Question

The dipole moment of $NaCl$ molecule is $8.5D$ and its bond length is $2\stackrel{o}{A}$. The percentage ionic character in this molecule will be:

• A. $50.25\%$
• B. $60.30\%$
• C. $88.50\%$
• D. $100\%$

Answer 1

The correct option is C $88.50\%$

${\mu }_{theo}=e\times d=(4.8\times {10}^{-10})\times (2\times {10}^{-8})esucm=9.6\times {10}^{-18}esu\times cm=9.6D\%\text{Ionic character}=\frac{{\mu }_{exp}}{{\mu }_{theo}}\times 100=\frac{8.5}{9.6}\times 100=88.5\%$