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Hybridization affects bond angle in perhaps too many ways to explain clearly. The most familiar is how, based on sp, sp2, or sp3 hybrization, bond angle is either 180 degrees (linear), 120 degrees (trigonal planar), or 109.47 degrees (tetrahedral). Those are optimal, theoretical values, and they just reflect the way that sp hybridization generates two hybrid orbitals for bonding, and that means that it's bonding to two groups, and the most distant way to spread out two groups is to put them on opposite sides of a central atom. Make sense?

All of this falls apart when you start thinking about atoms being bonded to groups of different electronegativities (including lone pairs--a lone pair is like a bond to an infinitely electropositive group). Because, you see, a central atom's orbitals will hybridize to give a lot of s-character to very lone-pair-like bonds (this is Bent's rule, approximately). So now, we don't have precisely equivalent hybrids! This is why H2S has a bond angle of around ninety degrees (also, hyperconjugation of lone pairs donating into antibonding orbitals, but whatever).

Anyway, you can compute bond angles, based on the percent s and p character of the hybrids, via Coulson's theorem.

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Q: How hybridization effect bond angle?
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