Animated Solution for Chemistry - Coordination Compounds: The correct order of intensity of colors of the compound is

Transition metal complexes are famous for their brilliant and diverse colors. But what determines how intense or deep these colors are? Let's dive into a fascinating problem that explores the relationship between ligand strength and color intensity.

We are given three nickel complexes: $[Ni(CN{)}_4{]}^{2-}$, $[Ni(H_{2}O{)}_6{]}^{2+}$, and $[NiC{l}_4{]}^{2-}$.

The first step is to identify the state of our central metal atom. In all three of these complexes, Nickel is in a $+2$ oxidation state, meaning it has a $3d^8$ electron configuration. Since the metal ion is identical across the board, any difference in their properties must stem entirely from the ligands attached to them.

When ligands approach a central metal ion, they create a crystal field that splits the degenerate d-orbitals into different energy levels. The energy gap between these levels is known as the crystal field splitting energy ($\Delta$).

The magnitude of this splitting depends heavily on the nature of the ligand. We can rank ligands based on their splitting power using the spectrochemical series. For our specific ligands, the order of strength is:

Because $C{N}^{-}$ is a strong field ligand, it causes a massive splitting of the d-orbitals. Conversely, $C{l}^{-}$ is a weak field ligand and causes only a small energy gap.

In many standard textbook problems, a useful heuristic is applied: The intensity of the color of a complex is inversely proportional to the crystal field splitting energy ($\Delta$).

Why? A larger splitting energy often means the complex absorbs higher-energy light (like violet or UV), which can sometimes lead to paler complementary colors or transitions that are less probable.

Following this logic, since the splitting energy order is:

The order of color intensity will be the exact reverse:

Laporte Selection Rules

While the inverse relationship heuristic gets us the right answer here, the true physical mechanism behind color intensity is governed by quantum mechanics, specifically the Laporte Selection Rules.

The Laporte rule states that in a molecule with a center of symmetry (like a perfect octahedron), transitions between orbitals of the same parity (e.g., $d\to d$) are forbidden. Because they are forbidden, octahedral complexes like $[Ni(H_{2}O{)}_6{]}^{2+}$ usually have pale colors.

However, tetrahedral complexes like $[NiC{l}_4{]}^{2-}$ lack a center of symmetry. This allows for $p-d$ orbital mixing, making the $d\to d$ transitions partially allowed. As a result, tetrahedral complexes are generally much more intensely colored (often deep blue or green) compared to their octahedral counterparts. Meanwhile, square planar complexes like $[Ni(CN{)}_4{]}^{2-}$ are often diamagnetic and can be very pale or even colorless depending on the specific transitions involved.