Question

If dipole moment of A-B molecule is 4.8 D and the bond length is 167 pm then the percentage ionic character of A-B bond is approximately :
40%
80%
70%
60%​

Answer 1

Given - Distance and bond length

Find - Percentage ionic character

Solution - Percentage ionic character is 76.9%.

Percentage ionic character can be calculated by the formula - observed dipole moment/theoretical dipole moment*100

Now calculating theoretical dipole moment by the formula - charge * bond length.

Bond length in centimetres = 1.6*10-⁸ centimetres

Theoretical dipole moment = 4.8*10-¹⁰*1.6*10-⁸

Theoretical dipole moment = 6.24*10-¹⁸ esu cm

Now, 10-¹⁸ esu cm is 1 Debye or 1 D

So, 6.24*10-¹⁸ esu cm will be 6.24 D.

Now, calculating percentage ionic character -

Percentage ionic character = 4.8/6.24*100

Percentage ionic character = 76.9%

Thus, percentage ionic character is 76.9%.

Answer 2

Given :

Dipole moment of A-B molecule = 4.8 D

Bond Length = 167 pm

To Find :

The percentage ionic character of A-B bond

Solution :

Theoretical dipole moment (μ) = Charge (q) × Distance (d)

                                                   = 1.6 × 10⁻¹⁹ × 167 × 10⁻¹²

                                                   =  267.2 × 10⁻³¹ Cm

Given, Observed Dipole moment = 4.8 D

Taking 1 D = 3.33564 × 10⁻³⁰ Cm  

So, Observed Dipole moment = 4.8 × 3.33564 × 10⁻³⁰ Cm  

Percentage ionic character = $\frac{Observed Dipole Moment}{Theoretical Dipole Moment} * 100%$

                                               =  $\frac{4.8 * 3.33564 * 10^-^3^0}{267.2 * 10^-^3^1} *100 %$

                                               = 60%

Therefore, the percentage ionic character of A-B bond is 60%

Hence, option (d) is correct.