Question

The dipole moment of $HBr$ is $1.6\times 10^{- 30}$ Coloumb-metre and inter-atomic spacing is $1\mathring{A} .$ The $\%$ ionic character of $HBr$ is (nearest integer)

Answer 1

Charge of electron $=1.6 \times 10^{-19} C$
Dipole moment of $HBr =1.6 \times 10^{-30} C \times m$
Inter-atomic space (distance) $=1\, \mathring{A}$
$=1 \times 10^{-10} m$
Percentage of ionic character in $HBr$
$=\frac{\text { Dipole moment of } HBr \times 100}{\text { inter-atomic distance } \times q} $
$=\frac{1.6 \times 10^{-30}}{1.6 \times 10^{-19} \times 10^{-10}} \times 100$
$=10^{-30} \times 10^{29} \times 100$
$=10^{-1} \times 100 $
$=0.1 \times 100$
$=10 \%$