[jailwall]

I understand that H2S has no hybridization so the bond angle is closer to 90 than 109.5 degrees and that's the reason why H2S bond angle is less than H2O.

However, assuming hybridization occurs why would H2S have a smaller bond angle? In H2S the sp3 hybrid orbitals are bigger than H2O and so the bond pair-bond pair repulsion decreases so with just this statement, the bond pairs should be able to come closer to each other. However, the lone pair-bond pair repulsions are also smaller now so with just this statement shouldn't the bond pairs also be able to move further away from each other?

So eventually how do we know which factor outweighs the other factor? We know that H2S has a smaller bond angle so that would mean the decrease in repulsion between the two bond pairs are more significant than the decrease of repulsion between the lone pair and bond pair - but why is this so and not the other way around?

Thanks in advance for the help

[Irlanur]

So eventually how do we know which factor outweighs the other factor?

Experiment. that's THE difference between science and theology.

[Rutherford]

The bond pair-bond pair is smaller in magnitude, still the bond pair-bond pair repulsion (steric effect) between the hydrogen in water is bigger than in hydrogen sulfide because oxygen is smaller. Going down the group the bond pair-bond pair repulsion decreases in magnitude much faster than lone pair-bond pair, therefore the smaller angle. Bigger size, in this case wins over smaller electronegativity (which would imply the opposite). If the central atom becomes big enough, the bond angle won't change much downwards.

[bharath reddy]

 In H2O and H2S both central atoms have two lone pairs. If size is the factor there would be large difference in bond angles of all these molecules.
OH2>>SH2>SeH2>TeH2
104.5>92.2>91.0>89.5

The trend is not much different in other molecules expect in water. There large electronegativity difference in between oxygen and hydrogen. This leads to form hydergen bonds in water. so that two lone pairs of oxygen in water engaged in hydrogen bond. Formation of hyderogen bonds reduce the lone pair - bond pair repulsion. Thats why bond angle is greater in water when compare to hydrogen sulfide.

[Irlanur]

This leads to form hydergen bonds in water. so that two lone pairs of oxygen in water engaged in hydrogen bond. Formation of hyderogen bonds reduce the lone pair - bond pair repulsion. Thats why bond angle is greater in water when compare to hydrogen sulfide.

I think it's the same in the gas-phase (where most bond angles are acutally measured...)