Animated Solution for Chemistry - Coordination Compounds: The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is

Welcome to the fascinating world of coordination chemistry! Imagine a central metal ion surrounded by six ligands forming a perfect octahedral geometry. As these negatively charged ligands approach the metal, their electron clouds repel the electrons in the metal's $d$-orbitals. However, because of the spatial orientation of the $d$-orbitals, this repulsion is not uniform.

The $d$-orbitals split into two distinct energy levels: the lower energy $t_{2g}$ set (three orbitals) and the higher energy $e_g$ set (two orbitals). The energy gap between these two sets is known as the octahedral crystal field splitting energy, denoted by ${\Delta }_o$.

Splitting Energy vs. Pairing Energy

When it comes to filling these orbitals with electrons, a fierce battle ensues between two opposing forces: the splitting energy (${\Delta }_o$) and the pairing energy ($P$).

If the approaching ligands are weak field ligands, the energy gap ${\Delta }_o$ is relatively small (${\Delta }_o<P$). In this scenario, electrons find it easier to jump to the higher $e_g$ level rather than pairing up in the lower $t_{2g}$ level. This maximizes the number of unpaired electrons, creating a High Spin complex.

Conversely, if the ligands are strong field ligands, the energy gap ${\Delta }_o$ is massive (${\Delta }_o>P$). The energy required to jump the gap is too high, so electrons are forced to pair up in the lower $t_{2g}$ level first. This minimizes the number of unpaired electrons, resulting in a Low Spin complex.

Our mission is to find the metal ion where the difference in the number of unpaired electrons between its high-spin and low-spin states is exactly two. Let's systematically build up the $d$-electron configurations.

1. Manganese(II): $3d^5$ With five electrons to place, a high-spin configuration ($t_{2g}^3{e}_g^2$) spreads them out singly across all five orbitals, giving us 5 unpaired electrons. In a low-spin configuration ($t_{2g}^5{e}_g^0$), the large gap forces them to pair up below, leaving only 1 unpaired electron. The difference is $5-1=4$. Not our winner.

2. Iron(II): $3d^6$ Adding a sixth electron, the high-spin state ($t_{2g}^4{e}_g^2$) pairs one electron in the bottom level, leaving 4 unpaired electrons. The low-spin state ($t_{2g}^6{e}_g^0$) completely fills the bottom level, leaving 0 unpaired electrons. The difference is $4-0=4$. Still not the one.

3. Cobalt(II): $3d^7$ Now for the seventh electron! In the high-spin state ($t_{2g}^5{e}_g^2$), we have two pairs at the bottom and three single electrons, meaning 3 unpaired electrons. In the low-spin state ($t_{2g}^6{e}_g^1$), the bottom level is full, so the seventh electron must jump to the top, leaving exactly 1 unpaired electron. The difference is $3-1=2$. We have found our match!

4. Nickel(II): $3d^8$ Just to be thorough, let's look at the eighth electron. In both high-spin ($t_{2g}^6{e}_g^2$) and low-spin ($t_{2g}^6{e}_g^2$) states, the bottom level is completely full, and two electrons sit unpaired at the top. The configurations are identical, meaning the difference is $2-2=0$. In fact, ${\text{Ni}}^{2+}$ does not even form low-spin octahedral complexes because there is no energetic advantage to doing so.

By systematically applying the rules of Crystal Field Theory, we can clearly see how the interplay between ${\Delta }_o$ and $P$ dictates the electronic structure. The only ion that yields a difference of exactly two unpaired electrons between its high-spin and low-spin states is ${\text{Co}}^{2+}$.