Question

The dipole moment of $HBr$ is $0.78\times {10}^{-18}$ esu cm and interatomic spacing is $1.41\mathring{A}$ The $\%$ ionic character of $HBr$ is?

• A. $7.5$
• B. $11.$
• C. $15$
• D. $27$

Answer 1

The correct option is B $11.$

$D=$ charge $\times$ distance

$=1.6\times {10}^{-19}\times 1.41\times {10}^{-8}$

$=\frac{2.256\times {10}^{-27}}{3.33\times {10}^{-30}}$

$=6.76\times {10}^2$

% of ionic charge= $0.78\times {10}^{-18}esu$

$=0.78$

$=\frac{0.78}{6.76}=11$% .