Platinum Orbitals in Cisplatin Forming Bonds with LCAOs of Chlorine My inorganic chemistry professor placed two questions on the exam review that he did not cover in class; I haven't been able to figure out how to answer them and could really use some help in learning how to think about this problem.

The two questions are:



  Which orbitals of $\ce{Pt}$ can form a $\sigma$ bond with the LCAO $\left[\Psi((\ce{Cl}')_{\mathrm p_z})+\Psi((\ce{Cl}'')_{\mathrm p_z})\right]$ of the two $\ce{Cl}$ atoms in cisplatin?


  Which $\mathrm d$ orbital of $\ce{Pt}$ can form a $\pi$ bond with the LCAO $\left[\Psi((\mathrm{Cl}')_{\mathrm p_z})-\Psi((\mathrm{Cl}'')_{\mathrm p_z})\right]$ of the $\mathrm p_{z}$ orbitals of the two $\ce{Cl}$ atoms in cisplatin?



How do I arrive at the corresponding oritals?
I figured it out!

We have to start by looking at d orbital shapes:



To form bonds, the d orbitals of platinum must have the same symmetry (i.e. shape) as the molecular orbital that's approximated by the LCAO. 

We start with the bonding orbital, which looks like this: (sorry about the pencil-and-paper phone picture). 



We can see we need a d orbital with exactly one lobe in the zx plane (x is pointing to the right in the above). The only d orbital that fits the requirement is the $\textrm{dz}^{2}$.

Moving on  to the anti-bonding orbital, which I can't post because I don't have at least 10 reputation, we see that it has four lobes with two nodal planes. It looks like a $\textrm{d}_{xz}$ orbital, in fact.

We can see we need a d orbital in the zx plane that has two lobes. Clearly, this is the $\textrm{d}_{xz}$ orbital.

Easier than I thought once I got the concepts down!