Chemical Bonding - Result Question 23
23. The dipole moment of KCl is $3.336 \times 10^{-29} \mathrm{C}$ -m which indicates that it is a highly polar molecule. The interatomic distance between $\mathrm{K}^{+}$ and $\mathrm{Cl}^{-}$ in this molecule is $2.6 \times 10^{-10} \mathrm{m}$. 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, 2M)
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Answer:
Correct Answer: 23. $80.2 %$
Solution:
- Dipole moment is calculated theoretically as
$ \mu=q \cdot d $
$ \text { Here, } q=1.6 \times 10^{-19} \mathrm{C} \text { and } d=2.6 \times 10^{-10} \mathrm{~m} $
$ \mu_{\text {Theo }}=1.6 \times 10^{-19} \times 2.6 \times 10^{-10}=4.16 \times 10^{-29} \mathrm{~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 $
$ =80.2 %$
BETA
