Which combinations of orbitals lead to pi bonds according to Valence Bond Theory?

Question

How many of the following combination of the orbitals will lead to formation of $\pi$-bonds with $z$ axis being the internuclear axis: $$p_x+p_x,\,p_z+p_z,\,p_y+p_y,\,d_{zx}+p_x,\,d_{zy}+p_y,\,s+p_y,\,d_{yz}+d_{zy},\,d_{zx}+d_{zx},\,d_{z^2}+s$$

My attempt/Understanding:

According to Valence Bond Theory, any orbital having any relations with $z$-axis or $z$ related planes should form $\sigma$-bonds. This is because $z$ is internuclear axis and these orbitals will have direct overlapping, forming $\sigma$-bonds.

Rest orbitals will form $\pi$-bonds by lateral overlapping, so I counted them as answer $(4)$: $$p_x+p_x,\,p_y+p_y,\,s+p_y,\,d_{xy}+d_{xy}$$ But answer given is $(6)$, and it is not mentioned which are orbitals are counted. So, please tell which orbitals are to be counted for, and tell any flaws in my understandings, if any.

Answer 1

Here are the requirements for a $\pi$ bond:

1. Each contributing orbital has to have exactly one nodal plane containing the bond axis. There can be other nodal planes that might cut through the axis, but don't contain it so those planes don't count. Two nodal planes containing the axis, as with a $d_{xy}$ orbital, also fail because that's too many.

2. The nodal planes identified in (1) that contain the bond axis must line up, so they end up as a single nodal plane going through the bond. Draw, let us say, the $p_x-p_x$ pair, which is easy to see it's a $\pi$ bond, to see what I mean.

If you sketch the orbital combinations for each of the nine given cases you see immediately that three fail test (1) (the $s$ orbital is no-go for a nodal plane, and $p_z$ has its nodal plane oriented wrong), but the others pass both tests (1) and (2) and they are the six combinations identified by the book.

The hidden lesson here is that $d$ orbitals can form $\pi$ bonds too. You will see them do so when you study transition-metal complexes.