------·--

----------

8 Periodic Properties and Chemical Bonding

Periodic Properties

Ionisation potential ofH-atom Ioni8ation energy ofH-atom

I. Ionisation Energy (IE) or Ionisation Potentlal (IP)

=

1. It is the minimum amount of energy required to remove an

•1 . . :. omsation energy

electron from the valence shell of an isolated gaseous atom or cation.

(An

M(g)----..+ M+(g)+e; IE1 = E(M+) - E(M) M+(g) ----..+ M2+(g)+e;

M2+(g)

----+

M3+(g)+e;

Mf;._IE1 Ml=IE2 Ml=IE3

effectively pulled towards nucleus and requires more energy. That is why for an element.

IE3 > IE2

> IE1

3. Ionisation energy is expressed in eV/atom or kcal/g-atom or kJ/g-atom or mol.

4. Ionisation energy values are endoergic in nature. 5. Ionisation energy values are derived from Born-Haber cycle or these are measured either by spectroscopic techniques or by passing an electric current of �ually increasing intensity through the vapours of the elements in a discharge

13.6 xL602x10-19 J

L602x10-19 (":

=

. eV 136

1.602x10-19 J = 1 eV)

2. The IE,. is highest for elements close to Helium and is lowest for elements close to cesium.

of elements.

(i) EtYective nuclear charge:

1. Ionisation energy eV/atom is equal to ionisation potential in volt i.e., potential difference which supplies energy to ionise

More is the effective

nuclear charge, more tightly are held electrons with nucleus and thus more is the energy required to remove electron which results in higher ionisation energy. Effective nuclear charge (Zeff) is given by: Zeff

=

Z - a, where Z is nuclear charge and

a is screening constant. Screening constant (a) may be calculated by

Slater's rule.

The phenomenon of repulsion of valency el�trons by the electrons in penultimate shell to reduce effective nuclear charge is known as shielding effect or screening effect. 1

-�e

tube.

an atom, e.g.,

13.6 V

13.6 VxL602x10-19 C

6. Following are the factors which influence ionisation energy

2. The removal of 2nd electron from a cation is relatively more difficult because remaining electrons in cation are more

Note :

=

=

13.6xL602xl0-19

isolated atom refers for the atom of an element

without any bonding with other atoms.)

=

Nucleus

Fig.

1

•·-------

�� �.'

Oe AttracliOn -1

Shielding of electrons of 2nd shell by

1st

shell

_/

" 380

Concepts of Physical Chemistry cr

The evaluation of shielding constant (cr) can be made by using following rules: (a) The electronic configuration of element under investigation is written according to aufbau principle in the

=

[2x0.85] + [ 4x0.35]=3.1

.. Effective nuclear charge=Z -cr

=

7 - 3 . 1=3 9 .

(ls2)(2s22 p 4)

0-atom :

following grouping (ls), (2s.2.p), (3 s 3 p)(3d ), (4 s 4 p)(4d ) (4j), (Ss Sp), (Sd), (Sf ) ...

(b) Electrc.ns present in any group [i.e., in ( )]contribute nothing to shielding constant.

cr =[2 x0.85) + [5x0.35]=3.45

(c) All the electrons excluding test electron in the (ns, np)

group shield the valence electrons to the extent of

z• = z - cr=8 -3.45=4.55

0.35 each

except for ls-orbital which acts as ns· in He for which a

0:30 has been preferred. (d) All the electrons in the (n -1) shell shield the valence electron to the extent of 0.85 each. (e) All the electrons in (n -2) or lower shells shield completely the valence electrons to the extent of 1.00 each.

contribution

Shielding constant for Ne is 4.15. Calculate the

EXAMPLE 2

effective nuclear charge on Na+ and F - using only this value.

10Ne: (ls2)(2s22 p6)

Solution

11 Na+ :(ls2)(2s22 p6) 9F- :(ls2)(2s2 2 p6)

(f ) When electrons shielded is present in (nd) or (nf)

(b) and (c) are same but rule (d) and (e) are ' changed. Here ali the electrons lying to the left of (nd) or (nf) group contributes 1.00 each.

group, rule

[0.35 xNumber of electrons in nth shell excluding test or valence electron] + [0.85 x Number of electrons in (n -1) th shell] + [1.0 x Number of electrons in Thus cr :;;

inner shells] e.g.,

Zn :ls2, (2s22 p6) , (Inner shell) l 0 electrons

(3 s23 p6)(3d10), 4s2

(n-1) 18 electrons

The effective nuclear charge on 4s test electron is given as: cr = [1.00x10] + [0.85x18] + [0. 35x1] cr=25.65

z• EXAMPLE 1

=

z - cr

=

Z�e :;; ZNe -cr=10 - 4.15= 5.85

z�.

+

=ZN

+

a

-cr = 1·1 - 4.15=6.85

zF• = Z -
The ionization energy of Li is

EXAMPLE 3

Calculate the effective nuclear charge at the

the formation of anion. Also calculate the effective nuclear charge of N-atom and 0-atom.

ionization energy of His 13.6 eV, then calculate the effective

For Li, electronic configuration is Is2, 2s1, so given

Solution

ionization energy value is for (n=2). We know that:

z

En=-X ;tr EI n2

En =Energy of charge,

Addition of extra electron ----Test electron

nth

level,

Zerr=n·

Given

E

1

= -13.6 eV;

Z=2x �

�

En=-5.39eV and n=2 5 39 · =1.26 1 3 .6

Thus, effective nuclear charge is

cr =[2x0.85] + [5x0.35]= 3.45

Effective nuclear charge

z• =Z - cr = 7 - 3.4,!i = 3.55

is shielded by ls2-electrons.

Also from Slater's rule cr Li

-Test electron

1.26 because 2s-electron

=2x0.85=1.70

z�i =3 -1.10= 1.Jo

..

In N-atom

Zeff=Effective nuclear

E1 =Energy of first orbit of H-atom.

or

In N--ion

5.39 eV. If

charge acting upon outermost elect1¥.Jn of Lt.

30 - 25.65= 4.35

periphery of nitrogen atom when an extra electron is added in

Solution

Thus, shielding constant cr is same for all these.

Ionisation energy generally increases along the period with some disorders shown in blocks. Il period

3Li

4Be

IE1 in eV

5.4

9.3

lsBl 6C � 11.26

7N

lsOl 9F 14.54 � 17.42

10Ne 21.56

Periodic Properties and Chemical Bonding

381 effective nuclear charge decreases which results in decrease in

Table 1 : Successive Values of Ionization Energies,

IE for some Elements (kJ/mol) 0

Element

IEi

IEz

IE3

IE4

IE5

IE6

ionisation energy.

IE1

Na

496

4560

Mg

738

1450

7730

Al

577

1816

2744. 11,600

Si

786

1577

3228

4354

16,100

p

1060

1890

2905

4950

6270

21,200

s

999

2260

3375

4565

6950

8490

27,000

Cl

1256

2295

3850

5160

6560

9360

11,000

Ar

1520

2665

3945

5770

7230

8780

12,000

I group

IEi in eV

5.1

EXA,..PLE 6

(Inner-shell electrons)

37Rb

11Na

ssCs 3.9

4.2

Why the first ionisation energy for Cu is higher

than that for potassium whereas the second ionisation energy values are in reverse order?

In Cu, effective nuclear charge is greater than K and

Ans.

thus, IE1 ofCu> IE1 ofK. But in Cu+, the electron is lost from

Although the ionization energies are given here in units of

3d10-orbital whereas in K+, the electron is lost from 3p6 . orbital. Thus, removal of elec tron from 3d10 is easier or IE of 2 Cu IE1 of 0 but IE of 0> IE2 of N; 2

382

Concepts of Phys;cal Chemistry 1N :

Ans.

ls2 ' 2s22 3 P

4 2 2 80: l s ,2 s 2p

Removal of 'e' from 2phalf filled

Removal of 'e' from --. ------. 2pnormal

a lower energy when it is attached to a

N+

difference E(X)-E(X-) is positive Thus,

+ o

Note:

1. A high electron affinity (EA) means a lot of energy is released when an electron attaches to a gas phase atom. 2. A negative electron affinity means that energy must be supplied to push an electron onto an atom. This convention ·

Removal of'e' from 2pnonnal

matches the everyday meaning of the term 'affinity'. 3. We have defined electron affinity in such a way that a positive EA is associated with an exothermic process. Thus, the more positive is the value for EA, the greater the attraction for electrons. 4. However in thermodynamics, energy given off during addition of electron should be taken as - ve and thus, the term electron affinity is replaced by electron gain enthalpy (Ml = Ega = - ve)

Removal of 'e' 3 ---o --- 2+: ls 2,2s22p 2 from 2phalffilled

IE of0 >IE of N 2 2 EXAMPLE 10 IE1 for inert gases is maximum. Explain. :.

Ans. Inert gases possess completely filled orbitals, possess octet of electron which is stable state.

i.e.,

Why. IE for transition metals increases and then becomes constant along the period? Due to much more shielding by innerd-orbitals when

5 electrons are paired up ind-orbitals after d configuration.

(v) Following are some important observations derived

from the general trend of ionisation energy.

M(g)

�

Lower is ionisation energy ofan element easier is removal

metallic or electropositive character.

In reference that define Ega in this second way, the more negative the E a value, the greater the attraction for electron. 8

2. The addition of2nd electron in an anion is more difficult

and energy is needed to overpower the forces of repulsions between negatively charged atomic sphere and the test electron. This energy being greater than the energy released

during gaining up of electron. That is why higher

M+(g)+e

of electron and thus: (a) More is the tendency to lose electron,

Ml= Ega

X(g) + e ----+ x-(g);

EXAMPLE 11

Ans.

X(g)+e � x-(g); Energy released = EA

IE1 ofN > IE1 ofO

o+: l s2, 2s22 p

X(g) atom, the

i.e., more is

(b) More is the tendency to form cation and thus, more is

the tendeacy to show ionic bonding. Therefore, tendency of metals to show ionic bonding decreases along the period but

(i.e.,EA2, EA3

.

•

) are endoergic in nature. x (g) + e � x2-(g); -

or

3. EA1

EA values

•

2

EA2 =E(X-)-E(X -)=-ve

!:Jl=E 0 =+ve 1 values are exoergic and rest all are endoergic in nature.

4. EA values are expressed in eV/atom or kcal/g-atom or

more is reducing nature. Therefore, reducing power decreases along the period but increases down the group. However, Li in

kJ/g-atom or mol. 5. EA values are also derived by Born-Haber cycle or by spectroscopic studies. 6. Following are the factors which influence the EA values. (i) Effective nuclear charge: More is the effective

small size and thus, has more tendency for hydration leading to

electron

increases down the group. (c) More is the tendency to get it self oxidized and thus,

I group is the most powreful reductant, which is due to its

higher hydration energy. EXAMPLE 12

Li is more powerful reductant than Na.

Explain. Ans.

Due to high beat ofhydration ofLi than·Na on account

ofits smaller size, !:JI for oxidation ofLi is more -ve than Na Li( s)+aq � Li + ( aq)+e Na(g)+aq � Na + (aq)+e

II� Electron Affinity and Electron Gain Enthal.JW

nuclear charge, more is the attraction of nucleus towards test

(i.e., electron to be added) and thus, more will be

electron affinity.

That is why on moving along the period

increase with some disorders shown in blocks: II period

Li

EA1in eV

0.54

1. Electron affinity is the amount of energy released during

X(g)+e � x-(g);

EA =E(X)-E(X-)

where E(X)and E(X-)are the energy ofa gas phase atom X

and gas phase anion X - respectively. Because the electron bas

F 3.62

INel LJU

(ii) Size of the atom : On moving down the group, effective nuclear charge decreases and therefore, EA values decrease with an exception of Cl in VII group as well as for S in VI group and for P in V group. F

the addition of an electron in the valence shell of an isolated gaseous atom.

ts;-i lil C INl 0 LJU� 1.13 � J.48

EA values

EA1in eV Note :

3.65

ICll �

Br

I

3.56

3.28

1. Electron affinity of anions are endoergic. 2. First electron affinity of alkaline

earth

metals is also

endoergic due to stable ns2 configuration. 3. EA1ofF < EA1 ofCl; as well as EA1ofO < EA1 ofS and EA1of

Periodic Properties and Chemical Bonding

383

EA1 of P. This may be explained as follows. In all such as we proceed from fluorine to. iodine, the added electron is going irito a p-orbital of increasing principal

N


p> d> f

(for a shell)

Why EA1 for B< EA1 for Li? Due to ellipticity. (iv) Nature of coaftguration: (a) Half filled sub-shells are extra stable and thus, oppose the addition of electron which leads to lower EA values.

EXAMPLE 13

Hint :

EA1 for C > EA1 for N, Why? 2

2

2 Addition of'e' in 2pnonnal

2

2

3

6C : b , 2s 2p

7N : b

' 2s 2 p

1 of Cu is -123 k.J mor whereas

Ega,

of

Explain.

In case of Cu, test electron occupies half filled 4s 16 sub-shell giving rise to a stable arrangement of 3 d 4s2 configuration and the process shows exothennic nature. In case of Zn, the test electron is to be added in 4p sub-shell and this requires work to be done and thus, process in endothennic. Following are some important observations derived from the general trend of electron gain enthalpy. Ans.

Cl

(ill) Ellipticity of sub-shells: More closer are sub-shells of valence shells to the nucleus, eaaier is . the add ition of electron in that sub-shell and thus, EA for addition of electron show the order

EXAMPLE 14

Ega,

1 Zn is + 87 k.J mor •

- 300

·

Why, EA

Ans.. After addition of one electron, halogen acquires nearest

Na+

-100

for C > EA1 for N

(b) Completely filled sub-shells have no space to accommodate the new electron and thus, show zero values, e.g., EA for inert gases is zero.

EXAMPLE 17

0

�

� •

Ca

..

. and thus, easter

Addition of 'e' in 2p half filled and thus. more difficult

·

A(g) + e

----+

A.-(g);

Ml=Eaa

Higher is Ega of an element, easier is addition of electron and thus, (a) More is the tendency to gain electron, i.e., more is non-metallic nature and therefore, non-metallic nature increases along the period but decreases down the group. (b) More is the tendency to anion formation, i.e., more is the teDdcncy to show ionic bonding. Thus, tendency of non­ metals to show ionic bonding increases along the period but decreases down the group. (c) More is the tendency to get itself reduced and thus, more is oxidant nature and therefore, oxidizing power increases along the period but decreases down the group. Oxidizing power: F2>Cl 2>Br2> 12•

Concepts of Physical Chemistry

384

F is more powerful oxidant than Cl . Explain. i i

EXAMPLE 19 Ans.

Due to high heat of hydration of F than Cl on account of

its smaller size,

:. X A -X B

=

Also

0.208 [EA-B -(EA-AxEB-'-B )112]11i ...(3)

...(4)

X A -X8=0.1017./X

1 where I:!. is in kJ mo1- •

F(g)+ e+aq --t F- ( aq);

X� -X 8 =0.1011[E A-B -(EA-A xE8_8)11i]112

or

Cl�) + e+ aq --t Cl ( aq) ;

...(5)

-

q >q due to smaller size of F. i 1 EXAMPLE 20 Why formation of Cr is exothermic but

formation ofoi- is endothermic?

Formation of Cl- from Cl(g) is exothermic because

Ans.

EA is negative. 1

Cl(g)+ e --t Cl - (g) ;

The formation of oi involves two processes: O(g) + e --t o (g); o-(g)+e --t oi (g);

Ega1 Ega1

Ega2

.. The E

gaz

is positive and E00 is negative as well as E002 is o I

endothermic (i.e., +ve) in nature.

Ill. Electronegatlvlty 1. It is the power of an element to attract the shared pair of

electron between two covalently bonded atoms.

2. The electronegativity is a measure of the tendency of an atom to attract shared pair of electron towards itself in combined state whereas electron affinity is the energy released during addition of an electron in uncombined state. 3. There are around 20 different scales to assign electronegativity values based on different theoretical assumptions, some of them have been given to express electronegativity as: (i) Pauling scale: According to Pauling, bond formed between two atoms A and B. (A -B) must be stronger than the A- A or B�B bond since, .bond energy ofA-B will be more than the average of single bond energies A-A or B-B. The electronegativity difference (X A - X8) betw�en two atoms A and B is given by: where I:!. is extra bond energy in kcal moi-1•

...(1)

I:!.= Actual bond energy (A-B)- bond energy

for 100% covalent bond ( A-B)

Bond energy for 100% covalent bond= geometric mean of

covalent bond A- A and Ii-B. • •

or

xE _ _B--B I:!. = EA- B - �E � -�- --A-

!J.=EA-B -(EA - A xEa-a>11i

i

Calculate

electronegativity of carbon

...(2)

on

83.1, 98.8 kcal mor1 respectively. A'so electronegativity of

hydrogen atom is 2. 1. Solution According to Eq. (3)

.

.

1 Xe -XH =0.208[Ec-H -(EH-H xEc-e) 1i]112 x e -2.l=0.208 [98.8-[104.2x83.l] 11iJ1'i

..

=0.498

Xe =2.598

(ii) Mulliken

According

scale:

t°' Mulliken,

·

the

electronegativity of an atom is the average of the two values, i.e., electron affinity and ionisation energy.

. .

Electronegattv1ty X M =

(IE+EA)

...(6)

2

---

where IE and EA are in electron volts. The numerical values of electronegativity of an element obtained on Mulliken scale were found 2.8 times greater than Pauling scale.

... (7)

XM =2.8xXp _IE+EA Xp 5.6

or or

...(8)

-

-

If IE and EA are being measured in kJ moi 1 , then

_ JE+EA XM 2x96.48

(': l

and

Xp =

...(9)

1 eV/molecule= 96.48 kJ mol- )

IE+EA

= 2 x2.8x96.48

IE+EA 540

...(lO)

EXAMPLE 22 Electronegativity of.fluorine on Pauling scale is 4.0. Calculate its valw on Mulliken scale. Solution

According to Eq. (7)

XM =2.8xXp

X M =4.0x2.8 = U.2

EXAMPLE 23 Ionisation potential and electron gain entlialpy offluorine lll'fJ 17.42 ·w -3.45 eY rmpecttvely. Calculate electronegativity of fluorine on Mulliken and Pauling scale. Solution

·: EA1

=-Eg =3.45eV "1.

According to Eq. (6)

where EA-B, EA _ A and E8 8 are bond energy of molecules 1 AB, Ai and B respectively in kcal moi- • _

EXAMPLE 21

Pauling scale. Given that EH-H ,Ee-e and Ee-H are 104.2,

o·

greater than Ego, and thus, Ml for the formation of oi is

X A - X B =0.208./X

where EA-B• EA-A and E8_8 are in kJ mot-1•

x M =IE+EA=1 2 By Eq. (8) Xp =

7.42

3 45 + . =10.435 2

17.42 3.45 + =3.73 5.6

Periodic Properties and Chemical Bonding (iii) Allred-Rochow scale: According to this concept, electronegativity of an atom is simple electrostatic force of attraction between atoms and an electron separated from the

nucleus by the covalent radius, i.e., electronegativity XA

R

=

( .3:;z )+ 0

*

0.744

... (11)

where Z* is effective nuclear charge and r is covalent radius of atom in A.

EXAMPLE 24

Calculate electronegativity of 14 Si

using

Allred-Rochow scale. The covalent radius of Si is 1.175 Solution

A.

AR

Z* - Z

]

-

[ 14Si: (ls2 ) (2s22 p6) (3s23 p2 )]

cr

R

0.359x 4.15 (1.175)2


Electronegativity of Fe +

Ans.

(a) If !Ji. (electronegativity) = O; Bond is 100% non:.

polar, if !Ji. (electronegativity) =less; Bond is· more covalent,

atom,

smaller

is

size

of

orbital,

more

is

electronegativity. Thus, electronegativity order for different hybridized state of carbon follow: sp>sp2 >sp3

X 8 )+3 5 (X A .

-

X 8 )2 ] . . .

(12 )

X A and X8 are electronegativities of A and B.

EXAMPLE 26 Calculate the % ionic character in HCI electronegativity of Hand Cl are 2.1 and 3.0 respectively. Solution

if

According to Eq. (12)

%ionic character=[16(Xci

2 -X H)+3.5 (Xc1 -X H ) ]

= (16 x 0.9)+3.5x (0.9)2

(Xc1 -XH -3.0-2.1=0.9} =14.4+2.835 =

17.235

(c) To predict effect of electronegativity on bond angles

(a) Bond

angles

are

generally

explained

by

electronegativity or size arguments when outer atoms are (b) Molecules

having

large

difference

in

electronegativities of central atom and atoms attached to it have smaller bond angles because highly electronegative atom reduces the electron density on central atom by pulling shared

and so, e.g., angles are small. Compounds containing halogens

(a) To predict nature of bond: . There is no hard and

covalent

bonding.

However, if the electronegativity difference (XA - X8) in A and B atoms of molecule A - B is:

-

electron pair away from the central atom. Thus, repulsive effect of these bonding electrons (bp-bp repulsion) is reduced

5. Applications fast dividing line between ionic and

It is used to calculate %ionic

different and central atom is same.

(iv) s-character: . More is the s-character in hybridized

of

Ionic

T hus, greater the difference in electronegativity, the

On moving down the group, effective nuclear charge decreases and therefore, electronegativity decreases. F Br Cl I

state

3.0

Polar covalent

more polar the bond.

(ii) Size of the atom:

(iii) Oxidation state:

LiF

1.8

Non-polar

%ionic character=[16(XA

nuclear charge of an atom more is its power to attract the

2.8

HF

0

Hanny-Smith equation :

shared pair of electron and thus more is electronegativity.

3.0

F2

:

character in a bond A-B.

4. Factors influencing electronegatlvity

4.0

ionic and less covalent

more

The bond A - B is less ionic and more covalent.

(b) To predict % ionic character

+0.744= l.S2

(i) Effective nuclear charge:

>

The bond A - Bis 500/o covalent and 50% ionic or po-

less covalent.

14-9.85 = 4.15

XA =

XA - X8

I. 7

The bond A-Bis 100% covalent ornon -polar bond.

less ionic, if !Ji. (electronegativity)=more; Bond is more ionic,

=14 -[(l x 2)+ (0.85x 8)+ (0.35x 3) ] =

XA - X8

=

0

State the kind of bond formed when atoms are : (a) With zero electronegativity difference (b) With small electronegativity difference (c) With large electronegativity difference

_I0.359Z* +0.744 r2

-l

XA - X8

=

EXAMPLE 25

According to Eq. (11)

x

XA - X8

have the following order of bond angles F'

EXAMPLE 30

The ionic radii of S 2- and Te2- are 1.84 and

2.21 A respectively. What may be the ionic radius ofSe2- and

p3-?

Ans. Ionic radius of Se2- is average of s2- and Te 2- , i.e., 1 4 +2 21 · =2.02Aas16th gp has 0, S, Se, Te. .s 2 p3- has ionic radius should be larger (i.e.,>2.02 A) than S 2- because the radii of isoelectronics decreases with increase in atomic number. EXAMPLE 31

K+ and p- have identical radius about 1.34 A.

What should be the atomic radius of K'and F atoms?

Ans.

Atomic radius of K > Ionic radius of K+ > 1.34 A

Atomic radius of F < Ionic radius of p­ < 1.34 A

Cations are always smaller than their parent atom; anions are always larger than their parent atom. EXAMPLE 32

Mg2+ > Al3+

An increase in atomic numbedn isoelectronic ions brings �n increase in effective nuclear charge to show a decrease in

have same electronic configuration. Explain. Ans. No doubt Mg2+ and 02- both

The atomic radius decreases along the period

but inert gases has maximum atomic radius in a period. Ans.

has

Atomic radii of inert gases are van der Waals' radius

whereas for rest all atoms covalent radii are determined

I

10 electrons but

has 8 protons)

exerts higher effective nuclear charge and thus, shells are more effectively pulled in Mg 2+ resulting to smaller size than 02-. EXAMPLE 33

Atomic and monovalent anion radius of

fluorine are 72 and 136 pm in gaseous state. Calculate the % increases in volume during conversion of F (g) to p- (g). Solution

Volume of F(g) = � 1t r3 3 = � x 3.14 x (72 ) 3 3

ionic radius.

EXAMPLe 27

.

Mg2+ is smaller than 0 2- in size, though both

Mg2+ having 12 protons in its nucleus (02-

02-> o- > 0

and thus, effective nuclear charge decreases. The valence shells are less tightly held by �ucleus and orient away from the

(vii)

in the case of the most strongly electronegative elements F, 0

Which oxide of N is isoelectronic with C0 ? 2 Ans. N 20 is isoelectronic with C02.

HCl, HBr and HI are altogether different.

l.33 0.89 0.80

between unlike atoms and the difference is ..;ery much marked

EXAMPLE 29

Covalent radii of H in

II period

The experimentally determined N-F bond length in

NF is greater than the sum of single bond covalent radii ofN 3 and F. This is due to the partial ionic character of the bonds

andN.

(ii) Percentage ionic character:

Atomic radius in A

The experimentally determined N-F bond

0.77 A

0.67 A

1.20 A

HC;;;;C ;;: H

EXAMPLE 28

length in NF is greater than the sum ofsingle bond covalent 3 radii of N and F. Ans.

5. Factors influencing atomic radii

(i) Multiplicity of bond:

experimentally. van der Waals' radii should not be compared

with covalent radii.

.

=1.56x106 ( pm)3

4 Volume of p- (g) = - 7t r3 3

Concepts of Physical Chemistry

388

together by the electrostatic attraction between these ions (e.g.,

=�x3.14x(l36)3 3

Na +c1-). This attraction results for ionic bond. If the lowest energy is achieved by sharing electrons, then the atoms link

=10.54x106 (pm)3

through a covalent bond and discrete molecules (e.g.,

:.Increase in volume =[10.54 -1.56]x106

are formed.

=8.98x106 ( pm)3 or

0/ l'o

mcrease =

•

8.98x106 l.56x 106

=

EXAMPLE 34

Types of Bonding I. Electrovalent bonding or ionic bonding-the strongest bonding among all.

x100

II. Covalent bonding.

S.7Sx 102

III. Co-ordinate bonding.

Atomic radius of Li is 1.23 A and ionic radius

of Li+ is 0.76 A. Calculate the percentage. of volume occupied by single valence electron in Li.

Solution

3

(Li:ts2,2s1)

=7.79(A)3

IV.· Metallic bonding.

V. Hydrogen bonding-the weakest among all.

I. Electrovalent or Ionic Bonding (Kossel and Lewis)

Volume of Li =�x3.14x(1.23)3

1.

an atom into the valence shell of other atom so that each

3

=l.84 (A)3

2.

:. Volume occupied by 2s sub-shell

7.79-1.84

=

.3.95 (A)3

3.

:. % volume occupied by single valence electron i.e., 2s

surrounded by anions and there is large negative (energy lowering) contribution from the attraction of opposite charges. Beyond those nearest neighbours, there are cations that contribute a positive (repulsive, energy

According to-Pauling, a chemical bond existing between two

raising) contribution to the total potential energy of the

atoms is defmed as the binding force between them of such a

central cation. There is also a negative contribution from

strength which gives rise to an aggregate of sufficient stability

the anions beyond their cations, a positive contribution from the cations beyond them, and so on, to the edge of the

warranting their consideration as independent molecular species.

solid.

These

repulsions

and

attractions

increases but the net outcome of all these contribution is

redistribution of electrons, so that each atom involved in

lowering of energy. Thus, the net result of these coulombic

bonding acquires stable configuration and attains lower energy

forces between oppositely charged ions give rise to

level in order to gain stability is known as chemical bonding. The combination of two atoms involves a decrease in

4.

evolution of energy, so that resultant molecule acquires lower energy level and gains stability. Ionic bonding is usually noticed between non-metals and metals, particularly metals in s-block, e.g.,

Ans. Chemical reactions occur by breaking and making of bonds. The strength of these bonds play important role to

K

;.� .. CI: = K+

(2,8,8, I)

�

••

..

[ :c1: r (2,8,8) ••

or

K+ Cl-

decide the nature and rate of reaction. EXAMPLE 36

Ans.

Why do atoms combine to form molecule?

Two atoms combine if the resulting arrangement of two

nuclei and their electrons has lower energy than the total energy of the separate atoms. If the lowest energy is achieved by the complete transfer of one or more electrons from one atom to another, then ions formed and resultant ions are held

becomes

progressively weaker as the distance from the central ion

The phenomenon of union of two or more atoms involving

bonds between atoms in a molecule?

Each ion in a solid experiences electrostatic attraction from all the other like-charged ions. The total potential

Chemical Bonding

Why is it important to know about chemical

Atom .losing electron forms cation and the other gaining electrons forms anion.

energy is the sum of all these contributions. Each cation is

5 95 = · X100 =76.4o/o 7.79

EXAMPLE 35

atom involved in bonding acquires nearest noble gas configuration to gain stability.

from all the other oppositely charged ions and repulsion

electron

potential energy as a result of net attraction in between them.

The union of two or more atoins involving complete transfer of one or more electrons from the valence shell of

Volume ofLi+ =�x3.14x(0.76)3

=

NH3)

(2,7) (2,8,2) Na

�·s. · Na + • •

Nav·

(2,8,l) (2,8,6)

.

Na+

(2,8)

(2,8)

(2,8)

•• cxxs•.1-••

or

(2,8)

(Na + ) 2

s--

·

Periodic Properties and Chemical Bonding

5.

The arrows indicate the transfer of an electron from more electropositive atom to more electronegative atom. Conditions for ionic bonding: Consider an ionic

-

B(g)+e

----+

A+(g)+e;

----+

B-(g);

A+(g)+B-(g)----+ A+B-(s);

Afl=Ega1 Afl=EL

• • •

A(g)+B(g)

-4

. ..(iii)

as the amount ofenergy released when the requisite number of isolated gaseous cations and anions are condensed into crystal to form one mole of ionic compound. Eq. (iv) reveals that more -ve is Ml, more is stability to ionic bond as well as more is the possibility for ionic bonding. Therefore, it may be derived that ionic bonding is favoured by: (i) Low IE values of atoms forming cation. (ii) High E a values of atoms forming anion. (iii) Higher 1attice energy (EL ) of ionic compound. The capacity of an element to form electrovalent or ionic bond is termed as electrovalency or in other words it is equal to the number of electrons lost by an atom of the element or gained by an atom of the element to acquire nearest noble gas configuratior.. There is no fixed direction in space among the ions, i.e., electrovalency is non-directional. An electrovalent bond cannot be formed between atoms of same electronegativity.

What type of atoms generally form ionic

linkage? Ans.

Atoms forming cation with low IE and atoms forming

anion with high Ega generally form ionic bonding.

It interrelates the· various energy terms involved during formation of an ionic compound. The fundamental is based upon the fact that the formation of an ionic crystal may occur either by direct combination of the elements or by an alternate process in which: (i) The reactants are vaporized to convert in gaseous state. (ii) The gaseous atoms are converted into ions. (iii) The gaseous ions are combined to form ionic lattice of molecule. Thus, for NaCl:

U

l

(-) I

2

Cl2(g)

HS

-----+

l HD/2 Na(g) + -Cl2 -----+ Na(g) + Cl(g) 2

E

Na+c1-(s) 1

l rl

°1

+Ei

1. The lattice energy of an ionic compound is the energy

required to separate 1 mole of solid ionic substance

completely into gaseous ions. NaCl(s)

j

Na+(g) + c1-(g

-----+

Lattice energy

= + 788 kJ mol-

1

Alternatively lattice energy of the solid is the difference in

energy between the ions packed together in a solid and the

ions widely separated as gas. The lattice energy is always

positive; a high lattice energy indicates that the ions interact strongly with one another to give a tightly bonded solid.

2. It is the sum of the electrostatic interaction energies between ions in a crystal and thus, measures the strength of the

crystals ionic bond. Lattice energy

( U)

is expressed as:

-Ei ocF xd oc

-

E

l

=

(·:

:

z 2 xd

A·N·�-� d

(In C.G.S.)

.

(N 1s

F= Av.

;;)

No.)

(In M.K.S.)

or

where d is distance of separation of ion, Z1 and Z2 are charges on ions arid F is Coulombic forces of attractions among ions.

Lower is the value ofd, more is lattice energy. A is Madelung value ofMadelung constant (A) is 1.748, 1.763, 2.519, 2.408 respectively.

Lattice energy ofsome ionic solids (kJ mol-1) Anions

Cation F-

er

er-

1-

2-

Lt

1036

853

807

757

2925

Na+

923

787

747

704

2695

IC'·

821

715

682

649

2360

Bell-

3505

3020

2914

2800

4443

Mg21-

2957

2524

2440

2327

3791

Na+(g)+ eCT(g)

llCa

2630

2258

2176

2074

3401

Al3+-

5215

5492

5361

5218

5916

U =HS +-HD + IE1 +E ga1 +Ei 2

U =HSea +HDc12 +IE1 +IE2 +2E8

Note :

+Ei

2(S)

,_,

IE; l

Similarly for Na2S:

constant (value of A depends upon geometry of crystal) e.g., for rocksalt (NaCl), CsCI, fluorite and rutile structures, the

Born-Haber cycle

Na (s) +

E L =lattice energy of Na +er

For CaC12:

. .. (iv)

where EL is lattice energy ofionic compound A+ B- defined

EXAMPLE 37

Ega1 = electron gain enthalpy of Cl

U =2HSNa +HSs +2IE1O: ••

N

0

( Y')

5

5

6

sub-shell

Non-bonding electrons in Lewis

2

2

6

overlapping.

I -x6=3 2

I -x6=3 2

0

(iii)

structure (L)

Between two sub-shells of same energy level, the more

0

(L � ) +

5

s-p

head on overlapping (a-bond)

overlap

� Fig.

Formal charge on carbonate ion -I

-0-c-011 0

The Valence Bond Theory : A Modem Approach for Covalence

-

c::>I - 2 > 2-2 > 2-3 for different overlapping. However, bond energy ofF2 (2p-2p) < B. E. ofCl 2 (3p-3p).

N

2

the

(b) cancellation or attraction between spins of antispin

0

I

••

electrons in Lewis

and

(b) nature of overlapping.

Atom

_! x Bonding

nuclei

electrons.

(i)

S

Valence electrons in free atom

between

accumulated electron cloud. 6

(ii) (Case II)

attraction

Paullng and Slater extension

(L � ) +

(v)

0

electrons in Lewis structure) or I -xS 2 Formal charge= V

As a result of overlapping, a new localized· bond

orbital is formed, in which probability for finding electron pair

••

xx

Bond

6

p-p

energy

head on overlapping (a-bond)

order:

2s - 2s< 2s - 2p< 2p-2p

for

different head on overlapping.

(iv) s -orbitals are spherically symmetrical and thus, show only head on overlapping. On the other hand p-orbitals are directionally concentrated and thus, show either head on overlapping or lateral overlapping.

Heltler and London concept

(i)

For a covalent bond to form, two atomnnust come

closer to each other so that orbitals of one overlap the other.

(ii)

Overlapping orbitals must have:

(a) half filled nature,

i.e.,

must have unpaired electron.

p

p

Fig. 7

(b) antispin electrons.

(iii)

Properly oriented atomic orbitals of differentatoms

can overlap together.

p-p overlapping

(v)

p-p

n-bond orbitals

lateral overlapping

Head on overlapping is more stronger than lateral or

sideways overlapping.

Periodic Properties and Chemical Bonding :. B.E.:

p-p

399

< s-s < s -p < p- p

Lateral overlapping

=

=

Head o n overlapping of same shell

4. S.

Head on overlapping and sigma bond

6.

(i) Formation of a sigma bond (cr) involves head on overlapping, i.e., overlapping oforbitals along their inter nuclear axis, giving rise to maximum electron density on the axis. (ii) The characteristic property of a a-bond is the cylindrical symmetry it has about the internuclear axis. There 1s not a nodal plane coincident with the internuclear axis. (iii) Formation ofcr-bonds can be noticed in the examples given below:

(J)+O-®

H2

Is HCI

CD Is

+

CI> L.P. - B.P. > B.P. - B.P. It can be illustrated in case of NH3 and H20 satisfactorily.

In F2B- C==C-BF2, the B and C-atoms are again

linear on account of sp-hybridized carbon. No doubt B-atoms 2 are sp -hybridized and are trigonal. No set of orientation of the fluorine atoms with respect to each other is expected, since,

N: 2 s2 2p

3

(G.S.)@ 1111111

Concepts of Physical Chemistry

406 Since, N has three unpaired electrons and th�, it can form three er-bonds. Also, excitation of electrons in higher d-sub-shell is not possible since 2nd shell does not have d-orbitals. That is why N does not show +5 covalence like its

greater repulsive force on adjacent electron pairs than do single bonds. Geometry of molecules or ions involving hybridization; havin2 only bond pair electrons

other group members P, As, etc. Furthermore, if all the three p-orbitals are involved in sigma bond formation with ls-orbitals of hydrogen, the bond angles will be at 90° whereas NH3 has bond angle of106°S1'. 3

Thus, it has been proposed that N-atom in NH3 is

sp -hybridized to show pyramidal geometry or tetrahedral nature with one position occupied by lone pair of electron. Presence of the lone pair of electron on N-atom contracts the bond angle from 109"28' to 106°51'.

H20 0:

(G.S.)

2 2s 2 p4

I

S=O

(b)

Thus, bond energies are different.

Ans.

F

than regular P-0 bond.

-

Ans.

I

I

formal charge arguments result in a double bond for P=O. The actual P-0 distance is (143 pm), considerably shorter

3 by overlapping of3sp -3 p. 3 In PCl5, P is sp d-hybridized and P-Cl bond is formed 3 by the overlapping of3sp d 3 p

EXAMPLE 91

II

(a) F-P -F � F-P-F

Octet rule results in single P-0 and P-F bond while

j

In PC13, P is sp -hybridized and P-Cl bond is formed

EXAMPLE 90

0

Why bond energy of P-Cl bond is different in

PC13 and PC15?

(c) S03F-

(b) SOF4

F

sp2 EXAMPLE 89

Discuss the structure of the following:

(a) POF3

for conduction of current. B: 2s2zp 1

(if any)

In I02F:Z

:

F · 3

I has sp d-hybridization with one of the

equatorial position unshared leading to trigonal bipyramid geometry and see-saw shape. 0

� AsC13 102°

98.4 °

(vii) .The bond angle in (CH3 )JN and (SiH3 )JN are

different due to lone pair-bond pair repulsion in (CH 3) 3 N and not in (SiH3 h N due to non-availability of lone pair of

electron in (SiH3 h N because of pn-d7t bonding.

Can the molecule of NH 3 be assumed to have

the shape of regular triangle?

EXAMPLE 107

The small size of F-atom leads to p're--

XeF2 is linear inspite of the fact that Xe

Ether and water have same hybridization on

o;cygen atom but their bond angles are different.

Oxygen in each has sp3-hybridization with two lone

pair of electrons. In H20 lone pair-lone pair repulsion contracts bond angles from 109°28 ' to I 04 °31'. In ether due to . mutual repulsion between alkyl groups counter balances lone

pair-lone specie pair repulsion and bond angles are greater than 109°28' and becomes 1 10°. Thus, bond angle slighly 3 increases from 109°28' (for sp -hybridization) due to steric hinderance of b�lky groups.

EXAMPLE 108

Repulsion between non-bonded orbitals are

greater than between the bonded orbitals.

Ans.

Non-bonded orbitals occupy relatively more space

compared to bonded orbitals (overlapping occurs) and thus, repulsion is greater.

Concepts of Physical Chemistry

408 EXAMPLE 109

Interpret the non-linear shape of H2S molecule and non-planar shape of PC� using valence shell electron pair repulsion (VSEPR) theory. 3 Ans. In H2S; S is in sp -hybridized state having two lone pairs on it as shown below. The angle is contracted due to lone pair effect to produce V-shaped structure. On the other hand, in PC13• P-atom is also

sp3-hybridized having one lone pair on

it. This time the shape from tetrahedron structure is distorted to produce pyramidal shape due to lone pair effect.

Explain which one is most probable structure and why? Ans.

The most stable structure will be one with lowest energy or with minimum steric repulsion. The order of repulsion between different kinds of electrons is :

Ip - Ip> Ip - bp> bp - bp

EXAMPLE 110

Using VSEPR theory identify the type of hybridization and draw the structure of OF2• What are the oxidation state of0 and F? 3 Ana. The structure of OF2 (similar to H20) involves sp -

also the less is separation, more is repulsion. Thus, repulsion between two electron pairs at 90° will be more than when they are at 120°. . Considering the above two facts, structure (C) will be most probable because: (i) in structure A, there are 6 repulsions between lp-bp. �·· (ii) in structure B, there is one 90° Ip-Ip repulsion, and three 90° lp-bp repulsions. (iii) in structure C, there are four 90° lp-bp repulsions and no 90° Ip-Ip repulsion and therefore, structure C is most probable. Note :

In trigonal bipyramidal structure, lone pair of electrons occupy equatorial positions rather than axial positions to minimise re­ pulsive forces.

hybridization on 0-atom having V-shape. Oxidation number of oxygen Oxidation number of fluorine

EXAMPLE 111 and B rF5.

+2

=

-1

�

Using VSEPR theory, draw the shape of PC15

sp3d-hybridization 3 shape with five 3sp d-3p-bonds

Ans.

=

PCI 5 :

having trigonal by pyramid

EXAMPLE 113

F-N-F

less than H-N-H angle in

Ans. Highly electronegative F-atoms pulls the bonding electron pair away from N than in NH3• Thus, repulsion between bond pairs is less in NF3 than in NH3•

EXAMPLE 114

li

Ans.

Draw the geometry ofr; ion.

is formed as : [I-It-Ir

.

!�--,Cl c11l Cl� l ''

c1

BrF5 :

2 sp3d -hyl;>ridizatio� having

distorted octahedral shape

with.one position occupied by lone pair (or square pyramidal) 3 2 with five 4s p d -2p-bonds

��/! .

Lone pair F

F ,-II

r

:

I

l �F

1 F --- F

In CIF3, central chlorine atom is sp3d­ hybridized and theoretically three structures are possible.

EXAMPLE 112

bond angle in NF3(102°30') is Explain.

NH3{1 07°48'}

In this ion, central I-atom has ten electrons and is

sp3d-hybridized. To minimise repulsive

forces, lone pair of electrons occupy equatorial positions and I-atoms, apical (or axial) positions. Thus, ion is linear in shape, yet hybridization is

sp3d.

EXAMPLE 115 Using the VSEPR model, predict the molecular geometries of the following:

(a) SnCI), Ans.

/ {a) Cl

t°

�;-

'

Cl

••

(b)

03,

(b)

(c)

SF4,

/ 1 - Cl

c

(d)

IF5•

s·l�Cl

C:l

••

o�o -- ci°'o

Pyramidal ••

�

�o

Angular

[ .ti-=f ]-

Periodic Properties and Chemical Bonding

See-saw

409

•.

(c)

3 EXAMPLE 116 Both Cl03 and C/04 have sp -hybridization but czo:; is pyramidal and c104 is tetrahedral. 2 Ans. Cl : 3s 3p5 (J) lllJl 11

Ana. (a) HCP : (H--O=P). (b)

�1Il!L!)

er: 3s 23p6

EXAMPLE 119 Discuss the structures of the following molecules: (a) HCP; (b) IOF4-; (c) Se0Cl2•

3 (sp )

In c103 three oxygen atoms can bond to the chloride ion

utilizing fully occupied Cl-orbitals with no electron8 furnished by the oxygen atom leaving one lone pair and thus, pyramidal. In Cl04 four oxygen atoms are bonded to the chloride ion in. same manner and thus, tetrahedral.

EXAMPLE 117 Discuss the structure of the following: (a) SbF4-; (b) SF 5-; (c) SeF3+ Ans. (a) SbFi : SbFi has a lone pair on Sb and so its

structure may be compared with SF4• Lone pair of electron occupies equatorial position and causes considerable distortion giving F-Sb-F (axial) angle 155° and F-Sb-F (equatorial) angle 90°.

IOFi

Ans.

.)F

••

r

(c) SeF� :

octahedron. Structure is comparable toIF5• .

Due to presence of lone pair on Se,

F-Se-F angle is

r l·r \.jl

V

e

EXAMPLE 118 Discuss the structure of following Ions.

[ ·r ·]-

(a) NH2; An•.

(b) NHt;

itl'(a)

_

(c) 13;

(d) PCf6.

..

I

F

(a)

r

F

,_sd'92.3° O �

(b)

�'

Cl

LF-Cl-F < 90° LF-Cl-0 > 90°

LCl-S-Cl = 96° LCl-:--S-0 > 106°

l

CI

,1n ,_s�Cl.2°

�

T

r

;s 98.2°r

O O OSF2