Question

The dipole moment of HBr is $1.6\times {10}^{-30}$ c.m. and inter atomic spacing is $1\stackrel{o}{A}$. The percentage ionic character of HBr is:

• A. $10$
• B. $0$
• C. $15$
• D. $27$

Answer 1

The correct option is A $10$

Percentage of ionic character = $\frac{Observeddipolemoment}{Calculateddipolemoment}\times 100$

Observed dipole moment $=1.6\times {10}^{-30}cm$

Calculated dipole moment = electron charge $\times$ interatomic distance

$=1.6\times {10}^{-19}\times {10}^{-10}=1.6\times {10}^{-29}cm$

% of ionic character $=\frac{1.6\times {10}^{-30}}{1.6\times {10}^{-29}}\times 100=10$

$\therefore$ Percentage of ionic character in $HBr$ is $10$