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Q.
The dipole moment of $HBr$ is $2.60 \times 10^{-30} \,cm$ and the interatomic spacing is $1.41 \,\mathring{A} $. What is the per cent ionic character of $HBr$ ?
Bihar CECEBihar CECE 2013
Solution:
Theoretical value of dipole moment of $100 \%$ ionic character $=e \times d$
$=\left(1.6 \times 10^{-19} C \right)\left(1.41 \times 10^{-10} m \right)$
$=2.26 \times 10^{-29} \,Cm$
Observed value of dipole moment
$=2.60 \times 10^{-30} \,Cm$
$\therefore $ Percent ionic character
$=\frac{\text { observed value }}{\text { theoretical value }} \times 100 $
$=\frac{2.60 \times 10^{-30}}{2.26 \times 10^{-29}} \times 100 $
$=11.5 \%$