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What is the percentage ionic character of $KCl$ given the following: The dipole moment of $KCl$ is $3.336 \times 10^{-29} \text{coulomb-meter}$. The interatomic distance between $K^{+}$ and $Cl^{-}$ is $2.6 \times 10^{-10}\;m$.

$\begin{array}{1 1} 75.32 \% \\ 47.96 \% \\ 77.93 \\ 80.09 \% \end{array}$

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If we know a molecule’s dipole moment and bond length, then we can assess the degree of charge separation (ionic character) across the bond: $\delta = \large\frac{\mu}{ed}$
Given $\mu$ dipole moment for $KCl = 3.336 \times 10^{-29} \text{coulomb-meter}$,
Interatomic distance $d = 2.6 \times 10^{-10}\;\text{m}$ and for complete separation of unit charge $e = 1.602 \times 10^{-19} \text{C}$
Therefore % ionic charge, $\delta = \large\frac{\mu}{ed}$$ = 100 \% \times$$ \large\frac{3.336 \times 10^{-29}}{1.60 \times 10^{-19} \times 2.6 \times 10^{-10}}$$ = 80.09 \%$
answered Mar 22, 2014 by balaji.thirumalai
 

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