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
- 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
