Question

The observed dipole moment for a molecule AB is 1.45 D and its bond is 1.654 $\mathring{A}$. Calculate the percentage of ionic character in the bond.

Answer 1

$\%ofIC=\frac{{\mu }_{Ovserved}}{{\mu }_{ionic}}\times 100{\mu }_{Ovserved}=1.45D=1.45\times {10}^{-30}cm{\mu }_{ionic}=1.654\times {10}^{-10}m=1.654\times {10}^{-8}cm{\mu }_{ionic}c=q\times d=1.9\times {10}^{19}\times 1.65\times {10}^{-8}=3.1426\times {10}^{-27}cm$