Dipole moment of $\mathrm{HCl}=1.03 \mathrm{D}, \mathrm{HI}=0.38 \mathrm{D}$. Bond length of $\mathrm{HCl}=1…
Dipole moment of $\mathrm{HCl}=1.03 \mathrm{D}, \mathrm{HI}=0.38 \mathrm{D}$. Bond length of $\mathrm{HCl}=1.3 Ã…$ and $\mathrm{HI}=1.6 Ã…$. The ratio of fraction of electric charge, $\delta$, existing on each atom in $\mathrm{HCl}$ and $\mathrm{HI}$ is
$12: 1$
$2.7: 1$
$3.3: 1$
$1: 3.3$
Solution
From the definition of dipole moment,
$\mu=\delta \times d$
where, $\delta=$ magnitude of electric charge
$d=$ distance between particles (here bond length)
$\begin{array}{rlrl}\therefore & \delta =\frac{\mu}{d} \\ \text { or, } \quad \frac{\delta_{\mathrm{HCl}}}{\delta_{\mathrm{HI}}} & =\frac{\mu_{\mathrm{HCl}}}{d_{\mathrm{HCl}}} \times \frac{d_{\mathrm{HI}}}{\mu_{\mathrm{HI}}} \\ & =\frac{1.03 \times 1.6}{1.3 \times 0.38}=3.3: 1\end{array}$