Chemical Bonding - Result Question 23
####23. The dipole moment of $KCl$ is $3.336 \times 10^{-29} C-m$ which indicates that it is a highly polar molecule. The interatomic distance between $K^{+}$and $Cl^{-}$in this molecule is $2.6 \times 10^{-10} m$. Calculate the dipole moment of $KCl$ molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of $KCl$.
$(1993,2 M)$
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Solution:
- Dipole moment is calculated theoretically as
$$ \begin{aligned} & \mu=q \cdot d \ & \text { Here, } q=1.6 \times 10^{-19} C \text { and } d=2.6 \times 10^{-10} m \ & \quad \mu _{\text {Theo }}=1.6 \times 10^{-19} \times 2.6 \times 10^{-10}=4.16 \times 10^{-29} cm \ & % \text { ionic character }=\frac{\mu _{\text {obs }}}{\mu _{\text {Theo }}} \times 100=\frac{3.336 \times 10^{-29}}{4.16 \times 10^{-29}} \times 100 \end{aligned} $$
$=80.2 %$
