MO diagram for pentaamminechlorocobalt(III)?
Let's start at the bottom. There will be three a1 bonding orbitals. First, we can have the metal $4s$ orbital interact with all six ligands in phase. Let's call this $1a_1$. Next, the metal $d_{z^2}$ can interact with all six ligands as well, but the axial ligands are now opposite phase from the equatorial. Let's call that $2a_1$. Both will of course have antibonding counterparts. The third $a_1$ orbital is an interaction of the metal $p_z$ with out of phase axial ligand orbitals. The equatorial ligands do not contribute to this orbital because they are on the nodal plane. Let's call this $3a_1$. 

Mixed in with these a1 orbitals will be the $b_1$ orbital comprised of the equatorial ligand orbitals interacting with $d_{x^2-y^2}$. That will be $1b_1$. It is lower in energy than $3a_1$ but likely higher than $1a_1$ and $2a_1$. (This $1b_1$ and the $2a_1$ correspond to the $1e_g$ orbitals of an $O_h$ complex.)

Near $3a_1$ we have the two $e$ orbitals comprised of the metal $p_x$ and $p_y$, each interacting with a pair of equatorial ligands. Let's call these $1e$. (These along with $3a_1$ correspond to the $1t_{1u}$ orbitals of an $O_h$ complex).

If we are only considering sigma interactions, we next have the $e$ and $b_1$ orbitals representing metal $d_{xy}$, $d_{xz}$, and $d_{yz}$, all of which are nonbonding for sigma. However, the pi interactions with Cl mean that the $e$ orbitals are stabilized and have antibonding destabilized counterparts. Let's call those $2e$. The $b_1$ orbital ($d_{xy}$) remains nonbonding. 

Getting the ordering exactly correct is tough in this sort of qualitative analysis, but that should get you a pretty good start.