Estimate the percentage ionic character of the $HBr$ molecule, given that the dipole moment $\left(\mu \right)$ is $0.72D$ and $HBr$ bond length $150pm$ 

We know,
Theoretical dipole moment $\left(\mu \right)=q\times d$ 
$=4.8\times {10}^{-10}esu\times 1.50\times {10}^{-8}\text{cm}$
$=7.2\times {10}^{-18}esucm$
Again,
$1\times {10}^{-18}esucm=1\text{Debye}$
$\therefore 7.2\times {10}^{-18}esucm=7.2D$
Percentage ionic character = $\frac{\text{observed dipole moment}}{\text{theoretical dipole moment}}\times 100=\frac{0.72}{7.2}\times 100=10\%$

Chemistry