Question

The correct statement(s) is/are with respect to chromyl chloride test.
A.formation of lead chromate
B.formation of chromyl chloride
C.liberation of chloride
D.formation of reddish brown vapours

Answer 1

Hint:In the qualitative analysis, various kinds of tests are done to detect the type of ions present in the given. There are two types of ions present in the inorganic salt i.e. cation and anion e.g. $\text{N}{{\text{a}}^{+}}\text{C}{{\text{l}}^{-}}$contains$\text{N}{{\text{a}}^{+}}$and$\text{C}{{\text{l}}^{-}}$.

Complete answer:
Chromyl chloride test is used for the conformation of $\text{C}{{\text{l}}^{-}}$ion present in the inorganic salt.
Chromyl Chloride test: When the given sample is treated with potassium dichromate ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$and conc.${{\text{H}}_{2}}\text{S}{{\text{o}}_{4}}$, reddish orange vapours of chromyl chloride are obtained.
${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}+\text{Kcl}+\text{6}{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\to \text{2Cr }{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}+\text{6KHS}{{\text{O}}_{4}}+3{{\text{H}}_{2}}\text{O}$
Potassium Dichromate Potassium Chloride And Sulphuric acid Chromyl Chloride
Here the chromyl chloride is mixture of $\text{Cr}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}$ &$\text{KHS}{{\text{O}}_{4}}$
 conformation of chromyl chloride test, the red vapour formed dissolves in a solution of sodium hydroxide (NaOH) The solution turns yellow (due to$\text{N}{{\text{a}}_{2}}\text{Cr}{{\text{O}}_{4}}$).
$\text{Cr}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}+\text{NaOH}\to \text{Na2Cr}{{\text{O}}_{4}}+\text{NaCl}+{{\text{H}}_{2}}\text{O}$
Reacts this solution further with the lead acetate and diluted acetic acid (CH3COOH) produced by the yellow precipitate.
$\text{Cr}{{\text{O}}^{42-}}+\text{Pb}{{\left( \text{C}{{\text{H}}_{3}}\text{COOH} \right)}^{2}}\to \text{PbCr}{{\text{O}}_{4}}+\text{C}{{\text{H}}_{3}}\text{COONa}$
$\text{PbCr}{{\text{O}}_{4}}$ is the yellow precipitate of chloride and hence the test is accurate.

Hence, Option (A) (B) and (D) is the correct option i.e. formation of chromyl chloride takes place, formation of lead chromate and formation of chromyl chloride takes place.
Note:
Chromyl Chloride and Its Properties: Chromyl chloride is a chemical compound and its chemical formula is given as$\text{Cr}{{\text{O}}_{2}}\text{C}{{\text{l}}_{12}}$. Chromyl chloride is dark red blood colour liquid where the molecules of chromyl chloride are tetrahedral. They are mostly chromium (IV) derivatives${{\left( \text{Cr}{{\text{O}}_{4}} \right)}^{2+}}$.
Chemical Property: It is often used as an oxidizing agent. The chromyl chloride can be used for the oxidation of toluene to Benzaldehyde. This reaction is followed by two steps:
1. On reacting chromyl chloride with benzene a chromyl compound is formed.
${{\text{C}}_{6}}{{\text{H}}_{5}}\text{C}{{\text{H}}_{3}}+2\text{Cr}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}\to {{\text{C}}_{6}}{{\text{H}}_{5}}\text{CH}{{\left( \text{Cr}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}} \right)}^{2}}$
2. This chromyl compound is hydrolyzed with water to give the Benzaldehyde.
${{\text{C}}_{6}}{{\text{H}}_{5}}\text{CH}{{\left( \text{Cr}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}} \right)}^{2}}+2{{\text{H}}_{2}}\text{O}\to {{\text{C}}_{6}}{{\text{H}}_{5}}\text{CH}=\text{O}+\text{Cr}{{\text{O}}_{3}}+4\text{HCl}$
This reaction is known as Etard’s Reaction and the chromyl chloride that is used here is a mild oxidizing agent which is excellent in preparation of aldehydes.
When Chromyl chloride reacts with water it results in formation of chromic acid and hydrochloric acid. This reaction is exothermic.
$\text{Cr}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}+2{{\text{H}}_{2}}\text{O}\to {{\text{H}}_{2}}\text{Cr}{{\text{O}}_{4}}+2\text{HCl}$