Q13 / 120 · VITEEE 2009 · Chemistry
Dipole moment of $\mathrm{HCl}=1.03 \mathrm{D}, \mathrm{HI}=0.38 \mathrm{D}$.
Bond length of $\mathrm{HCl}=1.3 Å$ and $\mathrm{HI}=1.6 Å$. The ratio of fraction of electric charge, $\delta$, existing on each atom in $\mathrm{HCl}$ and $\mathrm{HI}$ is
- A$12: 1$
- B$2.7: 1$
- Correct answerC$3.3: 1$
- D$1: 3.3$
Solution
Dipole moment,
$(\mu)=\delta \times d$
where, $\delta=$ magnitude of electric charge $d=$ distance between particles (or bond length)
$$
\begin{array}{l}
\therefore \delta=\frac{\mu}{d} \\
\begin{aligned}
\text { or } & \frac{\delta_{\mathrm{HCl}}}{\delta_{\mathrm{HI}}}=\frac{\mu_{\mathrm{HCl}}}{d_{\mathrm{HCl}}} \times \frac{d_{\mathrm{HI}}}{\mu_{\mathrm{HI}}} \\
&=\frac{1.03 \times 1.6}{1.3 \times 0.38}=3.3: 1
\end{aligned}
\end{array}
$$
Hence, the correct answer is option: 'C'.