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I'm really struggling with finding the magnetic moment of this complex:

enter image description here

In the solutions, it says that it is 2.83 BM (which corresponds to 2 unpaired electrons). However, in drawing out the d orbital diagram (using crystal field theory) I found that there are actually 4 unpaired electrons (since there are 6 d electrons and the complex will be octahedral and high spin). Am I missing something or is the solution wrong?

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    $\begingroup$ If it's high-spin, you'd be correct. I wouldn't be entirely surprised if it's low-spin (hence 0 unpaired electrons), though: Ni(IV) is a pretty high oxidation state, and d6 low spin is nicely stabilised by CFSE (or LFSE, whatever you prefer to call it), just like in Co(III) which is essentially exclusively low-spin... $\endgroup$
    – orthocresol
    Commented Jun 28, 2020 at 13:01
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    $\begingroup$ (Are you or the question writer sure that it's not meant to be $\ce{K4[Ni(ox)3]}$?) $\endgroup$
    – orthocresol
    Commented Jun 28, 2020 at 14:28
  • $\begingroup$ Thanks very much...yes I suspect it should be K4! Thanks for confirming my thinking about the d6 configuration though :) $\endgroup$
    – MatH
    Commented Jun 29, 2020 at 9:13
  • $\begingroup$ Just to reinforce orthocresol's point with the exception of some Fe2+ complexes, and one or two fluoro complexes of Co 3+ all d6 is low spin $\endgroup$
    – Ian Bush
    Commented Jun 29, 2020 at 10:21

1 Answer 1

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If the compound is really nickel(IV), then there are only two choices: a high-spin $\mathrm d^6$ or a low-spin $\mathrm d^6$ configuration, corresponding either to 4 or 0 unpaired electrons. As higher oxidation states make low-spin configurations more likely (due to the lower energy of metal orbitals which means the energy is more similar to the corresponding ligand orbitals leading to greater stabilisation and destabilisation of bonding/antibonding orbitals; $\mathrm{t_{2g}}$ is nonbonding while $\mathrm{e_g^*}$ is antibonding if π effects are ignored, so the energy difference increases), I would expect a nickel(IV) compound to be diamagnetic rather than paramagnetic.

However, oxalate is easily oxidised to carbon dioxide and likewise transition metals in unusually high oxidation states (which is true for nickel(IV)) are typically very good oxidation agents as their reduction is preferred. Therefore, I highly doubt this complex would be stable and suggest it immediately decomposes to carbon dioxide and a nickel(II) or nickel(III) complex. To quote Wikipedia:

Ni(IV) remains a rare oxidation state of nickel and very few compounds are known to date.

Thus, I am inclined to suspect some kind of typo. Two unpaired electrons would correspond to the well-known nickel(II) $\mathrm d^8$ system and tris(oxalato)nickelate(II) would seem much more likely.

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  • $\begingroup$ I agree that it's chemically unlikely, but been wrong enough times to not want to say it. :-) $\endgroup$
    – orthocresol
    Commented Jun 29, 2020 at 10:13
  • $\begingroup$ @orthocresol At least this absolutely terrible source gives $\pu{-0.386V}$ to $\pu{-0.49V}$ for the oxalate oxidation. Finding a number for nickel(IV) is hard but $$\begin{align}\ce{NiO2(s) + 2 H+ + 2 e- &<=> Ni^2+ + 2 OH-}&E^0 = \pu{+1.59V}\end{align}$$ (Wikipedia) That seems to prove my hunch that oxalate and nickel(IV) can’t be in the same solution. $\endgroup$
    – Jan
    Commented Jun 29, 2020 at 11:22

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