1. Determine the metal ion configuration

In $[Cr(H_2O)_6]^{3+}$, water is a neutral ligand, so the oxidation state of chromium is $+3$.

Chromium: $Z=24$

Neutral chromium configuration: $[Ar]3d^54s^1$

For $Cr^{3+}$, remove three electrons:

• one from $4s$
• two from $3d$

So, $Cr^{3+} = [Ar]3d^3$

2. Geometry and crystal field splitting

$[Cr(H_2O)_6]^{3+}$ is a hexaaqua complex with coordination number $6$, so it has octahedral geometry.

In an octahedral field, the five $d$-orbitals split into:

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

Thus, orbitals within each set are degenerate.

3. Check the options

• A: $d_{x^2-y^2}$ and $d_{xz}$
  These belong to different sets ($e_g$ and $t_{2g}$), so not degenerate.

• B: $d_{xz}$ and $d_{yz}$
  Both belong to the $t_{2g}$ set, so they are degenerate.

• C: $d_{yz}$ and $d_{z^2}$
  These belong to different sets, so not degenerate.

• D: $d_{x^2-y^2}$ and $d_{xy}$
  These belong to different sets, so not degenerate.

4. Conclusion

The degenerate orbitals are: $d_{xz} \text{ and } d_{yz}$

So the correct option is B.