Find the oxidation state and d-count of Ru in $[Ru(en)_3]Cl_2$

• The complex ion is $[Ru(en)_3]^{2+}$ because there are $2Cl^-$ outside the coordination sphere.
• Ethylenediamine $(en)$ is a neutral ligand.
• So, oxidation state of Ru is: $x+3(0)=+2 \Rightarrow x=+2$
• Ruthenium has atomic number $44$.
• Ground-state valence configuration of Ru is approximately $4d^7 5s^1$.
• On forming $Ru^{2+}$, two electrons are removed, giving: $Ru^{2+} : 4d^6$

Decide the crystal field splitting for $[Ru(en)_3]^{2+}$

• This is an octahedral complex.
• Ru is a 4d-series metal, and $4d/5d$ metals generally form low-spin octahedral complexes because $\Delta_o$ is large.
• Also, $en$ is a relatively strong-field ligand.
• Therefore, for $d^6$ low spin: $t_{2g}^6 e_g^0$

Find the oxidation state and d-count of Fe in $[Fe(H_2O)_6]Cl_2$

• The complex ion is $[Fe(H_2O)_6]^{2+}$ because there are $2Cl^-$ outside.
• Water is a neutral ligand.
• So, oxidation state of Fe is: $x+6(0)=+2 \Rightarrow x=+2$
• Iron has atomic number $26$.
• Ground-state configuration of Fe is: $[Ar]3d^6 4s^2$
• For $Fe^{2+}$, remove the two $4s$ electrons: $Fe^{2+} : 3d^6$

Decide the crystal field splitting for $[Fe(H_2O)_6]^{2+}$

• This is also octahedral.
• $H_2O$ is a weak-field ligand relative to pairing energy for $3d$ metals.
• Hence $Fe^{2+}$ $(3d^6)$ forms a high-spin complex.
• High-spin octahedral $d^6$ configuration is: $t_{2g}^4 e_g^2$

Match with the options

• $[Ru(en)_3]Cl_2 : \; t_{2g}^6 e_g^0$
• $[Fe(H_2O)_6]Cl_2 : \; t_{2g}^4 e_g^2$

Therefore, the correct option is: $\boxed{\text{C}}$

Comparison with stored correct answer

• Stored correct answer: C
• Derived answer: C
• Hence, they agree.