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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?

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What causes orbital hybridization?

Hybridization only occurs in response to bond formation, that is an atom will not just spontaneously hybridize on its own. Further, an atom will only hybridize if hybridization leads to the formation of a more stable (lower energy) molecule.

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.

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  • $\begingroup$ I'm sorry, I should have been more clear in my question. I (unjustifiably) presumed readers would regard the disposition of atoms to configure themselves most stably as common knowledge. I've edited my question. $\endgroup$ – Hal Jun 1 '15 at 0:09
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    $\begingroup$ 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 ($\ce{PH3}$). $\endgroup$ – ron Jun 1 '15 at 0:40
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    $\begingroup$ Whichever configuration (hybridized or unhybridized) has the strongest bonds, the smallest steric repulsion, etc. (e.g. the lowest energy) is the configuration that will be adopted. $\endgroup$ – ron Jun 1 '15 at 0:51

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