# Why does hybridization produce a more stable configuration?

My text, and the other sources I've checked, include information on the effects, and kinds, of hybridized orbitals; however, they do not explain what properties of hybridized bonds conduce greater stability.

Why does hybridization produce a (lower energy) more stable configuration?

Nature is very efficient at exploring the energy surface and finding whatever energy minimum is available. For example, ammonia ($\ce{NH3}$) is roughly a tetrahedron with $\ce{H-N-H}$ bond angles around 109.5° (the tetrahedral angle) using (roughly) $\mathrm{sp^3}$ hybridized orbitals. For ammonia, mixing (hybridizing) it's $\mathrm{2s}$ and three $\mathrm{2p}$ orbitals to form four $\mathrm{sp^3}$ hybridized orbitals that are used to form three $\ce{N-H}$ bonds and one lone pair orbital produces a more stable molecule than if it were not hybridized. On the other hand, the analogue phosphine ($\ce{PH3}$) chooses to remain roughly unhybridized, using p-orbitals to form its $\ce{P-H}$ bonds with a resultant $\ce{H-P-H}$ bond angle around 90°. Bond strengths, steric factors, etc. all play into determining whether a hybridized or unhybridized structure will be the more stable.
• As I tried to explain in my answer, hybridization does not always produce a more stable configuration - some atoms, when they combine to form a bond, do not hybridize ($\ce{PH3}$). – ron Jun 1 '15 at 0:40