I'd like to count the total valence electron of following neutral complex in the ionic counting (=donor-pair):
Electron count:
$$\begin{align} \ce{Cp-} : \ce{6e-} \\ \ce{CH3-}: \ce{2e-} \\ \ce{PPh3} : \ce{2e-} \end{align}$$
which gives $\ce{10e-}$ from these three ligands.
1. Question: How to decide the charge of NO
a) if I set $\ce{NO+}$ the charge of Re must be +, therefore Re(I): d6
b) if I set $\ce{NO-}$ the charge of Re must be 2+, therefore Re(III): d4
Probably only a) gives tetragonal geometry whereas b) gives square-planar geometry, correct?
2. Question: How many electrons for NO (linear/bent)
a) I expect linear NO is in ionic counting way a $\ce{2e-}$ ligand (in covalent counting (=neutral ligand method) a $\ce{3e-}$ donor).
b) I expect bent NO is in ionic counting way a $\ce{2e-}$ ligand (in covalent counting (=neutral ligand method) a 1e donor).
3. Question: Total valence electron number
I remember: When we go from linear to bent the VE number decreases with -2, therefore I expect:
a) linear goes along with $\ce{NO+}$, therefore: 10 + 6 + 2 = 18 VE
b) bent goes along with $\ce{NO-}$, therefore: 10 + 4 + 2 = 16 VE
Especially with 1. Question I feel pretty uncomfortable.