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Bond angle in \( H_{2} O\left(104.5^{\circ}\right) \) is higher than the bond angle of \( H_{2} S\left(92.1^{\circ}\right) . \) The difference is due to:

1. 0 is diatomic and \( S \) is tetra-atomic
2. difference in electronegativity of S and 0
3. difference in oxidation states of S and 0
4. difference in shapes of hybrid orbitals of S and

Correct answer: difference in electronegativity of S and 0

Solution

Oxygen is more electronegative than sulfur, so in H2O the bonding pairs are drawn closer to O, increasing lone-pair repulsion and giving a larger bond angle than in H2S. Sulfur’s lower electronegativity leads to less electron-pair repulsion and a smaller angle.

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