Animated Solution for Chemistry - Coordination Compounds: The complex that has highest crystal field splitting energy ($\Delta$), is

Imagine you are looking at a lineup of coordination complexes, each with its own unique geometry and set of ligands. Our goal is to find the one with the highest crystal field splitting energy, denoted by $\Delta$.

The first step is to identify the geometry of each complex. We do this by looking at the coordination number, which is the number of ligands directly attached to the central metal ion.

Options (a), (b), and (c) all have a coordination number of 6. This tells us they are octahedral complexes. On the other hand, option (d), $K_2[CoC{l}_4]$, has a coordination number of 4, making it a tetrahedral complex.

Octahedral vs. Tetrahedral

Why does geometry matter? Because the crystal field splitting energy is highly dependent on the arrangement of ligands around the central metal ion.

In an octahedral field, the ligands approach directly along the axes where the $e_g$ d-orbitals are located, causing significant repulsion and a large splitting energy, ${\Delta }_o$.

In a tetrahedral field, the ligands approach between the axes, leading to less direct repulsion. Mathematically, the splitting in a tetrahedral field (${\Delta }_t$) is roughly four-ninths of the splitting in an octahedral field (${\Delta }_o$).

Because ${\Delta }_t$ is significantly smaller than ${\Delta }_o$, we can immediately eliminate the tetrahedral complex, option (d), from our race for the highest splitting energy.

Now we are left with three octahedral complexes. Since their geometries are the same, the deciding factor is the strength of the ligands.

The magnitude of the octahedral splitting energy, ${\Delta }_o$, is directly proportional to the ligand strength. To compare them, we rely on the spectrochemical series, which ranks ligands based on their ability to split d-orbitals.

Looking at the ligands in our remaining options, we have chloride ($C{l}^{-}$), water ($H_{2}O$), ammonia ($N{H}_3$), and cyanide ($C{N}^{-}$). According to the spectrochemical series, the order of increasing ligand strength is:

Cyanide ($C{N}^{-}$) is a strong field ligand, primarily because it is a strong $\pi$-acceptor. It causes a massive splitting of the d-orbitals.

Since $K_3[Co(CN{)}_6]$ contains six of these powerful cyanide ligands, it will experience the maximum crystal field splitting. Therefore, option (c) is our correct answer.

As a bonus thought, because this splitting is so large, the electrons will prefer to pair up in the lower energy $t_{2g}$ orbitals rather than jumping the large energy gap to the $e_g$ orbitals. This makes $K_3[Co(CN{)}_6]$ a low-spin, diamagnetic complex. Always remember to connect the splitting energy to the physical properties of the molecule!