Bond angle greater in water or hydrogen sulphide? Explain me the reason.?

Question

Bond angle greater in water or hydrogen sulphide? Explain me the reason.?

Answer 1

always_nan has given you the correct experimental structural data, but hasn't offered an explanation.  Experimental result, no theoretical rationale.

jyoti makes no reference to the actual experimental values, but makes a blind prediction from his theory and calls the result "obvious".  It's correct in this case, but there are plenty of examples where his "obvious" reasoning would give him the wrong answer.  Theory without experiment.

It's also not a very good explanation.

The problems with jyoti's answer is this: if greater electronegativity attracts the bond pairs, it should also increase bond-pair bond-pair repulsion.  If the BP-BP repulsion is greater, then the angle should be larger!  It's not a useful theory if you can use exactly the same reasoning to explain exactly the opposite result.

So.  Which theory do you want to use, because we can explain this lots of different ways.

Lewis structures of H2O and H2S have eight electrons at O or S, four in bond pairs, four in lone pairs. VSEPR designation: AX2E2, predicts bent geometry, C2v symmetry, exactly as observed. The usual VSEPR explanation for the narrower angle in H2S is similar to that of jyoti's, but with more explicit consideration of the lone pair: In H2S the H atoms are farther away from each other because a S-H bonds is longer because S is a larger atom, and the S lone pairs are also much bigger because S is a larger atom, which means the BP-BP repulsions are less and the BP-LP repulsions are *greater*, and so the bonds get pushed closer together.

You asked a qualitative question, there's a perfectly good qualitative answer.

More interesting are the quantitative values: H2O is 104º, H2S is 92º, H2Se is 91º, H2Te also 91ºish. Why the sudden and dramatic jump from almost tetrahedral angle to almost orthogonal? Why does the angle never drop below 90? Can't explain that with Lewis.

Valence bond theory explains bond angles in terms of hybridization of orbitals. O has a small energy difference between the 2s and 2p orbitals, so it's not a great energy penalty to hybridize and form sp3 hybrids to both hold the lone pair and form the bonds -- indeed, the overlap is stronger with the hybrids, the bonds are stronger, so you get a big benefit by doing so. But not true with S. S-H is weaker than O-H, so the benefit to the bonds you can get from hybridization is reduced. S is bigger, Hs farther apart, steric benefit by avoiding steric repulsion is also reduced. The s-p gap is much much greater in S, so the energy penalty to hybridize is greater -- moving the lone pair out of a pure s into an ap3 hybrid isn't worth it. So you don't hybridize. H2S uses two atomic p orbitals for bonds, keeps a (nearly) 90º angle, and puts one of the LPs in a pure atomic s orbital, keeping it at as low an energy as possible.  Se and Te same argument but moreso.

There's a different argument using molecular orbitals, but it invokes symmetry mixing and a Walsh diagram. Boils down to the s-p gap and ease of mixing the s and the pz AOs to form the 2a1 HOMO. Most people stop caring long before you finish that one, though.

Answer 2

The bond angle of water is greater than the bond angle of hydrogen sulphide.

Water : 104.5 degrees

Hydrogen sulphide : 92.1 degrees

Answer 3

obviously of H2O.

oxygen is more electronegative so it attracts the bond pair. thus more lone pair -- bond pair repulsion.

while sulphur is less  electronegative.