Question

If the bond length and dipole
moment of a diatomic molecule are
1.25A and 1.0D respectively the
percent ionic character of the bond​

Answer 1

Answer: The  percent ionic character of the bond​ = 16.69%.

Explanation:

Given information in the question,

Bond Length = 1.25 Ang = d =  1.25 X $10^{-10}$ m

Observed dipole Moment = $u_{obs}$ = 1.0 D

dipole moment = $u$ = 3.335 X $10^{-30}$ coulomb metre.

Also, for complete separation of unit charge, e = 1.6 X $10^{-19}$ C.

Ionic dipole moment $u_{ionic}$ = $\frac{e.d}{dipole moment}$  coulomb metre

$u_{ionic}$ =  $\frac{1.25 X 10^{-10} X 1.6 X 10^{-19} }{3.335 X 10^{-30} }$ coulomb metre

$u_{ionic}$ = 5.99 D

%ionic character = $\frac{100 X u_{obs} }{u_{ionic} }$ = $\frac{100 X 1}{5.99}$ = 16.69 %

Hence, The  percent ionic character of the bond​ = 16.69%.

Answer 2

Thus the percentage of ionic character is 16.69 %

Explanation:

We are given that:

• Bond length of molecule = 1.25 A
• Dipole moment of diatomic length = 1.0 D

To find: Percent ionic character = ?

μionic = (1.6 × 10^−19 C)(1.25 × 10^−10 m)  / 3.335 × 10^−30 C.m / D        

μionic = 5.99 D

%  ionic character = 100 × μobs / μ ionic

% ionic character = 100 × 15.99=

% ionic character = 16.69%

Thus the percentage of ionic character is 16.69 %