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The electronic configuration of the central atom of $\ce{[MnO4]^2-}$ if given out to be $e^1\,t_2^0$

I want to know how to identify which one of the d-orbitals does this configuration utilizes because when I applied Crystal Field Theory, my answer came out to be $t_2^1\,e^0$

I already found out the oxidation number for $\ce{Mn}$ to be $+6$ and using the structure of the compound (which is tetrahedral) determined the electronic configuration as stated above but could not determine the priority order for which orbitals are to be correctly filled.

Where am I going wrong?

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    $\begingroup$ And in tetrahedral symmetry which of the orbitals has lowest energy? Is that the same as in an octahedral field? $\endgroup$ – Ian Bush Jun 3 at 10:41
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In tetrahedral geometry, considering CFT, the ligands are ions that approach from positions that are not axial in nature. This means that the $\mathrm{d}$-orbitals that are aligned along the axis are at a lower energy level compared to the orbitals that are aligned between the axes. Therefore, we notice two degenerate energy levels as shown below for tetrahedral symmetry of ligands.

From Chemistry Libretexts − Non-octahedral Complexes

The splitting of the energies of the orbitals in a tetrahedral complex ($Δ_\mathrm t$) is much smaller than that for an octahedral complex ($Δ_o$), however, for two reasons: first, the d orbitals interact less strongly with the ligands in a tetrahedral arrangement; second, there are only four negatively-charged regions rather than six, which decreases the electrostatic interactions

enter image description here

Therefore, the orbitals will be filled in the order e and then in the t2 orbitals.

Therefore since we need to fill one electron into these orbitals, we get $e^1\,t_2^0$ and not $t_2^1\,e^0$

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