Discuss the chemistry of Lassaigne's test.

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(N/A) Nitrogen $(N)$,Sulphur $(S)$,Halogens $(Cl, Br, I)$,and Phosphorus $(P)$ present in an organic compound are detected by "Lassaigne's test".
$(a)$ The elements $(N, X, S)$ present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. The following reactions take place:
$(i)$ $Na + C + N \xrightarrow{\Delta} NaCN$
$(ii)$ $2Na + S \xrightarrow{\Delta} Na_{2}S$
$(iii)$ $Na + X \xrightarrow{\Delta} NaX$ (where $X = Cl, Br, \text{ or } I$)
Cyanide,sulphide,and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.
$(b)$ Test for Nitrogen:
Procedure: The sodium fusion extract is boiled with iron $(II)$ sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron $(II)$ sulphate and forms sodium hexacyanidoferrate $(II)$. On heating with concentrated sulphuric acid,some iron $(II)$ ions are oxidised to iron $(III)$ ions which react with sodium hexacyanidoferrate $(II)$ to produce iron $(III)$ hexacyanidoferrate $(II)$ (ferriferrocyanide),which is Prussian blue in colour.
$6CN_{(aq)}^{-} + Fe_{(aq)}^{2+} \rightarrow [Fe(CN)_{6}]_{(aq)}^{4-}$
$3[Fe(CN)_{6}]_{(aq)}^{4-} + 4Fe^{3+} \xrightarrow{xH_{2}O} Fe_{4}[Fe(CN)_{6}]_{3} \cdot xH_{2}O$ (Prussian blue)
$(c)$ Test for Sulphur: Following two tests occur for detection of sulphur:
$(i)$ The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. $A$ black precipitate of lead sulphide indicates the presence of sulphur.
$S_{(aq)}^{2-} + Pb_{(aq)}^{2+} \rightarrow PbS_{(s)}$ (Black precipitate)
$(ii)$ On treating sodium fusion extract with sodium nitroprusside,the appearance of a violet colour further indicates the presence of sulphur.

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Similar Questions

Match the compounds in Column $I$ with their characteristic test$(s)$ reaction$(s)$ given in Column $II$.
Column $I$ Column $II$
$A. H_2N-NH_3^+Cl^-$ $p. \text{Sodium fusion extract of the compound gives Prussian blue colour with } FeSO_4$
$B. HO-C_6H_4-CH(NH_3^+)COOH \text{ (with } I^- \text{ counterion)}$ $q. \text{Gives positive } FeCl_3 \text{ test}$
$C. HO-C_6H_4-NH_3^+Cl^-$ $r. \text{Gives white precipitate with } AgNO_3$
$D. (NO_2)_2C_6H_3-NH-NH_3^+Br^-$ $s. \text{Reacts with aldehydes to form the corresponding hydrazone derivative}$

In the detection of halogens,what is formed by adding $AgNO_3$? What is its colour?

Appearance of blood red colour,on treatment of the sodium fusion extract of an organic compound with $FeSO_4$ in presence of concentrated $H_2SO_4$ indicates the presence of element/s

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 $FeSO_4$ and dilute sulphuric acid to a part of the 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.

Identify the binary mixture$(s)$ that can be separated into individual compounds by differential extraction,as shown in the given scheme.
$(A)$ $C_6H_5OH$ and $C_6H_5COOH$
$(B)$ $C_6H_5COOH$ and $C_6H_5CH_2OH$
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