I am doing some chemistry problems with given answers, and supposedly the $107^{\circ}$ and $92^{\circ}$ bond angles in $\ce{H_2O}$ and $\ce{H_2S}$, respectively, is due to the fact that"$\ce{O}$ uses $\ce{sp^3}$ hybrid orbitals for bonding, $S$ uses its $\ce{3p}$ orbitals."

There is quite a lot of stuff on the Internet pertaining to this exact problem. For example, the first answer on this page says that sulfur doesn't have as much electron repulsion because it's bigger, so it doesn't need to hybridize to lower its nergy. That answer assumes that hybridization in oxygen and sulfur atoms requires energy like it does in oft-used carbon, but I think that is wrong: since oxygen and sulfur don't need to promote any $ns$ electrons to $np$ orbitals (they already have filled $p$ orbitals), they can simply "mix" what they already have. Thus, why doesn't S just hybridize anyway to save a little bit of energy, and consequently have tetrahedral bond angles?

Another answer involves lone pair repulsion: perhaps both atoms hybridize, but the lone pairs on $S$ repel the $H-S-H$ bond more. I simply don't think this can have such a large effect. Even more, the other $H-X-H$ bond angles for Group 6 elements (selenium and tellurium) are around $90^{\circ}$, so it cannot be due to increased lone-pair size, as Se and Te have even bigger ones.

Because of the anomaly in $\ce{H_2O}$'s angle ($107^{\circ}$) versus the other ones (all around $90^{\circ}$), I think that the non-hybridization is more likely to be correct, so why don't sulfur atoms hybridize?

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    $\begingroup$ consider $\ce{SO4^{2-}}$ $\endgroup$ – MaxW Apr 16 '16 at 15:52
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    $\begingroup$ In principle, hybridisation is only a mathematical tool to understand bonding better. Hybridisation always results from a geometry. It is never cause for a certain geometry. $\endgroup$ – Martin - マーチン Apr 16 '16 at 16:03
  • $\begingroup$ chemistry.stackexchange.com/q/14981/16683 $\endgroup$ – orthocresol Apr 16 '16 at 16:33
  • $\begingroup$ see: chemistry.stackexchange.com/questions/14087/… $\endgroup$ – Swedish Architect Apr 16 '16 at 16:33
  • $\begingroup$ @Martin-マーチン I agree with you absolutely on that and that is why we cannot predict the hybridisation state of atoms in molecules without first observing geometry and other molecular characteristics. However, it is possible to deduce whether orbitals can hybridise. According to Mingos, an indicator for whether different orbitals can hybridise is the difference in their values of the most probable radius (r max). For the period 2 elements, Mingos says that it is due to the similarity in r max values of the 2s and 2p orbitals which allow for hybridisation. $\endgroup$ – Tan Yong Boon Dec 15 '17 at 16:59

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