Question

The dipole moment of $NaCl$ molecule is $8.5\text{D}$ and its bond length is $2.36A^o$. The percentage ionic character in this molecule will be

• A. $50.25\%$
• B. $60.30\%$
• C. $75.03\%$
• D. $100\%$

Answer 1

The correct option is C $75.03\%$

${\mu }_{exp}=e\times d=(4.8\times {10}^{-10})\times (2.36\times {10}^{-8})\text{e.s.u cm}$
$=11.328\times {10}^{-18}\text{e.s.u}\times cm$
$=11.328\text{D}$
% ionic character $=\frac{{\mu }_{obs}}{{\mu }_{exp}}\times 100=\frac{8.5}{11.328}\times 100$
$=75.03\%$