Question

The dipole moment of $HBr$ is $1.6\times {10}^{-30}$ cm and inter atomic spacing is $1\mathring{A}$. The $\%$ ionic character of $HBr$ is:

• A. $7$
• B. $10$
• C. $25$
• D. $27$

Answer 1

Answer: (B)

$10$

Charge of electron(q)$=1.6\times {10}^{-19}C$
Dipole moment of $HBr$$=1.6\times {10}^{-30}$ m
Inter-atomic spacing $=1\stackrel{o}{A}=1\times {10}^{-10}$ m

$\%$ of ionic character in $HBr$

$=\frac{dipolemomentofHBr\times 100}{inter-spacingdistance\times q}$

$=\frac{1.6\times {10}^{-30}\times 100}{1.6\times {10}^{-19}\times {10}^{-10}}$

$=10\%$.

Hence, option B is correct.