Animated Solution for Chemistry - Coordination Compounds: Which one of the following complexes is an outer orbital complex? (At. no. of Mn = 25, Fe = 26, Co = 27, Ni = 28)

Coordination compounds are like architectural marvels at the atomic level. Imagine the central metal ion as a gracious host and the incoming ligands as guests. For the guests to settle in, the host must provide empty rooms, which in the quantum world, are empty atomic orbitals. Depending on which rooms the host offers, the resulting structure can be classified as an inner orbital complex or an outer orbital complex.

If the metal ion uses its inner $(n-1)d$ orbitals for hybridization (forming a $d^{2}s{p}^3$ hybrid), it creates an inner orbital complex. Conversely, if those inner rooms are occupied and the metal is forced to open up its outer $nd$ orbitals (forming an $s{p}^3{d}^2$ hybrid), it creates an outer orbital complex. Let's analyze our given options to see which host is forced to use its outer rooms.

Let's start with $[Fe(CN{)}_6{]}^{4-}$. Iron is in the $+2$ oxidation state, giving it a $3d^6$ configuration. The cyanide ion ($C{N}^{-}$) is a notoriously strong field ligand. It acts like a strict manager, forcing the six $d$-electrons to pair up against Hund's rule. This pairing neatly packs the electrons into just three $3d$ orbitals, leaving exactly two $3d$ orbitals completely empty. These two empty inner orbitals are perfect for $d^{2}s{p}^3$ hybridization, making it an inner orbital complex.

A similar story unfolds for $[Mn(CN{)}_6{]}^{4-}$. Manganese is in the $+2$ state, meaning a $3d^5$ configuration. The strong $C{N}^{-}$ ligand forces pairing, packing the five electrons into three orbitals and again leaving two $3d$ orbitals vacant. Thus, it also forms a $d^{2}s{p}^3$ inner orbital complex.

What about $[Co(N{H}_3{)}_6{]}^{3+}$? Cobalt is in the $+3$ state ($3d^6$). While ammonia ($N{H}_3$) is generally a moderate ligand, it acts as a strong field ligand when paired with the highly charged $C{o}^{3+}$ ion. Just like in the iron complex, the six electrons pair up, freeing two $3d$ orbitals. Once again, we get a $d^{2}s{p}^3$ inner orbital complex.

Now we arrive at our final candidate: $[Ni(N{H}_3{)}_6{]}^{2+}$. Nickel is in the $+2$ oxidation state, which corresponds to a $3d^8$ configuration. This is where things get interesting.

Even though $N{H}_3$ is a strong field ligand, let's try to pair up those eight electrons. If we pack them as tightly as possible, they will occupy four out of the five $3d$ orbitals. This leaves only one $3d$ orbital empty. However, to form an octahedral complex, the metal needs two empty $d$-orbitals to mix with the $s$ and $p$ orbitals.

Since the inner $3d$ subshell simply cannot provide the required two empty rooms, the $N{i}^{2+}$ ion has no choice but to look outward. It utilizes its outer $4d$ orbitals instead. The six $N{H}_3$ ligands donate their electron pairs into one $4s$, three $4p$, and two $4d$ orbitals. This results in $s{p}^3{d}^2$ hybridization.

Because the complex relies on the outer $d$-orbitals for bonding, $[Ni(N{H}_3{)}_6{]}^{2+}$ is definitively classified as an outer orbital complex. The inability of $d^8$ systems to form inner octahedral complexes is a classic and highly testable concept in coordination chemistry!