Chemistry - Why are pi bonds only formed when sigma bonds are formed?

Solution 1:

A $\pi$ bond has a plane of symmetry along the bond axis. It cannot be formed by s-orbitals; it needs at least p-orbitals to be created. $90\,\%$ of all bonds described some time or another are somehow involving carbon, nitrogen or oxygen. (In fact, I probably underestimated). But these elements can only use p-orbitals to create $\pi$ bonds. To do that, one needs a p-orbital that is ortohogonal to the bond axis. So you run into the problem that you have an orbital pointing in one direction, but want to bond into another direction — hardly optimal, especially since there likely is already another orbital pointing in the direction you need to give a $\sigma$ bond.

Transition metals can use d-orbitals for $\pi$ bonding. They can actually point towards the atom they want to bond with so there is a greater chance of using them due to higher overlap. However, there will usually also be a different orbital pointing directly in the bonding direction which again will bond earlier and would give a $\sigma$ bond.

Solution 2:

$\sigma$-bonding MOs tend to have lower energy than $\pi$-bonding MOs so they will be formed first. One explanation is that the $\sigma$-bonding MOs have a lot of $s$-AO character and $s$-AOs have lower energy than $p$-AOs.

I don't know of any double bond that is purely $\pi$-bonding. Basic MO theory suggest that $\ce{C2}$ should have a double bond made of two $\pi$-bonding MOs because the $\sigma$-anti-bonding MO is lower in the energy than the $\pi$-bonding MOs. However, experiment tells us that $\ce{C2}$ has two unpaired electrons so this simple picture cannot be right. This is explained in greater detail here and here.