Question

The bond length of $NaCl$ bond is $2.36\times {10}^{-10}m$. The measured dipole moment for $NaCl$ is $7.552\times {10}^{-30}C.m$.The percentage ionic character of $NaCl$ is:

Answer 1

${\mu }_{cal}=e\times d$

= $1.6\times {10}^{-19}\times 2.36\times {10}^{-10}$

= $3.776\times {10}^{-29}C.m$

$\therefore \%$ ionic character = $\frac{{\mu }_{exp}}{{\mu }_{cal}}\times 100$

= $\frac{7.552\times {10}^{-30}}{3.776\times {10}^{-29}}\times 100=20\%$