Animated Solution for Chemistry - Coordination Compounds: The correct order of magnetic moments (spin only values in BM) among the following is (At. no of Mn = 25, Fe = 26, Co = 27)

When dealing with coordination compounds, one of the most fascinating properties we can calculate is the spin-only magnetic moment. This property gives us a direct window into the electronic structure of the central metal ion. The formula is beautifully simple:

Here, $n$ represents the number of unpaired electrons, and the result is measured in Bohr Magnetons (BM). To find $n$, we must act as molecular detectives, analyzing the oxidation state, the ligand strength, and the resulting crystal field splitting.

Let's start with $[Fe(CN{)}_6{]}^{4-}$. The overall charge is $-4$, and with six cyanide ligands (each $-1$), the iron must be in the $+2$ oxidation state. Iron(II) has a $3d^6$ electronic configuration.

Now, cyanide ($C{N}^{-}$) is a notoriously strong field ligand. In an octahedral field, it causes a massive splitting between the lower $t_{2g}$ and upper $e_g$ orbitals. Because the splitting energy (${\Delta }_o$) is greater than the pairing energy ($P$), the electrons prefer to pair up in the lower energy level. All six electrons fill the $t_{2g}$ orbitals perfectly, leaving zero unpaired electrons ($n=0$). Consequently, the magnetic moment is $0$ BM.

Next up is $[MnC{l}_4{]}^{2-}$. Manganese is also in the $+2$ oxidation state, giving it a $3d^5$ configuration. However, the environment is completely different. We have four chloride ligands, which are weak field ligands, leading to a tetrahedral geometry.

In a tetrahedral field, the splitting is inverted and much smaller (${\Delta }_t<P$). The electrons will singly occupy all five d-orbitals before any pairing occurs. Thus, we have five unpaired electrons ($n=5$). Plugging this into our formula:

Finally, we examine $[CoC{l}_4{]}^{2-}$. Cobalt is in the $+2$ state, which means a $3d^7$ configuration. Like the manganese complex, this is a tetrahedral complex with weak field chloride ligands.

The seven electrons will fill the orbitals following Hund's rule for a high-spin complex: first singly occupying the lower $e$ and upper $t_2$ sets, and then pairing up in the lower $e$ set. This results in an $e^4{t}_2^3$ configuration, leaving exactly three unpaired electrons ($n=3$).

With our detective work complete, we can easily rank the magnetic moments:

This perfectly matches the order: $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>[Fe(CN{)}_6{]}^{4-}$. The beauty of crystal field theory lies in how it elegantly predicts these physical properties from simple electronic principles!