(a) In octahedral crystal field, energies of which d-orbitals will be raised when ligands approach the central metal atom/ion? Give reason in support of your answer.

Q: (a) In octahedral crystal field, energies of which d-orbitals will be raised when ligands approach the central metal atom/ion? Give reason in support of your answer.

In an octahedral field the energies of the **$e_g$ orbitals — $d(x^2-y^2)$ and $d(z^2)$** — are raised. **Reason** - The six ligands approach the central metal ion **along the x, y and z axes**. - The $e_g$ orbitals, $d(x^2-y^2)$ and $d(z^2)$, have their lobes pointing **directly at the approaching ligands**, so they suffer greater electrostatic repulsion and their energy is **raised** (by $+0.6\,\Delta_0$ above the barycentre). - The $t_{2g}$ orbitals — $d(xy)$, $d(yz)$, $d(zx)$ — lie **between the axes**, experience less repulsion, and are therefore **lowered** in energy (by $-0.4\,\Delta_0$).

Q: Which d-orbitals form the $e_g$ set?

The $e_g$ set consists of $d(x^2-y^2)$ and $d(z^2)$. In an octahedral field these point directly at the ligands and are raised in energy, while the $t_{2g}$ set ($d_{xy}$, $d_{yz}$, $d_{zx}$) is lowered.

Coordination Compounds

Crystal field splitting of d-orbitals in an octahedral field

Official Answer

In an octahedral field the energies of the $e_g$ orbitals — $d(x^2-y^2)$ and $d(z^2)$ — are raised.

Reason

The six ligands approach the central metal ion along the x, y and z axes.
The $e_g$ orbitals, $d(x^2-y^2)$ and $d(z^2)$ , have their lobes pointing directly at the approaching ligands, so they suffer greater electrostatic repulsion and their energy is raised (by $+0.6\,\Delta_0$ above the barycentre).
The $t_{2g}$ orbitals — $d(xy)$ , $d(yz)$ , $d(zx)$ — lie between the axes, experience less repulsion, and are therefore lowered in energy (by $-0.4\,\Delta_0$ ).

octahedral crystal fieldeg orbitalsd(x²–y²)d(z²)t2g orbitalscrystal field splittingligand repulsionΔ₀

Marking Scheme

11 mark: correctly naming the $e_g$ orbitals $d(x^2-y^2)$ and $d(z^2)$ as the ones whose energy is raised.
21 mark: reason — these orbitals point directly at the axially approaching ligands, so greater ligand–electron repulsion raises their energy ( $t_{2g}$ lie between the axes and are lowered).

Hint

Ligands come in along the axes — the d-orbitals lying ON the axes get pushed up.

Quick Oral Answer

In an octahedral field the $e_g$ orbitals — $d(x^2-y^2)$ and $d(z^2)$ — are raised because they point straight at the ligands approaching along the axes, so they feel more repulsion, while the $t_{2g}$ orbitals between the axes are lowered.

Analysis & Explanation

Concept

Crystal Field Theory treats ligands as point negative charges. In the free ion all five d-orbitals are degenerate, but an octahedral arrangement of ligand charges lifts that degeneracy.

Orbitals pointing at the ligands ( $e_g$ : $d(x^2-y^2)$ , $d(z^2)$ ) are destabilised — energy raised.
Orbitals pointing between the ligands ( $t_{2g}$ : $d(xy)$ , $d(yz)$ , $d(zx)$ ) are relatively stabilised — energy lowered.

The energy gap between the two sets is the crystal field splitting energy, $\Delta_0$ .

Exam trap

A common slip is to name the wrong pair. Remember: 'along the axes → raised ( $e_g$ )'. The subscript in $d(x^2-y^2)$ and $d(z^2)$ is the memory hook because these expressions contain the axes directly.

Real-world

This splitting explains the colour of transition-metal complexes (d–d transitions across $\Delta_0$ ) and their magnetic behaviour (high-spin vs low-spin).

Common Mistakes

1Naming $t_{2g}$ orbitals ( $d_{xy}$ , $d_{yz}$ , $d_{zx}$ ) as the raised set — these are actually lowered in an octahedral field.
2Stating that all five d-orbitals are raised equally — degeneracy is lifted into two sets of different energy.
3Giving no reason; the mark for the reason needs the 'orbitals point along the axes toward ligands' argument.

Interesting Facts

The label ' $e_g$ ' means the pair is doubly degenerate and 'g' (gerade) indicates symmetry with respect to the centre of inversion in the octahedron.

In a tetrahedral field the pattern inverts: the e set is lowered and the $t_2$ set is raised, and the splitting $\Delta_t$ is only about $\frac{4}{9}$ of $\Delta_0$ .

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Which d-orbitals form the $e_g$ set?

The $e_g$ set consists of $d(x^2-y^2)$ and $d(z^2)$ . In an octahedral field these point directly at the ligands and are raised in energy, while the $t_{2g}$ set ( $d_{xy}$ , $d_{yz}$ , $d_{zx}$ ) is lowered.

What is $\Delta_0$ ?

$\Delta_0$ is the crystal field splitting energy in an octahedral field — the energy gap between the raised $e_g$ orbitals and the lowered $t_{2g}$ orbitals. Its size decides whether a complex is high-spin or low-spin.