Question

The Lassaigne’s extract is boiled with dil. ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ before testing for halogens because________.

A. AgCN is soluble in ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$

B. Silver halide are soluble in ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$

C. NaCN and ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}$ are decomposed by ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$

D. ${\rm{A}}{{\rm{g}}_{\rm{2}}}{\rm{S}}$ is soluble in ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$

Answer 1

Hint: We know that, Lassaigne test is used in detection of nitrogen, halogen (Fluorine, chlorine, Bromine, iodine) and sulphur in an organic compound. In this test, at first Lassaigne’s extract is prepared and then tested for nitrogen, sulphur or halogen.

Complete step by step answer:

Let's first understand how a Lassaigne's extract is prepared. To prepare this, heating a small piece of sodium is gently done in a fusion tube till the time it melts to result in a shining globule. At this stage, organic substance in a smaller amount is added and the tube is strongly heated for 2 to 3 minutes. The tube which is very hot is quickly transferred into distilled water present in a china dish. The contents of the dish is boiled for a couple of minutes, cooled and filtered. The filtrate is the Lassaigne's extract.

The reaction of formation of extract is shown as below.

${\rm{Na}} + {\rm{C}} + {\rm{N}} \to {\rm{NaCN}}$

$2{\rm{Na}} + {\rm{S}} \to {\rm{N}}{{\rm{a}}_2}{\rm{S}}$

${\rm{Na}} + {\rm{X}} \to {\rm{NaX}}$ (X is any halogen)

Now, come to the question. The reason is the boiling of Lassaigne’s extract with dil. ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ before testing for halogens needs to be determined.

We know an organic compound may contain nitrogen and sulphur along with halogen. In such case, the Lassaigne’s extract will contain sodium sulphide $\left({{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}} \right)$ and sodium cyanide (NaCN) along with sodium halide. So, the reaction of Lassaigne’s extract with nitric acid decomposes sodium cyanide and sodium sulphide to form vapours of HCN and ${{\rm{H}}_{\rm{2}}}{\rm{S}}$ which escaped ultimately.

${\rm{NaCN}} + {\rm{HN}}{{\rm{O}}_{\rm{3}}} \to {\rm{NaN}}{{\rm{O}}_{\rm{3}}} + {\rm{HCN}}\left( g \right)$

${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}} + {\rm{2HN}}{{\rm{O}}_3} \to 2{\rm{NaN}}{{\rm{O}}_{\rm{3}}} + {{\rm{H}}_{\rm{2}}}{\rm{S}}\left( g \right)$

If the decomposition of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}$ and NaCN will not occur, they undergo reaction with silver nitrate to form precipitate that will interfere with the colours of different silver halides.

Therefore, reason of reacting extracting with nitric acid is to decompose $\left({{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}} \right)$ and NaCN.

Hence, the correct answer is option C.

Note:
After reacting sodium extract with nitric acid, silver nitrate is added to the solution. The formation of colored precipitate indicates the presence of halogen. If precipitate is white colored, the halogen present in the compound is chlorine. If ppt is pale yellow colored, halogen present is bromine and if ppt. is yellow colored, halogen present in the organic compound is iodine.

${\rm{NaCl}} + {\rm{AgN}}{{\rm{O}}_{\rm{3}}} \to {\rm{AgCl(white ppt}}{\rm{.)}} + {\rm{NaN}}{{\rm{O}}_3}$

${\rm{NaBr}} + {\rm{AgN}}{{\rm{O}}_{\rm{3}}} \to {\rm{AgBr(pale yellow ppt}}{\rm{.)}} + {\rm{NaN}}{{\rm{O}}_3}$

${\rm{NaI}} + {\rm{AgN}}{{\rm{O}}_{\rm{3}}} \to {\rm{AgI(yellow ppt}}{\rm{.)}} + {\rm{NaN}}{{\rm{O}}_3}$