Question

The dipole moment of $NaCl$ molecule is $8.4\text{D}$ and its bond length is $2.1\stackrel{\circ }{\text{A}}$. The percentage ionic character in this molecule will be:

• A. $50.25\%$
• B. $60.30\%$
• C. $83.33\%$
• D. $100\%$

Answer 1

The correct option is C $83.33\%$

Given,
${\mu }_{\text{observed}}=8.4D$
Bond length $=2.1\stackrel{\circ }{\text{A}}$
$\text{Theoretical dipole moment},{\mu }_{\text{theoratical}}=e\times \text{distance}=(4.8\times {10}^{-10})\times (2.1\times {10}^{-8})\text{e.s.u cm}$
$=10.08\times {10}^{-18}\text{e.s.u}\times \text{cm}$
$=10.08\text{D}$
Percentage ionic character $=\frac{{\mu }_{\text{observed}}}{{\mu }_{\text{theoratical}}}\times 100=\frac{8.4}{10.08}\times 100$
$=83.33\%$