[MIA6]

In the absence of hybridization, the H-O-H bond angle in water would be
a) 90  b)  109.5  c) 104.5
I know the shape of H2O is bent, but how do you figure out the angle for non-hybridized H2O? My friend told me that because H2O is derived from tetrahedral (109.5 degrees), but since there are two lone pairs, so it pushes down more to 104.5, but that's after hybridized. If without hybridization, it's 90, which is a, and it's correct. But I still don't get why it's 90, not 109.5  Thanks.

[lancenti]

If you don't hybridize water, your electronic configuration for oxygen is

1s²,2s²,2p⁴

Basically, the 2s orbital and one of the 2p orbitals are filled up, leaving only two of the 2p orbitals with one electron each (let's call them x and y). To form the H-O bond, you can only use the x and y orbitals because hydrogen also has one electron. If you use the 2s or the filled 2p, you'll get either 3 electrons in one orbital which is not possible.

Since the x and y orbitals are oriented perpendicular to each other, then the two O-H bonds will be 90 degrees apart.

[MIA6]

your first sentence makes it sound like if I DO hybridize water, oxygen's electron configuration is no longer 2s2, 2p4?

[orgoclear]

hybridisation is a theoretical process that involves the functions defining the atomic orbitals. This occurs due to increase in stabilisation (decrease in potential energy)

If O in water was not hybridized, its electronic configuration would have been 1s2 2s2 2p4. Then see how you can bond the two hydrogens with the oxygen. There are two singly paired 2p electrons in the outermost shell of the oxygen (assume they are px and py) . So the O-H bond that are formed must be along x-axis and y-axis respectively so the bond angle becomes 90 degrees

If it is hybridised the configuration of O does not remain the same. It becomes 1s2 2sp(3) [6] ..