Animated Solution for Chemistry - Coordination Compounds: The values of the crystal field stabilisation energies for a high spin $d^6$ metal ion in octahedral and tetrahedral fields respectively, are

A Tale of Two Fields

Imagine a metal ion with a $d^6$ configuration. In isolation, its five d-orbitals are perfectly degenerate, meaning they all share the exact same energy level. However, the moment ligands approach to form a coordination complex, this peaceful symmetry is shattered. The incoming negative charges of the ligands repel the electrons in the d-orbitals, causing them to split into different energy levels.

The way these orbitals split depends entirely on the geometry of the approaching ligands. In this problem, we are tasked with calculating the Crystal Field Stabilisation Energy (CFSE) for this $d^6$ ion in two distinct environments: an octahedral field and a tetrahedral field. Furthermore, we are told the complex is "high spin," which is a crucial piece of the puzzle.

Breaking the Symmetry

When six ligands approach to form an octahedral complex, they come in directly along the x, y, and z axes. The d-orbitals that point directly along these axes ($d_{x^2-y^2}$ and $d_{z^2}$) experience the most repulsion and are pushed to a higher energy level. We call this the $e_g$ set. The other three orbitals ($d_{xy}$, $d_{yz}$, $d_{zx}$) point between the axes, experience less repulsion, and drop to a lower energy level. This is the $t_{2g}$ set.

Because the problem specifies a high spin complex, we know that the splitting energy (${\Delta }_0$) is relatively small compared to the pairing energy ($P$). The electrons would rather jump to the higher $e_g$ orbitals than pair up in the lower $t_{2g}$ orbitals.

So, how do we distribute our 6 electrons? We place one in each of the three $t_{2g}$ orbitals, then one in each of the two $e_g$ orbitals. That accounts for 5 electrons. The 6th electron has no choice but to pair up in one of the $t_{2g}$ orbitals. This leaves us with $n_{t_{2g}}=4$ and $n_{e_g}=2$.

To find the CFSE, we use the principle that the "barycenter" (average energy) must remain zero. Each electron in the lower $t_{2g}$ set stabilizes the system by $-0.4{\Delta }_0$, while each electron in the higher $e_g$ set destabilizes it by $+0.6{\Delta }_0$.

Substituting our electron counts:

A Reversed Reality

Now, let's shift our perspective to a tetrahedral field. Here, four ligands approach between the axes. This completely flips the script! The orbitals pointing between the axes ($d_{xy}$, $d_{yz}$, $d_{zx}$) now experience more repulsion and are pushed higher in energy, forming the $t_2$ set. The orbitals pointing along the axes ($d_{x^2-y^2}$ and $d_{z^2}$) experience less repulsion and drop lower, forming the $e$ set.

Tetrahedral complexes are almost universally high spin because the splitting energy (${\Delta }_t$) is inherently small (roughly $\frac{4}{9}$ of ${\Delta }_0$).

Distributing our 6 electrons in this high spin scenario: we place one in each of the two $e$ orbitals, then one in each of the three $t_2$ orbitals. The 6th electron pairs up in the lower $e$ set. This gives us $n_e=3$ and $n_{t_2}=3$.

Again, we calculate the net stabilization. In the tetrahedral case, each electron in the lower $e$ set stabilizes the system by $-0.6{\Delta }_t$, and each electron in the higher $t_2$ set destabilizes it by $+0.4{\Delta }_t$.

Substituting our electron counts:

By carefully analyzing the splitting patterns and electron distributions for a high spin $d^6$ ion, we have determined the stabilization energies. For the octahedral field, the CFSE is $-0.4{\Delta }_0$. For the tetrahedral field, the CFSE is $-0.6{\Delta }_t$.

Looking at our options, this perfectly aligns with option (d). It is a beautiful demonstration of how geometry dictates the quantum mechanical behavior of transition metals!