# Sp and sp2 orbital hybridisation, constructive and destructive.

Hello!

When an s and a p orbital are hybridised to give and sp orbital, two orbitals are produced. According to my chemistry book, there are two orbitals due to in-phase and out-of-phase combinations of the the s and p orbitals. When we progress to sp2 hybridisation, there are three orbitals, but no textbook I have read explains what happended to the in- and out-of-phase interactions.

I played around with different ideas. I know that two p-orbitals on different atoms don't mix when they are not aligned along the same axis because the orbitals have a constructive and destructive interaction - e.g. a px and a py orbital - so I thought that maybe this needed to be considered; otherwise, I don't see why there is not every combination. Take the s, py and pz orbitals, the combinations seem to be: the in-phase interaction of s and py and out-of-phase pz; all in-phase; s and out-of-phase px and in-phase pz; or all out-of-phase.

See what I mean? For sp, there is an in-phase or out-of-phase interaction, for the sp2 hybridisation this seems to disappear.

Thanks,
Nobahar.

chemisttree
Homework Helper
Gold Member
Maybe you should expand your search to include the term, 'antibonding orbital'.

DrDu
Maybe you should expand your search to include the term, 'antibonding orbital'.
I fear that won't help. Bonding and anti-bonding orbitals are concepts from MO theory, while he is asking about VB.

DrDu
Roughly the construction of the sp2 orbitals proceeds like this:
the tree orbitals - I call them o1, o2, o3 shall be linear combinations of one s and two p orbitals so that they are
1) orthogonal to each other
2) each orbital contains the same fraction of s and p contribution
3) the hybrid orbitals are as directional as possible.

Hence we search for an orthogonal matrix A so that (o1,o2,o3)=(s,p_x,p_y) A. The first column of the matrix is (a, b, 0)^T, as the first orbital may be taken to show into x direction. The second column is then (a, c, d)^T with a^2+bc=0 (from orthogonality) and c^2+d^2=b^2 (as the amount of p character should be the same). The third one is of the form (a,f,g)^T with f^2+d^2=b^2 and a^2+bf=0 and a^2+cf+dg=0. All in all we have one s orbital, so 3a^2=1 and two p orbitals so 3b^2=2. This already fixes the first orbital as a and b may be chosen positive. With a and b known you can solve a^2+bc=0 for c and then c^2+d^2=b^2 for d. The same way you may solve for f and d.

Condition 3) is not needed in this simple example.

Thanks Dr.Du. I'm not sure if I follow it at the moment. But when I get more time (which is never, it seems), I'll try to work through it.
Once again, many thanks.

DrDu
Sorry if my reply was to technical. The sp2 case is not the simplest one due to the trigonal symmetry. Easier is the case of four dsp2 hybrids pointing into the directions of a square.
The orbitals which make up the hybrids are s, p_x, p_y and d_x2-y2 (I will only call it "d" in the following).
Try to draw the orbitals!
Then you can convince yourself that (assuming some specific normalization of the orbitals)
s+p_x+d
s-p_x+d
s+p_y-d
s-p_y-d
are the hybrid orbitals pointing into the directions +x,-x,+y and -y, respectively. You can see that the p and d orbitals appear in deed in in phase and out of phase combination with the s orbital.