The degenerate orbitals of [Cr(H2O)6]3+ are:

🧠 Octahedral d-Orbital Splitting

In octahedral $[\mathrm{Cr(H_2O)_6}]^{3+}$, the five d-orbitals split into two sets:

• $t_{2g}$ (lower, triply degenerate): $d_{xy}, d_{xz}, d_{yz}$.
• $e_g$ (upper, doubly degenerate): $d_{x^2-y^2}, d_{z^2}$.

Within each set, the orbitals are degenerate (same energy).

🗺️ Match to Options

| Option | Orbitals | Same set? | |---|---|---| | (1) $d_{x^2-y^2}$ and $d_{xy}$ | $e_g$ vs $t_{2g}$ | No | | (2) $d_{z^2}$ and $d_{xz}$ | $e_g$ vs $t_{2g}$ | No | | (3) $d_{yz}$ and $d_z$ | $t_{2g}$ vs (notation issue: $d_z$ probably $d_{z^2}$) | No | | (4) $d_{xz}$ and $d_{yz}$ | both $t_{2g}$ | Yes ✓ |

So $d_{xz}$ and $d_{yz}$ — both in the $t_{2g}$ set — are degenerate.

⚡ Why $t_{2g}$ Has 3 Members

$d_{xy}, d_{xz}, d_{yz}$ all have lobes pointing between the cartesian axes (not along them). In an octahedral field (ligands along ±x, ±y, ±z), these three orbitals experience identical ligand repulsion → triply degenerate.

$d_{x^2-y^2}$ and $d_{z^2}$ have lobes pointing along the axes → directly at ligands → higher energy, but their lobe-pattern symmetry makes them mutually degenerate at $e_g$.

⚠️ Don't Mix $e_g$ and $t_{2g}$

A pair like $d_{x^2-y^2}$ + $d_{xy}$ may look similar (both have lobes in the xy-plane), but $d_{x^2-y^2}$ points at axes and $d_{xy}$ points between — different sets, different energies. Always check axis-pointing.

$\boxed{\text{Answer: (4) } d_{xz} \text{ and } d_{yz}}$