Well, MO's are typically calculated from all the electrons present in the entire molecule and hence span the entire molecule. You would really have to calculate them. However, there is the concept of LCAO, linear combination of atomic orbitals that you’re likely referring to and which eases up the process:
For that, you consider each atom to provide whichever atomic orbitals you decided to hybridise and donate them into bonds with neighboring atoms. It will usually be two neighbouring atoms of which each donates one orbital. Maths tells you, though, that the number of orbitals must stay the same after the linear combination step; we’ll come back to that in a second.
First, what is linear combination? It combines the vectors *a* and *b* to a resulting vector *c* by the following means: *c* = *k* · *a* + *l* · *b* Translated into orbitals that means that you need to sum up (or subtract) the wave functions creating ‘more’ or ‘stronger’ waves.
So take to orbitals which can overlap and combine them to form a bonding orbital. However, two orbitals need to create two orbitals, not just one. Hence you need to combine them with reverse signs on one AO to give the antibonding orbital.
Hybridisation can come a priori, i.e. I know I want to create a double bond, so my carbon needs to be sp², or a posteriori, i.e. I can reduce these MO's back to single bond AO's and they look like this carbon has sp². You would usually use the former method.
Did that semi-answer your question?
Somebody tell me why the equation up there doesn’t contain italic variables …
@FISOCPP Think of the $\sigma ^*$ orbital like a p-orbital: That has two lobes of different phase although it is only populated by one or two electrons (and two only if their spins differ). It’s just a nodal plane through the bond axis. Or think of it like a 2nd order vibrating guitar string: There’s a spot in the middle where it doesn’t vibrate = in orbital terms, amplitude is zero = no chance of finding an electon = not part of the drawing.
So you take two s-orbitals to create one bonding $\sigma$ orbital and because you need to end up with two orbitals, you create another one, which, due to maths, has to have the antibonding shape with the nodal plane. But it still is one further orbital.
Every one of these orbitals can be populated by up to two electrons; if there are two, they must have antiparallel spins.
@M.A.Ramezani And markdown is working again. Weird²
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The Periodic Table
The Periodic Table