Question

A certain diatomic molecule, $AB$ has dipole moment $1.6D$ and the inter-nuclear distance is $100pm$. The percentage of an electronegative atom is:

• A. $33\%$
• B. $25\%$
• C. $50\%$
• D. $10\%$

Answer 1

The correct option is A $33\%$

Solution:- (A) $33$ %

${\mu }_{observed}=1.6D=1.6\times {10}^{-30}cm(\because 1D={10}^{-30}cm)$

${\mu }_{ionic}=q\times d$

where,

q = amount of charge at either end of dipole = $4.8\times {10}^{-10}esu$

$\Rightarrow q=4.8\times {10}^{8}D=4.8\times {10}^{-22}cm(\because 1esu={10}^{18}D)$

d = distance between the two ends of dipole = $100pm={10}^{-8}cm(\because 1pm={10}^{-10}cm)$

$\therefore {\mu }_{ionic}=(4.8\times {10}^{-22})\times \left({10}^{-8}\right)$

$\Rightarrow {\mu }_{ionic}=4.8\times {10}^{-30}$

$\therefore$ percentage ionic character = $\frac{{\mu }_{observed}}{{\mu }_{ionic}}\times 100=\frac{1.6\times {10}^{-30}}{4.8\times {10}^{-30}}\times 100=33.33\%\approx 33\%$