3
$\begingroup$

I have recently studied ligand field theory but there is one thing I do not understand. Consider an octaedral geometry of ligands around a transition metal with a $t_{2g}$ - $e_g$ splitting of atomic $d$-orbitals due to bonding with the ligands. Suppose that the gap between the two is very large and electrons can only populate $t_{2g}$ orbitals. My question is: how do I understand how many electrons populate the $t_{2g}$ orbitals knowing the number of electrons in the atomic d orbitals? Originally I thought they were just the same, but apparently they are not! For example, in this paper https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.107.256401 the authors consider e.g. SrMnO$_3$ and they say that in this case the $t_{2g}$ is half filled with 3 electrons. However Mn has 5 valence electrons. Why is that?

I am a layman in chemistry, so please be patient if this is a silly question!

$\endgroup$

1 Answer 1

3
$\begingroup$

There is more to this question than meets the eye. First I answer with the OP's assumption that the metal ions are "low spin" — meaning because all the $t_{2g}$ orbitals are fully occupied before any of the hugher $e_g$ orbitals start to be populated. Then I will assess the accuracy of this assumption. (Hint: it's about as accurate as my typing. I had to make a lot of corrections and probably missed some.) In this discussion I assume octahedral coordination, which covers at least most oxides. I'll let the reader work out the corresponding situation with tetrahedral coordination, if that comes up.

There are three $t_{2g}$ orbitals, and in the low-spin case they would hold the first six $d$ electrons from the metal ion. Thus we render the count of $t_{2g}$ electrons based on the number of valence $d$ electrons in the metal ion:

$d^0$-$d^6$: $t_{2g}$ electrons = total $d$ electrons

$d^7$-$d^{10}$: $t_{2g}$ electrons = $6$

For instance, with $\ce{Fe^{2+}}$ we have the valence electron configuration $3d^6$, so all six of yhose $d$ electrons go into the $t_{2g}$ orbitals, filling them completely and leaving none for the $e_g$ orbitals. Thereby all electrons are paired, and there is zero spin — as low-spin as you can get.

Now how does that compare with the real world? If iron(II) were really low-spin in its oxide $\ce{FeO}$, then the oxide would be diamagnetic with no unpaired electrons, or at most only weakly paramagenetic (actual "$\ce{FeO}$" is not quite stoichiometric and contains a littke iron(III)).

Not so: Wikipedia renders a magnetic moment of $+7200×10^{−6}$ cm$^3$/mol, making it powerfully paramagnetic. Such a result is consistent instead with the "high-spin" model, where both $t_{2g}$ and $e_g$ get the first electron before any orbital gets another (thus maximizing unpaired electrons and total spin). So, at least for $3d$-valence metals, we don't get any more than the first three electrons into the $t_{2g}$ orbitals until the $e_g$ ones get a share:

$d^0$-$d^3$: $t_{2g}$ electrons = total $d$ electrons

$d^4$-$d^5$: $t_{2g}$ electrons = $3$ as these two electrons go into the $e_g$ orbitals

$d^6$-$d^8$: $t_{2g}$ electrons = total $d$ electrons minus the two that went into $e_g$

$d^9$-$d^{10}$: $t_{2g}$ electrons = $6$

We should expect this in $3d$ oxides because the pi-donor character of the oxide ligands tends to lower the $t_{2g}–e_g$ splitting.

Now we calculate that $\ce{Fe(II)}$ with six total $d$ electrons in $\ce{FeO}$ has only four electrons in the three $t_{2g}$ orbitals and two in the two $e_g$ orbitals — giving four unpaired electrons that match well with the observed paramagnetism of the oxide.

Thus still isn't the whole story. Ionic $3d$ transition metal oxides are indeed high-spin undrr the ambient pressure conditions in our everyday experience, but they may turn into low-spin under the gigapascal pressures deep inside Earth, with significant implications for our planetary interior and thus our planet's geological activity. See Sherman[1] for more details.

Reference

  1. Sherman, D.M. (1988). "High-Spin to Low-Spin Transition of Iron(II) Oxides at High Pressures: Possible Effects on the Physics and Chemistry of the Lower Mantle". In: Ghose, S., Coey, J.M.D., Salje, E. (eds) Structural and Magnetic Phase Transitions in Minerals. Advances in Physical Geochemistry, vol 7. Springer, New York, NY. https://doi.org/10.1007/978-1-4612-3862-1_6
$\endgroup$
5
  • 1
    $\begingroup$ Thanks for your very clear answer :) With this logic I expect SrVO$_3$ to be $t_{2g}^3$, but apparently they say it's $t_{2g}^1$, how is that possible? $\endgroup$
    – Matteo
    Commented Oct 31, 2023 at 1:17
  • 1
    $\begingroup$ Vanadium works out to an oxidation state of $+4$. That leaves only one $d$ electron with vanadium. So you can have only $t_{2g}^1$. $\endgroup$ Commented Oct 31, 2023 at 1:33
  • 1
    $\begingroup$ This is where I get confused: 1. the population of $t_{2g}$ orbitals also depends on the oxidation number of the metal, is that correct? 2. You say "that leaves only 1 $d$ electron with vanadium": that's because you are also counting the two electrons in the $4s^2$ shell, correct? $\endgroup$
    – Matteo
    Commented Oct 31, 2023 at 8:49
  • 1
    $\begingroup$ Vanadium starts out with only five valence electroms. $+4$ means only one is left, period. $\endgroup$ Commented Oct 31, 2023 at 9:06
  • $\begingroup$ Ok, I think I got it now. Thanks again $\endgroup$
    – Matteo
    Commented Oct 31, 2023 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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