Animated Solution for Chemistry - Coordination Compounds: Arrange the following cobalt complexes in the order of increasing crystal field stabilisation energy (CFSE) value. $[Co{F}_6{]}^{3-}\text{ (A)},[Co(H_{2}O{)}_6{]}^{3+}\text{ (B)},[Co(N{H}_3{)}_6{]}^{3+}\text{ (C)}\text{ and }[Co(en{)}_3{]}^{3+}\text{ (D)}$ Choose the correct option.

## The Power of Ligands: Decoding Crystal Field Splitting

Have you ever wondered why some coordination compounds are brilliantly colored while others are pale, or why some are strongly attracted to magnets while others are repelled? The secret lies in a fascinating phenomenon called Crystal Field Splitting. Let's dive into a classic JEE problem to unravel how different ligands manipulate the energy levels of a central metal ion.

A Cobalt Canvas

In our problem, we are presented with four distinct cobalt complexes: $[Co{F}_6{]}^{3-}$, $[Co(H_{2}O{)}_6{]}^{3+}$, $[Co(N{H}_3{)}_6{]}^{3+}$, and $[Co(en{)}_3{]}^{3+}$.

The first step in analyzing any coordination complex is to identify the central metal ion and its oxidation state. If we calculate the oxidation state of cobalt in each of these complexes, we find a common thread: it is $+3$ in all four cases.

Because the central metal ion ($C{o}^{3+}$) is identical across the board, the differences in their physical properties—specifically their Crystal Field Stabilization Energy (CFSE)—must stem entirely from the nature of the ligands surrounding them.

The Rulebook of Ligands

Imagine the five degenerate (equal energy) d-orbitals of a free $C{o}^{3+}$ ion. As ligands approach this central ion to form a complex, their electron clouds repel the electrons residing in the metal's d-orbitals. This repulsion shatters the degeneracy, splitting the d-orbitals into two distinct energy levels: a lower energy $t_{2g}$ set and a higher energy $e_g$ set.

The energy gap between these two sets is denoted as ${\Delta }_o$ (Delta naught) for octahedral complexes. The magnitude of ${\Delta }_o$ is directly proportional to the strength of the approaching ligands.

To determine which ligand is stronger, chemists rely on the Spectrochemical Series—an experimentally derived sequence that ranks ligands based on their ability to split d-orbitals. A simplified version of this series is:

Let's evaluate the ligands from our specific problem using the spectrochemical series:

1. Fluoride ($F^{-}$): Being a halide, it is a weak field ligand and causes minimal splitting. 2. Water ($H_{2}O$): An oxygen donor, it is slightly stronger than the halide. 3. Ammonia ($N{H}_3$): A nitrogen donor, it acts as a strong field ligand, causing significant splitting. 4. Ethylenediamine ($en$): Also a nitrogen donor, but with a twist. It is a bidentate ligand, meaning it attaches to the metal ion at two points, forming a stable ring structure. This 'chelating effect' makes it an exceptionally strong field ligand, even stronger than ammonia.

Therefore, the order of ligand strength is:

Since the Crystal Field Stabilization Energy (CFSE) is directly proportional to the ligand strength, the order of CFSE for the complexes perfectly mirrors the order of their ligands.

The complex with the weakest ligand, $[Co{F}_6{]}^{3-}$ (A), will have the lowest CFSE. This is followed by $[Co(H_{2}O{)}_6{]}^{3+}$ (B), then $[Co(N{H}_3{)}_6{]}^{3+}$ (C), and finally, the complex with the strongest ligand, $[Co(en{)}_3{]}^{3+}$ (D), will boast the highest CFSE.

Final Order: $A<B<C<D$

Understanding this concept doesn't just solve one MCQ; it unlocks the ability to predict magnetic behavior (high-spin vs. low-spin) and the vibrant colors of coordination compounds!