Question

The dipole moment of a bond $AB$ is $5D$ and its bond distance is $150pm$. The percentage ionic charactor of this bond is?

• A. $78.5\%$
• B. $83\%$
• C. $24\%$
• D. $69.5\%$

Answer 1

The correct option is D $69.5\%$

Dipole moment= charge $\times$ distance between the dipoles

$=1.6\times {10}^{-19}\times 150\times {10}^{-12}$ ($1$ pm= ${10}^{-12}$m)

$=\frac{240\times {10}^{-31}}{3.336}=7.190$ ($10=3.336\times {10}^{-30}$cm)

% of ionic character= $\frac{5}{7.19}\times 100=69.54$%