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 $1.596\mathring{A}$. What is the per cent ionic character in LiH?

• A. 75
• B. 76.8
• C. 79.8
• D. 100

Answer 1

The correct option is B 76.8

The practical dipole moment= $1.964\times {10}^{-29}$ cm

Theoretical dipole moment= = (1 electronic charge) $\times$ (interatomic distance)

= $1.602\times {10}^{-19}\times 1.596\times {10}^{-10}=2.557\times {10}^{-29}$ cm

Now the fraction of practical to theoretical dipole moment will give us the value of percentage ionic character in $LiH$

$\frac{Exp.valueofdipolemoment}{Theoreticalvalueofdipolemoment}=\frac{1.964\times {10}^{-29}}{2.557\times {10}^{-29}}$ $=0.768$ or $76.8$