Molecule AB has a bond length of 1.617A⁰ and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is (e₀= 4.802 × 10^–10 esu)

μ=dxδ
0.38 x 10^-18 =1.617 x 10^-8cm x δ
$\mathrm{\delta }=\frac{0.38\times {10}^{-18}\text{ esu }\mathrm{cm}}{1.617\times {10}^{-8} \mathrm{cm}}=0.235\mathrm{esu}\times {10}^{-10}$

4.802 x 10^-10 esu ------100% ionic
0.235x10^-10esu  -------?
$\frac{0.235\times {10}^{-10}\text{ esu }\mathrm{cm}}{4.802\times {10}^{-10} \mathrm{cm}}\times 100=0.048\times 100=4.8\%$

= 0.05