Question

The dipole moment of a molecule $AB$ is $1.6\times {10}^{-30}$ C.m . If intermolecular distance is $2.0\times {10}^{-10}$ m, then the percentage ionic character of $AB$ is :

Answer 1

The dipole moment for $100$% ionic character will be:

${\mu }_{ionic}=q\times d=1.6\times {10}^{-19}C\times 2.0\times {10}^{-10}m$

${\mu }_{ionic}=3.2\times {10}^{-29}Cm$

The percentage ionic character is $=\frac{\text{Observed dipole moment}}{\text{Dipole moment for complete ionic character}}\times 100$

$=\frac{1.6\times {10}^{-30}}{3.2\times {10}^{-29}}\times 100=5$%.

Hence, the percentage ionic character is $5$%.

Therefore, the answer is $5$.