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Atomic radius is the distance from the centre of the nucleus to the point where the electron density is effectively zero. a. Homonuclear diatomic molecules b. Heteronuclear diatomic molecule

Atomic radius is the distance from the centre of the nucleus to the point  where the electron density is effectively zero.

a. Homonuclear diatomic molecules

In case of homonuclear diatomic molecules of A2 type (e.g. F2, Cl2, Br2, I2 ... etc.) the bond length, d(A-A) is given by

d(A - A) =   r(A) + r(A)

d(A - A) =   2 r(A)

r(A) =  d(A-A) / 2

The above equation shows that in the case of homonuclear diatomic molecule of A2 type, the covalent radius of an atom A, r(A) is equal to one half of the inter- nuclear distance, d(A-A). Therefore, the covalent radius of an atom in a homonuclear diatomic molecule can be obtained by dividing the internuclear distance by two.

Example

1. Cl2 molecule

The value of Cl-Cl bond distance as found experimentally is 1.98�. Thus

r(Cl)= d(Cl- Cl) /  2 = 1.98/2 = 0.99�

2. Diamond

The value of d(C-C) distance as found experimentally in a variety of saturated  hydrocarbons is 1.54�.

r(C) = d(C - C) / 2 = 1.54 / 2 = 0.77 �

b. Heteronuclear diatomic molecule

In case of heteronuclear diatomic molecule of AB type, bond length

d(A - B) is given by

d(A - B) = r(A) + r(B)

r(A) and r(B) are the covalent radii of A and B atoms.

Example

i)  CCl4 molecule

The experimental value of d(C - Cl) is 1.76 �

Thus    d(C-Cl) = r(C) + r(Cl)

r(C) = d(C - Cl) - r(Cl)

= 1.76 - r(Cl)

Thus the covalent radius of carbon atom can be calculated by subtracting the covalent radius of Cl atom from d(C-Cl) bond length. The covalent radius of Cl atom can also be obtained, provided that covalent radius of C atom is known.

ii) Sic

The experimental value of d(Si-C) is 1.93 �. Thus,

d(Si - C) =   r(Si) + r(C)

r(Si)  = d(Si - C) - r(C)  = 1.93 - r(C)  = 1.93 - 0.77 = 1.16 �

The experimental values of covalent bond length for some common homonuclear diatomic molecules are given below.

Molecule  : Bond  : Bond length (�)

H2 : H-H : 0.74

F2 :  F-F  : 1.44

Cl: Cl-Cl  : 1.98

Br2 : Br-Br  : 2.28

H3C-CH3  : C-C  : 1.54

Pauling's Method

Pauling has calculated the radii of the ions on the basis of the observed internuclear distances in four crystals namely NaF, KCl, RbBr and CsI. In each ionic crystal the cations and anions are isoelectronic with inert gas configuration.

NaF crystal : Na+ - 2, 8   F- - 2, 8   Ne  type configuration

KCl crystal : K+ - 2, 8, 8    Cl-   - 2, 8, 8   Ar  type configuration

Further the following two assumptions are made to assign the ionic radii.

i) The cations and anions of an ionic crystal are assumed to be in contact

with each other and hence the sum of their radii will be equal to the inter nuclear distance between them.

r(C+) + r(A-) = d (C+-A-)

where

r(C+) - radius of cation, C+ r(A-)

d(C+-A-) - internuclear distance between C+ and A- ions in C+A- ionic crystal

ii) For a given noble gas configuration, the radius of an ion is inversely proportional to its effective nuclear charge. i.e.

r(C+ ) � = 1/ Z (C+ )

r(A- ) � = 1/ Z (A- )

where,

Z*(C+) & Z*(A-) are the effective nuclear charges of cation (C+) and anion (A-) respectively. On combining

r(C+ ) / r(A- ) = Z*(A- ) / Z*(C+ )

Hence the above two equations (1) & (4) can be used to evaluate the

values of r(C+) and r(A-) provided that the values of d(C+-A-), Z*(C+) and Z*(A-) are known.

Slater rules

The value of screening constant (S) and effective nuclear charge (Z*) can

be calculated by using Slater's rules. According to these rules the value of "S" for a given electron is estimated as follows.

i) Write down the complete electronic configuration of the element and divide the electrons into the following orbital groups starting from the inside of the atom.

(1s) :   (2s, 2p) : (3s, 3p) : (3d) : (4s, 4p) :

(4d) :   (4f)      : (5s, 5p) : (6s, 6p) .......etc.

ii) Select the electron for which the value of S is to be calculated. For this calculation add up the contributions to S for the other electrons according to the following rules.

Contribution to S for each electron of this type

Type of electron

1.           All electrons in groups outside the electron chosen  - 0

2.           All other electrons in the same group as the chosen one (n) - 0.35 (or 0.30 for 1s electron)

3.           All electrons in shell immediately inside (n-1)  - 0.85

4.           All electrons further inside  -  1.00  Ionisation potential

Ionisation energy of an element is defined as the amount of energy required to remove the most loosely bound electron from isolated neutral gaseous atom in its lowest energy state. The process is represented as

M(g) + Energy supplied - I1-- > M+(g) + e-

Ionisation energy is measured in electron volts per atom (eV/atom), kilo calories

per mole (kcal/mole) or kilo joules per mole (kJ/mole).

Successive ionisation potentials

In addition to first ionisation potential (I1) defined above, second, third. etc.

ionisation potentials are also known. Second ionisation potential (I2) is the energy required to remove one more electron from the gaseous cation, M+(g) to get the

doubly positively charged gaseous cation, M2+(g), i.e.,

M+(g) + I2 --- > M2+(g) + e-

Similarly, third ionisation potential (I3) is the energy required to remove still one more electron from M2+(g) cation to get M3+(g) cation, i.e.

M2+(g) + I3 - -- > M3+(g) + e-

Similarly ionisation potentials of higher and higher grades are also known.

Each successive ionization potential or energy is greater than the previous one, since the electron must be removed against the net positive charge on the ion.

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