Question

A certain diatomic molecule, $AB$ has dipole moment $1.6D$ and the internuclear distance is $100pm$. The percentage of electronic charge existing on more electronegative atom is

Answer 1

${\mu }_{observed}=1.6D=1.6\times {10}^{-18}esucm(\because 1D={10}^{-18}esucm){\mu }_{ionic}=q\times d$
where,
q = amount of charge at either end of dipole $=4.8\times {10}^{-10}esu$

d = distance between the two ends of dipole $=100pm={10}^{-8}cm(\because 1pm={10}^{-10}cm)$
$\therefore {\mu }_{ionic}=(4.8\times {10}^{-10})\times \left({10}^{-8}\right)\Rightarrow {\mu }_{ionic}=4.8\times {10}^{-18}=4.8D$
$\therefore$ percentage ionic character $=\frac{{\mu }_{observed}}{{\mu }_{ionic}}\times 100=\frac{1.6}{4.8}\times 100=33.33\%$