Instability of heavy hydrides

Question:

Why do hydrogen compounds from the third period down (that are electron deficient or have completed octets) readily decompose?

Thoughts:

Apparently, heavier group 14 elements form hydrides which decompose readily because they have valence orbitals with which $\ce{O_2}$ and $\ce{H_2O}$ can coordinate. But I've tried and tried and cannot think of why this makes $\ce{CH_4}$ more stable than $\ce{SiH_4}$. What is the difference between the two (except for the fact that silane contains a larger central atom)?

As for the remaining groups on the periodic table, I cannot fathom how this trend holds.

• What do you mean by 'electron deficient or precise'??
– bon
Sep 10, 2015 at 10:40
• I was referring to the number of electrons required by the central atom. Methane, for instance, is electron precise, because the central carbon has a completed octet. Monomeric borane, on the other hand, is electron deficient, because the central boron only has 6 electrons surrounding it. Sep 10, 2015 at 10:46
• Ok I guessed that was the case but precise is a weird word to describe it.
– bon
Sep 10, 2015 at 10:59

1 Answer

There are two factors at work here. One is that going down within a group of the periodic table, say from C to Si to Ge, gives you more electropositive (metallic) elements, so they behave more like metal hydrides and are easier to hydrolyze. A second factor is that with increasing atomic radius as you go down to heavier elements, the overlap between hydrogen's valence orbital(s) and those of the element will be poorer, and the resulting bonds weaker.