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If the bond length and dipole moment of a diatomic molecule are $1.25 \mathrm{~A}$ and $1.0 \mathrm{D}$ respectively, what is the per cent ionic character of the bond?
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The correct answer is:
16.66
$\mu$ experimental $=$ Dipole moment $\times 10^{-18}$
$\mu$ theoretical $=$ Bond length $\times 4.8 \times 10^{-10} \times 10^{-8} \mathrm{esu} \times \mathrm{cm}$
Per cent ionic character $=\frac{\mu_{\text {experimental }}}{\mu_{\text {theoretical }}} \times 100$
$=\frac{1.0 \times 10^{-18} \times 100}{1.25 \times 4.8 \times 10^{-10} \times 10^{-8}}=16.66 \%$
$\mu$ theoretical $=$ Bond length $\times 4.8 \times 10^{-10} \times 10^{-8} \mathrm{esu} \times \mathrm{cm}$
Per cent ionic character $=\frac{\mu_{\text {experimental }}}{\mu_{\text {theoretical }}} \times 100$
$=\frac{1.0 \times 10^{-18} \times 100}{1.25 \times 4.8 \times 10^{-10} \times 10^{-8}}=16.66 \%$
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