ccb – 1 chemical bonding and molecular …einsteinclasses.com/bluetooth folder/(4)ch_bond.pdfq....

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Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

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CHEMICAL BONDING AND MOLECULAR STRUCTURE

4.1 Kossel-Lewis Approach to Chemical Bonding :

Q. According to Kossel and Lewis why do atoms combine to form molecule.

Solution : The atoms of different elements combine with each other in order to complete their respectiveoctets (i.e., 8 electrons in their outermost shell) or duplet (i.e., outermost shell having 2 electrons) in case ofH, Li and Be to attain stable nearest noble gas configuration.

Q. Define electrovalent bond or ionic bond.

Solution : The bond formed, as a result of the electrostatic attraction between the positive and negativeions was termed as the electrovalent bond. The electrovalence is thus equal to the number of unit charge(s)on the ion.

Q. Define Octet rule.

Solution : According to this, atoms can combine either by transfer of valence electrons from one atom toanother (gaining or losing) or by sharing of valence electrons in order to have an octet in their valenceshells. This is known as octet rule.

Q. Draw the Lewis Dot structure for following molecules : Cl2, H

2O, CCl

4, CO

2, C

2H

4, N

2, C

2H

2.

Solution :

Cl2

H2O

CCl4

CO2

C2H

4

N2

C2H

2

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Q. Draw the Lewis representation of O2, O

3, NF

3, CO

32–, HNO

3.

Solution : Molecule/Ion Lewis Representation

O2

O3

NF3

CO3

2–

HNO3

Q. Write the Lewis dot structure of CO molecule. [NCERT Solved Example 4.1]

Solution :

Q. Write the Lewis structure of the nitrile ion, NO2–. [NCERT Solved Example 4.2]

Solution : or or

Q. Write the method to find the formal charge of an atom in Lewis structure.

Solution : Formal charge (F.C.) on an atom in a Lewis structure = [total number of valence electrons in thefree atom] – [total number of non bonding (lone pair) electrons] – ½ [total number of bonding (shared)electrons]

Q. Find the formal charge on each oxygen atom in Lewis dot structure of O3.

Solution : The atoms have been numbered as 1, 2 and 3. The formal charge on :

The central O atom marked 1 = 1)6(2

126

The end O atom marked 2 = 0)4(2

146

The end O atom marked 3 = 1)2(2

166

Hence, we represent O3 along with the formal charges as follows :

Q. What are the limitation of Octet rule ?

Solution : There are three types of exceptions to the octet rule :

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(a) The incomplete octet of the central atom : In some compounds, the number of electrons surrounding thecentral atom is less than eight. This is especially the case with elements having less than four valenceelectrons. Examples are LiCl, BeH

2 and BCl

3.

(b) Odd-electron molecules : In molecules with an odd number of electrons like nitric oxide, NO andnitrogen dioxide, NO

2, the octet rule is not satisfied for all the atoms

(c) The expanded octet : Elements in and beyond the third period of the periodic table have, apart from 3sand 3p orbitals, 3d orbitals also available for bonding. In a number of compounds of these elements thereare more than eight valence electrons around the central atom. This is termed as the expanded octet.Obviously the octet rule does not apply in such cases. Some of the examples of such compounds are : PF

5,

SF6, H

2SO

4.

(d) It is clear that octet rule is based upon the chemical inertness of noble gases. However, some noble gases(for example xenon and krypton) also combine with oxygen and fluorine to form a number of compoundslike XeF

2, KrF

2, XeOF

2 etc.

(e) This theory does not account for the shape of molecules.

(f) It does not explain the relative stability of the molecules being totally silent about the energy of amolecule.

4.2 Ionic or Electrovalent Bond :

Q. How the ionic bond is formed ?

Solution : Ionic bonds will be formed more easily between elements with comparatively low ionizationenthalpies and elements with comparatively high negative value of electron gain enthalpy.

The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom andthat of the negative ion involves the addition of electron(s) to the neutral atom.

M(g) M+(g) + e– ; Ionization enthalpy

X(g) + e– X– (g) ; Electron gain enthalpy

M+(g) + X–(g) MX (s)

Q. Define Lattice Enthalpy.

Solution : The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separateone mole of a solid ionic compound into gaseous constituent ions.

4.3 Bond Parameters :

Q. Define Bond length.

Solution : Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in amolecule.

Q. Define Covalent radius.

Solution : The covalent radius is measured approximately as the radius of an atom’s core which is incontact with the core of an adjacent atom in a bonded situation. The covalent radius is half of the distancebetween two similar atoms joined by a covalent bond in the same molecule.

Q. Define vander Waals radius.

Solution : The vander Waals radius represents the overall size of the atom which includes its valence shellin a nonbonded situation. Further, then vander Waals radius is half of the distance between two similaratoms in separate molecules in a solid.

Q. Define Bond angle and Bond enthalpy.

Solution : Bond angle is defined as the angle between the orbitals containing bonding electron pairs aroundthe central atom in a molecule/complex ion. Bond angle is expressed in degree.

Bond enthalpy is defined as the amount of energy required to break one mole of bonds of a particular typebetween two atoms in a gaseous state. The unit of bond enthalpy is kJ mol–1.

for e.g., H2(g) H(g) + H(g);

aH0 = 435.8 kJ mol–1

It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule.

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Q. Define Bond order.

Solution : In the Lewis description of covalent bond, the bond order is given by the number of bondsbetween the two atoms in a molecules. Isoelectronic molecules and ions have identical bond orders; forexample, F

2 and O

22– have bond order 1. N

2, CO and NO+ have bond order 3 with increase of bond order,

bond enthalpy increases and bond length decreases.

Q. Define Resonance.

Solution : According to the concept of resonance, whenever a single Lewis structure cannot describe amolecule accurately, a number of structures with similar energy, positions of nuclei, bonding andnon-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes themolecule accurately.

Q. Draw the resonating structure of O3 and carbonate ion.

Solution :

Q. Explain the resonating structure of CO2 molecule.

Solution :

Q. Define polar and non-polar covalent bond.

Solution : In H2, there is no displacement of the electric charge due to same electron affinity of both

H-atoms and the bond is non-polar. In HCl, the Cl atom has a more electronegativity than does the H atom.Electronic charge distribution is shifted towards the Cl atom. The H-Cl bond is said to be polar.

Q. Define dipole moment.

Solution : The magnitude of the charge displacement in a polar covalent bond is measured through aquantity called the dipole moment µ. It is the product of the magnitude of charges () and the distanceseparating them (d). (Here the symbol () suggests a small magnitude of charge, less than the charge on anelectron). Dipole moment is the vector quantity and direction of vector is from less electronegative to more

electronegative atom, for e.g., .

µ = × d

If = 4.8 × 10–10 esu and d = 1Å = 1 × 10–8 cm then µ = 4.8 × 10–10 × 1 × 10–8 = 4.8 × 10–18 esu cm

In S.I. unit, 1 D = 3.33 × 10–30 coulomb meter (when charge = 3.33 × 10–20 C and d = 1 × 10–10 m).

In diatomic molecule, µ = × d.

Q. Why the dipole moment in case of BeF2 and BF

3 is zero ?

Solution : In BeF2 two equal bond dipoles point in opposite direction and cancel the effect of each other for

e.g,

In tetra-atomic molecule, for example in BF3, the dipole moment is zero although the B – F bonds are

oriented at an angle of 1200 to one another, the three bond moments give a net sum of zero as the resultant

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of any two is equal and opposite to the third, for e.g., .

Q. Which out of NH3 and NF

3 has higher dipole moment and why ?

Solution : NH3 has higher dipole moment than NF

3. This is because, in case of NH

3 the orbital dipole due

to lone pair is in the same direction as the resultant dipole moment of the N — H bonds, whereas in NF3 the

orbital is in the direction opposite to the resultant dipole moment of three N – F bonds. The orbital dipolebecause of lone pair decreases the effect of the resultant N – F bond moments, which results in the lawdipole moment of NF

3

Q. Explain briefly the Fajans rules.

Solution : Just as all the covalent bonds have some partial ionic character, the ionic bonds also have partialcovalent character. The partial covalent character of ionic bonds was discussed by Fajans in terms of thefollowing rules :

The smaller the size of the cation and the larger the size of the anion, the greater the covalent character ofan ionic bond.

The greater the charge on the cation, the greater the covalent character of the ionic bond.

For cations of the same size and charge, the one, with electronic configuration (n – 1)dnns0, typical oftransition metals, is more polarising than the one which a noble gas configuration, ns2np6, typical of alkaliand alkaline earth metal cations.

The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing theelectronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup ofelectron charge density between the nuclei. The polarising power of the cation, the polarisability of theanion and the extent of distortion (polarisation) of anion are the factors, which determine the per centcovalent character of the ionic bond.

4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory :

Q. Explain the main postulates of (VSEPR) theory.

Solution : (a) The shape of a molecule depends upon the number of valence shell electron pairs (bonded ornonbonded) around the central atom.

(b) Pairs of electrons in the valence shell repel one another since their electron clouds are negativelycharged.

(c) These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximisedistance between them.

(d) The valence shell is taken as a sphere with the electron pairs localising on the spherical surface atmaximum distance from one another.

(e) A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiplebond are treated as a single super pair.

(f) Where two or more resonance structures can represent a molecule, the VSEPR model is applicable toany such structures. The repulsive interaction of electron pairs decrease in the order : Line pair (1p) – Lonepair (1p) > Lone pair (1p) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp).

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Q. Draw the molecular geometry of following molecules. BeCl2, BF

3, CH

4, PCl

5, SF

6, SO

2, NH

3, H

2O,

SF4, ClF

3, BrF

5, XeF

4

Solution : BeCl2 :

0180angleBondLinear

ClBeCl BF3 :

CH4 : PCl

5 :

SF6 : SO

2 :

NH3 : H

2O :

SF4 : ClF

3 :

BrF5 : XeF

4 :

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4.5 Valence Bond Theory :

Q. Explain the formation of H2 molecule on the basis of valence bond theory.

Solution : A labelled potential energy curve for H2 molecule is shown in the figure :

The potential energy curve for the formation of H2 molecule as a function of internuclear distance of H

atoms. The minimum on the curve corresponds to the most stable state of H2. At point A when the two atoms

are far apart from each other, there are no attractive or repulsive forces between the hydrogen atoms. Atpoint B when the two atoms starts approaching each other, they start interacting with each other and thesystem starts losing its energy because attractive forces between the two atoms become more and moredominant. At point C, the two atoms acquire minimum energy since the attractive forces are balanced by therepulsive forces. Beyond point C, after intermolecular distance r

0, repulsive forces predominate and

molecule formed tends towards disability. That is why potential energy increases abruptly.

Q. How many types of covalent bond are possible on the basis of type of overlapping ?

Solution : They are of two types : (i) sigma() bond, and (ii) pi() bond.

(i) Sigma() bond : This type of covalent bond is formed by the end to end (head-on) overlap of bondingorbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be formedby any one of the following types of combinations of atomic orbitals.

(a) s-s overlapping : In this case, there is overlap of two half filled s-orbitals along the internuclear axis asshown below :

(b) s-p overlapping : This type of overlap occurs between half filled s-orbitals of one atom and half filledp-orbitals of another atom.

(c) p-p overlapping : This type of overlap takes place between half filled p-orbitals of the two approachingatoms.

(ii) pi() bond : In the formation of bond the atomic orbitals overlap in such a way that their axes remainparallel to each other and perpendicular to the internuclear axis. The orbitals formed due to sidewiseoverlapping consists of two saucer type charged clouds above and below the plane of the participatingatoms.

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Q. Write the difference between sigma and pi bond ?

Solution : Sigma bond : (i) This bond is formed by overlap of orbitals along their internuclear axis (end toend overlap). (ii) This is formed by overlapping between s-s, s-p or p-p orbitals. (iii) Overlapping is quitelarge and hence sigma bond is a strong bond. (iv) Electron cloud in this case is symmetrical about the linejoining the two nuclei. (v) Sigma bond consists of only one electron cloud, symmetrical about theinternuclear axis. (vi) Free rotation about a -bond is possible.

Pi Bond : (i) This is formed by sideway overlapping of orbitals (lateral overlapping). (ii) This is formed bythe overlap of p-p orbitals only. (iii) Overlapping is to a small extent. Hence, -bond is a weak bond.(iv) Electron cloud of -bond is unsymmetrical. (v) Pi bond consists of two electron clouds, one above theplane of atomic nuclei and the other below it. (vi) Free rotation about a -bond is not possible because onrotation, overlapping vanishes and so the bond breaks.

4.6 Hybridisation :

Q. Explain the process of hybridisation.

Solution : Hybridisation is defined as the concept of intermixing of orbitals of same energy or of slightlydifferent energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetricallydisposed in plane. New orbitals formed are called hybrid orbitals.

Q. Write the salient feature of hybridisation and the conditions for hybridisation.

Solution : Salient features : (i) The number of hybrid orbitals is equal to the number of the atomic orbitalsthat get hybridised. (ii) The hybridised orbitals are always equivalent in energy and shape. (iii) The hybridorbitals are more effective in forming stable bonds than the pure atomic orbitals. (iv) These hybrid orbitalsare directed in space in some preferred direction to have minimum repulsion between electron pairs andthus a stable arrangement. Therefore, the type of hybridisation indicates the geometry of the molecules.

Conditions for hybridisation : (i) The orbitals present in the valence shell of the atom are hybridised.(ii) The orbitals undergoing hybridisation should have almost equal energy. (iii) Promotion of electron isnot essential condition prior to hybridisation. (iv) It is not necessary that only half filled orbitals participatein hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Q. Draw the hybridisied structure of BeCl2, BCl

3, CH

4.

Solution : BeCl2 :

In BeCl2, Be is sp hybridised. The shape of the molecule is linear having bond angle 1800.

BCl3 :

In BCl3, B is sp2 hybridised.

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CH4 :

CH4 is a tetrahedral molecule with each H–C – H angle equal to 1090 28’. The four sp3 hybrid orbitals

overlap with the half-filled 1s orbitals of four H-atoms.

Q. Draw the shape of C2H

4 molecule and C

2H

2 molecule showing the and bond.

Solution : C2H

4 : or

C2H

2 :

Q. Find the hybridisation of the following molecules : PF5, PCl

5, BrF

5, SF

6.

Solution : Molecule Hybridisation Shape of molecule

PF5

sp3d Trigonal bipyramidal

PCl5

sp3d Trigonal bipyramidal

BrF5

sp3d2 Square pyramidal

SF6

sp3d2 Square bipyramidal

Q. Describe the hybridisation of PCl5 and SF

6.

Solution : Formation of PCl5 (sp3d hybridisation) : The ground state and the excited state outer electronic

configurations of phosphorus (Z=15) are represented below.

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Formation of SF6 (sp3d2 hybridisation) : In SF

6 the central sulphur atom has the ground state outer

electronic configuration 3s23p4. In the exited state the available six orbitals i.e., one s, three p and two d aresingly occupied by electrons. These orbitals hybridise to form six new sp3d2 hybrid orbitals, which areprojected towards the six corners of a regular octahedron in SF

6. These six sp3d2 hybrid orbitals overlap

with singly occupied orbitals of fluorine atoms to form six S–F sigma bonds. Thus SF6 molecule has a

regular octahedral geometry as shown in figure below :

4.7 Molecular Orbital Theory :

Q. Write the salient features of molecular orbital theory.

Solution : (i) The electrons in a molecule are present in the various molecular orbitals as the electrons ofatoms are present in the various atomic orbitals.

(ii) The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.

(iii) While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it isinfluenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomicorbital is monocentric while a molecular orbital is polycentric.

(iv) The number of molecular orbital formed is equal to the number of combining atomic orbitals. Whentwo atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecularorbital while the other is called antibonding molecular orbital.

(v) The bonding molecular orbital has lower energy and hence greater stability than the correspondingantibonding molecular orbital.

(vi) Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital,the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.

(vii) The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeyingthe Pauli’s exclusion principle ans the Hund’s rule.

Q. Write the formation of molecular orbital by the linear combination of atomic orbitals.

Solution : Linear combination of Atomic orbitals (LCAO) in case of H2+

(i) A linear combination of two atomic orbitals A and B leads to the formation of two molecular

orbitals and

(ii) The energy E+ of molecular orbital is lowe than either of E

A and E

B (energies of isolated atoms). It

is therefore designated as bonding molecular orbital (BMO).

(iii) The energy E– of molecular orbital is higher than either of E

A and E

B. It is therefore designated as

antibonding molecular orbital (ABMO)

(iv) The extent of lowering of energy of the bonding molecular orbital is equal to the extent of increase ofenergy of antibonding molecular orbital.

Energies of bonding and antibonding molecular orbitals.

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Bonding MO’S : (2s) (2px) (2py) (2pz)

Anti Bonding MO’S : (2s) (2px) (2py) (2pz)

The order of energy of molecular orbital for lighter elements like boron, carbon and nitrogen are asfollows :

(1s) < *(1s) < (2s) < *(2s) < 2py = 2pz < 2px < *2py = *2pz < *px

The order of energy of molecular orbitals for heavier elements after nitrogen are :

(1s) < *(1s) < (2s) < *(2s) < 2px < 2py = 2pz < *2py = *2pz < *2px

Q. Define bond order.

Solution : Bond order (B.O.) is defined as one half the difference between the number of electrons presentin the bonding and the antibonding orbitals i.e.,

B.O. = 2

1 [N

b – N

a] N

b number of bonding electrons

Na number of anti-bonding electrons

If B.O. = 0, 1, 2, 3 so on it means that no bond is formed, one bond is formed, two bonds or three bonds areformed between the atoms respectively.

(i) B.O, Bond dissociation energy.

(ii) B.O. LengthBond

1

Q. What are diamagnetic and paramagnetic molecules ?

Solution : Electronic configuration helps to predict the magnetic character of the molecule. If all theelectrons in a molecule are paired they are diamagnetic (repelled by magnetic field) and if unpaired electronis present they are paramagnetic (attracted by magnetic field) for e.g., O

2 molecule.

Q. Write the molecular orbital electronic configuration for He2, Be

2, Li

2, C

2, O

2 and also write their

bond order.

Solution : Helium molecule (He2) : Molecular orbital electronic configuration :

1s2) *1s2)

Bond order of (He2) : ½(2 – 2) = 0

He2 molecule is therefore unstable and does not exist.

Berilium molecule (Be2) : Electronic configuration of Be is 1s22s2. Therefore for Be

2 molecule

1s2) *1s2) 2s2) *2s2)

Bond order of (He2) : ½(4 – 4) = 0

Be2 molecule is therefore unstable and does not exist.

Lithium molecule (Li2) : Electronic configuration of Lithium is 1s2 2s1. Therefore molecular orbital

electronic configuration is Li2 : (1s)2 (*1s)2 (2s)2. The configuration is also written as KK(2s)2 where

KK represents the closed K shell structure (1s)2 (*1s)2.

Bond order of (Li2) is ½ (4 – 2) = 1. It means it is stable and has no unpaired electrons. Therefore it is

diamagnetic.

Carbon molecule (C2) : Electronic configuration of C is 1s22s22p2. Therefore for C

2 molecule

1s2) *1s2) 2s2) *2s2) (2px2 = 2p

y2)

Bond order of (C2) is ½ (8 – 4) = 2

C2 molecule is diamagnetic.

Oxygen molecule (O2) : Electronic configuration of O is 1s22s22p4. Therefore for O

2 molecule

1s2) *1s2) 2s2) *2s2) (2pz)2 (2p

x2 2p

y2) (*2p

x1 *2p

y1)

Bond order of (O2) is ½ (10 – 6) = 2

O2 molecule is paramagnetic.

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4.9 Hydrogen Bonding :

Q. Define hydrogen bond and what is its causes.

Solution : Hydrogen bond can be defined as the attractive force which binds hydrogen atom of onemolecule with the electronegative atom (F, O or N) of another molecule.

When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between thetwo atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes highlyelectropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards X, thehydrogen acquires fractional positive charge (+) while ‘X’ attain fractional negative charge (–). Thisresults in the formation of a polar molecule having electrostatic force of attraction which can berepresented as : H+ – X– – – – H+ – X– – – – – H+ – X–.

Q. How many types of hydrogen bond are there. Explain them briefly ?

Solution : There are two types of hydrogen bonds :

(i) Intramolecular H-Bonding :

This type of H-bonding occurs when polar H and electronegative atom are present in the same molecule.

(a) (b)

(ii) Intermolecular H-Bonding :

This type of H-bonding takes place between H and electronegative element present in the differentmolecules of the same substance (as in between H

2O and H

2O) or different substances (as in between H

2O

and NH3)

e.g. In water molecules :

Due to polar nature of H2O, there is association of water molecules giving a liquid state of abnormally high

b.p.

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NCERT EXERCISE

4.1 Explain the formation of a chemical bond.

4.2 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N.

4.3 Write Lewis symbols for the following atoms and ions :

S and S2–; Al and Al3+, H and H–

4.4 Draw the Lewis structures for the following molecules and ions :

H2S, SiCl

4, BeF

2, CO

32–, HCOOH

4.5 Define octet rule. Write the significance and limitations.

4.6 Write the favourable factors for the formation of ionic bond.

4.7 Discuss the shape of the following molecules using VSEPR model : BeCl2, BCl

3, SiCl

4, AsF

5, H

2S, PH

3

4.8 Although geometries of NH3 and H

2O molecules are distorted tetrahedral, bond angle in water is less

than of ammonia. Discuss.

4.9 How do you express the bond strength in terms of bond order ?

4.10 Define the bond length.

4.11 Explain the important aspects of resonance with reference to the CO32– ion.

4.12 H3PO

3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as

the canocical forms of the resonance hybrid representing H3PO

3 ? If not, given reasons for the same.

4.13 Write the resonance structures for SO3, NO

2 and NO

3–.

4.14 Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b)Ca and O (c) Al and N.

4.15 Although both CO2 and H

2O are triatomic molecules, the shape of H

2O molecule is bent while that of

CO2 is linear. Explain this on the basis of dipole moment.

4.16 Write the significance/applications of dipole moment.

4.17 Define electronegativity. How does it differ from electron gain enthalpy ?

4.18 Explain with the help of suitable example polar covalent bond.

4.19 Arrange the bonds in order of increasing ionic character of the molecules :

LiF, K2O, N

2, SO

2 and ClF

3

4.20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown

incorrectly. Write the correct Lewis structure for acetic acid.

4.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four

H atoms at the corners of the square and the C atoms at its centre. Explain why CH4 is not square

planar ?

4.22 Explain, why BeH2 molecule has a zero dipole moment although the Be – H bonds are polar.

4.23 Which out of NH3 and NF

3 has higher dipole moment and why ?

4.24 What is meany by hybridisation of atomic orbitals ? Describe the shapes of sp, sp2, sp3 hybridorbitals.

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4.25 Describe the change in hybridisation if any, of the Al atom in the following reaction :

AlCl3 + Cl– AlCl

4–

4.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction ?

BF3 + NH

3 F

3B.NH

3

4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms inC

2H

4 and C

2H

2 molecules.

4.28 What is the total number of sigma and pi bonds in the following molecules ?

(a) C2H

2(b) C

2H

4

4.29 Considering x-axis as the intermolecular axis which out of the following will not form a sigma bondand why ?

(a) 1s and 1s (b) 1s and 2px

(c) 2py and 2p

y(d) 1s and 2s

4.30 Which hybrid orbitals are used by carbon atoms in the following molecules ?

(a) CH3 – CH

3(b) CH

3 – CH = CH

2(c) CH

3 – CH

2 – OH

(d) CH3 – CHO (e) CH

3COOH

4.31 What do you understand by bond pairs and lone pairs of electrons ? Illustrate by giving oneexample of each type.

4.32 Distinghish between a sigma and pi bond.

4.33 Explain the formation of H2 molecule on the basis of valence bond theory.

4.34 Write the important conditions required for the linear combination of atomic orbitals to formmolecular orbitals.

4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist.

4.36 Compare the relative stability of the following species and indicate their magnetic properties :O

2, O

2+, O

2– (superoxide), O

22– (peroxide).

4.37 Write the significance of a plus and a minus sign shown in representing the orbitals.

4.38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial

bonds ?

4.39 Define hydrogen bond. Is the weaker or stronger than the van der Waals forces ?

4.40 What is meant by the term bond order ? Calculate the bond order of : N2, O

2, O

2+ and O

2–.

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ANSWERS

4.2 , , , ,

4.3 , ,

4.4 , , , ,

4.5 Lewis postulated that atoms achieve the stable octet when they are linked by chemical bonds and indoing so each atom attains a stable outer octet of electrons. Significance of octet rule : It helps toexplain why different atoms combine with each other to form ionic compounds or covalentcompounds. Limitations of octet rule : (i) In the formation of hydrogen molecule duplet is completedand not the octet. (ii) Also in the formation of BeCl

2, BF

3, AlCl

3 etc. octet is not completed

4.6 (i) Low ionization enthalpy of the metal atom (ii) High electron gain enthalpy of the non-metal atom(iii) High lattice enthalpy of the compound formed.

4.9 Greater the bond order, shorter the bond length and greater the bond strength4.10 Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a

molecule

4.11

4.13 ,

,

4.15 Studies have shown that the net dipole moment of CO2 molecule is zero. This is possible only when

CO2 molecule is linear i.e., O = C = O, such that dipole moments of C – O are equal and opposite and

hence cancel out. On the other hand, it is found that H2O molecule has a net dipole moment of 1.84

D, though it contains 2O – H bonds. This suggests that it has a bent structure

4.16 (i) In determining the polarities of bonds. As µ = q × d, greater is the magnitude of dipole momenthigher will be the polarity of bonds H – F > H – Cl > H – Br > HI. Further in case of non-polarmolecules like O

2 and N

2, the dipole is found to be zero (ii) In calculation of percentage ionic

character4.17 Electronegativity : (i) It is the tendency of an atom to attract shared pair of electrons. (ii) It is the

property of bonded atom. (iii) The elements with symmetrical configuration have specificelectronegativities. (iv) It has no units. Electron gain enthalpy : (i) It provides a measure of the easewith which an atom adds an electron to form an anion. (ii) It is the property of an isolated atom.(iii) The elements with symmetrical configuration have almost zero electron gain enthalpy. (iv) It hasunits of kJ mol–1

4.18 Polar character of covalent bonds – when two unlike atoms form a covalent bond, the electrons maynot be shared equally by both the atoms. In such a case, the electrons cloud is closer to the one atomthan to the other. One end of the bond, therefore, develops partial positive charge and the other

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partial negative charge. Such a bond is called a partial polar bond. This is represented by symbols +

and –, where d stands for a partial charge. Example : ClH

4.19 N2 < SO

2 < ClF

3 < K

2O < LiF

4.22 (3) In BeH2 two dipoles act in opposite direction and hence the two dipoles cancel each other

completely and hence has a zero dipole moment.4.23 NH

3 has higher dipole moment than NF

3. This is because, in case of NH

3 the orbital dipole due to lone

pair is in the same direction as the resultant dipole moment of the N — H bonds, whereas in NF3 the

orbital is in the direction opposite to the resultant dipole moment of three N – F bonds. The orbitaldipole because of lone pair decreases the effect of the resultant N – F bond moments, which results inthe law dipole moment of NF

3.

4.26 Initially BF3 is sp2 hybridised and NH

3 is sp3 hybridised. F

3B.NH

3 is sp3 hybridised. Thus, there is a

change in hybridisation of boron.4.28 (a) (3, 2) (b) (5, 1)4.29 Only (c) will not give a -bond because taking x-axis as the internuclear axis, there will be lateral

(sideway) overlap between two 2py orbitals forming a -bond.

4.30 (a) both carbon atoms are sp3 hybridised (b) C1 = sp3, C

2 = sp2, C

3 = sp3 (c) both carbon atoms are sp3

hybridised (d) C1 = sp3, C

2 = sp2 (e) C

1 = sp3, C

2 = sp2

4.33 A labelled potential energy curve for H2 molecule is shown in the figure :

The potential energy curve for the formation of H2 molecule as a function of internuclear distance of

H atoms. The minimum on the curve corresponds to the most stable state of H2. At point A when the

two atoms are far apart from each other, there are no attractive or repulsive forces between thehydrogen atoms. At point B when the two atoms starts approaching each other, they start interact-ing with each other and the system starts losing its energy because attractive forces between the twoatoms become more and more dominant. At point C, the two atoms acquire minimum energy sincethe attractive forces are balanced by the repulsive forces. Beyond point C, after intermoleculardistance r

0, repulsive forces predominate and molecule formed tends towards disability. That is why

potential energy increases abruptly.4.34 (a) The combining atomic orbitals should have comparable energies (b) The combining atomic

orbitals must have proper orientation e.g., same symmetry, so that they are able to overlap to aconsiderable extent (c) The extent of overlaping should be large. Greater the overlap, greater will bethe electron density between the nuclei.

4.35 Molecular orbital electronic configuration of Be2 = 1s2, *1s2, 2s2, *2s2. Bond order =

2

NN ab

= 0)44(2

1 . From molecular orbital theory bond order of Be

2 is 0, which shows that Be

2 does not

exist.

4.36 Bond order of O2 = 2, O

2+ =

2

12 , O

2– =

2

11 , O

22– = 1. We know that higher the bond order higher the

stability i.e., O22– < O

2– < O

2 < O

2+

4.38 they undergo sp3d hybridization to form five equivalent orbitals.4.39 Hydrogen bond is stronger than ver der Waals forces.

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ADDITIONAL QUESTIONS AND PROBLEMS

Q. Identify the compounds which do not obey the octet rule from among SO2, SF

2, SF

4, SF

6, AlCl

3,

MgCl2, PCl

3, PCl

5 and XeF

2.

Q. Why is it easier to remove an electron from an O2 molecule than from an N

2 molecule ?

Q. Using Lewis electron-dot symbols, show the formation of bonds in

(a) H2O (b) C

2H

4

Q. What is meant by formal charge ? Calculate the formal charges of all the atoms in the nitrite and thecarbonate ion.

Q. In the following pairs, which compound will be more ionic and why ?

(a) NaCl and KCl (b) NaCl and NaI

(c) NaCl and MgCl2

Q. Arrange the following in order of increasing bond length, giving reasons.

(a) H –F, H – Cl, H – Br and H – I

(b) Carbon-carbon bond length in C2H

6, C

2H

4 and C

2H

2

Q. Draw the resonance structures for

(a) SO4

2– (b) NO3

– (c) ClO4

–

Q. Both CO2 and SO

2 contain polar covalent bonds, yet the former is nonpolar, while the latter is polar.

Give reasons.

Q. The N – F bond is more polar than the N – H bond, yet NH3 has a higher dipole moment than NF

3.

Justify.

Q. Using the VSEPR theory, predict and draw the structures of

(a) IF7

(b) SF4

(c) PCl5

Q. How does the VSEPR theory explain the

(a) bent structure of water

(b) pyramidal structure of ammonia

Q. Why is a sigma bond stronger than a pi bond ? Explain.

Q. Consider the following reaction : BF3 + F– BF

4–

What is the hybridisation of boron in the reactant and the product ?

Q. Is there any change in the hybridisation of nitrogen in NH3 and NH

4+ ? Justify your answer.

Q. Describe the hybridisation of PCl5 and SF

6.

Q. Explain why OF6 is not known but SF

6 is known.

Q. Using the molecular orbital theory, explain why Be2 does not exist.

Q. Explain why oxygen is paramagnetic but the peroxide ion is diamagnetic.

Q. Which is more stable – the peroxide ion or the superoxide ion ? Justify your answer by giving asuitable example.

Q. Define bond order. How does it affect the stability of a molecule ?

Q. Which of three – H2+, H

2– and H

2 – are paramagnetic and why ?

Q. Which will have a higher boiling point and why – NH3 or PH

3 ?

Q. Give an example of each of the following

(a) A molecule with a trigonal bipyramidal shape

(b) A molecule with a pyramidal shape

(c) A molecule containing a coordinate bond

Q. Using the VSEPR theory, explain why

(a) CF4 and XeF

4 differ in structures, and

(b) the shapes of PF5 and BrF

5 are different

Q. What is meant by sigma and pi bonds. Taking a suitable example, show under what conditions and bonds are formed.

Q. (a) What is meant by bonding and antibonding molecular orbitals ? Explain by taking theexample of H

2.

(b) Using the molecular orbital theory, explain the magnetic behaviour of O2, B

2 and C

2.

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