Animated Solution for Chemistry - Coordination Compounds: LIST-I contains metal species and LIST-II contains their properties. [Given : Atomic number of $\text{Cr}=24$, $\text{Ru}=44$, $\text{Fe}=26$] Match each metal species in LIST-I with their properties in LIST-II.

The world of coordination chemistry is a fascinating interplay of electrons, ligands, and energy levels. In this problem, we are tasked with matching four distinct coordination complexes to their electronic and magnetic properties. To conquer this, we must rely on Crystal Field Theory (CFT), which beautifully explains how the $d$-orbitals of a central metal ion split in the presence of a ligand field.

Our first candidate is the hexacyanochromate(II) ion. The first step in analyzing any coordination complex is determining the oxidation state of the central metal. Let the oxidation state of Chromium be $x$. Since each cyanide (${\text{CN}}^{-}$) ligand carries a $-1$ charge, we can set up the equation:

Neutral Chromium has an anomalous electronic configuration of $[\text{Ar}]3d^{5}4s^1$. Removing two electrons (one from $4s$ and one from $3d$) gives us the ${\text{Cr}}^{2+}$ ion with a $3d^4$ configuration. This immediately tells us that it is a $d^4$ species, matching property (T).

Now, we must consider the ligand. Cyanide is a strong field ligand (SFL). In an octahedral field, the five degenerate $d$-orbitals split into a lower energy $t_{2g}$ set and a higher energy $e_g$ set. Because ${\text{CN}}^{-}$ is a strong field ligand, the crystal field splitting energy (${\Delta }_o$) is greater than the pairing energy ($P$). Consequently, the four electrons will pair up in the lower energy $t_{2g}$ orbitals rather than jumping to the $e_g$ orbitals.

The resulting configuration is $t_{2g}^4{e}_g^0$. Because the electrons are paired as much as possible, this is a low spin complex, matching property (R). Furthermore, we can clearly see that the $t_{2g}$ orbitals contain exactly 4 electrons, matching property (P).

Next, we examine the hexachlororuthenate(IV) ion. Let's find the oxidation state of Ruthenium: This means the metal ion is in a 4+ oxidation state, which perfectly matches property (S).

Ruthenium (atomic number 44) is located right below Iron in the periodic table, belonging to the $4d$ transition series. Its neutral configuration is $[\text{Kr}]4d^{7}5s^1$. Losing four electrons leaves it with a $4d^4$ configuration. Thus, it is also a $d^4$ species, matching property (T).

Here is where a crucial concept comes into play: Metals in the $4d$ and $5d$ transition series almost always form low spin complexes, regardless of the ligand's strength. This is because their larger $d$-orbitals interact more strongly with ligands, leading to a very high crystal field splitting energy (${\Delta }_o>P$). Therefore, even though chloride (${\text{Cl}}^{-}$) is typically a weak field ligand, the complex will be low spin.

The electrons pair up in the lower orbitals, giving a $t_{2g}^4{e}_g^0$ configuration. This confirms it is a low spin complex (matching R) and that the $t_{2g}$ orbitals contain 4 electrons (matching P).

Our third complex is the hexaaquachromium(II) ion. Just like in the first complex, Chromium is in a $+2$ oxidation state, meaning it is a $3d^4$ ion. This makes it a $d^4$ species, matching property (T).

However, the ligand here is water (${\text{H}}_2\text{O}$), which acts as a weak field ligand (WFL) for ${\text{Cr}}^{2+}$. The splitting energy is less than the pairing energy (${\Delta }_o<P$). Instead of pairing up in the $t_{2g}$ level, the fourth electron will jump to the higher energy $e_g$ level. The configuration is $t_{2g}^3{e}_g^1$.

Because the electrons are spread out, this is a high spin complex. We have exactly 4 unpaired electrons ($n=4$). We can calculate the spin-only magnetic moment ($\mu$) using the formula: This calculated magnetic moment perfectly matches property (Q).

Finally, we look at the hexaaquairon(II) ion. The oxidation state of Iron is $+2$. Neutral Iron has a configuration of $[\text{Ar}]3d^{6}4s^2$, so ${\text{Fe}}^{2+}$ has a $3d^6$ configuration.

Water is a weak field ligand, so the complex will be high spin. The six electrons will fill the orbitals following Hund's rule: the first three go into the $t_{2g}$ orbitals, the next two go into the $e_g$ orbitals, and the sixth electron pairs up in one of the $t_{2g}$ orbitals. The final configuration is $t_{2g}^4{e}_g^2$.

Notice that the $t_{2g}$ orbitals contain exactly 4 electrons, which matches property (P). Additionally, there are 4 unpaired electrons (two in $t_{2g}$ and two in $e_g$). Using our magnetic moment formula again: This matches property (Q).

By systematically applying Crystal Field Theory, we have successfully mapped each complex to its properties: I matches P, R, T II matches P, R, S, T III matches Q, T IV matches P, Q

This problem beautifully demonstrates how the interplay between the metal's identity, its oxidation state, and the nature of the ligand dictates the fundamental properties of coordination compounds.