The bond angle in $H_{2} S$ is 97° and its dipole. moment is 1.5 D. The S-H bond distance is 0.15 run. Therefore, approximate percentage ionic character of S-H bond is (neglect the effect of dipole moment of lone pair on sulphur atom in $H_{2} S$ ).
$(Given : [\cos 97^{\circ} = - 0.12] and \sqrt{0.88} = 0.94)$

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32%

16%

84%

10%

answer is B.

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Detailed Solution

$1.5 = \sqrt{μ^{2} + μ^{2} + 2 μ^{2} \cos 97^{\circ}} \Rightarrow μ = 1.13 D$

Percentage ionic character of S-H bond is, $\begin{array}{l} % Ionic character = \frac{1.13 \times 10^{- 18}}{4.8 \times 10^{- 10} \times 1.5 \times 10^{- 8}} \times 100 \\ = 16 % \end{array}$

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