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The dipole moment of $KCl$ is $3.336 \times 10^{-29}$ coulomb meter which indicates that it is a highly polar molecule. The inter atomic distance between $K^+$ and $Cl^{-}$ in this molecule is $2.6 \times 10^{-10}\;m$ Calculate the dipole moment of $KCl$ molecular if there were opposite charges of one fundamental unit located at each nucleus. Calculate percentage ionic character of $KCl$

$(A)\;4.1652 \times 10^{-27} \\ (B)\;4.1762 \times 10^{-29} \\(C)\;4.2345 \times 10^{-28} \\(D)\;4.1652 \times 10^{-29} $

1 Answer

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Dipole moment $\mu =S \times d$
$3.36 \times 10^{29}=S \times 2.6 \times 10^{-10}$
$S= \large\frac{3.336 \times 10^{29}}{2.6 \times 10^{-10}}$
$\qquad = 1. 283 \times 10^{-19}$ coulomb
% ionic character $=\large\frac{1.283 \times 10^{-19}}{1.602 \times 10^{-19}} $$ \times 100$
$\qquad= 80.09 \%$
If one unit charge then $S= 1.602 \times 10^{-19} C$
$\mu =1.602 \times 10^{-19} \times 2.6 \times 10^{-10}$
$\qquad= 4.1652 \times 10^{-29}\;Coulomb \;meter$
Hence D is the correct answer.
answered Apr 4, 2014 by meena.p

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