The internuclear distance of H-F molecule is `0.92 Å` and dipole moment is 2 debye . Calculate the percentage of ionic character of the molecule .

Explanation

If HF is 100% ionic , each atom would carry a charge equal to one unit , i.e., `4.8 xx 10^(-10)` esu. As bond length of HF is `0.92 Å` , its dipole moment for 100 % ionic character would be ,
`mu_("ionic") = 4.8 xx 10^(-10)` esu `xx 0.92 xx 10^(-8)` cm
`= 4.416 xx 10^(-18)` esu cm = 4.416 D
`therefore` % ionic character = `(2)/(4.416) xx 100 = 45.3`