The degenerate orbitals of [Cr(H$$_2$$O)$$_6$$]$$^{3+}$$ are:

We begin by recalling the crystal-field theory statement that in an octahedral field the five $$d$$ orbitals of the central metal ion do not remain equal in energy. The electrostatic interactions with the six surrounding ligands split them into two distinct energy sets.

1. The higher-energy set is labelled $$e_g$$ and consists of the two orbitals that point directly toward the ligands along the Cartesian axes. In symbols, those are $$d_{x^2-y^2}$$ and $$d_{z^2}$$. These two orbitals are degenerate, i.e. they have exactly the same energy within the octahedral field.

2. The lower-energy set is labelled $$t_{2g}$$ and consists of the three orbitals that lie between the axes. In symbols, those are $$d_{xy},\; d_{xz},\; d_{yz}$$. All three of these orbitals are also mutually degenerate.

For the given complex $$[ \text{Cr(H}_2\text{O})_6 ]^{3+}$$ we therefore know:

$$\text{In an octahedral field:}\quad e_g = \{\,d_{x^2-y^2},\; d_{z^2}\,\}, \qquad t_{2g} = \{\,d_{xy},\; d_{xz},\; d_{yz}\,\}$$

Now we look at each option and check whether the two orbitals listed belong to the same set (only then will they be degenerate):

A. $$d_{x^2-y^2}$$ (in $$e_g$$) and $$d_{xy}$$ (in $$t_{2g}$$) – different sets, not degenerate.

B. $$d_{z^2}$$ (in $$e_g$$) and $$d_{xz}$$ (in $$t_{2g}$$) – different sets, not degenerate.

C. $$d_{yz}$$ (in $$t_{2g}$$) and $$d_{z^2}$$ (in $$e_g$$) – different sets, not degenerate.

D. $$d_{xz}$$ and $$d_{yz}$$ – both are in the $$t_{2g}$$ set, so they are degenerate.

Only Option D lists two orbitals that reside in the same crystal-field energy level, therefore only Option D gives a pair of degenerate orbitals.

Hence, the correct answer is Option 4.