Question

The bond length of HCl molecule is $1.275\mathring{A}$ and its dipole moment is 1.03 D. The ionic character of the molecule (in percent) (charge of electron $4.8\times 10^{-10}$ esu) is

• A. 100
• B. 67.3
• C. 33.6
• D. 16.83

Answer 1

Answer: (D)

% of ionic character $=\frac{{\mu }_{experimental}}{{\mu }_{theoretical}}\times 100$
${\mu }_{theoretical}=e\times d=4.8\times 10^{-10}esu\times 1.275\times 10^{-8}cm$
$=6.12\times 10^{-18}esucm=6.12D$
${\mu }_{experimental}=1.03D$
$\therefore \%$ of ionic character $=\frac{1.03}{6.12}\times 100=16.83$