Sodium Nitroprusside Test for Sulphide Ion Ligand Conversion

In the sodium nitroprusside test for the detection of the sulphide ion ($\text{S}^{2-}$), sodium nitroprusside, $\text{Na}_2[\text{Fe}(\text{CN})_5\text{NO}]$, reacts with sulphide ions in alkaline medium to yield a characteristic purple/violet-colored complex, sodium thionitroprusside, $\text{Na}_4[\text{Fe}(\text{CN})_5(\text{NOS})]$.

1. Initial Complex: In sodium nitroprusside, $\text{Na}_2[\text{Fe}(\text{CN})_5\text{NO}]$, the complex anion is $[\text{Fe}(\text{CN})_5\text{NO}]^{2-}$, where the iron ion is in the $+2$ oxidation state ($\text{Fe}^{2+}$) and one of the coordinated ligands is the nitrosyl cation, $\text{NO}^+$.

2. Reaction with Sulphide Ion: When sulphide ion ($\text{S}^{2-}$) is introduced, it attacks the electrophilic nitrosyl ligand ($\text{NO}^+$): $[\text{Fe}(\text{CN})_5\text{NO}]^{2-} + \text{S}^{2-} \longrightarrow [\text{Fe}(\text{CN})_5(\text{NOS})]^{4-}$

3. Ligand Conversion: During this reaction, the nitrosyl ligand ($\text{NO}^+$) coordinated to the iron center coordinates with the sulphide ion ($\text{S}^{2-}$), forming the thionitrito ligand, $\text{NOS}^-$.

Thus, the ligand coordinated to the metal ion is converted to $\text{NOS}^-$.

Correct Answer: (A) $\text{NOS}^-$