Question

Give the test for \[C{l^ - }\] radical.

Answer 1

Hint: Chloride radical is the acidic radical of the second group in which concentrated sulphuric acid is used as a group reagent. \[HCl\] gas is liberated when conc. \[{H_2}S{O_4}\] is reacted with the salt containing \[C{l^ - }\] radical. The presence of this radical can also be detected through some confirmatory tests like silver nitrate test or lead acetate test.

Complete step by step answer:
\[C{l^ - }\] radical is detected through some identification and confirmatory tests.
Solubility test: In this test, the salt is dissolved in distilled water. If the salt is readily soluble in distilled water without heating then there is a possibility that \[C{l^ - }\] radical may be present in the given salt.
Concentrated Sulphuric Acid Test: When the given salt containing \[C{l^ - }\]radical is reacted with conc. \[{H_2}S{O_4}\] , the chloride ions get displaced forming sodium bisulfite and hydrochloric acid in gaseous form.
\[NaCl + H2SO4 \to NaHSO4 + HCL\]
The liberated \[HCl\] gas can be detected when it comes in contact with ammonium hydroxide, it will produce white fumes of ammonium chloride.
\[HCl + N{H_4}OH \to N{H_4}Cl + {H_2}O\]
Silver Nitrate Test: When the salt containing chloride radical is added to a solution of silver nitrate, formation of white precipitate of \[AgCl\] occurs due to low solubility of \[AgCl\].
\[NaCl + AgN{O_3} \to NaN{O_3} + AgCl\]
Lead Acetate Test: When the given salt containing chloride radical is reacted with a solution of lead acetate, formation of white precipitate of lead chloride occurs due to the low solubility of \[PbC{l_2}\].
\[NaCl + {\left( {C{H_3}COO} \right)_2}Pb \to 2C{H_3}COONa + PbC{l_2}\]
Chromyl Chloride Test: This is the specific confirmatory test for detection of chloride radical in the given salt. When a chloride containing salt is reacted with acidified potassium dichromate salt, the orange coloured vapours of chromyl chloride gas \[\left( {Cr{O_2}C{l_2}} \right)\] is released.
\[{K_2}C{r_2}{O_7} + 4NaCl + 6{H_2}S{O_4} \to 2KHS{O_4} + 4NaHSO_4 + Cr{O_2}C{l_2}\]

Note: It is necessary to follow step by step analysis for any acidic radical starting from general identification tests to specific confirmatory tests. \[C{l^ - }\] radical does not react with dilute sulphuric acid to produce a characteristic gas.