Animated Solution for Chemistry - Coordination Compounds: The hybridisation and magnetic nature of $[Mn(CN{)}_6{]}^{4-}$ and $[Fe(CN{)}_6{]}^{3-}$, respectively are

Coordination chemistry often feels like a puzzle where the central metal ion and its surrounding ligands play a delicate game of tug-of-war with electrons. In this problem, we are tasked with uncovering the hybridisation and magnetic nature of two seemingly different complexes: $[Mn(CN{)}_6{]}^{4-}$ and $[Fe(CN{)}_6{]}^{3-}$. Let's break down the logic step-by-step.

Oxidation States and Electron Count

The first step in analyzing any coordination complex is to determine the oxidation state of the central metal ion.

For the hexacyanomanganate(II) ion, $[Mn(CN{)}_6{]}^{4-}$, we know that each cyanide ligand carries a $-1$ charge. Setting up the equation: $x+6(-1)=-4⟹x=+2$. So, Manganese is in the $+2$ oxidation state. The neutral Manganese atom has an electronic configuration of $[Ar]4s^{2}3d^5$. Removing two electrons (from the outermost $4s$ orbital) leaves us with $M{n}^{2+}=[Ar]3d^5$.

Now, let's look at the hexacyanoferrate(III) ion, $[Fe(CN{)}_6{]}^{3-}$. Setting up the equation: $y+6(-1)=-3⟹y=+3$. Iron is in the $+3$ oxidation state. The neutral Iron atom is $[Ar]4s^{2}3d^6$. Removing three electrons (two from $4s$ and one from $3d$) gives us $F{e}^{3+}=[Ar]3d^5$.

Fascinatingly, both metal ions end up with the exact same valence electronic configuration: $3d^5$. This means our analysis for both complexes will follow the exact same path from here on out!

Crystal Field Splitting

The true character of a complex is dictated by its ligands. Here, we have the cyanide ion ($C{N}^{-}$), which is renowned as a strong field ligand in the spectrochemical series.

According to Crystal Field Theory (CFT), when six ligands approach the central metal ion to form an octahedral complex, they cause the five degenerate $3d$ orbitals to split into two distinct energy levels: a lower energy $t_{2g}$ set (three orbitals) and a higher energy $e_g$ set (two orbitals).

Because $C{N}^{-}$ is a strong field ligand, it induces a massive splitting. The crystal field splitting energy (${\Delta }_o$) is significantly greater than the pairing energy ($P$). Mathematically, ${\Delta }_o>P$.

When we distribute our five $3d$ electrons into these split orbitals, they face a choice: jump the massive energy gap to the $e_g$ level, or pair up in the lower $t_{2g}$ level. Since ${\Delta }_o>P$, it costs less energy to pair up. Thus, all five electrons crowd into the $t_{2g}$ orbitals, resulting in the configuration: $t_{2g}^5{e}_g^0$.

Hybridisation and Magnetism

Now, let's determine the hybridisation. Because the two $e_g$ orbitals (which are part of the inner $3d$ subshell) are completely empty, the metal ion can utilize them to accept electron pairs from the incoming cyanide ligands.

The metal ion mixes these two $3d$ orbitals with one $4s$ and three $4p$ orbitals to create six equivalent hybrid orbitals. This results in $d^{2}s{p}^3$ hybridisation, forming what we call an inner orbital complex.

Finally, we assess the magnetic nature. We look at our $t_{2g}^5$ configuration. We have two pairs of electrons and exactly one unpaired electron. In chemistry, the presence of even a single unpaired electron renders the entire complex paramagnetic (it will be weakly attracted to an external magnetic field).

Since both $[Mn(CN{)}_6{]}^{4-}$ and $[Fe(CN{)}_6{]}^{3-}$ share the $3d^5$ configuration and the strong $C{N}^{-}$ ligand, they both exhibit $d^{2}s{p}^3$ hybridisation and are paramagnetic. The mystery is elegantly solved!