Question

The dipole moment of $HBr$ is $2.6\times 10^{- 30}$ esv.cm and the interatomic spacing is $1.41\,\mathring{A}$ . The percentage of ionic character in HBr is

• A. 10.5
• B. 11.5
• C. 12.5
• D. 13.5

Answer 1

Answer: (B)

Percentage ionic character $=\frac{\text { Observed dipole moment }}{\text { Theoretical dipole moment }} \times 100$
Theoretical dipole moment of a $100 \%$ ionic character
$=e \times d=\left(1.6 \times 10^{-19} C\right) \times\left(1.41 \times 1(0)^{-10} m\right) $
$=2.256 \times 10^{-29}$
$ \%$ ionic character $=\frac{2.6 \times 10^{-30}}{2.256 \times 6 \times 10^{-29}} \times 100=11.5 \%$