Question

The experimental value of the dipole moment of $HCl$ is $1.03 \, D$ . The length of the $H-Cl$ bond is $1.275 \, Å$ . The percentage of ionic character in $HCl$ is:

[Given: $1\text{D}=10^{- 18}\text{esu} \, \text{cm}$ ]

• A. 43
• B. 21
• C. 17
• D. 7

Answer 1

Answer: (C)

$\%$ Ionic character $=\frac{\text { Experimental value of dipole moment }}{\text { Theoritical value of dipolemoment }} \times 100$

$=\frac{\mu_{\operatorname{cxp}}}{\mu_{ th }} \times 100$

Where:

$\mu_{\exp }=1.030$ and $\mu_{\text {theo }}=| q | \times d$

$d =1.275 AA = 1.275 \times 10^{-8} cm$

$| q |=4.8 \times 10^{-10}$ esu

$\mu_{\text {theo }}=4.8 \times 10^{-10} \times 1.275 \times 10^{-8}$ esu $- cm$

$=4.8 \times 1.275 \times 10^{-18} esu - cm$

$=4.8 \times 1.275 D$

$1 D =10^{-18} esu cm$

$\%$ Ionic character $=\frac{1.03}{4.8 \times 1.275} \times 100=17$