Why does SiH4 have unusually large bond angles?

The bond angles of $$n > 2$$ hydrides are usually around $$90^\circ$$, such as $$\ce{PH3, H2S},$$ $$\ce{H2Se},$$ $$\ce{H2Te}.$$ I understand that this is from increased $$\mathrm{p}$$ character in the bonding with these compounds.

However, for $$\ce{SiH4}$$ and $$\ce{GeH4}$$ the bond angles are actually quite close to $$109.5^\circ$$. Why do $$\ce{Si}$$ and $$\ce{Ge}$$ not typically use unhybridized $$\mathrm{p}$$ orbitals, and instead use closer to $$\mathrm{sp^3}$$ hybrid orbitals for bonding?

• Tech note: please don't use ^\text{o} in place of a degree sign ^\circ, don't enclose text in \ce{…}, and keep in mind that labels such as atomic orbitals are recommended to be upright (to avoid confusion with other variables). If you haven't seen yet, or need to refresh your memory, visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. – andselisk Jan 8 at 17:01
• Because $\ce{SiH4}$ is forced to use its $s$ orbital for four bonds and the other elements can keep that orbital relatively inert? – Oscar Lanzi Jan 8 at 17:06
• – Nilay Ghosh Jan 8 at 17:07
• – Nilay Ghosh Jan 8 at 17:08
• Basic symmetry suggests the bond in trivalent hydrides won't be the same as a perfect tetrahedron: there are three bonds and a lone pair around the central atom. But there are 4 equivalent bonds and nothing else to interfere with symmetry in silicon tetrahydride. Why would you expect them not to be the same? – matt_black Jan 8 at 22:27

In hydrides that contain two or three hydrogens connected to a central atom such as $$\ce{PH3}$$ or $$\ce{H2Se}$$, the number of bonding partners does not exceed the number of available p orbitals on the central atom. In an extreme application of Bent’s rule, the most ideal bonding situation would be to have pure p-type orbitals form the required bonds and at least one lone pair reside entirely in an s-type orbital. This allows stabilisation of the s lone pair (as an s orbital’s energy is lower than a p orbital’s one) and better bonding (as p orbitals are directional and thus offer greater overlap with hydrogen; furthermore, they are higher in energy which better matches hydrogen’s orbital energy and again allows for greater bonding energy).
This ideal situation already falls apart in the second period where the bond angles of ammonia and water are unusually large; this is because the central atoms are too small to accomodate two or three bonding partners at $$90^\circ$$ angles so the bond angles are increased in a delicate balance until the lowest overall energy is obtained. The results are hybrid orbitals with a nonzero s component.
In $$\ce{AH4}$$ molecules, the situation is different and worse yet again. There are not enough p orbitals to accomodate four bonds so the s orbital has to help. Symmetrically, the optimal arrangement is a tetrahedron and this just happens to almost perfectly match an $$\mathrm{sp^3}$$ arrangement. As silane (and methane, germane, etc.) has four identical substituents, there is no energy to be gained by reducing symmetry so the ultimate geometry is that of an ideal tetrahedron whose bond angles are $$109.5^\circ$$.
The geometry depends on the nature of the doublets around the central atom. If the doublets are all bonding, like in $$\ce{CH4},$$ $$\ce{SiH4}$$ and $$\ce{GeH4},$$ the geometry of the molecule is tetrahedral, and the angles between the bonds are 109.5°. If one or two of these doublets are non-bonding, they are more cumbersome than covalent bonds. They repel the bonding electrons, and the angles between these bonding electrons are smaller. That is the case for $$\ce{NH3}$$ and $$\ce{PH3}$$, which have one pair of non-bonding electrons. It is even worse for $$\ce{H2O},$$ $$\ce{H2S},$$ $$\ce{H2Se}$$ and $$\ce{H2Te},$$ which have two pairs of non-bonding electrons.