NO coordination linear vs bent - valence electron

I'd like to count the total valence electron of following neutral complex in the ionic counting (=donor-pair):


![enter image description here](https://i.stack.imgur.com/YIMMb.png)


Electron count:


$$\begin{align} \ce{Cl-} : \ce{6e-} \\ \ce{CH3-}: \ce{2e-} \\ \ce{PPh3} : \ce{2e-} \end{align}$$


which gives $\ce{10e-}$ from these three ligands.




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**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




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Especially with **1. Question** I feel pretty uncomfortable.


