# What is the electron count in this nickel complex?

Our class has started learning about electron counting using the ionic method. I was having a little difficulty, especially when there are two metals in one complex, so I looked at Wikipedia for help.

In $$\ce{(allyl)2Ni2Br2}$$ and $$\ce{(allyl)Ni(C5H5)}$$, nickel is assigned to oxidation number +2, and the electron counts are 16 and 18, respectively.
I understand that in $$\ce{(allyl)Ni(C5H5)}$$, the allyl and cyclopentadienyl ligands both have a $$-1$$ "charge", so the nickel has a +2 "charge". Thus, the electron count is: $$4 + 8 + 6 = 18$$.
However, I am struggling to calculate the electron count in $$\ce{(allyl)2Ni2Br2}$$. Using the same logic, $$\ce{Ni}$$ is in the +2 oxidation state. Since the structure (see below) does not have a metal-metal bond, we do not have to add $$1$$ when counting electrons. So, I think the electron count should be $$\frac{4*2 + 8*2 + 2*2}{2} = 14$$, not $$16$$. How does Wikipedia get that $$\ce{Ni}$$ has an electron count of $$16$$ in $$\ce{(allyl)2Ni2Br2}$$?
The bromide ions contribute an electron pair on each side of the square they form with the nickel centers (thus the bromine has a positive formal charge like a bridged bromonium ion), so in your fraction you need $$4×2$$ in the numerator where you have $$2×2$$. Then it will come out to $$16$$.