Jonathan Fong

**U.C.Berkeley**

M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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If you remember the orbital types, there are S, P, D and F. And so you could have different mixtures and so obviously S is not a hybrid, because it’s not a combination. But you can have Sp, Sp2 and then Sp3 it stops there. Because remember p can only have 3 orbitals and then so you can have like Sp3d etcetera.

Here are some tips and tricks for figuring out the hybridization. Say for example, if you have methane, which is CH4, so we draw it out. That’s a little structure for that. And here I have 4 bonds or you might know it as ligands. And we have 4 bonds around the carbon; 1, 2, 3, 4. And then we have no lone pairs on the carbon. We want to take a look at the hybridization of the central atoms. So if I add those two numbers then I get 4 electron pairs total, and that’s equal to the steric number.

The steric number is just the way of determining the number of electron pairs. So that’s 4 and so the hybridization of these orbitals is actually pretty easy to figure out. All I have to do is match steric number to the sum of the exponents. What that means is, I have Sp. So in Sp the sum of the exponents would be 2, and so the steric number of 2 would correspond to Sp. The steric number is 3, 1 plus 2 equals 3. So steric number of 3 would correspond to Sp2. Sp3, 1 plus 3 gives me a steric number of 4. Sp3d2, 1 plus 3 plus 1 is 5, so that would be a steric number of 5.

We take a look here and, since we have 4 for our steric number or the number of electron pairs, that means that the hybridization would be Sp3. So we would have Sp3 hybridization here. If you take a look, just to verify or double check, if we draw the orbital boxes, we would in this case since we have sp3, I have the exponents that’s 4, so I draw 4 boxes; 1, 2, 3, 4. And then since the carbon had 4 valence electrons to start with, then I would draw 4 orbitals. Remember they all need to be separated. So these represent the 4 places that they can bond. So I could have 1, 2, 3, 4, bonds. So that verifies that.

Let’s do one more example just to show you. And so say if I have Sulphur hexafluoride or SF6. I’ll draw the Sulphur and I’ll put 6 Fluorines around here. Sulphur breaks the octet rule because it’s in period 3 or below, and it has 6 valence electrons to start of with. So if you draw the new substructures for this, then we have this right here. Now we take inventory. So the sulphur which is our central atom, so we want to figure out the hybridization around the sulphur. So we count out 1, 2, 3, 4, 5, 6. So we have 6 bonds or ligands. And then around the sulphur, we have to know..........