The dipole moment of HF molecule is 1.91D and the bond distance is 0.92 Å. What is the fractional charge on H and F in the HF bond? (Electronic charge = `4.8 xx 10^(-10)` e.s.u.).

Explanation

Dipole moment = Charge x distance
`1.91 xx 10^(-18) e.s.u. cm ` = charge ` xx 0.92 xx 10^(-8) cm`
charge = `1.91 xx 10^(-18) e.s.u. cm // 0.92 xx 10^(-8) cm = 2.08 xx 10^(-10) e.s.u.`
Therefore , fraction of electron charge , d = charge/electronic charge
` = (2.08 xx 10^(-10) e.s.u)/(4.8 xx 10^(-10) e.s.u) = 0.43`
Thus `d_H^+ = + 0.43 and d_(Cl)^- = - 0.43`