The d-electron configuration of $$[Ru(en)_3]Cl_2$$, and $$[Fe(H_2O)_6]Cl_2$$ respectively are:

We start with the complex $$[Ru(en)_3]Cl_2.$$ The two chloride ions are simple counter-ions carrying a charge of $$-1$$ each, while the bidentate ligand $$en$$ (ethylenediamine) is neutral. Writing the total charge balance, we have

$$\big[Ru(en)_3\big]^{2+} + 2\,Cl^- \;=\; 0.$$

So the charge on the coordination sphere is $$+2,$$ and therefore

$$\text{Oxidation state of Ru}=+2.$$

The atomic number of ruthenium is $$44,$$ and its ground-state configuration is $$[Kr]\,4d^7\,5s^1.$$ In forming $$Ru^{2+}$$ we remove first the one $$5s$$ electron and then one $$4d$$ electron (because $$s$$ electrons are lost before $$d$$ electrons of the same principal shell). Thus

$$Ru^{2+} : [Kr]\,4d^{6},$$

which means the metal ion brings $$d^{6}$$ electrons into the crystal field.

Next, we recall the crystal-field splitting pattern in an octahedral environment:

$$d\; \longrightarrow\; t_{2g}^{\,\text{lower}} + e_g^{\,\text{higher}}.$$

Whether those six electrons will all pair into the lower set or some will occupy the upper set depends on the magnitude of the crystal-field splitting energy $$\Delta_0$$ compared with the pairing energy $$P.$$ The qualitative rule is:

• If $$\Delta_0 \gt P,$$ we obtain a low-spin (strong-field) configuration, all possible electrons pairing in $$t_{2g}.$$
• If $$\Delta_0 \lt P,$$ we obtain a high-spin (weak-field) configuration, electrons occupying $$e_g$$ before pairing.

Here the ligand $$en$$ lies far to the right in the spectrochemical series, i.e. it is a strong-field ligand. Moreover, ruthenium belongs to the 4d series, for which the metal-ligand overlap is greater, automatically making $$\Delta_0$$ large. Hence we confidently take $$\Delta_0 \gt P.$$ Therefore all six electrons pair in the lower set:

$$[Ru(en)_3]^{2+} :\; t_{2g}^{6}\,e_g^{0}.$$

Now we analyse $$[Fe(H_2O)_6]Cl_2.$$ The two chloride ions again contribute a total charge of $$-2,$$ while the six water molecules are neutral. Setting up the charge balance,

$$\big[Fe(H_2O)_6\big]^{2+} + 2\,Cl^- \;=\; 0,$$

so

$$\text{Oxidation state of Fe} = +2.$$

Iron has atomic number $$26$$ with ground-state configuration $$[Ar]\,3d^6\,4s^2.$$ On ionisation to $$Fe^{2+},$$ the two $$4s$$ electrons leave, giving

$$Fe^{2+} : [Ar]\,3d^{6},$$

again a $$d^{6}$$ situation.

Water, however, is only a moderately weak-field ligand, and for a 3d metal the crystal-field splitting $$\Delta_0$$ is smaller than the pairing energy. Hence $$\Delta_0 \lt P,$$ leading to the high-spin arrangement in which electrons occupy the higher $$e_g$$ set before pairing in $$t_{2g}.$$ Distributing the six electrons according to Hund’s rule we obtain

$$[Fe(H_2O)_6]^{2+} :\; t_{2g}^{4}\,e_g^{2}.$$

Summarising the two results:

$$[Ru(en)_3]^{2+}\;:\; t_{2g}^{6}\,e_g^{0}, \qquad [Fe(H_2O)_6]^{2+}\;:\; t_{2g}^{4}\,e_g^{2}.$$

These match exactly with Option C: $$t_{2g}^6e_g^0$$ and $$t_{2g}^4e_g^2$$ respectively.

Hence, the correct answer is Option C.