Q.The CFSE for octahedral [CoCl6]4− is 18,000 cm⁻¹. The CFSE for tetrahedral [CoCl4]2− will be :

Crystal Field Stabilization Energy (CFSE): Why the Formula Holds

Let's build this from first principles — not just memorizing numbers, but understanding why the energy changes happen.

1. The Core Idea: d-Orbitals Are Not All Equal in an Octahedral Field

In a free metal ion, all five d-orbitals have the same energy (degenerate). But when you place the ion inside an octahedral ligand field, something changes:

• Ligands (negative point charges or dipoles) approach along the $x$, $y$, and $z$ axes.
• Some d-orbitals point directly at the ligands → high repulsion → higher energy.
• Other d-orbitals point between the ligands → less repulsion → lower energy.

Which orbitals point where?

So the five d-orbitals split into two groups:

• $e_g$ set (higher energy): $d_{x^2-y^2}$, $d_{z^2}$
• $t_{2g}$ set (lower energy): $d_{xy}$, $d_{xz}$, $d_{yz}$

2. The Energy Splitting: Why $\Delta_o$ and the "Barycenter" Rule

The total energy of all five d-orbitals must be conserved — it's the same as in the free ion. This is the barycenter (center of gravity) rule.

Let:

• Energy of $t_{2g}$ orbitals = $-x$ (below barycenter)
• Energy of $e_g$ orbitals = $+y$ (above barycenter)
• The splitting energy between them = $\Delta_o$ (called $10 Dq$ in older texts)

So:

Conservation of energy:

• There are 3 $t_{2g}$ orbitals and 2 $e_g$ orbitals.
• Total energy shift = $3(-x) + 2(+y) = 0$

From this:

Substitute into $x + y = \Delta_o$:

Then:

Key result:

• Each $t_{2g}$ electron is stabilized by $\mathbf{-\frac{2}{5}\Delta_o}$
• Each $e_g$ electron is destabilized by $\mathbf{+\frac{3}{5}\Delta_o}$

3. The CFSE Formula for Octahedral Complexes

Let:

• $n_{t_{2g}}$ = number of electrons in $t_{2g}$ orbitals
• $n_{e_g}$ = number of electrons in $e_g$ orbitals

Then:

Why this is the stabilization energy:

• The negative sign means energy is lowered (stabilization).
• The positive term means energy is raised (destabilization).
• Net CFSE = how much more stable the complex is compared to the free ion.

4. The "Why" Behind Pairing Energy and High/Low Spin

When you add electrons beyond $d^3$, you face a choice:

Example: $d^4$ configuration

Option A (High spin):

• Put 4th electron in $e_g$ (higher energy)
• Cost: $+\frac{3}{5}\Delta_o$ (destabilization)
• Benefit: No pairing energy ($P$)

Option B (Low spin):

• Pair the 4th electron in $t_{2g}$
• Cost: Pairing energy $P$ (electrostatic repulsion between two electrons in same orbital)
• Benefit: Avoid $+\frac{3}{5}\Delta_o$ destabilization

The decision rule:

• If $\Delta_o > P$ → Low spin (pairing is cheaper than going to $e_g$)
• If $\Delta_o < P$ → High spin (going to $e_g$ is cheaper than pairing)

CFSE for low-spin $d^4$:

CFSE for high-spin $d^4$: …