Question

A diatomic molecule having a dipole moment of 1.92D and bond length of 2 A. Its percentage ionic character is​

Answer 1

Answer:

dipole moment = q x d

= 4.8 × 10 ^ (-10) esu x 2 x 10^ (-8) cm

9.8 x 10^(-18) esu.cm = 9.8 D

% ionic character is = 1.92 D x 100

-------------- =19.5%

9.8D

Answer 2

Percentage ionic character is​ 20%.

Calculations:

Given

• Bond length = 2 A = 2 x 10⁻⁸ cm
• Observed dipole moment (μ) =1.92 D

Percentage ionic character = μ observed/ μ calculate  x 100%

μ calculated = q x D

• q is charge= 4.8  x 10⁻¹⁰ esu
• D is the bond length

       μ = 4.8  x 10⁻¹⁰ x 2 x 10⁻⁸ cm

      μ = 9.6 x 10⁻¹⁸ esu cm

      μ calculated = 9.6 D

% ionic character = 1.92/9.6 x 100

                              = 20 %