The values of the crystal field stabilization energies for a high spin $$d^6$$ metal ion in octahedral and tetrahedral fields, respectively, are:

In crystal field theory we have two standard splitting patterns.

• For an octahedral field the three lower orbitals are designated $$t_{2g}$$ and lie at an energy of $$-0.4\,\Delta_o$$ each, while the two upper orbitals are designated $$e_g$$ and lie at $$+0.6\,\Delta_o$$ each. The crystal-field stabilisation energy (CFSE) is therefore written as

$$\text{CFSE}_\text{octa} = (-0.4\,\Delta_o)\;n(t_{2g}) + (+0.6\,\Delta_o)\;m(e_g),$$

where $$n(t_{2g})$$ and $$m(e_g)$$ are the numbers of electrons in the respective sets.

• For a tetrahedral field the relative ordering is reversed and the magnitudes are smaller. The two lower orbitals are labelled $$e$$ and lie at $$-0.6\,\Delta_t$$ each, while the three upper orbitals are labelled $$t_2$$ and lie at $$+0.4\,\Delta_t$$ each. Thus

$$\text{CFSE}_\text{tetra} = (-0.6\,\Delta_t)\;n(e) + (+0.4\,\Delta_t)\;m(t_2).$$

Now we deal with a high-spin $$d^6$$ ion.

1. Octahedral field

In a weak (high-spin) octahedral field the six electrons are filled according to Hund’s rule until every orbital of the lower set is singly occupied; only then does pairing begin:

$$t_{2g}: \uparrow\;\uparrow\;\uparrow \quad\Rightarrow\quad \uparrow\downarrow$$ $$e_g:\; \uparrow\;\uparrow$$

So we have $$t_{2g}^4\,e_g^2.$$

Substituting $$n(t_{2g}) = 4$$ and $$m(e_g) = 2$$ in the octahedral CFSE expression,

$$\text{CFSE}_\text{octa} = (-0.4\,\Delta_o)(4) + (+0.6\,\Delta_o)(2)$$ $$= -1.6\,\Delta_o + 1.2\,\Delta_o$$ $$= -0.4\,\Delta_o.$$

2. Tetrahedral field

In a weak (high-spin) tetrahedral field the lower $$e$$ set is filled first with one electron in each orbital before electrons enter the higher $$t_2$$ set; pairing is avoided until every orbital is singly occupied:

Stepwise filling gives $$e^2\,t_2^3$$ after five electrons; the sixth electron must now pair in the lower set, producing $$e^3\,t_2^3.$$

Thus $$n(e) = 3$$ and $$m(t_2) = 3.$$

Putting these numbers into the tetrahedral CFSE expression,

$$\text{CFSE}_\text{tetra} = (-0.6\,\Delta_t)(3) + (+0.4\,\Delta_t)(3)$$ $$= -1.8\,\Delta_t + 1.2\,\Delta_t$$ $$= -0.6\,\Delta_t.$$

We have therefore obtained $$\text{CFSE}_\text{octa} = -0.4\,\Delta_o$$ and $$\text{CFSE}_\text{tetra} = -0.6\,\Delta_t,$$ which exactly matches the pair of values listed in Option A.

Hence, the correct answer is Option A.