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NCERT Exemplar · Q39

Q.Three students, Manish, Ramesh and Rajni were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne's extract (L.E.) independently by the fusion of the compound with sodium metal. Then they added solid FeSO4 and dilute sulphuric acid to a part of Lassaigne's extract. Manish and Rajni obtained prussian blue colour but Ramesh got red colour. Ramesh repeated the test with the same Lassaigne's extract, but again got red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this observation. Also, write the chemical equations to explain the formation of compounds of different colours.

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The compound contained both nitrogen and sulphur. Ramesh's sodium fusion was not

carried out with excess sodium, so the two elements combined as sodium thiocyanate

(NaSCN\ce{NaSCN}); with FeX3+\ce{Fe^{3+}} this gives the blood-red ferric-thiocyanate

colour instead of Prussian blue. Manish and Rajni used excess sodium, which decomposes

any thiocyanate to cyanide and sulphide, so they got Prussian blue.

Why a red colour instead of Prussian blue

All three students fused the same compound. When a compound contains both N and S,

sodium fusion can give sodium thiocyanate:

Na+C+N+S→ΔNaSCN\ce{Na + C + N + S ->[\Delta] NaSCN}

The thiocyanate ion contains no free cyanide, so it cannot form Prussian blue. Instead,

the FeX3+\ce{Fe^{3+}} present (from air-oxidation of the added FeSOX4\ce{FeSO4} in dilute

HX2SOX4\ce{H2SO4}) reacts with SCNX−\ce{SCN-} to give a blood-red complex:

FeX3++3 SCNX−→Fe(SCN)X3 (blood red)\ce{Fe^{3+} + 3SCN- -> Fe(SCN)3}\ \text{(blood red)}

This is exactly what Ramesh observed — and it repeated because his extract genuinely

contained thiocyanate, not because of contamination.

Why Manish and Rajni got Prussian blue

If the fusion is done with excess sodium, any thiocyanate formed is decomposed to

free cyanide and sulphide:

NaSCN+2 Na→ΔNaCN+NaX2S\ce{NaSCN + 2Na ->[\Delta] NaCN + Na2S} …

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