Animated Solution for Chemistry - Coordination Compounds: The $d$-electron configuration of $[\text{Ru}(\text{en}{)}_3]{\text{Cl}}_2$ and $[\text{Fe}({\text{H}}_2\text{O}{)}_6]{\text{Cl}}_2$, respectively are

The Tale of $3d$ vs $4d$ Metals

When dealing with coordination compounds, determining the exact $d$-electron configuration is like solving a microscopic puzzle. You need to know the metal's oxidation state, its electron count, and the strength of the surrounding ligand field. Let's break down the two complexes given in our problem: $[\text{Ru}(\text{en}{)}_3]{\text{Cl}}_2$ and $[\text{Fe}({\text{H}}_2\text{O}{)}_6]{\text{Cl}}_2$.

The $4d$ Rule

First, let's look at $[\text{Ru}(\text{en}{)}_3]{\text{Cl}}_2$. The two chloride counter ions give the complex sphere a $+2$ charge. Since ethylenediamine (en) is a neutral ligand, the Ruthenium atom must be in a $+2$ oxidation state.

Ruthenium is a $4d$ transition metal, located right below Iron in the periodic table. Therefore, ${\text{Ru}}^{2+}$ has a $4d^6$ electron configuration. Now, 'en' is a strong field ligand, but there is a more dominant rule at play here. Metals from the $4d$ and $5d$ series always form low-spin complexes, regardless of the ligand. Their larger $d$-orbitals interact more strongly with ligands, resulting in a massive crystal field splitting energy (${\Delta }_o$).

Because ${\Delta }_o>P$ (pairing energy), all six electrons will pair up in the lower energy $t_{2g}$ orbitals, leaving the $e_g$ orbitals completely empty.

This gives us the configuration: $t_{2g}^6{e}_g^0$.

The Weak Field Case

Next, we analyze $[\text{Fe}({\text{H}}_2\text{O}{)}_6]{\text{Cl}}_2$. Water is a neutral ligand, and with two chloride ions outside, Iron is also in a $+2$ oxidation state. Iron is a $3d$ metal, so ${\text{Fe}}^{2+}$ has a $3d^6$ configuration.

Water is a classic weak field ligand. This means the splitting energy it causes is relatively small compared to the energy required to pair electrons up in the same orbital (${\Delta }_o<P$).

Following Hund's rule, the electrons will prefer to occupy all available orbitals singly before pairing. We place three electrons in the $t_{2g}$ orbitals, two in the $e_g$ orbitals, and the final sixth electron is forced to pair up in one of the $t_{2g}$ orbitals.

This results in a high-spin configuration: $t_{2g}^4{e}_g^2$.

Comparing our findings, the Ruthenium complex has a $t_{2g}^6{e}_g^0$ configuration, while the Iron complex has a $t_{2g}^4{e}_g^2$ configuration. This perfectly matches option (b).

Beyond just the configuration, this tells us a lot about the physical properties of these materials. The Ruthenium complex has zero unpaired electrons, making it diamagnetic. In stark contrast, the Iron complex has four unpaired electrons, making it highly paramagnetic. Always remember to check the metal's series—it can completely override the nature of the ligand!