Question

The diatomic molecule has a dipole moment of 1.98D and the bond length is 0.92Angstroms . then the value of %ionic character/112 is​

Answer 1

Given:

Dipole moment of the diatomic molecule = 1.98D

Bond length in Angstroms = 0.92

To find:

% of ionic character in the molecule.

Solution:

As we know that

Dipole moment (theoretical) μ = Q x d (coulomb-metre)

Q is the charge of electron = 1.6 x 10^-19 C

d is the bond length in meters

μ = 1.6 x 10^-19 x 0.92 x 10^-10 = 1.472 x 10^-29 C-m = 14.72 x 10^-30 C-m

(1 debye = 3.34 x 10^-30 C-m)

Therefore,

 μ = 4.4 D

% ionic character = actual dipole x 100/ theortical dipole

= 1.98 x 100/ 4.4 =  45%

Therefore the % ionic character is 45%.