Animated Solution for Chemistry - Coordination Compounds: The calculated magnetic moments (spin only value) for species $[{\text{FeCl}}_4{]}^{2-}$, $[\text{Co}({\text{C}}_2{\text{O}}_4{)}_3{]}^{3-}$ and ${\text{MnO}}_4^{2-}$ respectively are

Spin-Only Magnetic Moment

When dealing with transition metal complexes, one of the most revealing properties is their magnetic moment. The spin-only magnetic moment, denoted by $\mu$, is a direct window into the electronic structure of the central metal ion. It is calculated using the elegant formula:

Here, $n$ represents the number of unpaired electrons in the d-orbitals, and the result is expressed in Bohr Magnetons (BM). To find $n$, we must embark on a journey through Crystal Field Theory (CFT), determining the oxidation state, the geometry of the complex, and the strength of the ligands involved.

Decoding the Tetrachloroferrate Ion

Let's begin with our first candidate: the $[{\text{FeCl}}_4{]}^{2-}$ ion.

First, we determine the oxidation state of iron. Since each chloride ion carries a $-1$ charge, we set up the equation $x+4(-1)=-2$, which gives us $x=+2$. A neutral iron atom has the electronic configuration $[\text{Ar}]3d^{6}4s^2$. Stripping away two electrons to form ${\text{Fe}}^{2+}$ leaves us with a $3d^6$ configuration.

Next, we look at the geometry and the ligand. With four chloride ligands, the complex adopts a tetrahedral geometry. Chloride is notoriously known as a weak field ligand in the spectrochemical series. In a tetrahedral field, the d-orbitals split into a lower energy $e$ set and a higher energy $t_2$ set. Because the crystal field splitting energy (${\Delta }_t$) is small, the electrons prefer to occupy higher energy orbitals before pairing up.

Distributing the six electrons, we get the configuration $e^3{t}_2^3$. This arrangement leaves us with exactly 4 unpaired electrons ($n=4$). Plugging this into our formula:

The Oxalate Anomaly with Cobalt

Our second complex is the trioxalato cobaltate ion, $[\text{Co}({\text{C}}_2{\text{O}}_4{)}_3{]}^{3-}$.

Oxalate (${\text{C}}_2{\text{O}}_4^{2-}$) is a bidentate ligand with a $-2$ charge. Setting up the oxidation state equation: $x+3(-2)=-3$, we find that cobalt is in the $+3$ oxidation state. The ${\text{Co}}^{3+}$ ion also possesses a $3d^6$ configuration.

Here is where many students fall into a trap. While oxalate is generally considered a weak to moderate field ligand, when it coordinates with the highly charged ${\text{Co}}^{3+}$ ion, it acts as a strong field ligand. The high charge density of ${\text{Co}}^{3+}$ pulls the ligands closer, significantly increasing the crystal field splitting energy (${\Delta }_o$).

In this octahedral field, the splitting is so large that it is energetically more favorable for all six electrons to pair up in the lower energy $t_{2g}$ orbitals, leaving the $e_g$ orbitals completely empty. The configuration is $t_{2g}^6{e}_g^0$.

With 0 unpaired electrons ($n=0$), the complex is diamagnetic:

The Highly Oxidized Manganate Ion

Finally, we examine the manganate ion, ${\text{MnO}}_4^{2-}$.

Oxygen carries a $-2$ charge, so $x+4(-2)=-2$ reveals a staggering $+6$ oxidation state for manganese. A neutral manganese atom is $[\text{Ar}]3d^{5}4s^2$. Removing six electrons leaves the ${\text{Mn}}^{6+}$ ion with a solitary electron: a $3d^1$ configuration.

In this tetrahedral complex, the single electron will naturally occupy the lowest available energy level, which is one of the $e$ orbitals. The configuration is simply $e^1{t}_2^0$.

With exactly 1 unpaired electron ($n=1$), we calculate the magnetic moment:

By meticulously applying Crystal Field Theory to each complex, we have determined their magnetic moments to be 4.90 BM, 0 BM, and 1.73 BM respectively. This perfectly aligns with option (b), showcasing the predictive power of coordination chemistry.