Question

The dipole moment of HBr is $2.60\times {10}^{-30}$ C-m and the
interatomic spacing is 1.41. What is the percentage ionic character of HBr.

Answer 1

$Observeddipole=2.6\times {10}^{-30}a=1.41\mathring{A}=1.41\times {10}^{-10}mObservedcharge=\frac{P}{a}=\frac{2.6\times {10}^{-30}}{1.41\times {10}^{-10}}=1.84\times {10}^{-20}\%ionic=\frac{Observedcharge}{Original}\times 100=\frac{1.84\times {10}^{20}}{1.6\times {10}^{-19}}\times 100=11.5$