Which of these ions will have a zero crystal field splitting energy in an octahedral complex?

$\left(1\right) {\text{ Fe}}^{3 +}$ in a low-spin system (2)" Fe"^(3+ in a high-spin system $\left(3\right) {\text{ Cr}}^{3 +}$ in a low-spin system $\left(4\right) {\text{ Cr}}^{3 +}$ in a high-spin system

Nov 26, 2015

${\text{Fe}}^{3 +} \to$ high spin

Explanation:

Right from the start, you can eliminate options (1) and (3) because those ions form a low spin complex.

Now, I will not go into too much detail about crystal field theory in general because I assume you're familiar with it.

So, you know that transition metal ions placed in symmetric fields have degenerate d-orbitals that are higher in energy than they would have been in an isolated cation.

When you place such a cation in a field with octahedral symmetry, the five degenerate d-orbitals will split into two ${e}_{g}$ orbitals, which are higher in energy, and three ${t}_{2 g}$ orbitals, which are lower in energy. The crystal field stabilization energy, or CFSE, $\Delta$, is defined as the stability gained by the ion after placing it in a crystal field.

The idea here is that electrons placed in the ${t}_{2 g}$ orbitals will increase the stability of the ion because these orbitals are lower in energy than the degenerate d-orbitals.

On the other hand, electrons placed in the ${e}_{g}$ orbitals will reduce the stability of the ion because the orbitals are higher in energy than the degenerate d-orbitals.

A low field complex is characterized by the fact that the electrons found in the d-orbitals are all placed in the lower energy ${t}_{2 g}$ orbitals. This means that the CFSE for such an ion cannot be equal to zero, since the ion is gaining stability relative to the initial energy level of the d-orbitals. However, a high spin complex has the potential of having a zero CFSE if the increase in stability resulting from the electrons being placed in the ${t}_{2 g}$ orbitals is cancelled out by the decrease in stability resulting from electrons being placed in the ${e}_{g}$ orbitals.

So, your two candidates are ${\text{Fe}}^{3 +}$ and ${\text{Cr}}^{3 +}$, which have the following electron configurations

$\text{Fe"^(3+): ["Ar"] 3d^5 " }$ and " " "Cr"^(3+): ["Ar"] 3d^3

As you can see, the chromium(III) cation only has three electrons in its d-orbitals, which means that it cannot form a high-spin complex. All three electrons will thus be placed in the lower energy ${t}_{2 g}$ orbitals.

Let's do the calculations for the iron(III) cation to make sure that its CFSE will indeed be equal to zero.

So, the iron(III) cation will have three electrons in the lower energy ${t}_{2 g}$ orbitals and two electrons in the higher energy ${e}_{g}$ orbitals.

In an octahedral; field, for every electron placed in a ${t}_{2 g}$ orbital, you need to add $\frac{2}{5} \cdot {\Delta}_{\text{oct}}$ to the total. At the same time, for every electron placed in the ${e}_{g}$ orbitals, you need to subtract $\frac{3}{5} \cdot {\Delta}_{\text{oct}}$ from the total.

This means that you have

Delta = overbrace( 3 xx 2/5 Delta_"oct")^(color(blue)("3 electrons in " t_(2g))) - overbrace(2 xx 3/5Delta_"oct")^(color(red)("2 electrons in " e_g)) = 0

The answer will indeed be (2) ${\text{Fe}}^{3 +} \to$ high spin