Formation of Na4[Fe(CN)5NOS], a purple coloured complex formed by addition of sodium nitroprusside in sodium carbonate…

Step 1: Recall the Lassaigne Test for Sulphur

In the Lassaigne test, an organic compound is fused with sodium metal to convert covalently bonded elements into ionic form: $\ce{C + S + Na ->[ ext{fusion}] Na2S}$

Sodium sulphide ($\ce{Na2S}$) is formed from the sulphur in the compound.

Step 2: The Sodium Nitroprusside Test for Sulphide

The sodium extract (containing $\ce{Na2S}$) is treated with sodium nitroprusside $\ce{Na2[Fe(CN)5NO]}$:

$\ce{Na2S + Na2[Fe(CN)5NO] -> Na4[Fe(CN)5NOS]}$

The product $\ce{Na4[Fe(CN)5NOS]}$ is violet/purple in colour. This is the confirmatory test for sulphide ion ($\ce{S^{2-}}$).

Step 3: Evaluate Each Option

| Option | Ion | Sodium Nitroprusside Test | |---|---|---| | (a) Sodium ion ($\ce{Na+}$) | Already present in reagent | No specific colour test | | (b) Sulphate ion ($\ce{SO4^{2-}}$) | Detected by BaCl₂/HCl (white ppt) | Not detected by nitroprusside | | (c) Sulphide ion ($\ce{S^{2-}}$) ✅ | Purple/violet complex with nitroprusside | ✅ This is the correct answer | | (d) Sulphite ion ($\ce{SO3^{2-}}$) | Detected differently | Not by nitroprusside test |

Step 4: Conclusion

Answer: (c) Sulphide ion

Key Points to Remember:

• Lassaigne fusion converts S → $\ce{Na2S}$ (sulphide)
• Sodium nitroprusside test is specific for sulphide ($\ce{S^{2-}}$) → purple/violet colour
• Sulphate detected by: BaCl₂ + HCl → white ppt of BaSO₄
• Nitroprusside formula: $\ce{Na2[Fe(CN)5NO]}$; the S from $\ce{Na2S}$ replaces the NO ligand in the complex