Animated Solution for Chemistry - Coordination Compounds: The correct order of the spin only magnetic moments of the following complexes is (I) $\left[\text{Cr}({\text{H}}_2\text{O}{)}_6\right]{\text{Br}}_2$ (II) ${\text{Na}}_4\left[\text{Fe}(\text{CN}{)}_6\right]$ (III) ${\text{Na}}_3\left[\text{Fe}({\text{C}}_2{\text{O}}_4{)}_3\right]({\Delta }_0>P)$ (IV) $({\text{Et}}_4\text{N}{)}_2\left[{\text{CoCl}}_4\right]$

When dealing with coordination compounds, the magnetic moment is a direct window into the electronic soul of the central metal ion. The spin-only magnetic moment is calculated using the formula $\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. To find $n$, we must dive into Crystal Field Theory (CFT) and analyze the oxidation state, the geometry, and the strength of the ligands for each complex. Let's break down the four complexes given in the problem.

The Weak Field Octahedral

Our first candidate is $[\text{Cr}({\text{H}}_2\text{O}{)}_6]{\text{Br}}_2$. First, we determine the oxidation state of chromium. Since water is neutral and there are two bromide counterions, chromium must be in the $+2$ state. The electronic configuration of ${\text{Cr}}^{2+}$ is $3d^4$.

Water (${\text{H}}_2\text{O}$) acts as a weak field ligand here, meaning the crystal field splitting energy (${\Delta }_0$) is less than the pairing energy ($P$). In an octahedral geometry, the $d$-orbitals split into a lower $t_{2g}$ set and a higher $e_g$ set. Because it's a weak field, the fourth electron will jump to the $e_g$ level rather than pairing up. This gives us a configuration of $t_{2g}^3{e}_g^1$, resulting in $4$ unpaired electrons.

Plugging this into our formula: $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90\text{ BM}$

The Strong Field Octahedral

Next, we look at ${\text{Na}}_4[\text{Fe}(\text{CN}{)}_6]$. Sodium provides a $+4$ charge, and the six cyanide ligands provide a $-6$ charge, which means iron must be in the $+2$ state to balance the complex. The electronic configuration of ${\text{Fe}}^{2+}$ is $3d^6$.

Cyanide (${\text{CN}}^{-}$) is a notoriously strong field ligand. It causes a massive splitting between the $t_{2g}$ and $e_g$ levels. The electrons are forced to pair up in the lower energy $t_{2g}$ orbitals, giving a configuration of $t_{2g}^6{e}_g^0$. With $0$ unpaired electrons, the complex is diamagnetic.

The Forced Low-Spin

Our third complex is ${\text{Na}}_3[\text{Fe}({\text{C}}_2{\text{O}}_4{)}_3]$. Here, iron is in the $+3$ oxidation state, making it a $3d^5$ system. The problem explicitly gives us a massive hint: ${\Delta }_0>P$. This mathematical inequality tells us that the splitting energy is greater than the pairing energy, so the complex will adopt a low-spin configuration regardless of the oxalate ligand's typical moderate strength.

The five electrons will crowd into the lower $t_{2g}$ orbitals, pairing up as much as possible. The configuration becomes $t_{2g}^5{e}_g^0$. This leaves exactly $1$ unpaired electron.

The Tetrahedral Twist

Finally, we examine $({\text{Et}}_4\text{N}{)}_2[{\text{CoCl}}_4]$. The tetraethylammonium cation has a $+1$ charge, so two of them give $+2$. The four chlorides give $-4$. Thus, cobalt is in the $+2$ state, which corresponds to a $3d^7$ configuration.

Notice the coordination number! There are only four ligands, making this a tetrahedral complex. In a tetrahedral field, the orbital splitting is inverted: the $e$ set is lower in energy, and the $t_2$ set is higher. Chloride is a weak field ligand, so the splitting is small. The seven electrons fill as $e^4{t}_2^3$. This arrangement leaves $3$ unpaired electrons.

Now we simply compare the calculated magnetic moments: - Complex I: $\approx 4.90\text{ BM}$ - Complex IV: $\approx 3.87\text{ BM}$ - Complex III: $\approx 1.73\text{ BM}$ - Complex II: $0\text{ BM}$

The decreasing order is clearly (I) > (IV) > (III) > (II). This problem beautifully illustrates how geometry, oxidation state, and ligand strength all intertwine to dictate the physical properties of coordination compounds.