Single Choice

The percentage ionic character of a bond having $$1.257A^0$$ its length and $$1.03$$D its dipole moment is: ($$q = 4.8 \times 10^{-10}$$ esu)

A$$10\%$$
B$$15\%$$
C$$16.83\%$$
Correct Answer
D$$18.88\%$$

Solution

$$\mu_{ionic}(ionic\,dipole)=4.8\times 10^{-10}\times 1.275\times 10^{-8}$$
$$\mu_{ionic}=6.12\times10^{-18} esu\, cm$$=$$6.12\,D$$
Observed dipole moment $$\mu_{observed}=1.032\,D$$
% ionic character =$$\frac{\mu_{observed}}{\mu_{ionic}}\times 100$$
=$$\frac{1.032\,D}{6.12\,D}\times 100$$
=$$16.38\%$$

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