5
$\begingroup$

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

enter image description here

$\endgroup$
2
$\begingroup$

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 $16$.

$\endgroup$
2
  • $\begingroup$ "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? $\endgroup$ May 21 '20 at 16:41
  • 1
    $\begingroup$ Yes, bromine can contribute four electrons. In fact if you follow my link you see it doing that routinely in organic chemical reactions. $\endgroup$ May 21 '20 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.