4 chemical bonding and molecular structure · chemical bonding and molecular structure 4 synopsis...

Chemical Bonding and Molecular Structure

4

SYNOPSIS

1. Chemical Bond: It is the force which holds the atoms together in a molecule. The causes of formation of bond are

• Tendency of atoms to complete their octets or duplets by rearrangement of their valence electrons (octet theory). • The system acquires minimum energy when atoms are at some equilibrium distance where attractive forces dominate

over repulsive forces.

Chemical Bond

Ionic Bond Covalent Bond Metallic Bond Co-ordinate Bond

2. Ionic Bond: It is the electrostatic force of attraction between oppositely charged ions which are formed by complete transference of electrons from one atom to another to get complete octet or duplet.• Favorable conditions to form ionic bond :

• Formed between electro positive element (group 1, 2, 13) and electro negative element group (15, 16, 17)• ∆EN ≥ 2 · Lower I.E of one atom and more negative Eeg of second atom. Higher lattice energy.• Larger cations, smaller anions.

3. Covalent Bond: Covalent bond is formed by sharing of electrons by two atoms to complete their octet or duplet. (octet theory)• Type of covalent bond: Depending upon the type of overlapping, the covalent bonds are of 2 types:

• Sigma bond (σ): When a bond is formed between two atoms by the overlap of their atomic orbitals along the inter nucleus axis (end to end or head on overlap), the bond formed is called sigma bond .

• Pi-bond (π): It is formed by lateral or side ways overlapping of p-orbitals i.e. by overlapping of p-orbitals in a direction at right angle to the inter nucleus axis.

• Favorable conditions to form covalent bond :• Formed between two electronegative elements (group 14, 15, 16, 17 )• ∆EN < 1.9 · Small cation, larger anions (Fajan’s rule). High charges on cation and anion (Fajan’s rule)• Upon cooling NO2 & ClO2 doesn't form Dimer, so they remain paramgnetic. The reason being that the unpaired

electron is present in unhybrid molecular orbital whereas No & ClO3 forms Dimer, so they change from paramagnetic

55C

hemical B

onding and Molecular Structure

to diamagnetic. The unpaired electron in these molecular is located in hybrid molecular orbital.• Formal charge The charge present at each atom of a covalent compound (neutral, or ion) is called its formal charge,

which is

Formal charge at atom=

Total number of its

valence electrons

Total number of non

bonding electrons around it

Total number of

bonding electrons

1

2�

• Drawbacks of octet theory Octet theory cannot explain

• Nature of force between atoms in the molecules like H2, Cl2 etc.• Formation of BCl3 BeCl2 AlCl3 etc. (electron deficient compounds)• Formation of PCl5, SF6, IF7etc.• Formation of odd electron molecules like NO, NO2 etc.• Shapes of molecules

4. Bond Length: It is the average distance between the nuclei of two bonded atoms. It depends uponSize of the atom: Bond length increases with increase in the size of atom(HI > HBr > HCl > HF)Multiplicity of bond : Bond length decreases with the multiplicity of bond (C – C > C = C > C ≡ C)

Type of hybridizationMore s-character, shorter is the bond length, greater is the acidity with comparable compound.

5. Bond Energy: It is the energy required to break one mole of bonds of particular type of a substance in gaseous state.6. Bond Angle: It is the internal angle between the orbitals containing electron pairs in the valence shell of central atom in

a molecule.• More lone pairs on central atom, smaller is the bond angle due to lp – bp repulsion.• More electronegative central atom, more is the bond angle.• More electronegative surrounding atom, less is the bond angle.

7. Resonance: In some molecules all the properties cannot be explained by single structure and the molecule is supposed to have many structures, but none of these can explain all the properties of the molecule. These different possible structures are called resonating structures and the actual structure is in between of all these structures, called resonance hybrid. This phenomenon is called resonance. Main features of resonance are:• Different structures have same positions of atoms, almost equal energies, same number of shared and unshared

electrons.• The difference in the energy of the most stable resonating structure and resonance hybrid is called resonance energy.• Resonating structures are imaginary.• Greater the resonance energy, greater will be the stability of the molecule.• More the number of canonical forms of nearly same energy, greater is the stability of molecule.

8. Valence Bond Theory (VBT)Proposed by Heitler and London and extended by Pauling and Slater. The postulates are• Overlapping of atomic orbitals of valence shell of two atoms leads to formation of a covalent bond.• Half-filled orbital and opposite spin electrons are used.• σ - bond is formed by head on overlapping, π- bond is formed by lateral overlapping. σ - bond is formed by s – s, s – p

or p – p overlapping but in π – bond formation no s– orbital can take part.• Greater the overlapping, stronger is the bond.• σ - bond is stronger than p- bond and s- bond is directional while p- bond is non-directional.• The direction of bond is same as the direction of overlapping of orbitals.• The strength of σ - bond follows the order s – s > s – p > p – p.• Paired electrons are shifted to higher energy levels while forming a bond.• VBT cannot explain paramagnetic behavior of O2.

Aim4AIIMS - Chemistry Prep Guide56

9. Shapes of Molecules : Shapes of molecules can be determined by following two concepts • Valence shell electron pair repulsion (VSEPR) theory : Main postulates of this theory are

• The electron pair surrounding the central atom repel one another to the extent that there are no further repulsions between them. Thus the molecule will have minimum energy and maximum stability.

• If a central atom is bonded to similar atoms and contains no lone pair of electrons, the shape of molecule is symmetrical or regular. e. g. CH4 ,BF3etc.

• If central atom is bonded with different atoms or is surrounded by lone pairs of electrons also along with bond pairs, then the shape of molecule is unsymmetrical or it has irregular geometry.

• The order of repulsion between electron pairs is as. lone pair - lone pair > lone pair - bond pair > bond pair - bond pair.

The regular geometry

Bond pairsGeometry

2 3 4 5 6 7

Linear Triangularplanar

Tetrahedral

Trigonalbipyramidal

Octahedral Pentagonalbipyramidal

• Hybridization: It is the phenomenon of intermixing of a number of orbitals of an atom with slightly different energies and then redistribution of their energies to form new orbitals of identical shape and equal energy.• No. of hybrid orbitals = no. of atomic orbitals intermixed• Hybrid orbitals form s– bond on overlapping.• It does not take place in isolated atom. It occurs only during bond formation• Hybrid orbitals tend to remain far apart, therefore, repulsion order is

lp – lp > lp – bp > bp – bp.Total number of hybrid orbitals of central atom= Number of its s- bond pairs + Number of lone pair of electrons around central atom• Different types of hybridization are sp, sp2 , sp3 sp3d, sp3d2, sp3d3.• Hybrid orbitals are directional in nature.

10. Dipole Moment (mu):• It is equal to the product of magnitude of the charge and the distance between the centres of + ve and - ve charges of

a dipole. mu = e × d• Common in covalent compound when formed between two dissimilar atoms.• Used to predict the extent of polarity in a molecule.• A vector quantity so follows vector addition or subtraction rule in predicting net dipole moment of a molecule.

� � � � � �r

Increases with increase in E N i e HI HBr

� � �

�

1

2

2

2

1 22 cos

. . . .� �� HCl HF Ionic charactert

Theoritical

%exp�

�100

Can predict the shape of molecule,(H2O = bent, BeF2 = linear.)

• Can distinguish between cis and trans isomers.

• Can distinguish between o-, m-, p- isomers.

11. Molecular Orbital Theory (MOT)

The theory was developed by Hund & Mulliken. Basic postulates are• The atomic orbitals of comparable energies and proper symmetry of two atoms are mixed up to form molecular

orbitals.• The MO is the electron cloud surrounding more than one nucleus.• Total number of MO = Total number of A.O• Two A.O. combine to form two M.Os. One is anti bonding M.O and the other is bonding M.O. The energy of

bonding MO is less than anti bonding MO.

• The filling up of electrons in M.Os must follow Aufbau’s principle, Pauli exclusion principle and Hund’s rule.

57C

hemical B


Bond order � �N Nb a

2

Bond order Bond energyBond length

∝ ∝1

Nb → Number of electrons in bonding M.O.Na → Number of electrons in anti bonding M.O.

12. Co-ordinate Bond :One of the group or atom must have lone pair of electron whereas the other must have the incomplete octet or duplet.

Also known as dative bond. Shown by H3O+, NH4

+ , O3, SO2, SO3, H2SO4

13. Other Bonds:• H-bonding : It is weak electrostatic force of attraction between an electronegative atom and already covalently bonded

H- atom with some other electronegative atom. It is formed by the compounds in which H is attached to N, O, F.• It’s strength is about 1/10th of covalent bond.• The atom with small size and high electronegativity form strong H-bonding.• Of two types

• Intermolecular (between two molecules of same or different compounds) e.g in HF,H2 O,NH3etc.• Intramolecular (within a molecule) e. g in o - Nitrophenol, salicylaldehyde etc.

• Used to predict high b.p & density of water, solubility of alcohols and carboxylic acids in water, acidic character of o, p- nitro phenols etc.

• Dipole-dipole interactions :• These exist between molecules having permanent dipoles e.g. H – C1, I – F.

• Ion - dipole interactions:• Between an ion and a polar molecule e-g hydration of ions like Na+, Mg2+. • Smaller the ion, more the dipole moment, stronger will be these interactions.

• Ion-induced dipole interactions : Between an ion and a dipole induced by the ion in a non polar molecule.• Dipole-induced dipole interactions : Between a dipole and an induced dipole in a non polar molecule.• Dispersion forces : Between two non polar substances. The forces are between an instantaneous dipole and an induced

dipole e.g in O , N , He, Ne, Ar etc.Note : Dipole-dipole, dipole-induced dipole and dispersion forces are called van der waal’s forces.

CONCEPT

The Lewis theory: The tendency of atoms to achieve eight electrons in their outermost shell is known as Lewis octet rule.

1. Resonance:- This is a very important concept as far as the Example:- CO2

O

C

O O

::

: ::

:::

(–)

O

C

O O

::

: : :::(–)

:

(–) O

C

O O

::

: : : :

(–)

:

(–)

:

structure of different compounds is concerned. It involves several structures each of which is not a complete description of the actual structure, but each of it contributes to a certain extent to the actual structure. Different structures contribute to a different extent depending on their stability. What this means is that, the final structure will have the features (various charges, bond lengths) of each of the contributing structures, but will not be equivalent to any of them entirely. Since these “midway features” (like fractional charges, delocalized charge) cannot be represented by our traditional lewis structures, we employ the superposition of several contributing structures.

Bond order

Total no of bonds between two atoms in all the structures=

.

TTotal no of resonating structures.


Bond order

double bond single bond�

��

��

2

2 1

21 5.

2. Hybridization:- It is the phenomenon in which a new orbital is formed as a result of the modification of the old orbitals. The number of orbitals will be preserved and only the space orientation of the orbital will change. It is simply the rearrangement and shape-orientation modification of the existing orbitals into new hybridized orbitals.

How to determine type of hybridization

H V M C A� � � �� 1

2

Where H = Number of orbitals involved in hybridization viz. 2, 3, 4, 5, 6 and 7, hence nature of hybridization will be sp, sp2, sp3, sp3d, sp3d2, sp3d3 respectively.

V = Number electrons in valence shell of the central atom,

M = Number of monovalent atom

C = Charge on cation,

A = Charge on anionDipole moment

“The product of magnitude of negative or positive charge (q) and the distance (d) between the centres of positive and negative charges is called dipole moment”.

μ = Electric charge bond length

As q is in the order of 10–10 esu and d is in the order of 10–8 cm, μ is in the order of 10–18 esu cm. Dipole moment is measured in “Debye” (D) unit. 1D = 10–18 esu cm = 3.3 ×10–30 coulomb metre (In S.I. unit).

Dipole moment is indicated by an arrow having a symbol ( ) pointing towards the negative end. Dipole moment has both magnitude and direction and therefore it is a vector quantity.

Fajan’s ruleThe magnitude of polarization or increased covalent character depends upon a number of factors. These factors are,

a. Small size of cation: Smaller size of cation greater is its polarizing power i.e. greater will be the covalent nature of the bond.

b. Large size of anion: Larger the size of anion greater is its polarizing power i.e. greater will be the covalent nature of the bond.

c. Large charge on either of the two ions: As the charge on the ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases with the result its ability for forming the covalent bond increases.

d. Electronic configuration of the cation: For the two ions of the same size and charge, one with a pseudo noble gas configuration (i.e. 18 electrons in the outermost shell) will be more polarizing than a cation with noble gas configuration (i.e., 8 electron in outer most shell).

Valence bond theory or VBT

• To form a covalent bond, two atoms must come close to each other so that orbitals of one overlaps with the other.

• Orbitals having unpaired electrons of anti spin overlaps with each other.

• After overlapping a new localized bond orbital is formed which has maximum probability of finding electrons.

• Covalent bond is formed due to electrostatic attraction between radii and the accumulated electrons cloud and by attrac-tion between spins of anti spin electrons.

• Greater is the overlapping, lesser will be the bond length, more will be attraction and more will be bond energy and the stability of bond will also be high.

• The extent of overlapping depends upon: Nature of orbitals involved in overlapping, and nature of overlapping.

• More closer the valence shells are to the nucleus, more will be the overlapping and the bond energy will also be high.

• Between two sub shells of same energy level, the sub shell more directionally concentrated shows more overlapping. Bond energy: 2s – 2s < 2s – 2p < 2p – 2p.

59C

hemical B


• s-orbitals are spherically symmetrical and thus show only head on overlapping. On the other hand, p-orbitals are di-rectionally concentrated and thus show either head on overlapping or lateral overlapping. Overlapping of different type gives sigma (σ) and pi (π) bond.

Type of molecule

No. of bond pairs of electron

No. of lone pairs of

electrons

Hybridiza-tion

Bond angle Expected geometry

Actual ge-ometry

Examples

AX3 2 1 sp2 < 1200 Trigonal planar

V-shape, Bent, Angu-

lar

SO2, SnCl2,NO2

−

AX4 2 2 sp3 < 1090 28' Tetrahedral V-shape, Angular

H2O, H2S, SCl2, OF2,NN ClO2 2

− −, AX4 3 1 sp3 < 1090 28' Tetrahedral Pyramidal NH3, NF3,

PCl3, PH3, AsH2, ClO3

− ,H3O

+

AX5 4 1 sp3d <1090 28' Trigonal bipyramidal

Irregular tetrahedron

SF4, SCl4, TeCl4

AX5 3 2 sp3d 900 Trigonal bipyramidal

T-shaped ICl3, IF3, ClF3

AX5 2 3 sp3d 1800 Trigonal bipyramidal

Linear XeF2,I3 2

− −, ICl AX6 5 1 sp3d2 < 900 Octahedral Square pyra-

midalICl5, BrF5,

IF5

AX6 4 2 sp3d2 – Octahedral Square planar

XeF4, ICl4−

AX7 6 1 sp3d3 – Pentagonal pyramidal

Distorted octahedral

XeF6

Molecular orbitals1. When two atomic orbitals combine or overlap, they lose their identity and form new orbitals. The new orbitals thus

formed are called molecular orbitals.

2. Molecular orbitals are the energy states of a molecule in which the electrons of the molecule are filled just as atomic orbitals are the energy states of an atom in which the electrons of the atom are filled.

3. Only those atomic orbitals can combine to form molecular orbitals which have comparable energies and proper orientation.

4. The number of molecular orbitals formed is equal to the number of combining atomic orbitals.

5. When two atomic orbitals combine, they form two new orbitals called bonding molecular orbital and anti-bonding molecular orbital. The bonding molecular orbital has lower energy and hence greater stability than the corresponding anti-bonding molecular orbital. The bonding molecular orbitals are represented by σ, π etc, whereas the corresponding anti-bonding molecular orbitals are represented by σ*, π* etc.

6. The filling of molecular orbitals in a molecule takes place in accordance with Aufbau principle, Pauli's exclusion principle and Hund's rule. The general order of increasing energy among the molecular orbitals formed by the elements of second period and hydrogen and their general electronic configurations are given below.

7. Electrons are filled in the increasing energy of the MO which is in order

� � � � � � � �1 1 2 2 2 2 2 2s s s s p p p px y y x, , , , , ,* * * *

Increasing energy (for electrons > 14) � � � � � � �1 1 2 2 2 2 2s s s s p p py x z, , , , , ,* * *

Increasing energy (for electrons 14)

8. Number of bonds between two atoms is called bond order and is given by Bond order � ��

��

N NB A

2


9. Bond order ∝ Stability of molecule ∝ Dissociation energy ∝1

Bond length.

10. If all the electrons in a molecule are paired then the substance is a diamagnetic on the other hand if there are unpaired electrons in the molecule, then the substance is paramagnetic. More the number of unpaired electron in the molecule greater is the paramagnetism of the substance.

Hydrogen bonding

For the formation of H-bonding the molecule should contain an atom of high electronegativity such as F, O or N bonded to hydrogen atom and the size of the electronegative atom should be quite small.

• Intermolecular hydrogen bond: Intermolecular hydrogen bond is formed between two different molecules of the same or different substances.

• Intramolecular hydrogen bond (Chelation): Intramolecular hydrogen bond is formed between the hydrogen atom and the highly electronegative atom (F, O or N) present in the same molecule. Intramolecular hydrogen bond results in the cyclisation of the molecules and prevents their association.

Out Side NCERT

Explanation of lower density of ice than water and maximum density of water at 277K: In case of solid ice, the hydrogen bonding gives rise to a cage like structure of water molecules as shown in following figure. As a matter of fact, each water molecule is linked tetrahedrally to four other water molecules. Due to this structure ice has lower density than water at That is why ice floats on water. On heating, the hydrogen bonds start collapsing, obviously the molecules are not so closely packed as they are in the liquid state and thus the molecules start coming together resulting in the decrease of volume and hence increase of density. This goes on upto 277K. After 277 K, the increase in volume due to expansion of the liquid water becomes much more than the decrease in volume due to breaking of H-bonds. Thus, after , there is net increase of volume on heating which means decrease in density. Hence density of water is maximum .

61C

hemical B


Self Assessment Questions

1. Which of the following statement is not true?a. Ionic bonds are non-directional while covalent bonds

are directionalb. Formation of π-bond shortens the distance between the

two concerned atomsc. Ionic bond is possible between similar and dissimilar

atomsd. Linear overlapping of atomic p-orbitals leads to a

sigma bond 2. How many orbitals are singly occupied in O2 molecule?

a. 2 b. 1 c. 3 d. None of these 3. What are the exceptions of the octet rule?

a. The incomplete octet of central atom b. An odd number of electrons on central atomc. Expanded octet of the central atomd. All of these

4. A pair of electrons present between two identical non-metals:a. Is shifted to one of the atomsb. Is shared equally between themc. Undergoes addition reactions d. Have same spin

5. Oxygen molecule is formed by:a. One axial s-s overlap and one p-p axial overlapb. Two p-p axial overlapsc. Two p-p side wise overlapsd. One s-s axial and one p-p side wise overlap

6. In a covalent bond formation: a. Transfer of electrons takes placeb. Equal sharing of electrons between two atoms takes

placec. Electrons are shared by one atom onlyd. Electrons are donated by one atom and shared by both

atoms 7. Which of the following statements is not true

regarding molecular orbital theory?a. The atomic orbitals of comparable energies combine to

form molecular orbitals. b. An atomic orbital is homocentric while a molecular

orbital is polycentric c. Bonding molecular orbital has higher energy than

antibonding molecular orbital.d. Molecular orbitals like atomic orbitals obey Aufbau

principle for filling of electrons. 8. Which of the following is the most stable state when

two atoms come closer to each other to form a molecule?

I IIIIIa. (i), when the bond is formed, the energy is minimum.

b. (ii), when the atoms touch each other, the energy is zero

c. (iii), when the atoms are isolated, the energy is minimum

d. (ii), when the attractive forces are more than repulsive forces

9. 2s and 2p-atomic orbitals combine to give how many molecular orbitals?a. 2 b. 4 c. 8 d. 6

10. Which of the following statements is not true?a. Intermolecular hydrogen bonds are formed between

two different molecules of compounds b. Intermolecular hydrogen bonds are formed between

two different molecules of the same compoundc. Intermolecular hydrogen bonds are formed within the

same molecule.d. Hydrogen bonds have strong influence on the physical

properties of a compound 11. Which pair of elements can form multiple bond with

itself and oxygen?a. F, N b. N, Cl c. N, P d. N, C

12. 0.01 mole of H3POx is completely neutralized by 0.56 gram of KOH hence:a. x = 3 and given acid is dibasicb. x = 2 and given acid is monobasicc. x = 3 and given acid is monobasicd. x = 4 and given acid forms three series of salt

13. The correct order of increasing C—O bond strength

of CO, ,2–3CO CO2 is:

a. CO CO CO3

2

2

� � � b. 2

2 3CO CO CO−< <

c. 23 2CO CO CO−< <

d. 22 3CO CO CO −< <

14. Phosphorus pentachloride in the solid exists as:

a. PCl5 b. PCl Cl4

� � c. PCl PCl4 6

� � d. PCl5.Cl2

15. An ionic compound A+ B– is most likely to be formed when:a. The ionization energy of A high and electron affinity

of B is lowb. The ionization energy of A is low and electron affinity

of B is highc. Both, the ionization energy of A and electron affinity

of B are highd. Both, the ionization energy of A and electron affinity

of B are low

16. The state of hybridization of the central atom is not the same as in the others:a. B in BF3 b. O in H3O

+ c. N in NH3 d. P in PCl3


17. Which of the following statements is incorrect for PCl5?a. Its three P—Cl bond lengths are equalb. It involves sp3d hybridizationc. It has an regular geometry d. Its shape is trigonal bipyramidal

18. The number of sp2— s sigma bonds in benzene are:a. 3 b. 6 c. 12 d. None of these

19. The structure of the noble gas compound XeF4 is:a. Square planar b. Distorted tetrahedralc. Tetrahedral d. Octahedral

20. What is the state of hybridisation of Xe in cationic part of solid XeF6?a. sp3d3 b. sp3d2 c. sp3d d. sp3

21. The geometry of ammonia molecule can be best described as:a. Nitrogen at one vertex of a regular tetrahedron, the

other three vertices being occupied by three hydrogensb. Nitrogen at the centre of the tetrahedron, three of the

vertices's being occupied by three hydrogensc. Nitrogen at the centre of an equilateral triangle, three

corners being occupied by three hydrogensd. Nitrogen at the junction of a T, three open ends being

occupied by three hydrogens 22. Which molecular geometry is least likely to result

from a trigonal bipyramidal electron geometry?a. Trigonal planar b. See-sawc. Linear d. T-shaped

23. Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false:

(I) The order of repulsion between different pair of electrons is lp – lp – lp – bp > bp – bp

(II) In general, as the number of lone pair of electrons on central atom increases, value of bond angle from normal bond angle also increases

(III)The number of lone pair on 0 in H2O is 2 while on N in NH3 is 1

(IV)The structures of xenon fluorides and xenon oxyfluorides could not be explained on the basis of VSEPR theorya. TTTF b. TFTF c. TFTT d. TFFF

24. BF3 and NF3 both are covalent compounds but NF3 is polar whereas BF3 is non-polar. This is because:a. Nitrogen atom is smaller than boron atomb. N —F bond is more polar than B —F bondc. NF3 is pyramidal whereas BF3 is planar triangulard. BF3 is electron deficient whereas NF3 is not

25. Which of the following is an example of super octet

molecule?a. ClF3 b. PCl5 c. IF7 d. All the three

26. Which of the following species is neither hypervalent nor hypovalent?

a. 4ClO− b. BF3

c. 24SO − d. 2

3CO −

27. Which of the following compound is planar?a. XeO4 b. SF4 c. XeF4 d. CF4

28. Low melting point is expected for a solid:a. Ionic solid b. Metallic solidc. Molecular solid d. Covalent solid

29. Which molecule does not exist?a. OF2 b. OF4 c. SF2 d. SF4

30. Which species is planar?

a. 23CO − b. 2

3SO −

c. 3ClO− d. 4BF−

31. How many sigma bonds are in a molecule of diethyl ether, C2H5OC2H5?a. 14 b. 12 c. 8 d. 16

32. Iodine molecules are held in the solid lattice by: a. London forces b. Dipole-dipole interactionsc. Covalent bonds d. Coulombic force

33. The strength of bonds formed by 2s-2s, 2p-2p and 2p – 2s overlap has the order:a. s-s > p-p > p-s b. s-s > p-s > p-pc. p-p > p-s > .s-s d. p-p > s-s > p-s

34. Carbon dioxide is gas, while SiO2 is solid because:a. CO2 is a linear molecule, while SiO2 is angularb. van der Waals’ forces are very strong in SiO2

c. CO2 is covalent, while SiO2 is ionicd. Si cannot form stable bonds with O, hence Si has to form a 3D lattice

35. Which of the following statements is incorrect for sigma and pi-bonds formed between two carbon atoms?a. Sigma-bond is stronger than a pi-bondb. Bond energies of sigma and pi-bonds are of the order

of 264 kJ/mol and 347 kJ/molc. Free rotation of surrounding atoms about a sigma-

bond is allowed but not in case of a pi-bondd. Sigma-bond determines the direction between carbon

atoms but a pi-bond has no primary effect in this regard

63C

hemical B


Higher Order Questions

1. Which type of hybridization is shown by carbon atoms from left to right in the given compound: CH2 = CH – C ≡ N?a. sp2, sp2, sp b. sp2, sp, spc. sp, sp2, sp3 d. sp3, sp2, sp

2. According to molecular orbital theory, which of the following will not exist?a. H2

+ b. Be2 c. B2 d. C2

3. Which of the following molecules is paramagnetic in nature?a. H2 b. Li2 c. B2 d. N2

4. Which of the following shows dsp2 hybridisation and a square planar geometry?a. SF6 b. BrF5 c. PCl5 d. [Ni(CN)4]

2–

5. In a diatomic molecule the bond distance is 1 × 10–8 cm. Its dipole moment is 1.2 D. What is the fractional electronic charge on each atom?a. 0.50 b. 1.2 × 10–10 c. 0.25 d. 1.2

6. What is the order of stability of N2 and its ions?a. N N N N2 2 2 2

2� � ��

b. N N N N2 2 2 2

2� � ��

c. N N N N2 2 2 2

2� � ��

d. N N N N2

2

2 2 2

� � ��

7. Oxygen molecule is paramagnetic in nature. What is the paramagnetic content in terms of magnetic moment inO2

- :a. 1.732 b. 3 c. 1.5 d. 2.5

8. Which of the following pairs will have same bond order?

a. F2 and O2

2−

b. N2 and CO2

c. O2 and O2

− d. N2 and N2

+

9. Which of the following does show octahedral geometry?

a. SF6 b. IF5 c. Si F6

2− d. Sf4

10. In which of the following molecules / ions BF3, NO ,NH2

-2- and H2O

The central atom is sp2 hybridized?a. NH2

− and H2O b. NO2

− and H2Oc. BF3 and NO2

− d. NO2

− and NH2

−

11. In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three: a. SF4 b. I3

− c. SbCI5

2− d. PCI5

12. Some of the properties of the two species, NO-3 and

H3O+ are described below. Which one of them is

correct? a. Similar in hybridization for the central atom with

different structures

b. Dissimilar in hybridization for the central atom with different structures

c. Isostructural with same hybridization for the central atom

d. Isostructural with different hybridization for the central atom

13. Which one of the following pairs is isostructural (i.e., having the same shape and hybridization)?a. [BCI3 and BrCI–

3] b. [NH3 and NO–3]

c. [NF3 and BF3] d. [BF-4 and NH+

4] 14. The pair of species with the same bond order is:

a. O B2

2

2

− , b. O2

+ +, NO c. NO, CO d. N2, O2

15. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them: a. NO O C He� � ��

2 2

2

2 b. O NO C He2 2

2

2

� � �� c. C He O NO2

2

2 2

� � �� d. He O NO C2 2 2

2� � ��

16. Which one of the following is the correct statement?a. O2 molecule has bond order 2 and is diamagneticb. N2 molecule has bond order 3 and is diamagneticc. H2 molecule has bond order zero and is diamagneticd. C2 molecule has bond order 2 and is diamagnetic

17. The shape of stannous chloride molecule is:a. See-saw b. Square planarc. Trigonal pyramidal d. Bent

18. In which of the following species maximum atom can lie in same plane:a. XeF2O2 b. PCl5 c. AsH4

+ d. XeF4

19. The ratio of σ-bond and π-bond in tetracyano ethylene is:a. 2 : 1 b. 1 : 1 c. 1 : 2 d. None of these

20. Bonds present in N2O5 (nitrogen pentaoxide) are:a. Only ionic b. Only covalentc. Covalent and co-ordinate d. Covalent and ionic

21. Which one of the following is the correct set with respect to molecule, hybridization and shape?a. BeCl2, sp2, linear b. BeCl2, sp2, triangular planarc. BCl3, sp2, triangular planar d. BCl3, sp3, tetrahedral

22. The hybridization of the central atom in ICl2– is:

a. dsp2 b. sp c. sp2 d. sp3


23. The molecule exhibiting maximum number of non-bonding electron pairs (1.p.) around the central atom is:a. XeOF4 b. XeO2F2 c. XeF3

– d. XeO3

24. The shapes of XeF4 , XeF5– and SnCl2 are:

a. Octahedral, trigonal bipyramidal and bentb. Square pyramidal, pentagonal planar and linearc. Square planar, pentagonal planar and angulard. See-saw, T-shaped and linear

25. Amongst NO AsO CO ClO SO3 3

3

3

2

3 3

2− − − − −, , , , and BO3

2− , the non-planar species are: a. CO SO O3

2

3

2

3

3− − −, , B b.AsO ClO SO3

3

3 3

2− − −, , c. NO CO BO3 3

2

3

3− − −, , d. SO NO BO3

2

3 3

3− − −, ,

26. The correct order of H — M — H bonds angle is:a. NH3< PH3< SbH3< BiH3 b. AsH3< SbH3< PH3< NH3c. NH3< PH3< BiH3< SbH3 d. BiH3< SbH3< AsH3< PH3

27. The molecular size of ICI and Br2 is approximately same, but b.p. of ICl is about 40°C higher than that of Br2. It is because:a. ICI bond is stronger than Br—Br bond b. IE of iodine < IE of brominec. ICl is polar while Br2 is nonpolard. I has-larger size than Br

28. Which of the following molecules are expected to exhibit intermolecular H-bonding?

(I) Acetic acid (II) o-nitrophenol (III) m-nitrophenol (IV) o-boric acid Select correct alternate :

a. I, II, III b. I, II, IV c. I, III, IV d. II, III, IV

29. Which set of compounds in the following pair of ionic compounds has the higher lattice energy?

(i) KC1 or MgO (ii) LiF or LiBr (iii) Mg3N2 or NaCl

a. KCl, LiBr, Mg2N2 b. MgO, LiBr, Mg3N2c. MgO, LiF, NaCl d. MgO, LiF, Mg3N2

30. The hybridization of central iodine atom in IF5, I3 and I3

- are respectively:a. sp3d2, sp3d, sp3 b. sp3d, sp3d, sp3

c. sp3d2, sp3d2, sp3 d. sp3d, sp3d2, sp3

31. The shape of XeF3+ is:

a. Trigonal planar b. Pyramidalc. Bent T-shape d. See-saw

32. Which of the following species used both axial set of d-orbitals in hybridisation of central atom?

a. 4PBr+ b. 4PCl−

c. 4ICl− d. None of these 33. The incorrect order of lattice energy is:

a. AlF3> MgF2 b. Li3N > Li2Oc. NaCl > LiF d. TiC > ScN

34. In which of the following species central atom is not

surrounded by exactly 8 valence electrons?

a. 4BF− b. 3NCl c. 4PCl+ d. SF4

35. What is the geometry of the IBr2-� � ion?

a. Linearb. Bent shape with bond angle of about 900

c. Bent shape with bond angle of about 1090

d. Bent shape with bond angle of about 1200

36. What is the shape of the ClF3 molecule?a. Trigonal planar b. Trigonal pyramidalc. T-shaped d. Tetrahedral

37. The H-O-H bond angles in H3O+ are approximately

107°. The orbitals used by oxygen in these bonds are best described as:a. p-orbitals b. sp-hybrid orbitalsc. sp2-hybrid orbital d. sp3-hybrid orbital

38. Which of the following fact is directly explained by the statement ‘oxygen is a smaller atom than sulphur’?a. H2O boils at a much higher temperature than H2Sb. H2O undergoes intermolecular hydrogen bondingc. H2O is liquid and H2S is gas at room temperatured. S—H bond is longer than O—H bond

39. If two different non-axial d-orbitals having `xz’ nodal plane form π-bond by overlapping each other, then internuclear axis will be:a. x b. y c. z d. They don’t form π-bond

40. In which of the following molecular shape dz2 -orbital must not be involved in bonding?a. Pentagonal planar b. Trigonal planar c. Linear d. Square planar

41. In which of the following pairs, both the species have the same hybridisation?

(I) SF4, XeF4

(II) 3 2,I XeF−

(III) 4 4,ICl SiCl+

(IV) 33 4,ClO PO− −

a. I, II b. II, III c. II, IV d. I, II, III 42. Select pair of compounds in which both have different

hybridization but have same molecular geometry:

a. BF3, BrF3 b. 2 2,ICl BeClΘ

c. 3 3,BCl PCl d. 3 3 3, ,BF BCl BBr

43. Which of the following compound has the smallest bond angle (X — A — X) in each series respectively:a. OSF2, SbCl3 and PI3 b. OSBr2, SbI3 and PI3c. OSF2, SbI3 and PI3 d. OSF2, SbCl3 and SbI3

44. The species having no pπ – pπ bond but has bond order equal to that of O2:

a. 3ClO− b. 34PO −

c. 24SO − d. 3XeO

65C

hemical B


NCERT Exemplar Problems

1. Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs:a. [NF3 and BF3] b. 4 4[ and ]BF NH− +

c. [BCl3 and BrCl3] d. 3 3[ and ]NH NO−

2. Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?a. CO2 b. HI c. H2O d. SO

3. The types of hybrid orbitals of nitrogen in 2 3,NO NO+ − and 4NH+ respectively are expected to be:a. sp, sp3 and sp2 b. sp, sp2 and sp3

c. sp2, sp and sp3 d. sp2, sp3 and sp 4. Hydrogen bonds are formed in many compounds e.g.,

H2O, HF, NH3. The boiling point of such compounds depends to a extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points above compounds is:a. HF > H2O > NH3 b. H2O > HF > NH3

c. NH3 > HF > H2O d. NH3 > H2O > HF

5. In 34PO − ion the formal charge on the oxygen atom of

P–O bond is:a. + 1 b. –1 c. – 0.75 d. + 0.75

6. In 3NO− ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are:a. 2, 2 b. 3, 1 c. 1, 3 d. 4, 0

7. Which of the following species has tetrahedral geometry?a. 4BH− b. 2NH−

c. 23CO − d. 3H O+

8. Number of π bonds and σ bonds in the following structure is:

H

H

H H

H HH

H

a. 6, 19 b. 4, 20 c. 5, 19 d. 5, 20 9. Which molecule/ion out of the following does not

contain unpaired electrons?a. 2N+ b. 2O c. 2

2O − d. B2

10. In which of the following molecule/ion all the bonds are not equal?a. XeF4 b. 4BF−

c. C2H4 d. SiF4

11. If the electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d2 4s2, the four electrons involved in chemical bond formation will be:

a. 3p6 b. 3p6, 4s2

c. 3p6, 3d2 d. 3d2, 4s2

12. Which of the following angle corresponds to sp2 hybridization?a. 90° b. 120° c. 180° d. 109°

Direction(Q. Nos. 1-16) The electronic configurations of the elements A, B and C are given below. Answer the questions from 14 to 17 on the basis of these configurations.A 1s2 2s2 2p6

B 1s2 2s2 2p6 3s2 3p3

C s2 2s2 2p6 3s2 3p5

13. Stable form of A may be represented by the formula:a. A b. A2 c. A3 d. A4

14. Stable form of C may be represented by the formula:a. C b. C2 c. C3 d. C4

15. The molecular formula of the compound formed from B and C will be:a. BC b. B2C c. BC2 d. BC3

16. The bond between B and C will be:a. Ionic b. Covalent c. Hydrogen d. Coordinate

17. Which of the following order of energies of molecular orbitals of N2 is correct?a. * *( 2 ) ( 2 ) ( 2 ) ( 2 )y z x yp p p pπ < σ < π ≈ π

b. * *( 2 ) ( 2 ) ( 2 ) ( 2 )y z x yp p p pπ > σ < π ≈ π

c. * *( 2 ) ( 2 ) ( 2 ) ( 2 )y z x yp p p pπ < σ < π ≈ π

d. * *( 2 ) ( 2 ) ( 2 ) ( 2 )y z x yp p p pπ > σ < π ≈ π* * *2 ( 2 2 ) 2z x y zp p p p< σ < π ≈ π < σ

18. Which of the following statement is not correct from the view point molecular orbital theory?a. Be2 is not a stable moleculeb. He2 is not stable but He-2’ is expected to exist.c. Bond strength of N2 is maximum amongst the

homonuclear diatomic molecules belonging to the second period.

d. The order of energies of molecular orbitals in N2 molecule is

19. Which of the following options represent the correct bond order?a. 2 2 2O O O− +> > b. 2 2 2O O O− +< <

c. 2 2 2O O O− +> < d. 2 2 2O O O− +< >

20. The electronic configuration of the outer most shell of the electronegative elements is:a. 2s22p5 b. 3s2 3p5 c. 4s2 4p3 d. 5s2 5p5

21. Amongst the following elements whose electronic configuration are given below, the one having the highest ionization enthalpy is:a. [Ne] 3s2 3p1 b. [Ne] 3s2 3p3

c. [Ne] 3s2 3p2 d. [Ar] 3s10 4s2 4p3


Assertion & Reason

1. Assertion: CHCl3 and CH3OH are miscible.

Reason: One of them is polar.

2. Assertion: Bond order can assume any value number including zero.

Reason: Higher the bond order, shorter is bond length and greater is bond energy.

3. Assertion: The dipole moment helps to predict whether a molecule is polar or nonpolar.

Reason: The dipole moment helps to predict geometry of moleculars.

4. Assertion: Bond order in a molecule, which can assume any value positive or negative, integral or fractional including zero.

Reason: Bond order depends on the number of electrons in the bonding and antibonding orbitals.

5. Assertion: B2H6, Si2H6 are said to have similar structure.

Reason: They have same number of σ and π bonds.

6. Assertion: Sigma σ is a strong bond, while pi (π) is weak bond.

Reason: Atoms rotate freely about pi (π) bond.

7. Assertion: All F- S –F angle in SF4 is greater than 90o but less than 180o.

Reason: The lone pair-bond pair repulsion is weaker than bond pair-bond pair repulsion.

8. Assertion: Molecular nitrogen is less reactive than molecular oxygen.

Reason: The bond length of N2 is shorter than that of oxygen.

9. Assertion: Multiple bond between two bonded atoms can have more than three bonds.

Reason: Multiple bond between two bonded atoms can not have more than three bonds.

10. Assertion: 2nd period elements do not involve in excitation of electron.

Reason: 2nd period elements do not have vacant 2d-orbitals.

11. Assertion: In SO3 molecule bond dissociation energy of

all S = O bonds are not equivalent.

Reason: SO3 molecule is having two types of 2pr – 3pr and 2pr – 3dr pi-bonds.

12. Assertion: PH+4 ion is having tetrahedron geometry.

Reason: P-atom is unhybridised in PH+4 ion.

13. Assertion: All diatomic molecules with polar bond have dipole moment.

Reason: Dipole moment is a vector quantity.

14. Assertion: Water is a good solvent for ionic compounds but poor for covalent compounds.

Reason: Hydration energy of ions releases sufficient energy to overcome lattice energy and break hydrogen bonds in water while covalent compounds interact so weakly that even van der Walls’ forces between molecules of covalent compounds cannot be broken.

15. Assertion: Xe-atom in XeF2 assumes sp-hybrid state.

Reason: XeF2 molecule does not follow octet rule.

16. Assertion: The atoms in a covalent molecule are said to share electrons, yet some covalent molecules are polar.

Reason: In polar covalent molecule, the shared electrons spend more time on the average near one of the atoms.

17. Assertion: CCl4 is a non-polar molecule.

Reason: CCl4 has polar bonds.

18. Assertion: Geometry of ICI3 is tetrahedral.

Reason: Its shape is T-shape, due to the presence of two lone pairs.

19. Assertion: The covalency of carbon is four in excited state.

Reason: The four half-filled pure orbitals of carbon form same kind of bonds with an atom as those are with hybridised orbitals.

20. Assertion: The shape of XeF4 is square-planar.

Reason: In an octahedral geometry, a single lone pair can occupy any position but a second lone pair will occupy the opposite position to the first lone pair.

67C

hemical B


ANSWER KEY

EXPLANATIONS


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17c a d b d b c a c b d b a c b a c

18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34c a a b a b c d d c c b a a a c d

35b


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17a b c d c a a a d c c b d a d d d

18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34d b c c d c c b d c c d a c c c d35 36 37 38 39 40 41 42 43 44a c d d d b c b d d


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17b c b b c d a c c c d b a b d b a

18 19 20 21

d b a b

Assertion & Reason

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17b a a a d c a a d b b a d c b c d

18 19 20

b a c


1. (c) Ionic bond is only possible between similar atoms. 2. (a) 2 orbitals are singly occupied in O2 molecule →

π*2px1 = π*2py1

5. (d) Oxygen molecule is formed by one sigma bond between two s orbitals & one pi bond between two p orbitals.

6. (b) In a covalent bond equal sharing of electrons between two atoms takes place.

7. (c) Bonding molecular orbital has lows energy than antibonding molecular orbital.

8. (a) Whenever formation of bond takes place by overlap of orbitals, minimum energy state is attained.

9. (c) 2s & 2p atomic orbitals combine to giveσ2s, σ*2s, σ2pz, σ*2pz, π2px = π2px, π*2px = π*2pyTotal 8 orbitals.

10. (b) Inter molecular hydrogen bonds are formed between two different molecules of same or different compound.

11. (d) C,N shows the property of catenation i.e. tendency to form multiple bonds with itself.

12. (b) 0.01 mole of H3PO2 is completely neutralized by 0.56 gm of KOH.

H3PO2 is monobasic.

13. (a) CO C O

C

O

O

O

(–)

(–)

2

3CO �

2CO C O

O


Since CO has triple bond ∴ bond strength is greatest.

14. (c) PCl5 in solid state exist as PCl PCl4 6

� �

16. (a) BF3 has sp2 hybridization whereas PCl3, NH3, H3O+

have sp3 hybridization.

17. (c) PCl5 doesn't have regular geometry 3-Cl atoms occupy equatorial position whereas 2-Cl have axial position.

18. (c) 12

H

H

H

H

H

Hσ bonds 6 → CH bonds

6 → C C bonds{ {

19. (a) Square planar sp3d2

XeF

FF

F

20. (a) XeF6 → sp3d3

Xe

Fe

Fe

Fe

Fe

Fe

Fe

21. (b) N

HHH

NH3

sp3

pyramidal

23. (b) As the number of lone pair of electron on central atom increases, the value of bond angle from normal bonds decreases due to increased repulsion from lone pair of electron.

Structures of xenon fluorides & xenon any fluorides can be explained on the basis of VSEPR theory.

25. (d) P

Cl

Cl

ClCl

Cl

10e–

I

F12 electron

F

F F

F F

27. (c) XeF4 is planar

XeF

FF

FSquare planar

sp d3 2

30. (a) C

O

O

O

(–)

(–)

planar molecule

31. (a) CH3 – CH2 – O – CH2 – CH3

Number of σ bonds = 14 32. (a) Atoms of same element are field together by London

forces.


1. (a) CH2 = CH – C ≡ N sp2 sp2 sp

2. (b) Be2 → 8 electronsσ1s2 σ*1s2 σ2s2 σ*2s2

Bond order � �� 1

2N Nb a

� �� 1

210 4 3

Bond order 0 means unstable molecule. 3. (c) B2 total electrons = 10

σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2

π2px1 = π2px1

Since molecular orbital are singly occupied it is paramagnetic.

4. (d) [Ni(CN)4]2–

NiCN

CNCN

CN

6. (a) N2 → 14e– →σ1s2, σ*1s2 σ2s2 σ*2s2 π2px2 = π2py2

σ2pz2, bond order � �� 1

210 4 3

N e pz B O2

113 21

29 4 2 5� �� . .

N e pz px py

B O

2

2 115 2 2 2

1

210 5 2 5

� ��

� ��

� � �* * ,

. .

N e pz px py

B O

2

2 2 1 216 2 2 2

1

210 6 2

� ��

� ��

� � �* * ,

.

∴ order of stability

N N N N2 2 2 2

2

3 2 5 2 5 2� � �

� � �

. .

8. (a) Since F2 & O2

2− posses same number of electrons therefore they have same bond order.

9. (d) SF4 possess seasaw shape

S

F

F

F

F

10. (c) For neutral molecules:

No. of electron pairs = No. of atoms bonded to it +1/2 [Group number of central atom – Valency of the central atom].

For ions:

No. of electron pairs = No. of atoms bonded to it + 1/2 [Group no. of central atom – Valency of the central atom + no. of electrons]

69C

hemical B


On calculating no. of electron pairs in given molecules. We find that in the given molecules hybridization isBF3 → sp2

NO2

− → sp2

NH2

− → sp3

H2O → sp3

no. of hybrid orbitals = 6

11. (c) SbCl52− : * = 6 means sp3d2

12. (b) In NO3

2− , nitrogen is in sp2 hybridization, thus planner in shape. In H3O

+, oxygen is in sp3

hybridization, thus tetrahedral in shape. 13. (d) BF4

− , hybridization sp3, tetrahedral structure.

NH4

+ hybridization sp3, tetrahedral structure. 14. (a) O2

2− no. of electrons = 18BO = –110e– = B2 = +1

15. (d) He+2 = s(1s)2s*(1s), B.O = 0.5

O2– = KK s(2s)2s*(2s)2s(2pz)

2p(2px)2p(2py)

2

p*(2px)2p*(2py)

1, B.O = 1.5NO = KK s(2s)2s*(2s)2p(2px)

2p(2py)2

s(2pz)2p*(2px)1, B.O = 2.5

C2-2 = KK s(2s)2s*(2s)2p(2px)

2p(2py)2s(2pz)

2

B.O = 3.0 16. (d) Total electrons in C2 = 12

σ1s2 σ*1s2 σ*2s2 π2px2 = π2py2

B N Nb a.O � �� 1

2

1

28 4 2

17. (d) SnCl2

Sn

ClCl

Bent

18. (d) XeF4

Xe

F

Square planarF F

F

19. (b) C

N

C

N

C

C

NC

NC

σ bonds = 9π bonds = 9Ratio = 1 : 1

20. (c) N

O

O

N

O:::

::

::O:::

O::

∴ Covalent bond & coordinate bonds.

21. (c) Cl B

Cl

Cl

Triangular planar

23. (c) XeFF

F

2 pairs

24. (c) Xe

F

F F

F

XeF4

Square planar

pentagonal planar5XeF�

Xe

F

F F

FF F

angular

2SnClSe

ClCl

26. (d) With increase in the electronegativity of central atom, the attraction of lone pair towards it increase ∴ bond angle increase.

27. (c) ICl is polar due to different electronegativity of both elements.

28. (c) o-nitro phenol has intra molecular hydrogen bonding.

30. (a) I

F

F F

F

Fsp d

3 2

31. (c) Xe

F

F F

Bent T-shape

+

32. (c) I

Cl

ClCl

Cl 4

3 2

ICl

sp d

�

34. (d) SF4 is not surrounded by exact 8 electrons

SF

F

F

F

It is surroundedby 10 electrons

35. (a)

I

Br3 2

hybridesation

sp d

Br

� �2IBr

� linear molecule with


36. (c)

I

F

3

3ClF sp d�

F

F

37. (d) The orbitals used by oxygen in forming bonds H–O–H in case of H3O

+ are sp3 hybrid orbitals. 38. (d) Since oxygen atom is more electronegative than

sulphur atom. Therefore S–H bond is longer than O–H bond.

40. (b) In case of trigonal planar only sp2 hydridised orbital are involved.

41. (c) XeF2, I3− has sp3d hydridisation ClO3 4

3− −, PO has sp3.


1. (b) (a) NF3 is pyramidal whereas BF3 is planar triangular.

(b) 4BF− and 4NH+ ion both are tetrahedral and sp3 hybridisation.

(c) BCl3 is triangular planar whereas BrCl3 is T shaped.

(d) NH3 is pyramidal whereas 3NO− is triangular planar. 2. (c) CO2 being symmetrical has zero dipole moment

Among HI, SO2 and H2O dipole moment is highest for H2O as in it the central atom contains 2 lone pairs.

3. (b) The type of hybrid orbitals of nitrogen can be decided by using VSEPR theory counting by and as lp in

2 2 0/NO bp p+ = + = linear = sp hybridized

23 2 0/NO bp p sp− = + ⇒ hybridized

34 4 0/NH bp p sp+ = + ⇒ hybridized

4. (b) Strength of H-bond is in the order H....F > H…O > H…..N.But each H2O molecule is linked to four other H2O molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules. Hence, b.p of H2O > b.p of HF > b.p. of NH3

5. (c) In 34PO − ion, formal charge on each 0-atom of P–O

bond

total charge 3 0.75Number of O-atom 4

= = − = −

6. (d) In N-atom, number of valence electrons = 5 Due to the presence of one negative charge, number of valence electrons = 5 + 1 = 6 one O-atom forms two bond (= bond) and two O-atom shared with two electrons of N-atomThus, 3 O-atoms shared with 8 electrons of N-atom.∴ Number of bond pairs (or shared pairs) = 4Number of lone pairs = 0

7. (a) 4 4BH− ⇒ bond pairs + 0 lone pair 3sp⇒ hybridised = tetrahedral geometry

2 shapeNH V− = −

23 triangular planarCO − =

3 pyramidalH O+ =

8. (c) The given compound will have the correct structure as

There are 5 π-bonds and 8 C— H + 11 C—C σ-bonds, i.e., 19 σ-bonds are present in the above molecule.

9. (c) The electronic configuration of the given molecules are

2 2 2 2 2 2 2 12 1 , * 1 , 2 , * 2s , , 2x y zN s s s p p p+ = σ σ σ σ π = π σ

It has one unpaired electron.2 2 2 2 2 2 1

2 1 , * 1 , 2 , * 2s , 2 , 2 2z x yO s s s p p p= σ σ σ σ σ π ≈ π + O2 has two unpaired electrons.

2 2 2 2 2 2 2 2 2 22 1 , * 1 , 2 , * 2s , 2 , 2 2 , * 2 * 2z x y x yO s s s p p p p p− = σ σ σ σ σ π ≈ π π ≈ π

Thus, 22O − has no unpaired electrons.2 2 2 2 1 1

2 1 , * 1 , 2 , * 2s 2 2x yB s s s p p= σ σ σ σ π ≈ π T h u s , B2 has two unpaired electrons.

10. (c) 4 4 2/XeF bp p⇒ + ⇒ square planar ⇒ all bonds are equal

4 4 0/BF bp p− ⇒ + ⇒ tetrahedral (all bonds are equal)

4 0/SiF bp p⇒ + ⇒ tetrahedral (all bonds are equal) Thus, in C2H4 all the bonds are not equal.

11. (d) The given electronic configuration shows that an element is vanadium (Z = 22). It belongs to d-block of the periodic table. In transition elements i.e., d-block elements, electrons of ns and (n – 1) d subshell take part in bond formation

12. (b) For sp2 hybridisation, the geometry is generally triangular planar.Thus, bond angle is 120°.

13. (a) The given electronic configuration shows that A represents noble gas because the octet is complete. A is neon which has 10 atomic number

14. (b) The electronic configuration of C represent chlorine. Its stable form is dichlorine.

15. (d) The electronic configuration show that B represents phosphorus and C represents chlorine. The stable compound formed is PCl3 i.e., BC3.

16. (b) The bond between B and C will be covalent. Both B and C are non-metal atoms B represents phosphorus and C represent chlorine.

17. (a) The correct increasing order of energies of molecular orbitals of N2 is given below

* *1 1 2 2 ( 2 2 )x ys s s s p pσ < σ < σ < σ π ≈ π

18. (d) Existence of molecule bonding nature and energy order of molecular orbitals can be explained on the basis of molecular orbital theory as follows

71C

hemical B


(i) Molecules having zero bond order never exists while molecular having non-zero bond order is either exists or expected to exist.(ii) Higher the value of bond order, higher will be its bond strength.Electrons present in bonding molecular orbital are known as bonding electrons (Nb) and electrons present on anti-bonding molecular orbital are known as anti-bonding electrons (Na) and half of their difference is known as bond order i.e.,

(a) 2 * 2 2 * 22 (4 4 8) 1 , 1 , 1 , 2Be s s s s+ = = σ σ σ σ

Bond order (BO) 12

=

[Number of bonding electrons (Na) – Number of anti-bonding electrons Na]

4 4 02−

= =

Here, bond order of Be2 is zero. Thus, it does not exist.

(b) 2 * 22 (2 2 4) 1 , 1He s s+ = = σ σ

2 2 02

BO −= =

Here, bond order of Be2 is zero. Hence, it does not exist. 2 * 1

2 (2 2 1 3) 1 , 1He s s+ + − = = σ σ

2 1 0.52

BO −= =

Since, the bond order is not zero, this molecule is expected to exist.

2 * 2 2 * 2 2 2 2 22 (7 7 14) 1 , 1 , 2 , 2 , 2 , 2x y zN s s s s p p p+ = = σ σ σ σ π ≈ π σ

10 4 32

BO −= =

Thus, dinitrogen (N2) molecule contain triple bond and no any molecule of second period have more than double bond. Hence, bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.(d) It is incorrect. The correct order of energies of molecular orbitals in N2 molecule is

* 2 * * *2 2 ( 2 2 ) 2 2 2x x y zs s p py pz p p pσ < σ < π ≈ π < σ < π ≈ π < σ

19. (b) Electronic configuration of O2 (16 electrons)

2 * 2 2 * 2 2 2 2 1 * 11 , 1 , 2 , 2 , 2 , 2 2 ,2 2z x y x ys s s s p p p p p= σ σ σ σ σ π ≈ π ≈ π

Bond order 1 1( ) (10 6) 22 2b aN N= − = − =

Electronic configuration of 2O+ (15 electrons)

2 * 2 2 * 2 2 2 2 * 1 * 61 , 1 , 2 , 2 , 2 , 2 2 , 2 2z x y x ys s s s p p p p p= σ σ σ σ σ π ≈ π π ≈ π

Bond order

1 1( ) (10 5) 2.52 2b aN N= − = − =

Electronic configuration of 2O− (17 electrons)

2 * 2 2 * 2 2 2 2 * 2 * 11 , 1 , 2 , 2 , 2 , 2 2 , 2 2z x y x ys s s s p p p p p= σ σ σ σ σ π ≈ π π ≈ π

Bond order

1 1( ) (10 7) 1.52 2b aN N= − = − =

Thus, the order of bond order is 2 2 2O O O− +< <

20. (a) The electronic configuration represents

2s2 2p5 = fluorine =most electronegative element

3s2 3p5 = chlorine

4s2 4p5 = bromine

5s2 5p5 = iodine

21. (b) The electronic configuration of options (b) and (d) have exactly half-filled 3p orbitals (b) represents phosphorus and (c) represents arsenic but (b) is smaller in size than (d).

Hence, (b) has highest ionization enthalpy, ionization enthalpy increases left to right in the periodic table as the size decreases.

Assertion & Reason

1. (b) Both the assertion and reason are true but reason is not the correct explanation of assertion.

Bond order

��1

2

No of bonding No of antibonding

orbital electrons orbital ele

. .

cctrons

�

��

�

��

The bond order zero indicates that the bond does not exist.

2. (a) Both the assertion and reason are true and reason is the correct explanation of assertion.

The linear triatomic molecules of the type B–A–B e.g. CO2 have no dipole moment, not because the individual C = O bonds are non-polar but because the two bond moments (polarity) cancel each other. Therefore, geometry of molecules can be predicted by the value of dipole moment.


Bond order is the half of the difference between bonding and anti-bonding electrons. Bond order

no. of e in No. of e in1bonding M.O. antibonding M.O.2

= −

Higher the value of bond order, stronger is the bond.


CHCl3 and CH3+OH are miscible due to intermolecular van der Waal’s forces of attraction. However CH3OH is polar and CHCl3 is non-polar.


5. (d) Both assertion and reason are false.B2H6 is an electron deficient compound, B2H6 contain some unusual bonds which are called as 2-electron, 3-centre bonds.

6. (c) Assertion is true but reason is false.

Sigma bonding involves end to end overlapping of the two atomic orbitals e.g. between s and s, s and p and p orbitals.

Pi bonding (π) involves side wise overlapping of the two atomic orbitals e.g. between p and p orbitals.

Overlapping of orbitals is more effective in σ-bonding than in π-bonding. Due to the geometry of the overlapping orbitals, rotation of an atom is not possible around π-bond.


Due to greater lone pair – bond pair repulsions than bond pair – bond pair repulsions, the F – S – F bond angle decreases from 180o.


Bond order of N2 = 3 , Bond order of O2 = 2.

Higher the bond order, higher is the bond dissociation energy. i.e. higher stability or less reactivity. Thus, N2 is less reactive than O2.

Higher the bond order, shorter is the bond length. Shorter bond length of N2 shows its higher bond order.

9. (d) Both assertion and reason are false.

Multiple bonds between two bonded atoms cannot have more than three bonds.




P

HH

H

H +

Tetrahedral


All diatomic molecules with polar bonds may have dipole moment.



Xe

F

F

XeF2

sp d3



18. (b) Both the assertion and reason are true and reason is not the correct explanation of assertion.

I

ClICl3

sp d3

Cl

ClT-shape



XeF

XeF4

F

F

FSquare planar

4 chemical bonding and molecular structure · chemical bonding and molecular structure 4 synopsis...

Documents