During sodium nitroprusside test of sulphide ion in an aqueous solution, one of the ligands coordinated to the metal ion is converted to

The correct answer is:

${\mathrm{NOS}}^{-}$

Step-by-Step Explanation:

1. Nature of Sodium Nitroprusside:
Sodium nitroprusside is a coordination complex with the chemical formula ${\mathrm{Na}}_2\left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\mathrm{NO}]$. In this complex, the central iron atom is coordinated to five cyanide (${\mathrm{CN}}^{-}$) ligands and one nitrosyl (${\mathrm{NO}}^{+}$) ligand.

2. Reaction with Sulphide Ion (${\mathrm{S}}^{2-}$):
When sodium nitroprusside is added to an aqueous solution containing sulphide ions, a characteristic purple-violet color is formed. This color change is used as a standard confirmatory test for the presence of sulphide ions in qualitative analysis.

3. Ligand Conversion:
The sulphide ion acts as a nucleophile and attacks the nitrogen atom of the coordinated nitrosyl (${\mathrm{NO}}^{+}$) ligand. This reaction converts the coordinated nitrosyl ligand into a thionitrite (${\mathrm{NOS}}^{-}$) ligand.

The chemical reaction is represented as follows:

$\left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\left(\mathrm{NO}\right){]}^{2-}+{\mathrm{S}}^{2-}\to \left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\left(\mathrm{NOS}\right){]}^{4-}$

4. Conclusion:
The purple complex formed is $\left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\left(\mathrm{NOS}\right){]}^{4-}$, showing that the nitrosyl ligand has been converted to ${\mathrm{NOS}}^{-}$.