Question

Compare the partial positive charge on H-atoms of $H-Cl$ with $H-I$ respectively. Given the dipole moment of $HCl$ and $HI$ is 1.03 D and 0.38 D. Also $H-Cl$ and $H-I$ bond distance are 127 pm amd 161 pm respectively.
(Given: 1 D = ${10}^{-18}\text{esu}$)

Answer 1

Let $\delta$ be the magnitude of positive charge on H atom
For $HCl$
$\delta =\frac{{\mu }_{HCl}}{d}=\frac{1.03\times {10}^{-18}}{127\times {10}^{-10}}=8.11\times {10}^{-11}\text{esu}$
For $HI$
$\delta =\frac{{\mu }_{HI}}{d}=\frac{0.38\times {10}^{-18}}{161\times {10}^{-10}}=2.4\times {10}^{-11}\text{esu}$
Now
$\frac{{\delta }_{HCl}}{{\delta }_{HI}}=\frac{8.11\times {10}^{-11}\text{esu}}{2.4\times {10}^{-11}\text{esu}}=3.3$
${\delta }_{HCl}$ is 3.3 times of ${\delta }_{HI}.$