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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.

The Wikipedia page for organonickel complexes says:

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 $\ce{(allyl)2Ni2Br2}$?

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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 $16$.

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  • $\begingroup$ "bromide ions contribute an electron pair on each side of the square they form with the nickel centers". What do you mean by each side of the square? There are 2 sides of the square that connect to a nickel, and if each side counts as an electron pair, doesn't that mean 2*2 = 4 electrons? $\endgroup$
    – Cyclopropane
    May 21, 2020 at 16:41
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    $\begingroup$ Yes, bromine can contribute four electrons. In fact if you follow my link you see it doing that routinely in organic chemical reactions. $\endgroup$
    – Oscar Lanzi
    May 21, 2020 at 16:44

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