Animated Solution for Chemistry - Coordination Compounds: Which one of the following complexes is violet in colour?

When we step into the organic chemistry laboratory, one of the most fascinating aspects is qualitative analysis—the art of detecting which elements are hiding inside an unknown organic compound. Among the various tests, the detection of sulphur is particularly striking because it yields a brilliant, unmistakable color change.

To detect sulphur, we first prepare a Lassaigne's extract by fusing the organic compound with sodium metal. This converts any covalently bonded sulphur into ionic sodium sulphide ($N{a}_{2}S$). Once we have the sulphide ions ($S^{2-}$) floating in our aqueous extract, we are ready for the magic trick: the Sodium Nitroprusside Test.

When a few drops of sodium nitroprusside solution are added to the alkaline Lassaigne's extract containing sulphide ions, a deep, rich violet color instantly appears. But what exactly is happening at the molecular level to cause this beautiful transformation?

Sodium nitroprusside is a coordination complex with the formula $N{a}_2[Fe(CN{)}_{5}NO]$. In aqueous solution, it dissociates to give the nitroprusside ion, $[Fe(CN{)}_{5}NO{]}^{2-}$.

When the sulphide ion encounters this complex, it doesn't attack the central iron atom. Instead, it acts as a nucleophile and attacks the coordinated nitrosonium ligand ($N{O}^{+}$).

The product of this reaction is a new coordination complex: the thionitroprusside ion, $[Fe(CN{)}_{5}NOS{]}^{4-}$.

This specific complex absorbs light in the visible region in such a way that it transmits a vibrant violet color. The appearance of this violet color is the definitive confirmatory test for the presence of sulphur in the original organic compound. Therefore, among the given options, $[Fe(CN{)}_{5}NOS{]}^{4-}$ is the complex responsible for the violet color.

Here is a classic trap that examiners love to set: What is the oxidation state of iron in the reactant and the product?

Let's break it down. In the reactant $[Fe(CN{)}_{5}NO{]}^{2-}$, we have five cyanide ligands ($C{N}^{-}$) and one nitrosonium ligand ($N{O}^{+}$). Let the oxidation state of Fe be $x$. $x+5(-1)+(+1)=-2$ $x-4=-2⟹x=+2$

Now, let's look at the violet product, $[Fe(CN{)}_{5}NOS{]}^{4-}$. The sulphide ion ($S^{2-}$) has bonded with the $N{O}^{+}$ ligand to form the thionitrosyl ligand ($NO{S}^{-}$). Let the oxidation state of Fe be $y$. $y+5(-1)+(-1)=-4$ $y-6=-4⟹y=+2$

Surprisingly, the oxidation state of the central iron atom does not change during this reaction! It remains in the $+2$ state throughout. The entire chemical transformation happens on the ligand itself. Keeping this subtle detail in mind will save you from making silly mistakes in advanced coordination chemistry problems.