[RaceMer]

Okay so I'm supposed to find the hybridization of the underlined atoms in the pic above (I circled and numbered them. Hopefully it doesn't get too confusing).

The think I'm confused about is what to do with situation like the one I labeled "1". I first thought that there was 3 charge clouds, 2 for the lone pairs and one for the bond with the other atom. So that would mean a sp^2 hybridization yes? But the H in the OH confuses me. Does that count as a bond? Would it be sp^3?

The answers I got were: 1) sp^2, 2) sp^2, 3) sp, 4) sp^2. I have no idea if I'm just beyond confused though.

If the picture is too confusing, let me know.

[UG]

The think I'm confused about is what to do with situation like the one I labeled "1". I first thought that there was 3 charge clouds, 2 for the lone pairs and one for the bond with the other atom. So that would mean a sp^2 hybridization yes? But the H in the OH confuses me. Does that count as a bond? Would it be sp^3?

Yes, the H is bonded to the O atom so there are actually 2 lone pairs and 2 bonds.
Same thing for the nitrogen atom, it has 3 bonds and 1 lone pair. The other two are correct

[JustinCh3m]

If you're just trying to appoint atoms' hybridization as sp, sp2, sp3, dsp3, or d2sp3, it's relatively easy if you go by this method:

Look at the atom whose hybridization you're trying to designate.  Then "go around" the atom saying this to yourself for every bond (single,double,triple - doesn't matter - that's "one attachment") and for every lone-pair:  "s, p1, p2, p3, d1, d2"

As you're counting via: "s, p1, p2, p3, d1, d2", stop when you've counted all of the attachments.  for example: 2 attachments would be "s, p1" - WHICH MEANS: "sp hybridization."  Another example: 5 attachments would be "s, p1, p2, p3, d1" == "dsp3 hybridization"