Question

The dipole moment of $LiF$ was experimentally determined and was found to be $6.32D$. The percentage ionic character in $LiF$ molecule is
$Li-F$ bond length is $0.156$nm. (write the value to the nearest integer)

Answer 1

Dipole moment assuming 100 % ionic character would be the product of its charge and the inter atomic distance.

It would be $4.8\times {10}^{-10}esu\times 1.56\times {10}^{-8}cm=7.48\times {10}^{-18}esu.cm=7.48D$
as $1D=1\times {10}^{-18}esucm$
Percent ionic character is the ratio of the observed dipole moment to the dipole moment assuming 100% ionic character. This ratio is multiplied by $100$.

Percent ionic character $\frac{6.32}{7.48}\times 100=84.35$