Question

The dipole moment of $LiH$ is $1.964 \times 10^{-29}$ cm and the interatomic distance between $Li$ and $H$ in this molecule is $1.596\, \mathring{A}$. The percentage of ionic character in $LiH$ is

• A. $75.0$
• B. $76.8$
• C. $79.8$
• D. $100$

Answer 1

Answer: (B)

The dipole moment of $100 \%$ ionic molecule $= \left(1 {\text{electronic charge}}\right) \times\left({\text{interatomic distance}}\right) $
$= 1.602 \times10^{-19 }\times 1.596\times 10^{-10}$
$ = 2.557 \times 10^{-29} cm $
Fractional ionic character $= \frac{{\text{Exp. value of dipole moment}} }{{\text{Theoretical value of dipole moment}} } =\frac{ 1.964\times10^{-29}}{2.557\times10^{-29}}$
$ =0.768$
The bond in $LiH$ is $76.8\%$ ionic.