Which one of the following ions has zero CFSE (crystal field stabilisation energy) in octahedral field ? Options 1. Fe3+ (low spin) 2. Fe3+ (high spin) 3. Co2+ (low spin) 4. Cr+3 (high spin)

##"Fe"^(3+) ->## high spin

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_(2g)## 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_(2g)## 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_(2g)## 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_(2g)## orbitals is cancelled out by the decrease in stability resulting from electrons being placed in the ##e_g## orbitals.

So, your two candidates are ##"Fe"^(3+)## and ##"Cr"^(3+)##, which have the following electron configurations

##"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_(2g)## 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_(2g)## orbitals and two electrons in the higher energy ##e_(g)## orbitals.

In an octahedral; field, for every electron placed in a ##t_(2g)## orbital, you need to add ##2/5 * Delta_"oct"## to the total. At the same time, for every electron placed in the ##e_g## orbitals, you need to subtract ##3/5 * Delta_"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) ##"Fe"^(3+) ->## high spin