Q5 / 160 · TS EAMCET 2008 · Chemistry
The bond length of $\mathrm{HCl}$ molecule is $1.275 Å$ and its dipole moment is $1.03 \mathrm{D}$. The ionic character of the molecule (in percent) (charge of the electron $=4.8 \times 10^{-10} \mathrm{esu}$ ) is
- A$100$
- B$67.3$
- C$33.6$
- Correct answerD$16.83$
Solution
Given,
observed dipole moment $=1.03 \mathrm{D}$
Bond length of $\mathrm{HCl}$ molecule, $d=1.275 Å$
$$
=1.275 \times 10^{-8} \mathrm{~cm}
$$
Charge of electron, $\quad e^{-}=4.8 \times 10^{-10}$ esu
Percentage ionic character $=$ ?
Theoretical value of dipole moment $=e \times d$
$$
\begin{aligned}
& =4.8 \times 10^{-10} \times 1.275 \times 10^{-8} \mathrm{esu}-\mathrm{cm} \\
& =6.12 \times 10^{-18} \mathrm{esu}-\mathrm{cm} \\
& =6.12 \mathrm{D}
\end{aligned}
$$
Percentage ionic character
$$
\begin{aligned}
& =\frac{\text { observed dipole moment }}{\text { theoretical value of dipole moment }} \times 100 \\
& =\frac{1.03}{6.12} \times 100 \\
& =16.83 \%
\end{aligned}
$$
Hence, the correct answer is option: 'D'.