chemical bonding chemistry

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SUBJECT – CHEMISTRY

TOPIC – CHEMICAL BONDING

COURSE CODE – AISM-09/C/CMB

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Content :- Chemical Bonding

Introduction………………………………………………………………………………3 Electrovalency……………………………………………………………………………4 Energy change during the formation of ionic bond……………………6 Born Haber Cycle……………………………………………………………………….8 Covalency…………………………………………………………………………………12 Co-ordinate Covalency…………………………………………………………….15 Hybridization……………………………………………………………………………19 Maximum Covalency……………………………………………………………….34 VSEPR theory…………………………………………………………………………..35 Resonance……………………………………………………………………………….41 FAJAN’s Rule……………………………………………………………………………45 Dipole Moment……………………………………………………………………….49 Bond Characteristics……………………………………………………………….56 Hydrogen Bonding………………………………………………………………….59 Intermolecular Forces…………………………………………………………….62 Molecular Orbital Theory………………………………………………………..65 Miscellaneous Exercises………………………………………………………….72 Solved Problems…………………………………………………………………….77

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INTRODUCTION

A molecule is formed if it is more stable and has lower energy than the

individual atoms. Normally only electrons in the outermost shell of an atom

are involved in bond formation and in this process each atom attains a stable

electronic configuration of inert gas. Atoms may attain stable electronic

configuration in three different ways by loosing or gaining electrons by

sharing electrons. The attractive forces which hold various constituents

(atoms, ions etc) together in different chemical species are called chemical

bonds. Elements may be divided into three classes.

Electropositive elements, whose atoms give up one or more electrons

easily, they have low ionization potentials.

Electronegative elements, which can gain electrons. They have higher

value of electronegativity.

Elements which have little tendency to loose or gain electrons.

Three different types of bond may be formed depending on electropositive or

electronegative character of atoms involved.

Electropositive element + Electronegative element = Ionic bond

(electrovalent bond)

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Electronegative element + Electronegative element = Covalent bond

or less electro positive + Electronegative element = Covalent bond

Electropositive + Electropositive element = Metallic bond.

ELECTROVALENCY

This type of valency involves transfer of electrons from one atom to another,

whereby each atom may attain octet in their outermost shell. The resulting

ions that are formed by gain or loss of electrons are held together by

electrostatic force of attraction due to opposite nature of their charges. The

reaction between potassium and chlorine to form potassium chloride is an

example of this type of valency.

K Clxx

xx

xx

x K Clxx

xx

xxx or K Cl

Here potassium has one electron excess of it‘s octet and chlorine has one

deficit of octet. So potassium donates it‘s electron to chlorine forming an

ionic bond.

Ca++

O2–

(Ionic bond)

xx

CaO

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Here the oxygen accepts two electrons from calcium atom. It may be noted

that ionic bond is not a true bond as there is no proper overlap of orbitals.

Criteria for Ionic Bond:

One of the species must have electrons in excess of octet while the other

should be deficit of octet. Does this mean that all substance having surplus

electron and species having deficient electron would form ionic bond? The

answer is obviously no. Now you should ask why? The reasoning is that in an

ionic bond one of the species is cation and the other is anion. To form a

cation from a neutral atom energy must be supplied to remove the electron

and that energy is called ionization energy. Now it is obvious that lower the

ionization energy of the element the easier it is to remove the electron. To

form the anion, an electron adds up to a neutral atom and in this process

energy is released. This process is called electron affinity.

So for an ionic bond one of the species must have low ionization energy and

the other should have high electron affinity. Low ionization energy is mainly

exhibited by the alkali and alkaline earth metals and high electron affinity by

the halogen and chalcogens. Therefore this group of elements are

predominant in the field of ionic bonding.

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Energy Change During the Formation of Ionic Bond

The formation of ionic bond can be consider to proceed in three steps

(a) Formation of gaseous cations

+ -A(g)+I.E. A (g)+ e

The energy required for this step is called ionization energy (I.E)

(b) Formation of gaseous anions

- -X(g)+ e X (g)+E.A

The energy released from this step is called electron affinity (E.A.)

(c) Packing of ions of opposite charges to form ionic solid

+ -A (g)+ X (g) AX(s)+ energy

The energy released in this step is called lattice energy.

Now for stable ionic bonding the total energy released should be more than

the energy required.

From the above discussion we can develop the factors which favour

formation of ionic bond and also determine its strength. These factors have

been discussed below:

(a) Ionization energy: In the formation of ionic bond a metal atom loses

electron to form cation. This process required energy equal to the

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ionization energy. Lesser the value of ionization energy, greater is the

tendency of the atom to form cation. For example, alkali metals form

cations quite easily because of the low values of ionization energies.

(b) Electron affinity: Electron affinity is the energy released when

gaseous atom accepts electron to form a negative ion. Thus, the value

of electron affinity gives the tendency of an atom to form anion. Now

greater the value of electron affinity more is the tendency of an atom

to form anion. For example, halogens having highest electron affinities

within their respective periods to form ionic compounds with metals

very easily.

(c) Lattice energy: Once the gaseous ions are formed, the ions of

opposite charges come close together and pack up three dimensionally

in a definite geometric pattern to form ionic crystal.

Since the packing of ions of opposite charges takes place as a result of

attractive force between them, the process is accompanied with the

release of energy referred to as lattice energy. Lattice energy may be

defined as the amount of energy released when one mole of ionic solid

is formed by the close packing of gaseous ion.

In short, the conditions for the stable ionic bonding are:

(a) I.E. of cation forming atom should be low:

(b) E.A. of anion – forming atom should be high;

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(c) Lattice energy should be high.

Born Haber Cycle

Determination of lattice energy

The direct calculation of lattice enthalpy is quite difficult because the

required data is often not available. Therefore lattice enthalpy is determined

indirectly by the use of the Born – Haber cycle. The cycle uses ionization

enthalpies, electron gain enthalpies and other data for the calculation of

lattice enthalpies. The procedure is based on the Hess‘s law, which states

that the enthalpy of a reaction is the same, whether it takes place in a single

step or in more than one step. In order to understand it let us consider the

energy changes during the formation of sodium chloride from metallic

sodium and chlorine gas. The net energy change during the process is

represented by Hf.

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Example 1:

Calculate the lattice enthalpy of2MgBr . Given that

Enthalpy of formation of 2MgBr = -524 kJ -1mol

Some of first & second ionization enthalpy (IE1 + IE2 ) = 148 kJ mol 1

Sublimation energy of Mg = +2187 kJ -1mol

Vaporization energy of 2Br (I) = +31kJ -1mol

Dissociation energy of 2Br (g) = +193kJ -1mol

Electron gain enthalpy of Br(g) = -331 kJ -1mol

Solution:

f vapH S I.E H D 2 E.A. U

2 f

1Na(s) Cl (g) NaCl(s) ;Energychange ( H )

2

S1

D2

Na(g) Cl(g)

IEEA

Na (g) Cl (g)

e

2

S Sublimation of sodium

D Dissociation energyof Cl

IE IE of sodium(IE)

EA EA of chlorine(EA)

U Lattice energy of NaCl

U

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Or f vapU H S I.E H D 2 E.A.

Or U = 524 [2187 148 + 31 + 193 + 2 ( 331)]

= 524 1897= 12421kJ mol

Characteristics of ionic compounds:

The following are some of the general properties shown by these compounds

(i) Crystalline nature: These compounds are usually crystalline in nature

with constituent units as ions. Force of attraction between the ions is

non-directional and extends in all directions. Each ion is surrounded by a

number of oppositely charged ions and this number is called co-

ordination number. Hence they form three dimensional solid aggregates.

Since electrostatic forces of attraction act in all directions, therefore, the

ionic compounds do not posses directional characteristic and hence do

not show stereoisomerism.

(ii) Due to strong electrostatic attraction between these ions, the ionic

compounds have high melting and boiling points.

(iii) In solid state the ions are strongly attracted and hence are not free to

move. Therefore, in solid state, ionic compounds do not conduct

electricity. However, in fused state or in aqueous solution, the ions are

free to move and hence conduct electricity.

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(iv) Solubility: Ionic compounds are fairly soluble in polar solvents and

insoluble in non-polar solvents. This is because the polar solvents have

high values of dielectric constant which defined as the capacity of the

solvent to weaken the force of attraction between the electrical charges

immersed in that solvent. This is why water, having high value of

dielectric constant, is one of the best solvents.

The solubility in polar solvents like water can also be explained by the dipole

nature of water where the oxygen of water is the negative and hydrogen being

positive, water molecules pull the ions of the ionic compound from the crystal

lattice. These ions are then surrounded by water dipoles with the oppositely

charged ends directed towards them. These solvated ions lead an independent

existence and are thus dissolved in water. The electrovalent compound

dissolves in the solvent if the value of the salvation energy is higher than the

lattice energy of that compounds.

AB Lattice energy A B

These ions are surrounded by solvent molecules. This process is exothermic

and is called solvation.

x

A x solv. A solv. energy

y

B y solv. B solv. energy

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The value of solvation energy depend on the relative size of the ions. Smaller

the ions more is the solvation. The non-polar solvents do not solvate ions and

thus do not release energy due to which they do not dissolve ionic compounds.

(v) Ionic reactions: Ionic compound furnish ions in solutions. Chemical

reactions are due to the presence of these ions. For example

2

2 4 4Na SO 2Na SO

2

2BaCl Ba 2Cl

COVALENCY

This type of valency involves sharing of electrons between the concerned

atoms to attain the octet configuration with the sharing pair being

contributed by both species equally. The atoms are then held by this

common pair of electrons acting as a bond, known as covalent bond. If two

atoms share more than one pair then multiple bonds are formed. Some

examples of covalent bonds are

Sigma and Pi Bonding:

Cl Cl – Cl N N

Cl

N x x x

x x x

N x x

x x

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When two hydrogen atoms form a bond, their atomic orbitals overlap to

produce a greater density of electron cloud along the line connecting the two

nuclei. In the simplified representations of the formation of H2O and NH3

molecules, the O—H and N—H bonds are also formed in a similar manner,

the bonding electron cloud having its maximum density on the lines

connecting the two nuclei. Such bonds are called sigma bonds ( -bond).

A covalent bond established between two atoms having the maximum

density of the electron cloud along the line connecting the centre of the

bonded atoms is called a -bond. A -bond is thus said to possess a

cylindrical symmetry along the internuclear axis.

Let us now consider the combination of two nitrogen atoms. Of the three

singly occupied p-orbitals in each, only one p-orbital from each nitrogen

(say, the px may undergo ―head –on‖ overlap to form a -bond. The other

two p-orbitals on each can no longer enter into a direct overlap. But each

p-orbital may undergo lateral overlap with the corresponding p-orbital on the

neighbour atom. Thus we have two additional overlaps, one by the two py

orbitals, and the other by the two pz orbitals. These overlaps are different

from the type of overlap in a -bond. For each set of p-orbitals, the overlap

results in accumulation of charge cloud on two sides of the internuclear axis.

The bonding electron cloud does no more posses an axial symmetry as with

the -bond; instead, it possess a plane of symmetry. For the overlap of the

pz atomic orbital, the xy plane provides this plane of symmetry; for the

overlap of the py atomic orbitals, the zx plane serves the purpose. Bonds

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arising out of such orientation of the bonding electron cloud are designated

as -bonds. The bond formed by lateral overlap of two atomic orbitals having

maximum overlapping on both sides of the line connecting the centres of the

atoms is called a -bond. A -bond possess a plane of symmetry, often

referred to as the nodal plane.

-Bond

(a) s-s

overlapping

s s

(b) s-p

overlapping

+s p

(c) p-p

overlapping +

p p

- Bond:

This type of bond is formed by the sidewise or lateral overlapping of two half

filled atomic orbitals.

p p

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CO-ORDINATE COVALENCY

A covalent bond results from the sharing of pair of electrons between two

atoms where each atom contributes one electron to the bond. It is also

possible to have an electron pair bond where both electrons originate from

one atom and none from the other. Such bonds are called coordinate bond

or dative bonds. Since in coordinate bonds two electrons are shared by two

atoms, they differ from normal covalent-bond only in the way they are

formed and once formed they are identical to normal covalent –bond.

It is represented as [ ]

Atom/ion/molecule donating electron pair is called Donor or Lewis base.

Atom / ion / molecule accepting electron pair is called Acceptor or Lewis

acid [ ] points donor to acceptor

NH4+, NH3 has three (N – H) bond & one lone pair on N – atom. In NH4

+

formation this lone pair is donated to H+ (having no electron)

NH3 + H+ NH4+

Lewis base Lewis acid

NH

H

H

H+

NH

H

H

or H

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Properties of the coordinate compounds are intermediates of ionic and

covalent compounds.

Comparison of ionic, covalent & coordinate compounds

Property Ionic Covalent Coordinate

1. binding force Between ions strong

(coulombic)

Between molecules

smaller (Vander

Waal‘s)

in between

2. mp/bp High less than ionic in between

3. condition conductor of

electricity in fused

state & in aqueous

solution

bad conductor Greater than

covalent

4. solubility in

polar solvent

(H2O)

High Less in between

5. Solubility in non Low High in between

O S

O

O

SO3

P

ClCl

ClCl

Cl ClPCl 6

Sb

FF

FF

F FSbF6

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polar solvent

(ether)

6. Physical state generally solid liquid & gaseous solid, liquid

gas

Example 2:

Which of the following statement is/are not true for -bond.

1. It is formed by the overlapping of s s or s p orbitals

2. It is weaker than pi bond

3. It is formed when bond exists already.

4. It is resulted from partial overlapping of orbitals.

(A) 1, 2, 3, 4 (B) 2, 3 and 4

(C) 2 and 4 (D) 1, 2 and 4

Solution:

(B)

Example 3:

The types of bond present in ZnSO4.7H2O are only

(A) Electrovalent and covalent

(B) Electrovalent and co-ordinate

(C) Electrovalent, Covalent and coordinate

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(D) None of these

Solution:

(C)

Example 4:

Classify the following bonds as ionic, polar covalent or covalent and

give your answer

(a) Si Si bond in Cl3 SiSiCl3 (b) SiCl bond in Cl3SiSiCl3

(c) CaF bond in CaF2 (d) NH bond in NH3

Solution:

(a) Covalent due to identical electronegativity

(b) One electron pair is shared between Si & Cl and thus, covalent

bond is expected but electron negativity of Cl is greater than

that of Si & some polarity develops giving polar – covalent

nature

(c) Ionic since Ca completes its octet by transfer of two outershell

electrons thus, completing their octets

Ca [Ar]4s2, F[He)2s22p5

(d) Polar covalent, explanation as in (b)

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Example 5:

Arrange the bonds in order of increasing ionic character in the

molecules:

LiF, K2O, N2, SO2 and ClF3

Solution:

N2 < ClF3 < SO2 < K2O < LiF

Example 6:

Arrange the following in order of increasing ionic character:

C H, F H, Br H, Na I, K F and Li Cl

Solution:

C H < Br H < F H < Na I < Li Cl < K F

HYBRIDIZATION

The tetravalency shown by carbon is actually due to excited state of carbon

which is responsible for carbon bonding capacity.

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C excited state

1s 2sx2P y2P z2P

If the bond formed is by overlapping then all the bonds will not be

equivalent so a new concept known as hybridization is introduced which can

explain the equivalent character of bonds.

s and p orbital belonging to the same atom having slightly different energies

mix together to produce same number of new set of orbital called as hybrid

orbital and the phenomenon is called as hybridization.

Important characteristics of hybridization

(i) The number of hybridized orbital is equal to number of orbitals that

get hybridized.

(ii) The hybrid orbitals are always equivalent in energy and shape.

(iii) The hybrid orbitals form more stable bond than the pure atom orbital.

(iv) The hybrid orbitals are directed in space in same preferred direction to

have some stable arrangement and giving suitable geometry to the

molecule.

Depending upon the different combination of s and p orbitals, these types of

hybridization are known.

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(i) sp3 hybridization: In this case, one s and three p orbitals hybridise

to form four sp3 hybrid orbitals. These four sp3 hybrid orbitals are

oriented in a tetrahedral arrangement.

For example in methane CH4

109028'

2s

2px

2py2pz

(ii) sp2 hybridization: In this case one s and two p orbitals mix together

to form three sp2 hybrid orbitals and are oriented in a trigonal planar

geometry.

1200

2s2px 2py

The remaining p orbital if required form side ways overlapping with the

other unhybridized p orbital of other C atom and leads to formation of

bond as in H2C = CH2

(iii) sp hybridization: In this case, one s and one p orbital mix together

to form two sp hybrid orbitals and are oriented in a linear shape.

2s 2ps

1800

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The remaining two unhybridised p orbitals overlap with another

unhybridised p orbital leading to the formation of triple bond as in

HC CH.

Shape Hybridisation

Linear sp

Trigonal planar sp2

Tetrahedral sp3

Trigonal bipyramidal sp3d

Octahedral sp3d2

Pentagonal bipyramidal sp3d3

Example 7:

Which of the following molecule has trigonal planer geometry?

(A) CO2 (B) PCl5

(C) BF3 (D) H2O

Solution:

BF3 has trigonal planer geometry (sp2 - hybridized Boron).

Hence (A) is correct.

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Rule for determination of total number of hybrid orbitals

Detect the central atom along with the peripheral atoms.

Count the valence electrons of the central atom and the peripheral

atoms.

Divide the above value by 8. Then the quotient gives the number of

bonds and the remainder gives the non-bonded electrons. So number

of lone pair =non bonded electrons

2.

The number of bonds and the lone pair gives the total number of

hybrid orbitals.

An Example Will Make This Method Clear

SF4 Central atom S, Peripheral atom F

total number of valence electrons = 6 + (4 7) = 34

Now 34/8 = 4 2

8

Number of hybrid orbitals = 4 bonds + 1 lone pair)

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So, five hybrid orbitals are necessary and hybridization mode is sp3d and it

is trigonal bipyramidal (TBP).

Note:

Whenever there are lone pairs in TBP geometry they should be placed

in equatorial position so that repulsion is minimum.

1. NCl3 Total valence electrons = 26

Requirement = 3 bonds + 1 lone pair

Hybridization = sp3

Shape = pyramidal

2. BBr3 Total valence electron = 24

Requirement = 3 bonds

Hybridization = sp2

Shape = planar trigonal

3. SiCl4 Total valence electrons = 32


Hybridization = sp3

Shape = Tetrahedral

F

F

F

S

F

S

F

F

F

F

(A) (B)

N

ClCl

Cl

B

Br Br

Br

Si

Cl Cl Cl

Cl

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4. CI4 Total valence electron = 32

Requirements = 4 bonds

Hybridization = sp3

Shape = Tetrahedral

5. SF6 Total valence electrons = 48


Hybridization = sp3d2

Shape = octahedral/square bipyramidal

6. BeF2 Total valence electrons : 16

Requirement : 2 bonds

Hybridization : sp

Shape : Linear

F – Be – F

7. ClF3 Total valence electrons : 28

Requirement: 3 bonds + 2 lone pairs

Hybridization : sp3d

Shape : T – shaped

We have already discussed that whenever there are lone pairs they should

be placed in equatorial positions. Now a question that may come to your

mind that though the hybridization is sp3d, so the shape should be T.B.P.

But when all the bonds are present the actual shape is TBP. But when

instead of bond there are lone pairs in TBP the actual geometry is

C

l

l l l

S

F

F

F F

F

F

Cl F

F

F

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determined by the bonds not by the lone pairs. Here in ClF3 the bond

present (2 in axial and 1 in equatorial) gives the impression of T shape.

8. PF5 Total valence electrons : 40



Shape : Trigonal bipyramidal (TBP)

9. XeF4 Total valence electrons : 36

Requirement:4 bonds+ 2 lone pairs

Hybridisation : sp3d

Shape : Square planar

Now three arrangements are possible out of which A and B are same. A and

B can be inter converted by simple rotation of molecule. The basic difference

of (B) and (C) is that in (B) the lone pair is present in the anti position which

minimizes the repulsion which is not possible in structure (C) where the lone

pairs are adjacent. So in a octahedral structure the lone pairs must be

placed at the anti positions to minimize repulsion. So both structure (A) and

(B) are correct.

P

F

F

F

F

F

Xe

F

F

F

F

(A)

Xe

F

F F

F

(B)

Xe

F

F

F

(C)

F

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10. XeF2 Total valence electrons : 22

Requirements : 2 bonds + 3 lone pairs

Hybridisation: sp3d

Shape : Linear

[ l.p. are present in equatorial position and ultimate shape is due to the

bonds that are formed]

11. PF2Br3 Total valence electrons : 40

Requirements : 5 bonds

Hybridisation: sp3d

Shape : trigonal bipyramidal

Here we see that fluorine is placed in axial position whereas bromine is

placed in equatorial position. It is the more electronegative element that is

placed in axial position and less electronegative element is placed in

equatorial position. Fluorine, being more electronegative pulls away bonded

electron towards itself more than that is done by bromine atom which results

in decrease in bp – bp repulsion and hence it is placed in axial position.

In this context it can also be noted that in T.B.P. shape the bond lengths are

not same. The equatorial bonds are smaller than axial bonds. But in square

bipyramidal shape, all bond lengths are same.

Xe

F

F

P

FBr

BrF

Br

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12. 3-

4PO Total valence electrons : 32


Hybridisation: sp3

Shape: tetrahedral

Here all the structures drawn are resonating structures with O– resonating

with double bonded oxygen.

13. NO2– Total valence electron: 18

Requirement : 2 bonds + 1 lone pair

Hybridisation: sp2

Shape: angular

14. CO32– Total valence electrons: 24


Hybrdisation = sp2

Shape: planar trigonal

But C has 4 valence electrons of these 3 form bonds the rest will form a

bond.

O

P

O O O

O–

P

O O O

O

P

O O O

etc.

N

O O

C

O O

O

C

O O

O–

C

O O

O–

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In the structure one bond is a double bond and the other 2 are single. The

position of the double bonds keeps changing in the figure. Since peripheral

atoms are isovalent, so contribution of the resonanting structures are equal.

Thus it is seen that none of the bonds are actually single or double. The

actual state is

15. CO2 Total valence electrons:16

Requirement: 2 bonds

Hybridisation: sp

Shape: linear

O = C = O

16. -

4BF Total valence electrons = 32

Requirement= 4 bonds

Hybridisation: sp3

Shape: Tetrahedral

17. -

3ClO Total valence electron = 26

Requirement = 3 bond + 1 lone pair

Hybridization: sp3

Shape: pyramidal

18. XeO2F2 Total valence electrons : 34

Requirement: 4 bonds +1 lone pairs

O

O

Xe

F

F

C

O–2/3

O–2/3

O–2/3

Bond order = 3/2 = 1.5

B

FF

F

F–

Cl

OO

O-

Cl

OO

- O

Cl

O-

OO

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Shape: Distorted TBP (sea-saw geometry)

19. XeO3 Total valence electrons : 26

Requirement: 3 bonds + 1 lone pair

Hybridization: sp3

Shape: Pyramidal

20. XeOF4 Total valence electrons : 42

Requirement: 5 bonds + 1 lone pair

Hybridization: sp3d2

Shape: square pyramidal.

Example 8:

Predict the hybridization for the central atom in 3POCl , 4OSF , 5OIF

Solution:

3POCl Total No. of V.E. = 5 6 21 32

48 8

So, hybridization = 3sp

4OSF = 6 6 28 40

58 8

So, hybridization of s = 3dsp

Xe

O OO

Xe

F

F

F

FO

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5OIF

6 7 35 486

8 8

So, hybridization of I = 2 3d sp

Example 9:

Out of the three molecules XeF4, SF4 and SiF4 one which has

tetrahedral structures is

(A) All of three (B) Only SiF4

(C) Both SF4 and XeF4 (D) Only SF4 and XeF4

Solution:

Hybridization of XeF4 = sp3d2, SF4 = sp3d, SiF4 = sp3

Hence (B) is correct.

Example 10:

Among the following compounds in which case central element uses d

orbital to make bonds with attached atom

(A) 2BeF (B) 2XeF

(C) 4SiF (D) 3BF

Solution:

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In 2XeF . Xe atom has 3sp d hybridisation.

Hence (B) is correct.

Example 11:

When NH3 is treated with HCl, state of hybridisation on central

nitrogen

(A) Changes from sp3 to sp2

(B) Remains unchanged

(C) Changes from sp3 to sp3d

(D) Changes from sp3 to sp

Solution:

On NH4+ state of hybridisation on central nitrogen atom is sp3 as

in NH3.

N

HH

H

H+

Hence (B) is the correct answer.

Exercise 4:

Among the following which has bond angle very near to 90°

(A) NH3 (B) XeF4

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(C) BF3 (D) H2O

Exercise 5:

Homolytic fission of C – C bond in hexafluoroethane gives an

intermediate in which hybridization state of carbon is

(A) sp2 (B) sp3

(C) sp (D) cannot be determined

Exercise 6:

A molecule XY2 contains two , two bonds and one lone pair of

electrons in valence shell of X. The arrangement of lone pair as well

as bond pairs is

(A) Square pyramidal (B) Linear

(C) Trigonal planar (D) Unpredictable

Exercise 7:

Draw the structure the following indicating the hybridisation of the

central atom.

(A) SOF2 (B) SO2

(C) POCl3 (D) I3–

Exercise 8:

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The type of hybridisation of orbitals employed in the formation of SF6

molecule is …………..

Exercise 9:

The angle between two covalent bonds is maximum for……………(CH4,

H2O, CO2)

Exercise 10:

The bond angle in 2

4SO ion is ………………..

Maximum Covalency

Elements which have vacant d-orbital can expand their octet by transferring

electrons, which arise after unpairing, to these vacant d-orbital e.g. in

sulphur.

↿⇂ ↿⇂ ↿ ↿

↿ ↿ ↿ ↿ ↿ ↿ In excited state

In ground state

3s 3p 3d

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In excited state sulphur has six unpaired electrons and shows a valency of

six e.g. in SF6. Thus an element can show a maximum covalency equal to its

group number e.g. chlorine shows maximum covalency of seven.

VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY

(SHAPES AND GEOMETRY OF MOLECULES)

Molecules exist in a variety of shapes. A number of physical and chemical

properties of molecules arise from and are affected by their shapes. For

example, the angular shape of the water molecules explains its many

characteristic properties while a linear shape does not.

The determination of the molecular geometry and the development of

theories for explaining the preferred geometrical shapes of molecules is an

integral part of chemical bonding. The VSEPR theory (model) is a simple

treatment for understanding the shapes of molecules.

Strictly speaking VSEPR theory is not a model of chemical bonding. It

provides a simple recipe for predicting the shapes of molecules. It is infact

an extension of the Lewis interpretation of bonding and is quite successful in

predicting the shapes of simple polyatomic molecules.

The basic assumptions of the VSEPR theory are that:

Pairs of electrons in the valence shell of a central atom repel each other

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1. These pairs of electrons tend to occupy position in space that minimize

repulsions and thus maximize distance between them.

2. The valence shell is taken as a sphere with the electron pairs localizing

on the spherical surface at maximum distance from one another.

3. A multiple bonds are treated as a single super pair.

4. Where two or more resonance structures can depict a molecule, the

VSEPR model is applicable to any such structures

For the prediction of geometrical shapes of molecules with the help of VSEPR

model, it is convenient to divide molecules into two categories

Regular Geometry

Molecules in which the central atom has no lone pairs

Irregular Geometry

Molecules in which the central atom has one or more lone pairs, the lone

pair of electrons in molecules occupy more space as compared to the

bonding pair electrons. This causes greater repulsion between lone pairs of

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electrons as compared to the bond pairs repulsions. The descending order of

repulsion

(lp – lp) > (lp – bp) > (bp – bp)

where lp-Lone pair; bp-bond pair

Regular Geometry

Number of

electron pairs

Arrangement of

electrons

Molecular geometry Examples

2

B – A – B

Linear

2 2BeCl ,HgCl

3

=

120

=

120

3 3BF ,AlCl

4

A

1090 28'

A

1090 28'

4 4 4CH ,NH SiF

A

180

Linear

A

Trigonal planar

B

B B

A

Trigonal planar

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5

trigonal bipyramidal

900

1200

trigonal bipyramidal

900

1200

5 5PCl , PF

6

6SF

Irregular Geometry

Molecul

e Type

No. of

Bonding

pairs

No. of

lone

pair

Arrangement of

electrons pairs

Shape

(Geometry

)

Examples

2AB E 2 1 A

B B

Trigonal planar

Bent 2 3SO ,O

3AB E 3 1

Trigonal

pyramidal 3NH

2 2AB E 2 2

Bent 2H O

A 90

AB

B

B

B

B

B

Octahedral

A

B BB

Tetrahedral

A

BB

Tetrahedral

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4AB E 4 1

See saw 4SF

3 2AB E 3 2

T – shaped 3CIF

5AB E 5 1

Square

pyramidal 5BrF

4 2AB E 4 2

Square

planar 4XeF

Example 13:

Why the bond angle of H – C – H in methane (CH4) is 109° 28‘ while

H – N – H bond angle in NH3 is 107° though both carbon and nitrogen

are sp3 hybridized

Solution:

In CH4 there are 4 bond pair of electrons while in NH3 are 3 bond pair

of electrons and 1 lone pair of electrons. Since bond pair bond pair

repulsion is less than lone pair bond pair repulsion, in NH3 bond angle

is reduced from 109°28‘ to 107°.

B

A

BB

B

Trigonal bi-pyramidal

B

AB

B

Trigonal bi-pyramidal

B

AB

B

B

B

Octahedral

AB

B

B

B

Octahedral

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Example 14:

Why bond angle in NH3 is 107° while in H2O it is 104.5°?

Solution:

In NH3, central nitrogen atom bears only one lone pair of electrons

whereas in H2O central oxygen atom bears two lone pair of electrons.

Since the repulsion between lone pair-lone pair and lone pair – bond

pair is more than that between bond pair-bond pair, the repulsion in

H2O is much greater than that in NH3 which results in contraction of

bond angle from 109°28‖ to 104.5° in water while in NH3 contraction is

less i.e. from 109°28‖ to 107°.

“If the electronegativity of the peripheral atoms is more, then the

bond angle will be less”.

For example if we consider NH3 and NF3, F – N – F bond angle will be lower

than H – N – H bond angle. This is because in NF3 the bond pair is displaced

more towards F and in NH3 it is displaced more towards N. So accordingly

the b.p. – b.p. interaction is less in NF3 and more in NH3.

Example 15:

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The bond angle of H2O is 104° while that that of F2O is 102°.

Solution:

Both H2O and F2O have a lone pair of electrons. But fluorine being

highly electronegative, the bond pair of electrons are drawn more

towards F in F2O, whereas in H2O it is drawn towards O. So in F2O the

bond pairs being displaced away from the central atom, has very little

tendency to open up the angle. But in H2O this opening up is more as

the bond pair electrons are closer to each other. So bond of F2O is

less than H2O.

RESONANCE

There may be many molecules and ions for which it is not possible to draw a

single Lewis structure. For example we can write two electronic structures

of O3.

O

H H

repulsion more O

F F

Repulsion less

O

O

O O

O

O

(A) (B)

(+) (+)

( ) ( )

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In (A) the oxygen oxygen bond on the left is a double bond and the

oxygen oxygen bond on the right is a single bond. In B the situation is just

opposite. Experiment shows however, that the two bonds are identical.

Therefore neither structure A nor B can be correct.

One of the bonding pairs in ozone is spread over the region of all the three

atom rather than associated with particular oxygen oxygen bond. This

delocalised bonding is a type of bonding in which bonding pair of electrons is

spread over a number of atoms rather than localised between two.

Structures (A) and (B) are called resonating or canonical structures and C is

the resonance hybrid. This phenomenon is called resonance, a situation in

which more than one plausible structure can be written for a species and in

which the true structure cannot be written at all.

Some other examples

(i) CO32– ion

O

O

O

(C)

O O

O

O O

O

O O

O

O O

O

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(ii) Carbon oxygen bond lengths in carboxylate ion are equal due to

resonance.

(iii) Benzene

(iv) Vinyl Chloride

Difference in the energies of the canonical forms and resonance hybrid is

called resonance stabilization energy and provides stability to species.

Rules for writing resonating structures

Only electrons (not atoms) may be shifted and they may be shifted

only to adjacent atoms or bond positions.

The number of unpaired electrons should be same in all the canonical

form.

R

O-

O

R

O

O-

R

O

O

CH2

Cl

H2C

Cl+

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The positive charge should reside as far as possible on less

electronegative atom and negative charge on more electronegative

atom.

Like charge should not reside on adjacent atom

The larger the number of the resonating structures greater the

stability of species.

Greater number of covalency add to the stability of the molecule.

Example 17:

Out of the following resonating structures for CO2 molecule, which are

important for describing the bonding in the molecule and why?

O C O O C O O C O O C O

(I) (II) (III) (IV)

2

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Solution:

Out of the structures listed above, the structure (III) is wrong since

the number of electron pairs on oxygen atoms are not permissible.

Similarly, the structures (II) has very little contribution towards the

hybrid because one of the oxygen atoms (electronegative) is show to

have positive charge. Carbon dioxide is best represented by structures

(I) and (IV).

FACTORS GOVERNING POLARIZATION AND POLARISABILITY

(FAJAN’S RULE)

Cation Size: Smaller is the cation more is the value of charge density

( ) and hence more its polarising power. As a result more covalent

character will develop. Let us take the example of the chlorides of the

alkaline earth metals. As we go down from Be to Ba the cation size

increases and the value of decreases which indicates that BaCl2 is

less covalent i.e. more ionic. This is well reflected in their melting

points. Melting points of BeCl2 = 405°C and BaCl2 = 960°C.

Cationic Charge: More is the charge on the cation, the higher is the

value of and higher is the polarising power. This can be well

illustrated by the example already given, NaBr and AlBr3. Here the

charge on Na is +1 while that on Al in +3, hence polarising power of Al

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is higher which in turn means a higher degree of covalency resulting in

a lowering of melting point of AlBr3 as compared to NaBr.

Noble Gas vs Pseudo Noble Gas Cation: A Pseudo noble gas cation

consists of a noble gas core surrounded by electron cloud due to filled

d-subshell. Since d-electrons provide inadequate shielding from the

nuclei charge due to relatively less penetration of orbitals into the

inner electron core, the effective nuclear charge (ENC) is relatively

larger than that of a noble gas cation of the same period. NaCl has got

a melting point of 800°C while CuCl has got melting point of 425°C.

The configuration of Cu+ = [Ar] 3d10 while that of Na+ = [Ne]. Due to

presence of d electrons ENC is more and therefore Cl– is more

polarised in CuCl leading to a higher degree of covalency and lower

melting point.

Anion Size: Larger is the anion, more is the polarisability and hence

more covalent character is expected. An e.g. of this is CaF2 and CaI2,

the former has melting point of 1400°C and latter has 575°C. The

larger size of I– ion compared to F– causes more polarization of the

molecule leading to a lowering of covalency and increasing in melting

point.

Anionic Charge: Larger is the anionic charge, the more is the

polarisability. A well illustrated example is the much higher degree of

covalency in magnesium nitride (3Mg++ 2N3–) compared to magnesium

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fluoride (Mg++ 2F—). This is due to higher charge of nitride compare to

fluoride. These five factors are collectively known as Fajan‘s Rule.

Example 18:

The melting point of KCl is higher than that of AgCl though the crystal

radii of Ag+ and K+ ions are almost same.

Solution:

Now whenever any comparison is asked about the melting point of the

compounds which are fully ionic from the electron transfer concept it

means that the compound having lower melting point has got lesser

amount of ionic character than the other one. To analyse such a

question first find out the difference between the 2 given compounds.

Here in both the compounds the anion is the same. So the deciding

factor would be the cation. Now if the cation is different, then the

answer should be from the variation of the cation. Now in the above

example, the difference of the cation is their electronic configuration.

K+ = [Ar]; Ag+ = [Kr] 4d10. This is now a comparison between a noble

gas core and pseudo noble gas core, the analysis of which we have

already done. So try to finish off this answer.

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Example 19:

AlF3 is ionic while AlCl3 is covalent.

Solution:

Since F– is smaller in size, its polarisability is less and therefore it is

having more ionic character. Whereas Cl being larger in size is having

more polarisability and hence more covalent character.

Example 20:

Which compound from each of the following pairs is more covalent and

why?

(a) CuO or CuS (b) AgCl or AgI ‗

(c) PbCl2 or PbCl4 (d) BeCl2 or MgCl2

Solution:

(a) CuS (b) AgI

(c) PbCl4 (d) BeCl2

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DIPOLE MOMENT

Difference in polarities of bonds is expressed on a numerical scale. The

polarity of a molecule is indicated in terms of dipole moment . To measure

dipole moment, a sample of the substance is placed between two electrically

charged plates. Polar molecules orient themselves in the electric field

causing the measured voltage between the plates to change.

The dipole moment is defined as the product of the distance separating

charges of equal magnitude and opposite sign, with the magnitude of the

charge. The distance between the positive and negative centres called the

bond length.

Thus, electric charge bond length= q d

As q is in the order of 1010 esu and d is in the order of 810 cm, is the order

of 1810 esu cm . Dipole moment is measured in ‗Debye‘ unit (D)

18 301D 10 esu cm 3.33 10 coulomb metre

Note:

(i) Generally as electronegativity difference increase in diatomic

molecules, polarity of bond between the atoms increases

therefore value of dipole moment increases.

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(ii) Dipole moment is a vector quantity

(iii) A symmetrical molecule is non- polar even though it contains

polar bonds. For example, 2 3 4CO , BF , CCl etc because summation of

all bond moments present in the molecules cancel each other.

(iv) Unsymmetrical non-linear polyatomic molecules have net value

of dipole moment. For example, 2 3 3H O, CH OH, NH etc.

Calculation of Resultant Bond Moments

Let AB and AC are two polar bonds inclined at an

angle their dipole moments are 1 and

2.

Resultant dipole moment may be calculated

using vectorial method.

2 2

R 1 2 1 22 cos

when = 0 the resultant is

maximum

R 1 2

when, = 180 , the resultant

is minimum

R 1 2

Example 21. The compound which has zero dipole moment is

Cl

C

Cl

Cl

Cl

A

B

C

1

R

2

180o

180o

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(A) CH2Cl2 (B) NF3

(C) PCl3F2(D) ClO2

Solution: (C)

Example 22. Sketch the bond moments and resultant dipole moment in

(i)2SO (ii)

2 2 2Cis C H Cl and (iii) 2 2 2trans C H Cl

Solution:

(iii)

C C

Cl

H Cl

H

Resultant = 0

Example 23. CO2 has got dipole moment of zero. Why?

Solutions: The structure of CO2 is . This is a highly

symmetrical structure with a plane of symmetry passing through

the carbon. The bond dipole of C–O is directed towards oxygen

as it is the negative end. Here two equal dipoles acting in

opposite direction cancel each other and therefore the dipole

moment is zero.

Example 24. Dipole moment of CCl4 is zero while that of CHCl3 is non

zero.

S

OO

(i)

C C

H

Cl Cl

H

(ii)

O = C = O

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Solution: Both CCl4 & CHCl3 have tetrahedral structure but CCl4 is

symmetrical while CHCl3 is non-symmetrical.

Due to the symmetrical structure of CCl4 the

resultant of bond dipoles comes out to be zero. But in case of

CHCl3 it is not possible as the presence of hydrogen introduces

some dissymmetry.

Example 25. Compare the dipole moment of H2O and F2O.

Solution: Let‘s draw the structure of both the compounds and then

analyse it.

In both H2O and F2O the structure is quite the same.

In H2O as O is more electronegative than hydrogen so the

resultant bond dipole is towards O, which means both the lone

pair and bond pair dipole are acting in the same direction and

dipole moment of H2O is high. In case of F2O the bond dipole is

acting towards fluorine, so in F2O the lone pair and bond pair

dipole are acting in opposition resulting in a low dipole.

In C-H, carbon being more electronegative the

dipole is projected towards C. Now the question comes whether

hybridization has anything to do with the dipole moment. The

Cl

C

ClCl

Cl

Symmetrical

H

C

ClCl

Cl

Non-Symmetrical

O

H H

O

FF

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answer is obviously yes. If yes, why? Depending on the

hybridization state the electronegativity of carbon changes and

therefore the dipole moment of C-H bond will change. As the s

character in the hybridized state increases, the electronegativity

of C increases due to which C attracts the electron pair of C-H

bond more towards itself resulting in a high bond dipoles.

Now as we have said about carbon hydrogen

bonds, the question that is coming to your mind is whether we

would be dealing with organic compounds or not. Yes we would

be dealing with the organic compounds.

For instance but -2- ene. It exists in two forms cis and

Trans.

The trans isomer is symmetrical with the 2 methyl

groups in anti position. So the bond dipoles the two Me– C bonds

acting in opposition, cancel each other results into a zero dipole.

Whereas in cis isomer the dipoles do not cancel each other

resulting in a net dipole.

Example 26. The molecule having largest dipole moment among the

following is

(A) CH4 (B) CHCl3

Cis

H CH3

H CH3

C

C

Trans

H CH3

HCH3

C

C

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(C) CCl4 (D) CHI3

Solution: (B)

Example 27. Compare the dipole moment of cis 1, 2 dichloroethylene

and trans 1, 2 dichloroethylene.

Solution:

In the trans compound the C-Cl bond dipoles are equal and

at the same time acting in opposition cancel each other while in

cis compound the dipoles do not cancel each other resulting in a

higher value.

Generally all Trans compounds have a lower dipole

moment corresponding to Cis isomer, when both the

substituents attached to carbon atom are either electron

releasing or electron withdrawing.

PERCENTAGE OF IONIC CHARACTER

Every ionic compound having some percentage of covalent character

according to Fajan‘s rule. The percentage of ionic character in a compound

having some covalent character can be calculated by the following equation.

The percent ionic character =

Observed dipole moment100

Calculated dipole moment assuming 100% ionic bond

Example 28. Dipole moment of KCl is 3.336 10–29 coulomb metre

which indicates that it is highly polar molecule. The interatomic

distance between K+ and Cl– is

2.6 10–10 m. Calculate the dipole moment of KCl molecule if

Cl

Cis

Cl

HH

C C

H

Trans

Cl

ClH

C C

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there were opposite charges of one fundamental unit located at

each nucleus. Calculate the percentage ionic character of KCl.

Solution: Dipole moment = e d coulomb metre

For KCl d = 2.6 10–10 m

For complete separation of unit charge

e = 1.602 10–19 C

Hence = 1.602 10–19 2.6 10–10 = 4.1652 10–29 Cm

KCl = 3.336 10–29 Cm

% ionic character of KCl = 29

29

3.336 10100

4.165 10 = 80.09%

Example 29. Calculate the % of ionic character of a bond having length

= 0.83 Å and 1.82 D as it’s observed dipole moment.

Solution: To calculate considering 100% ionic bond

= 4.8 10–10 0.83 10–8esu cm

= 4.8 0.83 10–18 esu cm = 3.984 D

% ionic character = 68.45100984.3

82.1

The example given above is of a very familiar compound called

HF. The % ionic character is nearly 43.25%, so the % covalent

character is (100 – 43.25) = 56.75%. But from the octet rule

HF should have been a purely covalent compound but actually it

has some amount of ionic character in it, which is due to the

electronegativity difference of H and F. Similarly knowing the

bond length and observed dipole moment of HCl, the % ionic

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character can be known. It was found that HCl has 17% ionic

character. Thus it can be clearly seen that although we call HCl

and HF as covalent compounds but it has got appreciable

amount of ionic character. So from now onwards we should call

a compound having more of ionic less of covalent and vice versa

rather than fully ionic or covalent.

BOND CHARACTERISTICS

1. Bond Length: The distance between the nuclei of two atoms bonded

together is termed as bond length or bond distance. It is expressed in

angstrom A

units or picometer (pm).

8 121A 10 cm;1pm 10 m

Bond length in ionic compound = c ar r

Similarly, in a covalent compound, bond length is obtained by

adding up the covalent (atomic) radii of two bonded atoms.

Bond length in covalent compound (AB) = A Br r

The factors such as resonance, electronegativity, hybridization, steric

effects, etc., which affect the radii of atoms, also apply to bond lengths.

Important features

(i) The bond length of the homonuclear diatomic molecules are twice the

covalent radii.

(ii) The lengths of double bonds are less than the lengths of single bonds

between the same two atoms, and triple bonds are even shorter than double

bonds.

Single bond > Double bond > Triple bond (decreasing bond length)

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(iii) Bond length decreases with increase in s-character since s-orbital is

smaller than a

p – orbital.

3sp C H 1.112A ; 2sp C H 1.103A ; spC H 1.08A ;

(25% s-character as in alkanes) (33.3% s-character as in alkenes) (50%

s-character as in alkynes)

(IV) BOND LENGTH OF POLAR BOND IS SMALLER THAN THE THEORETICAL

NON-POLAR BOND LENGTH.

2. BOND ENERGY OR BOND STRENGTH: BOND ENERGY OR BOND

STRENGTH IS DEFINED AS THE AMOUNT OF ENERGY REQUIRED TO BREAK

A BOND IN MOLECULE.

Important features

(I) THE MAGNITUDE OF THE BOND ENERGY DEPENDS ON THE TYPE OF

BONDING. MOST OF THE COVALENT BONDS HAVE ENERGY BETWEEN

50 TO 100 KCAL 1mol (200-400 KJ 1mol ). STRENGTH OF SIGMA BOND IS

MORE THAN THAT OF A -BOND.

(ii) A double bond in a diatomic molecules has a higher bond energy than a

single bond and a triple bond has a higher bond energy than a double

bond between the same atoms.

C C C C C C (decreasing bond length)

(iii) The magnitude of the bond energy depends on the size of the atoms

forming the bond, i.e. bond length. Shorter the bond length, higher is

the bond energy.

(iv) Resonance in the molecule affects the bond energy.

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(v) The bond energy decreases with increase in number of lone pairs on the

bonded atom. This is due to electrostatic repulsion of lone pairs of

electrons of the two bonded atoms.

(vi) Homolytic and heterolytic fission involve different amounts of energies.

Generally the values are low for homolytic fission of the bond in

comparison to heterolytic fission.

(vii) Bond energy decreases down the group in case of similar molecules.

(viii) Bond energy increase in the following order:

2 3s p sp sp sp

C C N N O O

No lone pair One lone pair Two lone pair

3. Bond angles: Angle between two adjacent bonds at an atom in a

molecule made up of three or more atoms is known as the bond angle.

Bond angles mainly depend on the following three factors:

(i) Hybridization: Bond angle depends on the state of hybridization of

the central atom

3 2

4 3 2

Hybridization sp sp sp

Bond angle 109 28 120 180

Example CH BCl BeCl

Generally s- character increase in the hybrid bond, the bond angle

increases.

(ii) Lone pair repulsion: Bond angle is affected by the presence of lone

pair of electrons at the central atom. A lone pair of electrons at the

central atom always tries to repel the shared pair (bonded pair) of

electrons. Due to this, the bonds are displaced slightly inside resulting in

a decrease of bond angle.

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(iii) Electronegativity: If the electronegativity of the central atom

decreases, bond angle decreases.

HYDROGEN BONDING

Introduction:

An atom of hydrogen linked covalently to a strongly electronegative atom

can establish an extra weak attachment to another electronegative atom in

the same or different molecules. This attachment is called a hydrogen bond.

To distinguish from a normal covalent bond, a hydrogen bond is represented

by a broken line eg X – H…Y where X & Y are two electronegative atoms. The

strength of hydrogen bond is quite low about 2-10 kcal mol–1 or 8.4–42 kJ

mol–1 as compared to a covalent bond strength 50–100 kcal mol–1 or 209 –

419 kJ mol–1

Conditions for Hydrogen Bonding:

Hydrogen should be linked to a highly electronegative element.

The size of the electronegative element must be small.

These two criteria are fulfilled by F, O, and N in the periodic

table. Greater the electronegativity and smaller the size, the stronger

is the hydrogen bond which is evident from the relative order of

energies of hydrogen bonds.

Types of Hydrogen Bonding:

Intermolecular hydrogen bonding: This type of bonding takes

place between two molecules of the same or different types. For

example,

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Intermolecular hydrogen bonding leads to molecular association in liquids

like water etc. Thus in water only a few percent of the water molecules

appear not to be hydrogen bonded even at 90°C. Breaking of those

hydrogen bonds throughout the entire liquid requires appreciable heat

energy. This is indicated in the relatively higher boiling points of hydrogen

bonded liquids. Crystalline hydrogen fluoride consists of the polymer (HF)n.

This has a zig-zag chain structure involving

H-bond.

Intramolecular hydrogen bonding: This type of bonding occurs

between atoms of the same molecule present on different sites.

Intramolecular hydrogen bonding gives rise to a closed ring structure

for which the term chelation is sometimes used. Examples are

o-nitrophenol, salicylaldehyde.

Effect of Hydrogen Bonding

Hydrogen bonding has got a very pronounced effects on certain properties of

the molecules. They have got effects on

State of the substance

Solubility of the substance

O—H----- O — H ------ O — H ------

H H H

F

F

H

H

HH

F

F

H

F

O

OOH

N

o-nitrophenol Salicaldehyde

H

OO

H

C

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Boiling point

Acidity of different isomers

These can be evident from the following examples.

Example 30. H2O is a liquid at ordinary temperature while H2S is a gas

although both O and S belong to the same group of the periodic

table.

Solution: H2O is capable of forming intermolecular hydrogen bonds.

This is possible due to high electronegativity and small size of

oxygen. Due to intermolecular H-bonding, molecular association

takes place. As a result the effective molecular weight increases

and hence the boiling point increases. So H2O is a liquid. But in

H2S no hydrogen bonding is possible due to large size and less

electronegativity of S. So it‘s boiling point is equal to that of an

isolated H2S molecule and therefore it is a gas.

Example 31. Ethyl alcohol (C2H5OH) has got a higher boiling point than

dimethyl ether

(CH3-O-CH3) although the molecular weight of both are same.

Solution: Though ethyl alcohol and dimethyl ether have the same

molecular weight but in ethyl alcohol the hydrogen of the O-H

groups forms intermolecular hydrogen bonding with the OH group

in another molecule. But in case of ether the hydrogen is linked

to C is not so electronegative to encourage the hydrogen to from

hydrogen bonding.

H O

C2H5

H O H O

C2H5C2H5

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Due to intermolecular H-bonding, ethyl alcohol remains in

the associated form and therefore boils at a higher temperature

compared to dimethyl ether.

Importance of Hydrogen Bonding in Biological Systems:

Hydrogen bonding plays a vital role in physiological systems. Proteins

contain chains of amino acids. The amino acid units are arranged in a spiral

form somewhat like a stretched coil spring (forming a helix). The N-H group

of each amino acid unit and the fourth C=O group following it along the

chain, establishes the N–H---O hydrogen bonds. These bonds are partly

responsible for the stability of the spiral structure. Double helix structure of

DNA also consists of two strands forming a double helix and are joined to

each other through hydrogen bond.

INTERMOLECULAR FORCES

We have enough reasons to believe that a net attractive force operates

between molecules of a gas. Though weak in nature, this force is ultimately

responsible for liquifaction and solidification of gases. But we cannot explain

the nature of this force from the ideas of ionic and covalent bond developed

so far, particularly when we think of saturated molecules like H2, CH4, He

etc. The existence of intermolecular attraction in gases was first recognised

by Vanderwaal‘s and accordingly intermolecular forces have been termed as

Vanderwaal‘s forces. It has been established that such forces are also

present in the solid and liquid states of many substances. Collectively they

have also been termed London forces since their nature was first explained

by London using wave mechanics.

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Nature of Vanderwaal’s Forces:

The Vanderwaal‘s forces are very weak in comparison to other chemical

forces. In solid NH3 it amount to about 39 KJ mol–1 (bond energy N-H bond

= 389 KJ mol–1). The forces are non directional. The strength of

Vanderwaal‘s force increases as the size of the units linked increases. When

other factors (like H-bonding is absent), this can be appreciated by

comparison of the melting or boiling points of similar compounds in a group.

Origin of Intermolecular Forces:

Intermolecular forces may have a wide variety of origin.

Dipole-dipole interaction: This force would exist in any molecule

having a permanent dipole e.g. HF, HCl, H2O etc.

Ion-dipole interaction: These interactions are operative in

solvation and dissolution of ionic compounds in polar solvents.

Induced dipole interaction: These generate from the polarisation

of a neutral molecule by a charge or ion.

Instantaneous dipole-induced dipole interaction: In non polar

molecules dipoles may generate due to temporary fluctuations in

electron density. These transient dipole can now induce dipole in

neighbouring molecules producing a weak temporary interaction.

METALLIC BONDING

Metals are characterised by bright, lustre, high electrical and thermal

conductivity, malleability, ductility and high tensile strength. A metallic

crystal consists of very large number of atoms arranged in a regular pattern.

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Different model have been proposed to explain the nature of metallic

bonding two most important modules are as follows

The electron sea Model

In this model a metal is assumed to consist of a lattice of positive ion (or

kernels) immersed in a sea of mobile valence electrons, which move freely

within the boundaries of a crystal. A positive kernel consists of the nucleus

of the atom together with its core on a kernel is, therefore, equal in

magnitude to the total valence electronic charge per atom. The free

electrons shield the positively charged ion cores from mutual electrostatic

repulsive forces which they would otherwise exert upon one another. In a

way these free electrons act as ‗glue‘ to hold the ion cores together.

The forces that hold the atoms together in a metal as a result of the

attraction between positive ions and surrounding freely mobile electrons are

known as metallic bonds.

Through the electron sea predated quantum mechanics it still satisfactorily

explains certain properties of the metals. The electrical and thermal

conductivity of metals for example, can be explained by the presence of

mobile electrons in metals. On applying an electron field, these mobile

electrons conduct electricity throughout the metals from one end to other.

Similarly, if one part of metal is heated, the mobile electrons in the part of

+

+

+

+

+

+

+

+

+

+

+

+

Positive Kernles

Mobile electrons

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the metals acquire a large amount of kinetic energy. Being free and mobile,

these electrons move rapidly throughout the metal and conduct heat to the

other part of the metal.

On the whole this model is not satisfactory.

Example 32. Sodium metal conducts electricity because it

(A) is a soft metal

(B) contains only one valence electron

(C) has mobile electron

(D) reacts with water to form H2 gas

Solution: (C)

MOLECULAR ORBITAL THEORY

In Molecular Orbital Theory (MOT) the atoms in a molecule are supposed to

loose their individual control over the electrons. The nuclei of the bonded

atoms are considered to be present at equilibrium inter-nuclear positions.

The orbitals where the probability of finding the electrons is maximum are

multicentred orbitals called molecular orbitals extending over two or more

nuclei.

In MOT the atomic orbitals loose their identity and the total number of

electrons present are placed in MO‘s according to increasing energy

sequence (Auf Bau Principle) with due reference to Pauli‘s Exclusion Principle

and Hund‘s Rule of Maximum Multiplicity.

When a pair of atomic orbitals combine they give rise to a pair of molecular

orbitals, the bonding and the anti-bonding. The number of molecular orbitals

produced must always be equal to the number of atomic orbitals involved.

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Electron density is increased for the bonding MO‘s in the inter-nuclear region

but decreased for the anti-bonding MO‘s, Shielding of the nuclei by increased

electron density in bonding MO‘s reduces inter nuclei repulsion and thus

stabilizes the molecule whereas lower electron density even as compared to

the individual atom in anti-bonding MO‘s increases the repulsion and

destabilizes the system.

In denotation of MO‘s, indicates head on overlap and represents side

ways overlap of orbitals. In simple homonuclear diatomic molecules the

order of MO‘s based on increasing energy is

This order is true except B2, C2 & N2. If the molecule contains unpaired

electrons in MO‘s it will be paramagnetic but if all the electrons are paired up

then the molecule will be diamagnetic.

Bond order

= no. of e s occupying bonding MO's no. of e soccupyingantibonding MO's

2

Application of MOT to homonuclear diatomic molecules.

H2 molecule : Total no. of electrons = 2

Arrangement : 2

1s

Bond order : ½ (2 – 0) = 1

2H molecule : Total no. of electrons = 1

Arrangement : 1

1s

Bond order : ½ (1 – 0) = 1/2

He2 molecule : Total no. of electrons = 4

y y

x x

zz

2p 2p

1s 1s 2s 2s 2p 2p

2p2p

** * *

*

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Arrangement : 2

1s

2

1s

*

Bond order : ½ (2 – 2) = 0

He2 molecule does not

exist.

2He molecule : Total no. of electrons = 3

Arrangement : 2

1s

1

1s

*

Bond order : ½ (2 – 1) = 1/2

So 2He exists and has been

detected in discharge tubes.

Li2 molecule : Total no. of electrons = 6

Arrangement : 2

1s

2

1s

* 2

2s

Bond order : ½ (4 – 2) = 1

No unpaired e‘s so

diamagnetic

Be2 molecule : Total no. of electrons = 8

Arrangement : 2

1s

2

1s

* 2

2s

2

2s

*

Bond order : ½ (4 – 4) = 0

No unpaired e–s so

diamagnetic

B2 molecule : Total no. of electrons = 10

Arrangement : 2

1s

2

1s

* 2

2s

2

2s

*x

2

2p

Bond order : ½ (6 – 4) = 1 diamagnetic

But observed Boron is

paramagnetic

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C2 molecule : Total no. of electrons = 12

Arrangement : 2

1s

2

1s

* 2

2s

2

2s

*x

2

2p

y

z

1

2p

1

2p

Bond order : ½ (4 – 0) = 2

It is paramagnetic

But observed C2 is

diamagnetic

N2 molecule : Total no. of electrons = 14

Arrangement : 2

1s

2

1s

* 2

2s

2

2s

* 2

2px

y

z

2

2p

2

2p

Bond order : ½ (6 – 0) = 3

It is diamagnetic

O2 molecule : Total no. of electrons = 16

Arrangement : 2

1s

2

1s

* 2

2s

2

2s

*x

2

2p

y

z

2

2p

2

2p

y

z

1

2p

1

2p

*

*

Bond order : ½ (6 – 2) = 2

It is paramagnetic

F2 molecule : Total no. of electrons = 18

Arrangement : 2

1s

2

1s

* 2

2s

2

2s

*x

2

2p

y

z

2

2p

2

2p

2

2py

2

2pz

*

* 2px

*

Bond order : ½ (6 – 4) = 1

It has been seen that in case of B2, C2 & N2 the order of filling the e‘s is

different from the normal sequence.

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B2 : 2

1s

2

1s

* 2

2s

2

2s

* y

z

1

2p

1

2p

It is paramagnetic

C2 : 2

1s

2

1s

* 2

2s

2

2s

* y

z

2

2p

2

2p

It is diamagnetic

N2 : 2

1s

2

1s

* 2

2s

*2

2s

2

2py 2

2px2

2pz

It is diamagnetic

Example 33. Compare the bond energies of O2, 2O & 2O

Solution: Higher the bond order greater will be the bond energy.

Now configuration of O2 = 2

1s

2

1s

* 2

2s

2

2s

*x

2

2p

y

z

2

2p

2

2p

y

z

1

2p

1

2p

*

*

Now formation of 2O means to remove an electron

from anti-bonding one, which means increase in B.O.

B.O. of 2O = ½ (6-1) = 2.5

2O means introduction of an e– in the anti-bonding

thereby reducing the bond order.

Bond order of 2O = ½ (6 – 3) = 1.5

So bond energy of 2O > O2 > 2O

Example 34. Give MO configuration and bond orders of H2, H2–, He2 and

He2–. Which species among the above are expected to have

same stabilities?

Solution: H2 = 1s2

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Bond order = 1

H2– = 1s

2*1s

1

Bond order = 0.5

He2 = 1s2

1s*2

Bond order = 0

He2– = 1s

2 *1s

22s

1

Bond order = (3 2)

2 = 0.5

H2– and He2

– are expected to have same stabilities.

Example 35. Which of the following species have the bond order same as2N

?

(A)CN (B) OH

(C)NO (D) CO

Solution. In 2N no. of bonding electrons and anti bonding electrons are 10

and 4 respectively. Therefore the bond order is 10 4

2 = 3. Out of

those given only CN is isoelectronic with2N . Therefore CN has

the same bond order as 2N .

Hence (A) is correct

1.1.1

1.1.2 M.O. of Some Diatomic Heteronuclei Molecules

The molecular orbitals of heteronuclei diatomic molecules should differ from

those of homonuclei species because of unequal contribution from the

participating atomic orbitals. Let‘s take the example of CO.

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The M.O. energy level diagram for CO should be similar to that of the

isoelectronic molecule N2. But C & O differ much in electronegativity and so

will their corresponding atomic orbitals. But the actual MO for this species is

very much complicated since it involves a hybridisation approach between

the orbital of oxygen and carbon.

HCl Molecule: Combination between the hydrogen 1s A.O‘s. and the

chlorine 1s, 2s, 2p & 3s orbitals can be ruled out because their energies are

too low. The combination of H 1s1 and 1

x3p gives both bonding and anti-

bonding orbitals, and the 2 electrons occupy the bonding M.O. leaving the

anti-bonding MO empty.

NO Molecule: The M.O. of NO is also quite complicated due to energy

difference of the atomic orbitals of N and O.

As the M.O.‘s of the heteronuclei species are quite complicated, so we should

concentrate in knowing the bond order and the magnetic behaviour.

Molecules/Ions Total No. of electrons Magnetic

behaviour

CO 14 Diamagnetic

NO 15 Paramagnetic

NO+ 14 Diamagnetic

NO– 16 Diamagnetic

CN 13 Paramagnetic

CN– 14 Diamagnetic

INERT PAIR EFFECT

Heavier p-block and d-block elements show two oxidation states. One is

equal to group number and second is group number minus two. For example

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Pb(5s25p2) shows two OS, +II and +IV. Here +II is more stable than +IV

which arises after loss of all four valence electrons. Reason given for more

stability of +II O.S. that 5s2 electrons are reluctant to participate in chemical

bonding because bond energy released after the bond formation is less than

that required to unpair these electrons (lead forms a weak covalent bond

because of greater bond length).

Example 36. Why does PbI4 not exist?

Solution: Pb(+IV) is less table than Pb(+II) due to inert pair effect and

therefore Pb(+IV) is reduced to Pb(+II) by I– which changes to

I2(I– is a good reducing agent)

MISCELLANEOUS EXERCISES

Exercise 1: Compare the bond energies of N2, +

2N & -

2N .

Exercise 2: Though the electronegativities of nitrogen and chlorine are

same, NH3 exists as liquid whereas HCl as gas. Why?

Exercise 3: Explain giving reason:

-

2ClF is linear, but the ion +

2ClF is bent

Exercise 4: Explain with reason:

Two different bond lengths are observed in PF5 but only one

bond length observed in SF6.

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Exercise 5: (a) Amongst BBr3 and BF3 which is a stronger acid and

why?

Exercise 6: Explain why the measured resultant dipole moment for FNO

is 1.81 D, is so much higher than the value for nitryl fluoride FNO2

(0.47 D).

Exercise7:

Why

O

H

OH

(I)

is liquid at room

temperature while

O

H

OH(II)

is a high melting

solid?

Exercise 8: (i) Among the compounds CH3COOH, NH3, HF and CH4 in which

maximum hydrogen bonding is present?

(ii) Which one of the following has strongest bond?

HF, HCl, HBr, HI

Exercise 9: Why BeF2 and BF3 are stable though Be and B have less

than 8 electrons? Which one is more stable?

Exercise10:

Ether R

O

R

and

water H

O

H

have same hybridisation at

oxygen. What angle would you

expect for them?

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ANSWER TO MISCELLANEOUS EXERCISES

Exercise 1: The configuration of N2 is

2

1s

2

1s

* 2

2s

*2

2s

y

z

2p 2

2px

2p

y

z

2p

2p

*

*

Now 2N means removal of an electron from a bonding M.O. This

will decrease the B.O.

B.O. of 2N = ½ (5 – 0) = 2.5

Now again for 2N bond order is ½ (6–1) = 2.5

So from the bond order it may seem that both 2N &

2N may have

the same bond energy. But removal of an electron from a

diatomic species tend to decrease the inter electronic repulsion

and thereby shortens the bond length. So the bond energy

becomes more than compared to 2N

N2 > 2N >

2N

Exercise 2: The size of nitrogen is less than the size of chlorine.

Therefore, electron density in nitrogen is more than that of

chlorine. So, nitrogen forms hydrogen bonding leading to

association of molecules. Hence, NH3 is liquid. Hydrogen bonding

is not possible with chlorine.

Exercise 3: In2ClF central chorine atom involves sp3d hybridisation, to

have minimum electronic repulsion three lone pairs should be in

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equitorial position as follows; giving linear shape to the ion.

Whereas, in case of 2ClF ion central atom Cl involves sp3

hybridisation having two lone pairs, resulting in bent shape for

the ion, (bond angle less than 109°28 due to repulsion of bond

pair by lone pair)

F

F

120°

Linear shape-bond angle-180°

Cl

F F

Bent shape

+

Exercise 4: PCl5 has trigonal pyramidal structure (sp3d hybridisation of

central atom) in which bond angles are 90° and 120°

respectively and there are two types of bond axial and

equatorial. In case of SF6 the structure is octahedral (sp3d2

hybridization of the central atom - S) resulting only one type of

bond, bond angle (90°) and one type of bond length.

(PF5)

(Bond angle 90° & 120°)

(SF6)

F

F F

F

F

F

Bond angle - 90°

P

F

F

F

F

F

Exercise 5: (a) BBr3 since back bonding is present in BF3.

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Exercise 6: Molecular symmetry (in terms of bond angles) leads to

lesser dipole moment.

Exercise 7: (i) shows intramolecular hydrogen bonding while

(ii) shows intermolecular hydrogen bonding.

Exercise 8: (i) HF due to maximum electronegativity of fluorine.

(ii) HF

Exercise 9: The stability is explained by symmetrical linear structure

for BeF2 and triangular planar structure for BF3. BeF2 is more

stable because its greater bond angle (180o).

Exercise 10: In H2O bond angle is less than 109o.28‘ due to lone pair

and bond pair repulsion. But in ether, due to strong mutual

repulsion between two alkyl groups bond angle becomes greater

than 109o.28‘.

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SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 2. Arrange the bonds in order of increasing ionic character in the

molecules:

LiF, K2O, N2, SO2 and ClF3

Sol. N2 < ClF3 < SO2 < K2O < LiF

Prob 3. Arrange the following in order of increasing ionic character:

C H, F H, Br H, Na I, K F and Li Cl

Sol. C H < Br H < F H < Na I < Li Cl < K F

IIT Level Questions

Prob 6. Predict the shapes of the following ions

(a) BeF3- (b) BF4

-

(c) IF4- (d) IBr2

-

Sol. (a) Triangular

(b) Tetrahedral

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(c) Square planar

(d) Linear

Prob 8. Arrange the following in increasing order of stability O2, O2+, O2

-,

O22-

Sol. O22- < O2

- < O2 < O2+

Calculate first the bond order which is as follows

O2 2, O2+ 2.5, O2

- 1.5, O22- 1 & then arrange according to

increasing bond order.

Prob 11. Arrange the following:

(i) N2, O2, F2, Cl2 in increasing order of bond dissociation energy.

(ii) Increasing strength of hydrogen bonding (X – H – X):

O, S, F, Cl, N

Sol. (i) F2 < Cl2 < O2 < N2 (ii) Cl < S < N < O < F

Prob 12. Explain the following

o - hydroxy benzaldehyde is liquid at room temperature while p -

hydroxy benzaldehyde is high melting solid.

Sol. There is intramolecular H bonding in o - hydroxy benzaldehyde while

intermolecular hydrogen bonding in p-hydroxy benzaldehyde

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Prob 14. Explain why ClF2- is linear but ClF2

+ is a bent molecular ion?

Sol. Chlorine atom lies in sp3d hybrid state. Three lone pairs are oriented

along the corners of triangular plane

F Cl F

2ClF

Chlorine atom lies in sp3 hybrid state. Two lone pairs are oriented

along two corners of tetrahedral

Prob 16. AlF3 is ionic while AlCl3 is covalent.

Sol. Since F– is smaller in size, its polarisability is less and therefore it is

having more ionic character. Whereas Cl being large in size is having

more polarisability and hence more covalent character.

CH=O---H

o-hydroxy benzaldehyde

CH=O---H—O

O—H----OCH

CHO

OH

O

Cl

FF[ClF

2

+]

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Prob 18. Write down the resonance structure of nitrous oxide 2N O

Sol.

Prob 20. Explain why BeH2 molecule has zero dipole moment although the

Be H bonds are polar.

Sol. BeH2 is a linear molecule (H Be H) with bond angle equal to 180o.

Although the Be H bonds are polar on account of electronegativity

difference between Be and H atoms, the bond polarities cancel with

each other. The molecule has resultant dipole moment of zero.

1. What are the factors influencing ionic bond formation?

2. Out of MgO and NaCl, whech has higher lattice energy and why?

3. BaSO4 being an electrovalent compound and still it does not pass into

solution state in water.

4. Why an ionic bond is formed between two elements having large

difference in their electronegativity?

N N O N N O

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COVALENT BOND :

6. Which compound has 3 carbon atoms, 4 double bonds and the total

no. of atoms are five?

7. Which compound has 3 carbon atoms. 4 and 4 bonds but not 4

double bonds and the total no. of atoms are 5. Writes its IUPAC

8. Element A has 3 electrons in the valency shell and its principal

quantum no. for the last electron is 3 and element B has 4 electrons in

the valency shell and its principal quantum to. For the last electrons is

2. Identify the compound and it‘s nature of bonding.

9. The compound X has 8 atoms, 4 bonds, no bonds, no. ionic

bonds, no coordinate bonds, no H-bond. Explain its structure.

10. The compound X has 8 atoms, 0 bonds, no ionic or bonds.

Explain its structure.

11. SnCl4 has melting point –15ºC where as SnCl2 has melting point

535ºC. Why?

12. Inorganic benzene is more reactive than organic benzene. Why?

FAJAN’S RULE :

13. SnCl2 is white but SnI2 is red. Why?

14. Explain the least melting point and highest solubility in H2O.

(i) LiCl NaCl KCl

(ii) NaCl NaBr Nal

(iii) LiCl BeCl2 BCl3

(iv) beSO4 MgSO4 CaSO4 SrSO4 BaSO4

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(v) CaF2 CaCl2 CaBr2

15. Which one has highest and lowest melting point and why?

NaCl KCl RbCl CsCl

16. LiOH and carbonates decomposes on heating in I-group. Other

hydroxides and carbonates of this group will not. Why?

LEWIS STRUCTURE AND FORMAL CHARGE :

17. Draw the Lewis structures of the following molecules and ions.

PH3, H2S, BeF2, SiCl4, N2O4, H2SO4, O22, IO6

5-.

V.B.T. & HYBRIDISATION :

20. Explain hydrbidisation in

(1) XeF2 (2) XeF6 (3) PCl3 (4) IF3

(5) IF5 (6) IF7 (7) CCl4 (8) SiCl4

(9) SlH4 (10) H2O

22. PH5 is not possible but PCI5 is possible. Why?

VSEPR THEORY :

24. Which one has highest and least bond angle in the following -

(1) NH3 PH3 AsH3 SbH3

(2) CH4 PH3 AsH3 SbH3

(3) H2O H2S H2Se H2Te

(4) CH4 PH3 AsH3 H2Te

(5) CH4 SiH4 CCl4 SiCl4

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(6) NCl3 PCl3 NBr3 PBr3

(7) PF3 PH3

(8) As3F3 AsH3

MOT :

29. Super oxide are coloured and paramagnetic why?

30. Write the important conditions required for the linear combination of

atomic orbitals to form molecular orbitals.

33. Exalain why NO+ is more stable towards dissociation into its atoms

than NO, where as CO+ is less stable than CO.

BACK BONDING AND HYDROLYSIS :

40. Arrange the following boron trihalides in the increasing order of their

ease of hydrolysis. Also give the reason for the same.

BF3, BCl3, BBr3

DIPOLE MOMENT :

53. Why NH3 is having more dipole moment than NF3.

54. Why is having more dipole moment than /

55. Why p-dichloro benzene in having zero dipole moment while

hydroquinone is having some dipole moment?

56. Why CH3 Cl is having high dipole moment than CH3F?

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57. Why ortho fluoro Phenol have greater dipole moment than ortho chloro

phenol?

58. Why trans–1, 2 sichloro ethene in having zero dipole moment than cis

form?

59. Write the order of dipole moments of 1, 2–1.3– and 1, 4–

dichlorobenzene.

60. Arrange in increasing order of dipole moment ; H2O, H2S, BF3.

61. BcF2 has zero dipole moment whereas H2O has a dipole moment?

62. CCl4 having zero dipole momnet but CHCl3 having some dipole

moment?

63. While down the resonance structure (S) for :

(i) N3– (ii) O3 (iii) CO2

(iv) N2O4 (v) SCO22–

1. Arrange the following in increasing order of property given-

(i) O, F, S, Cl, N strength of H-bonding (X-H_X)

(ii) N2, O2, F2, Cl2 bond dissociation energy

(iii) MCl, MCl2, MCl3 ionic nature

(iv) HI, HBr, HCl, HF dipole moment

(v) AsH3, PH3, NH3 bond angle

5. Explain the structure hybridisation and oxidation state of S i sulphuric

acid, marshall‘s acid, caro‘s acid, oleum.

6. Boric acid is monobasic acid. Why?

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7. Boron has exceptionally high melting point. Why?

8. BCl3 is more acidic than BF3. Why?

9. CCl4 is not dissolved in H2O but SiCl4 dissolves. Why?

10. Trimethylamine (CH3)3 N, is pyramidal but trisylyamine (SiH3)3 N is

planer. Why?

11. SnCl4 has melting point –15ºC Where as SnCl2 has melting point

535ºC.Why?

12. PbCl4 is possible but PbBr4 and Pbl4 are not. Why?

13. Pb+4, Bi+5 and Tl+3 act as oxidising agent. Why?

14. NCl3 & PCl3 on hydrolysis will give different products. Why?

15. ClO2 does not forms dimer but NO2 forms. Why?

16. How many and bonds are presents in hexacyanoethane and tetra

cyanoethylene?

17. Explain the structure of ClF3 on the basis of bent rule.

18. All bonds length of PCl5 are not equal but PF5 has same bond lengths.

Why?

19. The experimentally determined N – F bond length in NF3 is greater

than sum of single bond covalent radii of N and F.

24. The dipole moment of HBr is 7.95 debye and the intermolecular

separation is 1.94 × 10-10 m Find the % ionic character in HBr

molecule.

25. HBr has dipole moment 2.6 × 10-30 cm. If the ioninc character of the

bond is 11.5%, calculate the interatomic spacing.

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26. Dipole moment of LiF was experimentally determined and was found to

be 6.32 D. Calculate percentage ionic character in LiF molecule Li – F

bond length is 0.156 pm.

27. A diatomic molecule has a dipole moment of 1.2 D. If bond length is

1.0 A, what percentage of an electronic charge exists on each atom.

SECTION (B) FILL IN THE BLANKS :

1. Two atoms of similar electronegativity are expected to form

__________ compound.

2. When two atoms approach each other, potential energy ______ and a

___________ is formed between them.

3. Conversion of a neutral atom into a cation is _________ process.

4. The strongest hydrogen bond is formed between _________ and

hydrogen.

5. Low ionization potential of electropositive element and high electron

affinity of electronegative element favours the formation of

__________ bond.

6. NaCl is soluble in water due to its low ___________ energy.

7. The ________ value of lattice energy of a crystal favours the

formation of an ionic compound.

8. Solid NaCl ___________ conductor of electricity.

9. For dissolution of an ionic solid, ________ energy should be low and

hydration energy should be _________________.

10. ___________ cation and ___________ anion favour covalency.

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11. Anhydrous AlCl3 is a _____________ compound while hydrated AlCl3

is ___________.

12. Covalent compounds are generally _________ conductors of

electricity.

13. There are ___________ bonds in a nitroge molecule.

14. A double bond is shorter than ___________ bond.

15. Axial overlapping of half-filled atomic orbitals results in __________

bond.

16. ____________ and ___________ bonds are present in N2O5

molecule.

17. The angle between two covalent bonds is maximum in _________

(CH4, H2O, CO2).

18. ___________ hybrid orbital of nitroger atom are involved in the

formation of ammonium ion.

19. The hybridization state of oxygen in water molecule is ___________.

20. In the ion [Cu (H2O)4]++, copper is in dsp2 state of hybridization.

The shape of the ion is ________.

TRUE OR FALSE :

5. A diatomic molecule has a dipole moment of 1.2 D. If bond length is

1.0 a, what percentage of an electronic charge exists on each atom.

6. The size of negative ion decreases with increasing magnitude of

negative charge.

7. Ca2+ is smaller in size than K+ because the effective nuclear charge is

greater.

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8. The higher the latteice energy of an ionic solid, the greater will be its

stability.

9. Molten sodium chloride conducts electricity due to the presence of free

ions.

10. Linear overlap of two atomic p-rbitals leads to a -bond.

11. The H-N bond angle in NH3 is greater than the H-AS-H bond angle in

AsH3.

12. sp2 hybrid orbitals have equal s and p-character.

13. The tetrahedral geometry in SiF4 is due to sp3 hybridization of Si

atom.

14. SnCl2 is a non-linear molecule.

15. There are seven electron bond pairs in lF7 molecules.

16. Dipole moment of CHF3 is greater than CHCl3.

17. Dipole moment of NF3 is lesser than NH3.

18. Among HF, HCl, HBr and HI, HF has highest dipole moment.

19. All molecules with polar bonds have dipole moment.

20. The presence of polar bonds in a polyatomic molecule suggests that

the molecule has non zero dipole moment.

21. AgCl is more covalent than NaC.

REASONING AND ASSERTION :

Direction: These quaestions consist of two statements each printed as

Assertion and Reason.While answering these questions you are

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required to choose any one of the following four responses to encircle

(A, B, C, D) as per instructions given below:

(A) If both Assertion and Reason are true and Reason is a correct

explanation of Assertion.

(B) If both Assertion and Reason are true and Reason is not a correct

explanation of Assertion.

(C) If Assertion is true but Reason is false.

(D) If Assertion is false but Reason is true.

3. Assertion : Na2SO4 is soluble in water while BaSO4 is insoluble.

Reason : Lattice energy of BaSO4 exceeds its hydration energy.

(a) A (b) b (c) C (d) D

4. Assertion : py and pz cannot combine to give molecular orbital.

Reason : Both py and pz are dumb-bell shaped.

(a) A (b) b (c) C (d) D

7. Assertion : Although PF5, PCl5 and PBr5 are known, the pentahalides

of nitrogen have not been observed.

Reason : Phosphorus has lower electronegativity than nitrogen.

(a) A (b) b (c) C (d) D

9. Assertion : NO3– is planar while NH3 is pyramidal.

Reason : N in NO3– is sp2 and in NH3 it is sp3 hybridized.

(a) A (b) b (c) C (d) D

11. Assertion : s-orbital cannot accommodate more than two electrons.

Reason : s-orbitals are extremely poor shielders.

(a) A (b) b (c) C (d) D

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MATCH THE FOLLOWING :

1. Column - I Column - II

(A) Size of secondary layer of

hydrated ions (P)Maximum in

solid and minimum in gaseous

state.

(B) Magnitude of hydrogen

bonding (Q) Strength of ion -

dipole attraction

(C) Mobility of ions in water

(R) Inversely proportional to

the size of metal ion

(D) Degree of polarity of a

bond (S) Dipole moment

(T) Directly

proprotional to the size of

metal ion


(A) SO3 (gas) (P)Polar

with p - d bonds and

identical S – O bond, lengths.

(B) OSF4 (Q) One

lone pair and p – d bond.

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(C) SO3F– (R) non -

polar with p –p and p –

d bonds. Identical S–O bod

lengths.

(D) ClOF3 (S) Polar

with p – d bond.

(T) Hybridisation of

central atom in ClO2F3.


Molecule/ion

Hybridisation of central atom

(A) IO2F2– (P) sp3d

(B) F2SeO (Q) sp3

(C) ClOF3 (R) sp2

(D) XeF5+ (S) sp2

ONLY ONE ANSWER CORRECT:

1. In which molecule (s) is/are the vander waals force likely to be most

important in determining m.p. and b.p.

(a) ICI (b) Br2 (c) H2S (d) CO

2. Which is the most ionic

(a) LiF (b) Li2O (c) Li3N (d) All same

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3. The correct order of the increasing ionic character is-

(a) BeCl2 < MgCl2 < CaCl2 < BaCl2 (b) BeCl2 < MgCl2 < BaCl2 <

CaCl2

(c) BeCl2 < BaCL2 < MgCl2 < CaCl2 (d) BaCl2 < MgCl2 < CaCl2 <

BeCl2

4. An ionic bond A+ B– is most likely to be formed when :

(a) the ionization energy of A is high and the electron affinity of B is

low

(b) the ionization energy of A is high and the electron affinity of B is

low

(c) the ionization energy of A and the electron affinity of B is high

(d) the ionization energy of A and the electron affinity of B is low

5. Which of the following compounds of elements in group IV is expected

to the most ionic?

(a) PbCl2 (b) PbCl4 (c) CCl4 (d) SiCl4

6. The molecule BF3 borth are covalent compounds. But BF, is non,polar

and NF3 is polar. The reason is that-

(a) boron is a metal and nitrogen is a gas in uncombined state

(b) B – F bond have no dipole moment whereas N – F bond have

dipole moment

(c) atomic size of boron is smaller than that of nitrogen

(d) BF3 is symmetrical molecule where as NF3 is unsymmetrical

7. Least melting point is shown by the compound-

(a) PbCl2 (b) SnCl4 (c) NaCl (d) AlCl3

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8. Which of the following is in rder of increasing covalent character?

(a) CCl4 < BeCl2 < BCl3 < LiCl (b) LiCl < CCl4 < BeCl2 < BCl3

(c) LiCl < BeCl2 < BCl3 < CCl4 (d) LiCl < BeCl2 < CCl4 < BCl3

9. Which of the following compounds contain/s both ionic and covalent

bonds?

(a) NH4Cl (b) KCN (c) CuSO4.5H2O (d) NaOH

10. The tyes of bonds present in CuSO4.5H2O are

(a) electrovalent and covalent (b) electrovalent and coordinate

(c) covalent and coordinate (d) electrovalent, covalent and

coordinate

11. Three centre-two electron bonds exist in :

(a) B2H6 (b) Al2(CH3)6 (c) BeH2(s) (d) BeCl2(s)

Octet Rule :

12. Example of super octet molecule is :

(a) SF6 (b) PCl5 (c) IF7 (d) All of these

13. To which of the following species octet rule is not applicable :

(a) BrF5 (b) SF6 (c) IF7 (d) CO

14. NH3 and BF3 combine readily because of the formation of :

(a) a covalent bond (b) hydrogen bond (c) acoordinate bond

(d) an ionic bond

15. Which of the following species cotain covalent and ccordincate bond.

(a) AlCl3 (b) CO (c) [Fe(CN)6]4– (d) N–3

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1.1.2.1.1.1.1.1 Fajans Rule

16. Which of the following combination of ion will have highest polarisation

(a) Fe2+, Br– (b) Ni4+, Br– (c) Ni2+, Br– (d) Fe, Br–

17. Which of the following cannot be explained on the basis of Fajan‘s

Rules.

(a) Ag2S is much less soluble than Ag2O

(b) Fe(OH)3 is much less soluble than Fe(OH)2

(c) BaCO3 is much less soluble than MgCO3

(d) Melting point of AlCl3 is much less than that of NaCl

18. The correct order of decreasing polarizability of ion is :

(a) Cr–, Br–, I–, F– (b) F–, I–, Br–, Cl– (c) I–, Br–, Cl–, F–

(d) F–, Cl–, Br–, I–

19. Which ion has a higher polarising power

(a) Mg2+ (b) Al3+ (c) Ca2+ (d) Na+

20. Which combination will show maxinum polarising power & maxinum

polarisability

(a) Mn2+.F– (b) Mn7+, I– (c) Mn2+, I– (d) Mn7+, F–

LEWIS STRUCTURE :

21. The possible structure (s) of monothiocarbonate ion is :

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

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22. Pick out among the following species isoelectronic with CO2.

(a) N–3 (b) (CNO)– (c) (NCN)2– (d) NO–

2

V.B.T. & HYBRIDISATION :

23. The strength of bonds by s – s, p – p, s – p overlap is in the order :

(a) s – s < s – o < p – p (b) s – s < p – p < s – p

(c) s – p < s – s < p – p (d) p – p < s – s < s – p.

24. Number and ype of bonds between two carbon stoms in CaC2 are :

(a) one sigm3 () and one pi () bond (b) one and two bonds

(c) one and one and a half bond (d) one bond

25. In the context of carbon, which of the follwoing is arranged in the

correct order of electronegativity:

(a) sp > sp2 > sp3 (b) sp3 > sp2 > sp (c) sp2 > sp > sp3

(d) sp3 > sp > sp2

26. Which of the following oxyacids of sulphur contain ‗S – S bonds?

(a) H2S2O8 (b) H2S2O6 (c) H2S2O4 (d) H2S2O5

27. Which of the following represent the given mode of hybridisation sp2 –

sp2 – sp – sp from left to right

(a) H2C = C = c = CH2 (b) HC C – C CH

(c) H2C = CH – C N (d) H2C = CH – C CH

28. For BF3 molecule which of the following is true.

(a) B-atom is sp2 hybridised

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(b) There is a P –P back bonding in this molecule

(c) Observed B–F bond length is found to be less than the expected

bond length.

(d) All of these

29. Which of the following is isoelectronic as well as isostructura‘ with N2O

(a) N3H (b) H2O (c) NO2 (d) CO2

30. The hybridization state of B in B2H6 is –

(a) sp (b) sp2 (c) sp3 (d) sp3d

31. What is not true for SiH4 molecule –

(a) Tetrahedral hybridisation (b) 109º angle

(c) 4 bond (d) 4-lone pair of electrons

32. Which of the following has a geometry different from the other three

species (having the same geometry)?

(a) (b) (c) XeF4 (d)

33. In C – C bond C2H6 undergoes heterolytic fission, the hybridisation of

two resulting carbon atoms is/are

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

34. Which of the following statements are not correct?

(a) Hybridization is the mixing of atomic orbitals of large energy

difference.

(b) sp2 – hybrid orbitals are formed form two p – atomic orbitals and

one s-atomic orbitals

(c) dsp2 – hybrid orbitals are all at 90º to one another

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(d) d2sp3 – hybrid orbitals are directed towards the corners of a

regular octahedron

35. p – d back bonding occurs between oxygen and

(a) phosphorus in P4O10 (b) chlorine in HClO4

(c) nitrogen in N2O5 (d) carbon in CO2

36. In which of the following groups all the the members have linear shape

(a) NO2+, N3

–, H – C – H (b) N3–, I3

–, H – C – H

(c) XeF2, C2H2, SO2 (d) CO2, BeCl2, SnCl2

37. Consider the following molecules :

H2O H2S H2Se H2Te

I II III IV

Arrange these molecules in increasing order of bond angles.

(a) I < II < III < IV (b) IV < III < II < I (c) I < II < IV < III

(d) II < IV < III < I

38. Which has the smallest bond angle (X – O – X) in the given molecules?

(a) OSF2 (b) OSCl2 (c) OSBr2 (d) OSI2

39. Consider the following iodides :

PI3 AsI3 SbI3

102º 100.2º 99º

The bond angle is maximum in Pl3, which is

(a) due to small size of phosphorus (b) due to more bp–bp repulsion in

PI3

(c) due to less electronegativity of P (d) none of these

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BONDS ANGLES & BOND LENGTH :

40. The correct order of increasing X – O – X bond angle is (X = H, F or

Cl):

(a) H2O > Cl2 > F2O (b) Cl2O > H2O > F2O

(c) F2O > Cl2O > H2O (d) F2O > H2O > Cl2O

41. Which of the following compounds have bond angle as nearly 90º?

(a) NH3 (b) H2S (c) H2O (d) SF6

42. Number of non bonding electrons in N2 is :

(a) 4 (b) 10 (c) 12 (d) 14

43. Which of the following species is paramagnetic ?

(a) NO– (b) (c) CN– (d) CO

44. Tho bond order depends on the number of electrons in the bonding

and non bonding orbitals. Which of the following statements is/are

correct about bond order?

(a) Bond order cannot have a negative value

(b) It always has an integral value

(c) It is a nonzero quantity

(d) It can assume any value-positive or negative, integral or fractional,

including zero

45. In the formation of from N2, the electron is removed from :

(a) orbital (b) orbital (c) * orbital (d) * orbital

46. Which of the following has fractional bond order :

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

47. How many unpaired electrons are present in :

(a) 1 (b) 2 (c) 3 (d) 4

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48. Which of the following have identical bond order?

(a) (b) (c) NO (d)

49. Which of the following have identical bond order?

(a) (b) NO+ (c) CN– (d) CN+

50. Given the species : N2, CO, CN– and NO+, Which of the following

statements are true for these

(a) All species are paramagnetic (b) The species are isoelectronic

(c) All the species have dipole moment (d) All the species are linear

51. Which of the following are paramangetic?

(a) B2 (b) O2 (c) N2 (d) He2

52. Which of the following species have a bond order of 3?

(a) CO (b) CN– (c) NO+ (d)

53. Which of the following pairs have identical values of bond order?

(a) and (b) F2 and Ne2 (c) O2 and B2 (d) C2 and N2

54. Find out the bond order of :

(a) H2 (b) (c) He2 (d) Li2 (f) B2

55. The correct order of boiling point is :

(a) H2O < H2S < H2Se < H2Te (b) H2O > H2Se > H2Te > H2S

(c) H2O > H2S > H2Se > H2Te (d) H2O > H2Te > H2Se > H2S

56. Which of the following models best describes the bonding within a

layer of the graphite structure?

(a) metallic bonding (b) ionic bonding

(c) non-metallic covalent bonding (d) van der Waals forces

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57. The critical temperature of water is higher than that of O2 because the

H2O molecule has :

(a) fewer electrons than O2 (b) two covalet bonds

(c) V - shape (d) dipole moment

58. Arrange the following in order of decreasing boiling point :

(I) n-Butane (II) n-Butanol (III) n-Butyl chloride (IV)

Isobutane

(a) IV > III > II > I (b) IV > II > III > I (c) I > II > III > IV

(d) II > III > I > IV

59. For H2O2, H2S, H2O and HF, the correct order of increasing extent of

hydrogen bonding is :

(a) H2O > HF > H2O2 > H2S (b) H2O > HF > H2S > H2O2

(c) HF > H2O > H2O2 > H2S (d) H2O2 > H2O > HF > H2S

60. Which one of the following does not have intermolecular H-bonding?

(a) H2O (b) o-nitro phenol (c) HF (d) CH3COOH

61.

(a) has intermolecular H - bonding (b) has intramolecular H-bonding

(c) has low boilng point (d) is steam-volatile

62. Which of the following is/are observed in metallic bonds?

(a) Mobile valence electrons (b) Overlapping valence orbitals

(c) Highly directed bond (d) Delocalized electrons

63. Intermolecular hydrogen bonding increases the enthalpy of

vapourization of a liquid due to the

(a) decrease in the attraction betwwen molecules

(b) increase in the attraction between molecules

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(c) decrease in the molar mass of unassociated liquid molecules

(d) increase in the effective molar mass of hydrogen - bonded

molecules

64. Of the following molecules, the one, which has permanent diple

moment, is-

(a) SiF4 (b) BF3 (c) PF3 (d) PF5

65. The dipole moments of the given molecules are such that-

(a) BF3 > NF3 > NH3 (b) NF3 > BF3 > NH3 (c) NH3 > NF3 > BF3

(d) NH3 > BF3 > NF3

66. Which of the following has the least dipole moment

(a) NF3 (b) CO2 (c) SO2 (d) NH3

67. Which of the following compounds possesses zero dipole moment?

(a) Water (b) Benzene

(c) Carbon tetrachloride (d) Boron trifluoride

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