Animated Solution for Chemistry - Coordination Compounds: Which of the following compounds is not coloured yellow?

When we dive into the fascinating world of coordination chemistry, one of the most striking features is the vibrant colours of transition metal complexes. But where do these colours come from?

In most cases, the colour arises from d-d transitions. When ligands approach a central metal ion, they cause the degenerate d-orbitals to split into different energy levels. If visible light strikes the complex, an electron can absorb a specific wavelength and jump from a lower energy d-orbital to a higher one. The colour we see is the complementary colour of the light absorbed.

However, this isn't the only mechanism. Sometimes, even metal ions with no d-electrons (like ${\text{Cr}}^{6+}$) exhibit intense colours. This happens due to Charge Transfer, specifically Ligand-to-Metal Charge Transfer (LMCT), where an electron temporarily hops from a ligand orbital to an empty metal orbital.

Chromates, Fischer's Salt, and Arsenomolybdates

Let's analyze the compounds given in our problem.

First, we have Barium chromate, ${\text{BaCrO}}_4$. If you recall your qualitative salt analysis, the chromate ion (${\text{CrO}}_4^{2-}$) is famous for its bright yellow colour. Even though Chromium is in a $+6$ oxidation state with a $d^0$ configuration, it exhibits this intense yellow colour due to LMCT.

Next is ${\text{K}}_3[\text{Co}({\text{NO}}_2{)}_6]$, widely known as Fischer's salt. This is a classic yellow precipitate that forms when we test for the presence of potassium ions. The colour here is a result of d-d transitions within the Cobalt(III) complex.

Then we have $({\text{NH}}_4{)}_3[\text{As}({\text{Mo}}_3{\text{O}}_{10}{)}_4]$, which is ammonium arsenomolybdate. During the qualitative test for arsenic, the addition of ammonium molybdate yields this characteristic canary yellow precipitate.

Zinc Ferrocyanide

Finally, let's examine Zinc ferrocyanide, ${\text{Zn}}_2[\text{Fe}(\text{CN}{)}_6]$.

Here, the zinc ion (${\text{Zn}}^{2+}$) has a $d^{10}$ configuration. With completely filled d-orbitals, it cannot undergo d-d transitions and thus does not contribute to any colour.

But what about the iron? The iron is in a $+2$ oxidation state (${\text{Fe}}^{2+}$), which is a $d^6$ system. However, it is surrounded by cyanide ions (${\text{CN}}^{-}$), which are incredibly strong field ligands. According to the spectrochemical series, ${\text{CN}}^{-}$ causes a massive crystal field splitting (${\Delta }_o$).

Because the energy gap is so large, the energy required for an electron to transition is pushed out of the visible spectrum and into the ultraviolet (UV) region. Since it doesn't absorb visible light, the compound appears white.

By systematically evaluating the origin of colour and recalling our qualitative analysis tests, we can confidently conclude that Zinc ferrocyanide is the odd one out. While the other three compounds are famously yellow, ${\text{Zn}}_2[\text{Fe}(\text{CN}{)}_6]$ is a white precipitate.