Question

The dipole moment of $NaCl$ molecule is $9\text{D}$ and its bond length is $2.5\stackrel{\circ }{\text{A}}$. The percentage ionic character in this molecule will be

• A. $50\%$
• B. $60\%$
• C. $75\%$
• D. $100\%$

Answer 1

The correct option is C $75\%$

${\mu }_{exp}=e\times d=(4.8\times {10}^{-10})\times (2.5\times {10}^{-8})\text{e.s.u cm}$
$=12\times {10}^{-18}\text{e.s.u}\times cm$
$=12\text{D}$
Percentage of ionic character $=\frac{{\mu }_{obs}}{{\mu }_{exp}}\times 100=\frac{9}{12}\times 100$
$=75\%$