[jerm174]

The H2S bond angle is about 93°, H2O is 104.5°.
"Do not use electronegativity in your answer."

My thoughts are: Since the central atom, S, is larger, the less the H's have to spread out. Also, the H-S bonds are less polar than the H-O bonds so therefore there's less repulsion.
Is that an acceptable answer?

[ramboacid]

Which do you think are more significant, the repulsions between the lone pairs and the hydrogen atoms or the repulsions between the two hydrogen atoms? And how would an increase in atomic radius between oxygen and sulfur affect that?

[Sophia7X]

The H2S bond angle is about 93°, H2O is 104.5°.
"Do not use electronegativity in your answer."

My thoughts are: Since the central atom, S, is larger, the less the H's have to spread out. Also, the H-S bonds are less polar than the H-O bonds so therefore there's less repulsion.
Is that an acceptable answer?

In H2S the sulfur is not really hybridized hence the close to 90° bond angle (p orbitals are orthogonal or at 90° to each other). This has something to do with sulfur's size, you are on the right track. The bond angle actually decreases when you move down the periodic table. H2O>H2S>H2Se>H2Te

I suspect that sulfur has no need to hybridize due to its size.

This is because for the larger elements, the s and p orbitals are much too different in size; and accordingly, hybridization will not necessarily lead to greater overlap (and therefore, unnecessary).

[Babcock_Hall]

The H2S bond angle is about 93°, H2O is 104.5°.
"Do not use electronegativity in your answer."

Also, the H-S bonds are less polar than the H-O bonds so therefore there's less repulsion.
Is that an acceptable answer?

It seems to me that you are trying to slip electronegativity in via the back door with this argument.