Animated Solution for Chemistry - Coordination Compounds: The crystal field stabilisation energy (CFSE) of $[Co{F}_3(H_{2}O{)}_3]$ (${\Delta }_0<P$) is

Let's embark on a journey into the heart of coordination chemistry! We are given the complex $[Co{F}_3(H_{2}O{)}_3]$ and asked to find its Crystal Field Stabilisation Energy (CFSE). The problem also hands us a crucial piece of intel: ${\Delta }_0<P$.

First things first, we need to identify the central character of our story: the Cobalt ion. What is its oxidation state?

We know that fluoride ($F^{-}$) carries a $-1$ charge, and water ($H_{2}O$) is a neutral molecule. Since there are three fluorides, the total charge contributed by the ligands is $-3$. For the entire complex to be electrically neutral, the Cobalt ion must balance this out with a $+3$ charge.

So, we are dealing with $C{o}^{3+}$.

Now, let's look at the electronic configuration. The atomic number of Cobalt is $27$. Its ground state configuration is $[Ar]3d^{7}4s^2$.

When it loses three electrons to become $C{o}^{3+}$, it first loses the two electrons from the outermost $4s$ orbital, and then one electron from the $3d$ orbital. This leaves us with a $3d^6$ configuration.

Imagine these six electrons sitting in the five degenerate (equal energy) d-orbitals of the free ion. Following Hund's rule, five electrons will singly occupy the five orbitals, and the sixth electron will pair up. So, the free $C{o}^{3+}$ ion naturally has one pair of electrons.

When the six ligands approach the Cobalt ion to form an octahedral complex, the five 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 difference between them is ${\Delta }_0$.

Here is where the condition ${\Delta }_0<P$ becomes the star of the show. ${\Delta }_0$ is the crystal field splitting energy, and $P$ is the pairing energy (the energy required to force two negatively charged electrons into the same orbital).

Because ${\Delta }_0<P$, it takes less energy for an electron to jump up to the higher $e_g$ orbitals than it does to pair up in the lower $t_{2g}$ orbitals. This creates a high-spin complex.

Let's fill our six electrons: the first three go into the $t_{2g}$ orbitals. The next two jump up to the $e_g$ orbitals. The final, sixth electron has nowhere else to go but to pair up in one of the $t_{2g}$ orbitals.

Our final configuration is $t_{2g}^4{e}_g^2$.

Now for the grand finale: calculating the CFSE. The formula for an octahedral complex is:

Plugging in our numbers ($n_{t_{2g}}=4$ and $n_{e_g}=2$):

This is a classic trap! We only add a pairing energy term if there is a net increase in the number of electron pairs compared to the free metal ion. Remember how we established that the free $C{o}^{3+}$ ion already has one pair of electrons? Well, our complex configuration ($t_{2g}^4{e}_g^2$) also has exactly one pair.

Since the number of pairs didn't change, the pairing energy terms cancel out perfectly. Thus, our final, pristine answer is simply $-0.4{\Delta }_0$.