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

I was doing some research on Wikipedia and I came across the article on non-innocent ligands. 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.