# What's the configuration of chromium(III) in aqueous medium according to crystal field theory?

It has $\mathrm{3d^3}$ configuration, but how will we decide if they enter $\mathrm{t_{2g}}$ or $\mathrm{e_g}$ orbitals since we don't know which complex will form?

• The amount of ligands that can fit around a cation is guarded by sterical factors. The exact shape and spin of the complex formed is guarded by energy: a configuration with lowest energy is formed. (actually it is a bit more complex, read about kinetic and thermodinamic control) – permeakra Aug 28 '17 at 9:59

You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call.

In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they are lower in energy. (Look up any introduction on crystal field theory or check out the answers on this site.) In a tetrahedral complex, $\mathrm{e}$ — not $\mathrm{e_g}$, as there is no centre of symmetry, meaning the the $\mathrm{u/g}$ designators do not make sense, even though the orbitals are still often called $\mathrm{e_g}$ at introductory level — are the orbitals of choice to begin with. Again, just look at the scheme.

Thankfully, however, we can make an educated guess based on various factors.

• octahedral complexes are more common especially at introductory levels and for the ‘top middle’ of the d block elements (roughly from titanium to nickel)
• the LFSE (ligand field stabilisation enthalpy) can be calculated to $\pu{-12Dq}$ in the octahedral case and $\pu{-8Dq}$ in the tetrahedral case which twice prefers octahedral
• there is no compelling reason (size or charge balance) to consider tetrahedral to be preferable to octahedral.

Thus, the complex you are looking at will most likely be octahedral — and so it is: it is hexaaquachromium(III) $\ce{[Cr(H2O)6]^3+}$, which is octahedral.

• Btw, I think electronic spectra are good enough evidence that the complex is octahedral: since these have a centre of symmetry, the transitions are parity forbidden and tend to have extinction coefficients 1-2 orders of magnitude smaller than typical tetrahedral complexes. I'll see if I can find a spectrum – orthocresol Aug 28 '17 at 11:26
• Yes, I didn’t even go into spectroscopic details. That requires an experiment that OP maybe cannot make being restricted to paper chemistry. – Jan Aug 28 '17 at 11:31