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The lone pair - OH bond repulsion in water is greater than the OH bond- OH bond repulsion. In methane all of the bonds are the same so it has perfect tetrahedral symmetry. This is VSEPR theory

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11y ago
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11y ago

The HOH bond angle in water is 104.50, the HSH bond angle in hydrogen sulfide is less at 92.10. The O-H bond length is 95.84 pm and the S-H bond length is longer at 133.6 pm.

Two explanations, the most intuitive is the VSEPR:-

VSEPR theory- this predicts that the lone pair-bonding pair repulsion is greater than bonding pair-bonding pair. This explains why the HOH angle is decreased from the ideal tetrahedral shape. In H2S the reduced electronegativivty of the S atom leads to the lone pair bing further away from the atom and causing an even greater repulsion.

Hybridisation- as you go down group 16 the tendency is for s electrons to take less part in the bonding and the angles get closer to 900 as the p "character" increases.

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11y ago

Maybe because the flourines are the most electronegative atom in Periodic Table so they pull each other closer making F-O-F bond angle smaller then H-O-H bond angle.

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Q: Why bond angle in H2S is lower then H2O?
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