Question

The dipole moment of $KCl$ is $3.336\times {10}^{-29}$ coulomb metre which indicates that it is a highly polar molecule. The interatomic distance between $K^{+}$ and ${Cl}^{-}$ in this molecule is $2.6\times {10}^{-10}m$.Calculate percentage ionic character of $KCl$.

Answer 1

Dipole moment of bond ${\mu }_B=\delta \times d$
$\therefore 3.336\times {10}^{-29}=\delta \times 2.6\times {10}^{-10}$
$\therefore \delta =\frac{3.336\times {10}^{-29}}{2.6\times {10}^{-10}}=1.283\times {10}^{-19}coulomb$
$\because 1.602\times {10}^{-19}$ charge on each, % ionic character $=$ 100
$\therefore 1.283\times {10}^{-19}$ charge on each, % ionic character
$=\frac{1.283\times {10}^{-19}}{1.602\times {10}^{-19}}\times 100=80.09$%
If one unit charge, then $\delta =1.602\times {10}^{-19}C$
$\therefore \mu =1.602\times {10}^{-19}\times 2.6\times {10}^{-10}=4.1652\times {10}^{-29}couolombmetre$