Question

Sodium extract is heated with concentrated $HN{O_3}$ before testing for
halogens because:
(a) $A{g_2}S$ and $AgCN$ are soluble in acidic medium.
(b) Silver halides are totally insoluble in nitric acid.
(c) ${S^{2 - }}$ and $C{N^ - }$, if present, are decomposed by conc. $HN{O_3}$ and hence do not interfere in the test.
(d) $Ag$ reacts faster with halides in acidic medium.

Answer 1

The test for halogens is known as the Lassaigne’s test. Apart from halogen, Nitrogen and Sulphur is also detected by Lassaigne’s test. The test involves heating the samples containing Nitrogen, Sulphur or Halogen with a clean sodium metal, or else we can simply say fusing it with the sample.

Complete Step by step answer:
So, we know the test for halogen is known as Lassaigne’s test. The Lassaigne’s test follows the simple principle as the sodium metal will convert all the elements present in it to its ionic form. i.e., in form of reaction we can write
$Na + C + N \to NaCN$
Here the $C{N^ - }$ ion is formed due to the presence of nitrogen in it.
$2Na + S \to N{a_2}S$
Here, ${S^{2 - }}$ ion is formed due to the presence of sulphur in it.
And when they react with halogens
$Na + X \to NaX$
Where $X$ is $Cl,\;Br$ or $I$.
This formed ionic salt is extracted from the fused mass by boiling with distilled water which is then called the sodium extract.
So in order to make the test happen for halogens the sodium cyanide and sodium sulphide has to be removed. It may interfere with the reaction happening so we have to react then with conc. $HN{O_3}$
$NaCH + HN{O_3} \to NaN{O_3} + HCN$
$N{a_2}S + HN{O_3} \to 2NaN{O_3} + {H_2}S$
So by this the $C{N^ - }$ ions and ${S^{2 - }}$ ion escapes from the reaction.
Hence we can say that ${S^{2 - }}$ and $C{N^ - }$, if present, are decomposed by
conc. $HN{O_3}$ and hence do not interfere in the test.

Hence the option (c) is correct.

Note: If the sulphur or nitrogen group are still present in the sample, we can identify them by looking into the colour of the sample after the reaction. If colour formed is Prussian blue then it indicates that there is presence of nitrogen and if the colour formed is violet then it indicates the presence of sulphur. Hence we can identify whether this has interfered with the reaction while testing the presence of halogens.