What is the $\%$ ionic character of $\mathrm{HCl}$ if its observed dipole moment is $1.03 \mathrm{D}$,…
What is the $\%$ ionic character of $\mathrm{HCl}$ if its observed dipole moment is $1.03 \mathrm{D}$, electronic charge (q) is $4.8 \times 10^{-10}$ e.s.u. and distance (d) between atoms is $1.27 Å$ ?
$63 \%$ ionic
$17 \%$ ionic
$83 \%$ ionic
$89 \%$ ionic
Solution
$\begin{aligned} \mu_{\text {ionic }} &=\mathrm{q} \times \mathrm{d} \\ &=4.8 \times 10^{-10} \mathrm{esu} \times 1.27 \times 10^{-8} \mathrm{~cm} \\ & \Rightarrow 6.096 \times 10^{-18} \mathrm{esucm} \end{aligned}$
\% Ionic character
$=\frac{\text { Actual dipole moment of the bond } \times 100}{\text { dipole moment of a pure ionic bond }}$
$\Rightarrow \frac{1.03 \times 10^{-18}}{6.096 \times 10^{-18}} \times 100=17 \%$
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