Question

If the observed dipole moment of $LiF$ is $6.32D$, calculate percentage ionic character; given $Li-F$ bond length is $0.156$ nm.

• A. $84.4$
• B. $55$
• C. $64.6$
• D. $89$

Answer 1

The correct option is A $84.4$

${\mu }_{100\%}$= Interatomic spacing $\times$ electron charge

${\mu }_{100\%}=0.156\times {10}^{-9}\times 1.6\times {10}^{-19}C-m$
Using $1D=3.38\times {10}^{-30}{\mu }_{100\%}=1.6\ast 15.6/3.3=7.56D\%ioniccharacter=\frac{{\mu }_{observed}}{{\mu }_{100\%}}=.844$