%S character of bonding orbital of sulphur in H2S is (bond angle H-S-H=92°, cos 92° = -0.035)

cos⁡92∘=SS−1−0.035=SS−1⇒−0.035S+0.035=S0.035=S(1.035)S=0.0351.035=0.0338%S =3.38%

%S character of bonding orbital of sulphur in H2S is (bond angle H-S-H=92°, cos 92° = -0.035)

cos⁡92∘=SS−1−0.035=SS−1⇒−0.035S+0.035=S0.035=S(1.035)S=0.0351.035=0.0338%S =3.38%

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%S character of bonding orbital of sulphur in H₂S is (bond angle H-S-H=92°, cos 92° = -0.035)

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25%

33.33%

3.38%

20%

answer is B.

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$\begin{array}{l} \cos 92^{\circ} = \frac{S}{S - 1} \\ - 0 .035 = \frac{S}{S - 1} \Rightarrow - 0 .035 S + 0 .035 = S \\ 0 .035 = S (1 .035) \\ S = \frac{0 .035}{1 .035} = 0 .0338 \end{array}$

%S =3.38%

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%S character of bonding orbital of sulphur in H2S is (bond angle H-S-H=92°, cos 92° = -0.035)

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%S character of bonding orbital of sulphur in H₂S is (bond angle H-S-H=92°, cos 92° = -0.035)