3I is/are

8. Which among the following has bond order zero?

9. In which species the hybrid state of central atom is sp

10. KF combines with HF to form KHF2. The compound contains the species

(a) K+, F and H+

(c) K+ and [HF2]

one cation and one anion

11. Among the following species, identify the isostructural pairs

NF3, 3 NO,BF3, H3O+ , 3 N, 3I

(a)

[NF3, H3O+] and [ 3 NO,BF3]

12. Which of the following molecule involves sp2 hybridisation?

13. NH3 and BF3 form adduct readily

(a) Hybridization of NH3 remains same.

(b) through co-ordinate bond between B and N.

(c) Hybridization of NH3 changes from sp3 to sp2 .

(d) Hybridization of B changes from sp2 to sp3

14. Which of the following molecular species has unpaired electron(s)?

(a)

15. Which of the following species is/are superoctet molecule?

(a)

16. Which of the following statements are correct about sulphur hexafluoride?

(a) all SF bonds are equivalent

(b)

6SF is a planar molecule

(c) oxidation number of sulphur is the same as number of electrons of sulphur involved in bonding

(d) sulphur has acquired the electronic structure of the gas argon

17. Which of the following combination of bond pair (b.p.) and lone pair (l.p.) give same shape?

(i) 3 b.p.+1 l.p. (ii) 2 b.p. 2+ l.p. (iii) 2 b.p.+1 Lp. (iv) 2 b.p.+0 1.p.

(v) 3 b.p.+2 l.p. (vi) 2 b.p. 3+ l.p.

(a) (ii) and (iii) (b) (iv) and (v) (c) (iv) and (vi) (d) (iii) and (vi)

18. Select the true statement(s) among the following

(a) Pure overlapping of two xy d orbitals along x -axis results in the formation of -bond

(b) 232 NONONO +−−  is the correct order of bond angle as well as N o bond order

(c) 3333 NFNClNBrNI  is tho correct order of Lewis basic character as well as bond angle

(d) HFHClHBrHI  is the correct order of dipole moment as well as boiling point Valence Bond Theory

19. yp -orbital can not form -bond by lateral overlap with

(a) xz d -orbital (b) 22 xy d -orbital (c) xy d -orbital (d) sp -orbital

20. Which of the following statements is/are correct?

(a) All carbon to carbon bonds contain a sigma bond and one or more -bonds

(b) All carbon to carbon bonds are sigma bonds

(c) All oxygen to hydrogen bonds are hydrogen bonds

(d) All carbon to hydrogen bonds are sigma bonds

Matrix Match Type Questions

1. Match the column & choose the correct option among the options given.

(Note: Each statement in column I has only one match in column II.)

Column-I

I. SO2

II. SO3

III PCl5

IV. 4PCl

(a) I-A ; II-E ; III-C ; IV-D

(c) I-E ; II-A ; III-C ; IV-D

(A) sp2, angular

(B) sp3 , tetrahedral

Column-II

(C) sp3d, trigonal bipyramidal

(D) sp3d, see saw

(E) sp2, trigonal planar

(b) I-A ; II-E ; III-D ; IV- C

(d) I-E ; II-E ; III-D ; IV-B

2. Match the column & choose the correct option among the options given.

(Note: Each statement in column I has only one match in column II.)

Column-I

I. CH4 (A) tetrahedral

II. NH3

III HCl

Column-II

(B) hydrogen bonding

(C) see saw

IV. SF4 (D) linear

(E) trigonal planar

(a) I-C ; II-B; III-C; IV-A (b) I-A ; II-B; III-D; IV-C

(c) I-C ; II-E; III-B; IV-A

(d) I-A ; II-C; III-B; IV-C

3. Match the column & choose the correct option among the options given.

(Note: Each statement in column I has only one match in column II.)

Column-I

(Compound)

I. N2

II. CO2

III H2O

IV. XeF4

(a) I-A ; II-E ; III-C ; IV-C

(c) I-A ; II-B ; III-E ; IV-D

Column-II

(Shape/Property)

(A) zero dipole moment with non-polar bonds

(B) zero dipole moment with polar bonds

(C) tetrahedral

(D) square planar

(E) angular

(b) I-A ; II-B ; III-D ; IV-E

(d) I-A ; II-D ; III-E ; IV-B

4. Match the Column

(a) I2C6 (P) Allatomsaresp3 hybridised

(b) SiC (Q) have non planar structure

(c) Al2(CH3)6

(d) S3O9

(a) A-R, B-P, C-S, D-Q

(c) A-Q, B-R, C-S, D-P

5. Match the Column

have 3c - 4e bonds

have 3c – 2e bonds

(a) ICl3 (P) Hybridisation of central atom is similar in both dimer and monomer form.

(b) AlCl3 (Q) Both monomer and dimer forms are planar

(c) AlF3 (R) In dimer form all atoms are sp3 hybridised.

(d) NO2 (S) Does not exist in dimer form

6. Match the Column

Column-I Column-II

D-R

(a)

(d) py - orbital

dsp2 (square planar)

(a) A-QR,B-PQRS, C-QS, D-PQRS (b) A-QS,B-PSRQ, C-QS, D-PSRQ

(c) A-RQ,B-PSRQ, C-QS, D-SQRQ (d) A-RS,B-PSRQ, C-QR, D-PSRQ

7. Match the Column

Column-I

(i) Methane molecules (a) covalency

(ii) Sharing of electrons

(iii) H2O

Column-II

(b) tetrahedral molecules

(c) pyramidal molecules

(iv) NH3 (d) polar molecule

(a) (i)-(b); (ii)–(a); (iii)-(d); (iv)-(c) (b) (i)-(a); (ii)–(d); (iii)-(c); (iv)-(b)

(c) (i)-(c); (ii)–(b); (iii)-(c); (iv)-(d) (d) (i)-(c); (ii)–(d); (iii)-(a); (iv)-(c)

Integer Type Questions

8. Consider following compounds A to E (a)

(e) ( ) 2 4 XeF n+

If value of n is 4, then calculate value of “ pq  “ here, ‘ p ‘ is total number of bond pair and ‘ q ‘ is total number of lone pair on central atoms of compounds (A) to (E).

9. Consider the following species and find out total number of species which are polar and can act as Lewis acid

42233434 CCl,CO,SO,AlCl,HCHO,SO,SiCl,BCl,CF

10. Sum of  and -bonds in 4NH+ cation is ……

11. Calculate the value of XY , for 4 XeOF.(X = Number of  bond pair and Y = Number of lone pair on central atom)

12. How many right angle, bond angles are present in 5TeF molecular ion?

13. Consider the following species & find out how many are polar & can act as Lewis acid CO2, SO2, CCl4, AlCl3, HCHO, SO3, SiCl4, BCl3, CF4

14. In the compound PCIk F5-k possible values of k are 0 to 5 then sum of all possible value of k for the compounds having zero dipole moment, is :

15. Find the total number of non-linear species out of given species:

16. In cation of XeF6(s) total number of orbtials of Xe involved in hybridisation is:

17. Total number of covalent bonds in C₂O2, is ……….

18. In how many of the following cases ratio of σ-bond and π-bond is 1:1?

19. The difference of number of sigma bond and π bond in 1, 3, 5-tricynobenzene is

20. In the chromate dianion, how many Cr-O bonds are identical

Linked Comprehension Type Questions

Comprehension # 1

Ionic bond is defined as the electrostatic force of attraction holding the oppositely charged ions. Ionic compounds are mostly crystalline solids having high melting and boiling points, electrical conductivity in molten state, solubility in water etc. Covalent bond is defined as the force which binds atoms of same or different elements by mutual sharing of electrons in a covalent bond. Covalent compounds are solids, liquids or gases. They are low melting and boiling point compounds. They are more soluble in non-polar solvents.

21. The valence electrons not involved in formation of covalent bonds are called

(a) non-bonding electrons

(c) unshared pairs

(b) lone pairs

(d) all of these

22. The amount of energy released when one mole of ionic solid is formed by close packing of gaseous ion is called

(a) Ionisation energy (b) Solvation energy

(c) Lattice energy (d) Hydration energy

23. Examples of covalent compounds are

(a) Urea (b) Sugar

(c) Both (a) and (b) (d) Calcium fluoride

24. Which of the following gives a white ppt. with AgNO3?

(a)

Comprehension # 2

When hybridisation involving d-orbitals are considered then all the five d -orbitals are not degenerate, rather 2 2 , x

dyd−= and ,, xyyzxx ddd form two different sets of orbitals and orbitals of appropriate set is involved in the hybridisation.

25. In 32spd hybridisation, which sets of d -orbitals is involved?

26. In 33spd hybridisation, which orbitals are involved?

27. Molecule having trigonal bipyramidal geometry and 3 spd hybridization, d -orbitals involved is

28. Which of the following orbitals can-not undergo hybridization amongst themselves.

(a) only II (b) II & III

Comprehension # 3

II & IV

Hybridisation involves the mixing of orbitals having comparable energies of same atom. Hybridised orbitals perform efficient overlapping than overlapping by pure s, p or d orbitals.

29. Which of the following is not correctly match between given species and type of overlapping?

(a) 3XeO : Three (dπ – pπ) bonds (b) 24HSO : Two (dπ – pπ) bonds

(c) 3 SO: Three (dπ – pπ) bonds (d) 4HClO : Three (dπ – pπ) bonds

30. Consider the following compounds and select the incorrect statement from the following:

3322333742NH,PH,H S,SO,SO,BF,PCl,IF,P,H

(a) Six molecules out of given compounds involves hybridization

(b) Three molecules are hypervalent compounds

(c) Six molecules out of above compounds are non-planar in structure

(d) Two molecules out of given compounds involves (dπ – pπ) bonding as well as also involves (pπ – pπ) bonding

Comprehension # 4

A covalent bond, in which electrons are shared unequally and the bonded atoms acquire a partial positive and negative charge, is called a polar covalent bond. Bond polarity is described in terms of ionic character which usually increases with increasing difference in the electronegativity between bonded atoms. The polarity of a molecule is indicated in terms of dipole moment ().

Similarly in ionic bond, some covalent character is introduced because of the tendency of the cation to polarise the anion. The magnitude of covalent character in the ionic bond depends upon the extent of polarization caused by cations.

In general:

It’s a vector quantity and is defined as the product of distance between two heteronuclei and partial charge, developed any of the two heteronuclei atoms. For example,

Dipole moment  = x e

(i) Smaller the size of cation, larger is its polarizing power.

(ii) Larger the anion, more will be its polarisabilty.

(iii) Amongtwocationsofsimilarsize,thepolarizingpowerofcationswithpseudo-noblegasconfiguration (ns2np6nd10) is larger than cation with noble gas configuration (ns2np6) e.g. polarizing power of Ag+ is more than K+ .

Read the above passage carefully and answer the questions given below it.

31. The decreasing order of polarity of the bond in NH3, PH

is in the order

32. Which has maximum covalent character?

33. Which is highest melting point halide? (a)

34. The correct order of decreasing polarity is: (a)

Comprehension # 5

Hydrogen bonding originates from the dipole-dipole interaction between H-atom and any of the other atom like F, O, N and in some cases with Cl atom. There are two types of H-bonding like intermolecular and intramolecular H-bonding

35. Which of the following molecule does not consist of intramolecular H-bonding.

(a) Chloral

ortho hydroxy benzaldehyde

hydrate

36. Which of the following plot for  (density) v/s T (temperature) for liquid H2O is correct.

(a) (b)

0ºC T 0ºC T

(c) (d)

0ºC T 0ºC T

37. Which of following statement is incorrect

(a) Boiling point of H2O2 is greater that of H2O

(b) Ethylene glycol is less viscous than glycerol.

(c) o-nitrophenol can be separated from its meta and paraisomer using its steam volatile property.

(d) In ice each 'O' atom is tetrahedrally arranged by four H-atoms which are all equidistant.

Comprehension # 6

Phosphorusformsdifferenthalidecompounds withhalogens.In crystalline state some ofthem exist asion pairs.

38. What is hybridisation of central atom of anionic part of PBr5 in crystalline state.

(a) sp2 (b) sp3 (c) sp (d) not applicable

39. What is the difference between bond angles in cationic species of PCl5 and PBr5 in solid state.

(a) 60º (b) 109º28' (c) 0º (d) 90º

40. All possible bond angles in anionic part of PCl5 are

(a) 109º,28º (b) 90º,180º (c) 90º, 120º, 180º (d) 72º, 90º, 180º

Comprehension # 7

Hybridisation is the mixing of atomic orbital of comparable energy and the number of hybrid orbitals formed is equaltothe number of pure atomic orbitals mixed upandhybrid orbitals are occupied by -bond pairandlonepair.

41. Whichofthefollowing geometryismost likelytonot formfrom sp3dhybridisationof thecentralatom.

(a) Linear (b) Tetrahedral (c) T-Shaped (d) See-Saw

42. The orbital is not participated in sp3d2 hybridisation.

(a) px (b) dxy (c) dx2 –y2 (d) pz

43. "The hybrid orbitals are at angle of Xº to one another " this statement is not valid for which of the followinghybridisation.

(a) sp3 (b)sp2 (c)sp3d2 (d)sp

Assertions and Reasons Type Questions

These questions contains, Statement-I (assertion) and Statement-II (reason).

(A) Statement-1 is True, Statement-Il is True; Statement-II is a correct explanation for Statement-I

(B) Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I

(C) Statement-I is True, Statement-II is False.

(D) Statement-I is False, Statement-II is True.

44. Statement-I: N2O, CO2, & I3 - are isostructral.

Statement-II: All three have same hybridise central atom.

45. Statement-I: Al(OH)3 is amphoteric in nature.

Statement-II: Al-O and O-H bonds can be broken with equal case in Al(OH)3

46. Statement-I: Boron does not show univalent nature but unipositive nature of thallium is quite stable.

Inert pair effect predominates in thallium.

47. Statement-I: In NF3 molecule lone pair resides in sp3 hybrid orbital.

Statement-II: NF3 has pyramidal shape.

48. Statement-I: H2SO4 is more viscous than water. Statement-II: In H2SO4 ,S has highest oxidation state.

EXERCISE-5A

Single Correct Answer Type Questions

1. Among the following the maximum covalent character is shown by the compound- [AIEEE-2011]

(a) AICI3 (b) MgCl2 (c) FeCl2 (d) SnCl2

2. The structure of IF7 is [AIEEE-2011]

(a) octahedral

(c) square pyramid

(b) pentagonal bipyramid

(d) trigonal bipyramid

3. The hybridisation of orbitals of N atom in NO3 - , NO2 + and NH4 + are respectively: [AIEEE-2011]

(a) sp, sp3, sp2 (b) sp², sp³, sp (c) sp, sp2, sp3 (d) sp², sp,sp³

4. Which of the following has maximum number of lone pairs associated with Xe? [AIEEE-2011]

(a) Xeo3 (b) XeF4 (c) XeF6 (d) XeF2

5. The number of types of bonds between two carbon atoms in calcium carbide is :- [AIEEE-2011]

(a) One sigma, two pi

(c) Two sigma, one pi

(b) One sigma, one pi

(d) Two sigma, two pi

6. Ortho-Nitrophenol is less soluble in water than p- and m-Nitrophenols because: [AIEEE-2012]

(a) Melting point of o-Nitrophenol is lower than those of m- and p-isomers

(b) o-Nitrophenol is more volatile in steam than those of m- and p- isomers

(c) o-Nitrophenol shows Intramolecular H-bonding

(d) o-Nitrophenol shows Intermolecular H-bonding

7. Among the following species which two have trigonal bipyramidal shape? [AIEEE Online-2012]

(I) NI3

8. Among the following, the species having the smallest bond is:

9. Which of the following has the square planar structure:

(a)

BF4 –

10. The compound of Xenon with zero dipole moment is: [AIEEE-Online-2012]

(a) XeO3 (b) XeO

11. Among the following the molecule with the lowest dipole moment is :- [AIEEE-Online-2012]

(a)

12. Although CN– ion and N2 molecule are isoelectronic, yet N2 molecule is chemically inert because of: [AIEEE-Online-2012]

(a) Uneven electron distribution

(b) Absence of bond polarity

(c) Presence of more number of electrons in bonding orbitals

(d) Long bond energy

13. Bond order normally gives idea of stability of a molecular species. All the molecules viz. H2, Li2 and B2 have the same bond order yet they are not equally stable. Their stability order is: [JEE

14. Which has trigonal bipyramidal shape? [JEE Main-Online-2013]

(a) XeOF4

XeO3

XeO3F₂ (d) XeOF2

15. Which one of the following does not have a pyramidal shape? [JEE Main-Online-2014)

(a)

16. The geometry of XeOF4, by VSEPR theory is:-

(a) Trigonal bipyramidal (b) Octahedral

(c) Pentagonal planar (d) Square pyramidal

Main-Online-2015]

17. The correct order of thermal stability of hydroxides is: [JEE Main-Online-2015]

(a) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 < MG(OH)2

(b) MG(OH)2 < Sr(OH)2 < Ca(OH)2 < Ba(OH)2

(c) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2

(d) Ba(OH)2 < Ca(OH)2 < Sr(OH)2 < Mg(OH)2

18. Which among the following is the most reactive?

(a) Cl2 (b) Br2

[JEE Main 2015]

I2 (d) ICI

19. Which intermolecular force is most responsible in allowing xenon gas to liquefy?

[JEE Main-Online-2016]

(a) lonic (b) Instantaneous dipole-induced dipole

(c) Dipole-dipole (d) lon-dipole

20. The bond angle H-X-H is the greatest in the compound: [JEE Main-Online-2016]

(a)

21. Which of the following is paramagnetic?

Which one of the following is an oxide?

23. The correct sequence of decreasing number of π-bonds in the structures of H

is:-

24. The increasing order of the boiling points for the following compounds is:

Main-Online-2017]

25. The group having triangular planar structures is:-

26. Which of the following compounds contain(s) no covalent bond(s)?

27. Total number of lone pair of electrons in l3 - ion is:

28. In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic? [JEE Main-Online-2019]

(a) N₂ → N₂+ (b) NO → NO+ (c) O2 → O2 2- (d) 02 → 02 +

29. According to molecular orbital theory, which of the following is true with respect to Li2 + and Li2 - ?

[JEE Main - Online- 2019]

(a) Both are unstable (b) Li2 + is unstable and Li2 - is stable

(c) Li2 + is stable and Li2 - is unstable (d) Both are stabel

30. Two pi and half sigma bonds are present in:

[JEE Main-Online-2019]

(a) N+ ₂ (b) N₂ (c) O2 + (d) O2

31. The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF4, respectively, are:-

[JEE Main-Online-2019]

(a) sp³d and 1 (b) sp3d and 2 (c) sp³d2 and 1 (d) sp³d2 and 2

32. The amphoteric hydroxide is: [JEE Main-Online-2019]

(a) Ca(OH)2 (b) Be(OH)2 (c) Sr(OH)2 (d) Mg(OH)2

33. The ion that has sp3d2 hybridization for the central atom, is: [JEE Main-Online-2019]

(a) [ICI2]– (b) [IF6]– (c) [ICI4]– (d) [BrF2]–

34. The covalent alkaline earth metal halide [JEE Main-Online-2019]

(X = Cl, Br, I) is:

(a) CaX2 (b) SrX₂ (c) BeX

35. Among the following molecules/ions,

which one is diamagnetic and has the shortest bond length?

Main-Online-2019]

36. Among the following, the molecule expected to be stabilized by anion formation is: C2, O2, NO, F2

C₂

37. Among the following species, the diamagnetic molecule is

(a) O₂

NO

Main-Online-2019]

Main-Online-2019)

CO

38. The group number, number of valence electrons, and valency of an element with atomic number 15, respectively are [JEE Main-Online-2019]

(a) 16, 5 and 2 (b) 16, 6 and 3

15, 5 and 3 (d) 15, 6 and 2

39. The relative strength of interionic/intermolecular forces in decreasing order is: [JEE Main-Online-2020]

(a) ion-dipole > ion-ion > dipole-dipole

(c) ion-dipole > dipole-dipole > ion-ion

(b) dipole-dipole > ion-dipole > ion-ion

(d) ion-ion > ion-dipole > dipole-dipole

40. If the magnetic moment of a dioxygen species is 1.73 B.M, it may be [JEE Main - Online - 2020]

(a) O2 – or O2 + (b) O2, or O2 + (c) O2 or O2 – (d) O₂, O2 – or O2 +

41. The shape/structure of [XeF5]- and XeO3F2, respectively, are: [JEE Main-Online-2020]

(a) pentagonal planar and trigonal bipyramidal

(b) trigonal bipyramidal and pentagonal planar

(c) octahedral and square pyramidal

(d) trigonal bipyramidal and trigonal bipyramidal

42. The molecular geometry of SF6 is octahedral. What is the geometry of SF4 (including lone pair(s) of electrons, if any)? [JEE MAINS-Online-2020]

(a) Trigonal bipyramidal (b) Square planar

(c) Tetrahedral (d) Pyramidal

43. If the boiling point of H2O is 373 K, the boiling point of H2S will be: [JEE Main-Online-2020]

(a) Greater than 300 K but less than 373 K (b) Less than 300 K

(c) Equal to 373 K (d) More than 373 K

44. Of the species, NO, NO+, NO2+, NO- the one with minimum bond strength is:

[JEE Main - Online-2020]

(a) NO2+ (b) NO+

45. The structure of PCI5 in the solid state is

(a) square pyramidal

(c) NO (d) NO–

[JEE Main-Online-2020]

(b) tetrahedral [PCl4]+ and octahedral [PCl6]–

(c) square planar [PCl4]+ and octahedral [PCl6]– (d ) trigonal bipyramidal

46. The compound that has the largest H-M-H bond angle (M=N, O, S, C), is: [JEE Main-Online-2020]

(a) H₂O (b) CH4 (c) NH3 (d) H2S

47. The correct shape and I-I-I bond angles respectively in I3 - ion are [JEE Main-Online-2021]

(a) Distorted trigonal planar, 135° and 90° (b) T-shaped; 180° and 90

(c) Trigonal planar, 120° (d) Linear, 180°

48. According to molecular theory, the species among the following that does not exist is:

[JEE Main-Online-2021]

(a) He2 + (b) He2 –

(c) Be2 (d) 02 2–

49. Which among the following species has unequal bond lengths? [JEE Main-Online-2021]

(a) BF4 – (b) XeF4 (c) SF4 (d) SiF4

50. Given below are two statements: [JEE Main-Online-2021]

Statement I: o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding.

Statement II: 0-Nitrophenol has high melting due to hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below:

(a) Statement I is false but Statement II is true

(b) Both statement I and statement II are true

(c) Both statement I and statement II are false

(d) Statement I is true but statement II is false

51. In molecule, the hybridization of carbon 1,2,3 and 4 respectively are:

[JEE Main-Online-2021]

(a) sp3 , sp , sp3 , sp3

(b) sp2 , sp2 , sp² ,sp3

(c) sp², sp, sp², sp3 (d) sp2, sp3, sp², sp3

52. Which one of the following pairs of species have the same bond order?

[AIEEE–2008]

(a) CN– and NO+ (b) CN– and CN+ (c) 02 – and CN– (d) NO+ and CN+

53. Using MO theory predict which of the following species has the shortest bond length?

(a) 2O (b) 2 2O

[AIEEE–2009]

54. In which of the following pairs the two species are not isostructural ? [AIEEE–2012]

(a) AIF6 3

and SF6 (b) CO

55. The formation of molecular complex BF3–NH3, results in a change in hybridisation of boron :

(a) from sp³ to sp³d

(c) from sp³ to sp2

56. In which of the following ionization processes the bond energy has increased and also the magnetic behaviour has changed from paramagnetic to diamagnetic?

57. Which one of the following molecules is polar?

58. In which of the following sets, all the given species are isostructural?

2013] (a) BF3, NF3, PF3, AIF3

Which one of the following molecules is expected to exhibit diamagnetic behaviour?

In which of the following pairs of molecules/ions, both the species are not likely to exist?

Based on data provide the value of electron gain enthalpy of fluorine would be:

63. Stability of the species L. Li, and Li; Increases in the order of:– [JEE Main– 2013]

(a) 222 LiLiLi+−

(c) 222 LiLiLi−+

64. Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated?

(a) IF7 : Pentagonal bipyramid

(c) ICI3 : Planar dimeric

65. The correct order of bond dissociation energy among N

BrF5: Trigonal bipyramid

BrF3 : Planar T–shaped

is shown in which of the following arrange?

–2014]

66. Which one of the following molecules is paramagnetic?

67. Which one of the following properties is not shown by NO?

(a) It is diamagnetic in gaseous state

(b) It is a neutral oxide

(c) It combines with oxygen to form nitrogen dioxide

(d) It's bond order is 2.5

Main– Online–2014]

2014]

68. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules [JEE Main –2015]

(a) ion–ion interaction

(b) ion–dipole interaction

(c) London force (d) Hydrogen band

69. Molecule AB has a bond length of 1.617 Å and a dipole moment of 0.38 D. The fractional charge or atom (absolute magnitude) is : [JEE Main– Online–2015]

(e0 = 4.802x10–10 esi) :–(a) 0

70. The species in which the N atom is in a state of sp hybridization is: [JEE Main –2016]

(a) NO

71. The group of molecules having identical shape is:

(a) SF4, XeF4, CCI4,

(b) CIF3, XeOF2, XeF3 +

Main– Online–2016]

72. Assertion: Among the carbon allotropes, diamond is an insulator, whereas, graphite is a good conductor of electricity.

Reason: Hybridization of carbon in diamond and graphite are sp3 and sp², respectively.

[JEE Main– Online–2016]

(a) Assertion is incorrect statement, but the reason is correct.

(b) Both assertion and reason are correct, and the reason is the correct explanation for the assertion.

(c) Both assertion and reason are incorrect.

(d) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion.

73. sp3d2 hybridization is not displayed by: [JEE Main– Online–2017]

(a) [CrF6]3– (b) BrF5 (c) PF5 (d) SF6

74. Which of the following species is not paramagnetic? [JEE Main– 2017]

(a) B2 (b) NO

(c) CO (d) O2

75. According to molecular orbital theory, which of the following will not be a viable molecule?

[JEE Main– 2018]

(a) He2 2+ (b) He2 + (c) H2 – (d) H2 2–

76. The correct statement about ICI5, and 4ICI , is [JEE Main– Online–2019]

(a) ICI5 is trigonal bipyramidal and 4ICI , is tetrahedral.

(b) ICI5 is square pyramidal and 4ICI , is tetrahedral.

(c) ICI5, is square pyramidal and 4ICI , is square planar.

(d) Both are isostructural

77. HF has highest boiling point among hydrogen halides, because it has : [JEE Main– Online–2019]

(a) lowest dissociation enthalpy

(b) strongest van der Waals' Interactions

(c) strongest hydrogen bonding (d) lowest ionic character

78. During the change of O2 to 2O the incoming electron goes to the orbital: [JEE Main– Online–2019]

(a) σ*2Pz (b) π*2Py

*2Px

79. The correct statement among the following is [JEE Main– Online–2019]

(a) (SiH3)3 N is pyramidal and more basic than (CH3)3N

(b) (SiH3)3 N is planar and more basic than (CH3)3N

(c) (SiH3)3 Nis pyramidal and less basic than (CH3)3N

(d) (SiH3)3 Nis planar and less basic than (CH3)3N

80. The bond order and the magnetic characteristics of CN are: [JEE Main– Online–2020] (a)

81. The dipole moments of CCI4, CHCI3, and CH4, are in the order:

4

82. Arrange the following bonds according to their average bond energies in descending order:

C–CI, C–Br, C–F, C–I [JEE Main– Online–2020]

(a) C–1 > C–Br > C–CI > C–F (b) C–Br> C–1 > C–CI > C–F

(c) C–F > C–CI > C–Br> CI (d) C–CI > C–Br > CI > C–F

83. The predominant Intermolecular forces present in ethyl acetate, a liquid, are:

[JEE Main– Online–2020]

(a) hydrogen bonding and London dispersion

(b) Dipole–dipole and hydrogen bonding

(c) London dispersion and dipole–dipole

(d) London dispersion, dipole–dipole and hydrogen bonding

84. The number of sp2 hybrid orbitals in a molecule of benzene is : [JEE Main– Online–2020]

(a) 24 (b) 6 (c) 12 (d) 18

85. If AB4 molecule is a polar molecule, a possible geometry of AB4 is [JEE Main– Online–2020]

(a) Square pyramidal

(c) Square planar

(b) Tetrahedral

(d) Rectangular planer

86. Match the type of interaction in Column A with the distance dependence of their interaction energy in Colum B:

Column A

(i) ion – ion

(ii) dipole – dipole

(iii) London dispersion

(a) (I)–(a), (II)–(b). (III)–(c)

(c) (I)–(a), (II)–(b), (III)–(d)

[JEE Main– Online–2020]

Column B

(b) (I)–(a), (II)–(c), (III)–(d)

(d) (I)–(b), (II)–(d), (III)–(c)

87. The intermolecular potential energy for the molecules A, B, C and D given below suggests that:

[JEE Main– Online–2020]

(a) D is more electronegative than other atoms

(c) A–B has the stiffest bone

88. Which of the following are isostructural pairs?

(a) C and D only

(c) A and C only

(b) A–D has the shortest bond length

(d) A–A has the largest bond enthalpy

[JEE Main– Online–2021]

(d) B and C only

89. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:

[JEE Main– Online–2021]

Assertion A: The H–O–H bond angle in water molecule is 104.5°

Renson R: The lone pair–lone pair repulsion of electrons is higher than the bond pair–bond pair repulsion.

(a) A is false but R is true

(b) Both A and R are true, but R is not the correct explanation of

(c) A is true but R is false A

(d) Both A and R are true, and R is the correct explanation of A

90. A central atom in a molecule has two lone pairs of electrons and forms three single bands. The shape of this molecule is:

(a) see–saw

(c) T–shaped

[JEE Main– Online–2021]

(b) planar triangular

(d) trigonal pyramidal

EXERCISE-5B

Single Correct Answer Type Questions

1. The nitrogen oxide(s) that contain(s) N-N bond(s) is/are

(a) N2O

N2O3

2. The species having pyramidal shape is/are:

(a) SO3

3. In allene (C3H3), the type(s) of hybridisation of the carbon atoms is (are)

(a) sp and sp3 (b) sp and sp2 (c) only sp2

sp3

4. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen

(a) HNO3, NO, NH4CI, N2

HNO3, NH4CI, NO, N2

5. The shape of XeO2F2, molecule is:

(a) Trigonal bipyramidal

(b) Square planar

6. The total number of lone-pairs of electrons in melamine is.

see-saw

[JEE -2013]

7. Hydrogen bonding plays a central role in the following phenomena: [JEE -2014]

(a) Ice floats in water

(b) Higher Lewis basicity of primary amines than tertiary amines in aqueous solution

(c) Formic acid is more acidic than acetic acid.

(d) Dimerisation of acetic acid in benzene.

8. Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is

[JEE -2014]

9. Match the orbital overlap figures shown in List-I with the description given in List-Il using the code given below the lists. [JEE -2014]

10. The correct statement(s) regarding. (i) HCIO, (ii) HCIO2, (iii) HCIO3, and (iv) HCIO4, is (are)

[JEE -2015]

(a) The number of Cl = O bonds in (ii) and (ii) together is two

(b) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three

(c) The hybridization of Cl in (iv) is sp³

(d) Amongst (i) to (v), the strongest acid is (i)

11. Among the triatomic molecules/ions, BeCl2, N3-, N2O, NO2+, O3, SCI2, ICI2-, I3 - and XeF2, the total number linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the d-orbital(s) is [JEE -2015]

[Atomic number: S = 16, CI = 17, I = 153 and Xe = 54]

12. The total number of lone pairs of electrons in N₂O3, is [JEE -2015]

13. According to Molecular Orbital Theory, [JEE -2016]

(a) C2 2- is expected to be diamagnetic.

(b) O2 2+ is expected to have a longer bond length than O2.

(c) N2 + and N2 - have the same bond order.

(d) He2 + has the same energy as two isolated He atoms.

14. The compound(s) with TWO lone pairs of electrons on the central atom is(are) [JEE -2016]

(a) BrF6 (b) CIF3 (c) XeF4 (d) SF4

15. Among H2, He2 + , Li2, Be2, B2, C2, N2, O2, and F₂, the number of diamagnetic species is [JEE -2017]

(Atomic numbers: H=1, He =2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, F = 9)

16. The sum of the number of lone pairs of electrons on each central atom in the following species is [TeBr6]2-, [BrF2]+, SNF3, and [XeF3]- [JEE -2017]

(Atomic numbers: N=7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54)

17. Among the following, the correct statement(s) is (are) [JEE -2017]

(a) Al(CH3)3, has the three-centre two-electron bonds in its dimeric structure

(b) AICI3, has the three-centre two-electron bonds in its dimeric structure

(c) BH3, has the three-centre two-electron bonds in its dimeric structure

(d) The Lewis acidity of BCI3, is greater than that of AICI3

18. Each of the following options contains a set of four molecules. Identify the option(s) where all four molecules possess permanent dipole moment at room temperature. [JEE -2019]

19. Among

molecules containing cova bond between two atoms of the same kind is

20. Consider the following compounds in the liquid form:

the total number

5CI. When a charged comb is brought near their flowing stream, how many of them show deflection as per the following figure?

JEE-Main Level Questions