Question

The observed dipole moment of $HBr$ is $2.60\times {10}^{-30}$ and the interatomic spacing is $1.41\mathring{A}$. What is the percentage ionic character of $HBr$?

• A. 12%
• B. 15%
• C. 5%
• D. 10%

Answer 1

The correct option is A 12%

% ionic character in a molecule $=\frac{observeddipolemoment}{expecteddipolemoment}\times 100$

Common unit of interatomic spacing$=A°$ (Angstrom)

$H^{\delta +}-{Br}^{\delta -}\equiv H^{+}{Br}^{-}$

Expected dipole moment$=Q\times d$

Interatomic spacing$=1.41A°$

$=1.41\times {10}^{-8}cm$

$=141\times {10}^{-10}cm$

Expected dipole moment$=1.6\times {10}^{-19}\times 141\times {10}^{-10}$

$=255.6\times {10}^{-29}cm$

Observed dipole moment$=2.6\times {10}^{-30}cm$

% Ionic character$=\frac{2.6\times {10}^{-30}}{225.6\times {10}^{-29}}\times 100$

$=\frac{2.6}{225.6}\times 1000$

$=\frac{26}{225.6}=11.55$ % $\approx 12$ %.