What are the magnetic properties of [Ni(S2C2Ph2)2]^z's 3 oxidation states?

[![enter image description here](https://i.stack.imgur.com/je5ua.png)](https://i.stack.imgur.com/je5ua.png)


I was doing some research on Wikipedia and I came across the article on [non-innocent ligands](https://en.wikipedia.org/wiki/Non-innocent_ligand). It talked about the chemical $$[\ce{Ni(S\_2C\_2Ph\_2)\_2}]^z$$ as an example. It mentioned that the transition metal complex was diamagnetic when it was neutral(z=0). I decided to figure out how many unpaired electrons the central nickel atom would have at various formal oxidation states. I found the following: 


$$
\begin{array}{ccc}
\ce{Ni(II)}&(z=-2)&=2~\text{unpaired}\\
\ce{Ni(III)}&(z=-1)&=3~\text{unpaired}\\
\ce{Ni(IV)}&(z=0)&=4~\text{unpaired}\\
\end{array}$$
I don't understand why the complex is diamagnetic when its charge is 0. The radical ligands would be coupled with each other anti-ferromagneticly but there would still be 4 unpaired electrons in the central nickel atom. Wouldn't they make the complex paramagnetic? Could someone please explain to me why it is diamagneic and tell me whether the other 2 oxidation states are para or dia-magnetic.




$\ce{Ni^{2+}}$ is a $d^8$ metal. In a square planar orbital splitting, all $8$ electrons are paired, hence it is diamagnetic. Even when the complex is neutral, the electron loss is at the ligands (by definition, non-innocent), and they are antiferromagnetically coupled, which means that their spins "cancel" each other. Since there are two, the net moment is zero.


All complexes are $\ce{Ni(II)}$, so assuming all complexes are square planar, all $d$ electrons are paired up and the magnetic moment from these is 0. Both the $z=-2$ (no radicals) and $z=0$ (antiferromagnetic coupling) area diamagnetic. The $z=-1$ complex is paramagnetic. There is a single unpaired electron in one of the ligands that can't couple to the other one, where all of them are paired up.


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