Question

The bond length of $HCl$ molecule is $1.275\mathring{A}$ and its dipole moment is $1.03D$. The ionic character of the molecule (in percent) ( charge of the electron $=4.8\times {10}^{-10}esu$) is:

• A. $100$
• B. $67.3$
• C. $33.66$
• D. $16.83$

Answer 1

The correct option is D $16.83$

We know, $\mu =q\times l$

=$1.6\times {10}^{-19}C\times 1.275\left({10}^{-10}\right)m$

=$2.04\times {10}^{-29}Cm$

Actual $\mu$ of HCl = $1.03D$ = $1.03\times 3.336\times {10}^{-30}Cm$

=$0.34\times {10}^{-29}Cm$

So percentage of ionic character = $\frac{Actual\mu }{Calculated\mu }\times 100$

=$\frac{0.34\times {10}^{-29}}{2.04\times {10}^{-29}}\times 100$

=$16.667$% (approx)