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I assume you have come across VSEPR theory.

The central sulphur atom has 4 pairs of electrons around it (two pairs in the covalent bonds between S and H and two lone pairs that are sometimes drawn as rabbits ears!) If these four pairs repelled each other equally would form a tetrahedral angle of about 1090 ,this is the angle found in methane wheer the four pairs are identical and repel one another equally.

In H2S the two lone pairs repel more strongly and this pushes the hydrogen atoms closer together reducing the bond angle to 920

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Wiki User

11y ago
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Anonymous

Lvl 1
3y ago
The real answer is VSEPR is wrong. Lone pairs are not "repelling" each other - the difference between H2O and H2S' structure is that sulphur has larger p orbitals, so the H aren't bumping into each other as much as in water. Google water orbital diagram
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Anonymous

Lvl 1
3y ago
I should note that the lone pairs are "kind of" repelling. There are more electrons than will fit in bonding orbitals. The extra electrons are at lower energy in nonbonding orbitals (out of the way of the H). So the molecule organises to make these.

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Q: Why the bond angle of hydrogen sulphide is 92 degrees?
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