# 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 '15 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 '15 at 10:46
• Ok I guessed that was the case but precise is a weird word to describe it.
– bon
Sep 10 '15 at 10:59