Question

Explain the chromyl chloride test with an equation.

Answer 1

Hint: The chromyl chloride test is used to detect chloride ions in the qualitative analysis. If any chloride salt like sodium chloride is heated with acidified potassium dichromate it produces red colour fumes of chromyl chloride. It confirms the presence of chloride ions in that salt.

Complete step by step answer:
Now let’s start the understanding of chromyl chloride test in detail.
- A little amount of salt-containing chloride ion e.g. NaCl is mixed with an equal amount of solid potassium dichromate in a test tube along with concentrated sulphuric acid.
\[\underset{\text{Sodium Chloride}}{\mathop{4NaCl}}\,+\underset{\text{Potassium Chromate}}{\mathop{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}\,+\underset{\text{conc}\text{. Sulphuric Acid}}{\mathop{6{{H}_{2}}S{{O}_{4}}}}\,\to \underset{\text{Potassium bisulphate}}{\mathop{2KHS{{O}_{4}}}}\,+\underset{\text{Chromyl Chloride (Red fumes)}}{\mathop{2Cr{{O}_{2}}C{{l}_{2}}}}\,+\underset{\text{Sodium Bisulphate}}{\mathop{4NaHS{{O}_{4}}}}\,+\underset{\text{Water}}{\mathop{3{{H}_{2}}O}}\,\]

- Then it is heated for some time which leads to the evolution of red fumes of chromyl chloride ($Cr{{O}_{2}}C{{l}_{2}}$). When this gas is passed through a sodium hydroxide solution then a yellow solution is obtained due to the presence of sodium chromate ($N{{a}_{2}}Cr{{O}_{4}}$).
\[\underset{\text{Chromyl Chloride}}{\mathop{Cr{{O}_{2}}C{{l}_{2}}}}\,+\underset{\text{Sodium Hydroxide}}{\mathop{4NaOH}}\,\to \underset{\text{Sodium Chromate (Yellow)}}{\mathop{N{{a}_{2}}Cr{{O}_{4}}}}\,+\underset{\text{Sodium Chloride}}{\mathop{2NaCl}}\,+\underset{\text{Water}}{\mathop{2{{H}_{2}}O}}\,\]

- Then the solution is divided into two parts:
- When we take the first part and acidify it with acetic acid and lead acetate solution then there is a formation of a yellow precipitate of lead chromate (\[PbCr{{O}_{3}}\]) which confirms the presence of chloride ions in the salt this chromyl chloride test.
\[\underset{\text{Lead Acetate}}{\mathop{{{\left( C{{H}_{3}}COO \right)}_{2}}Pb}}\,+\underset{\text{Sodium Chromate}}{\mathop{N{{a}_{2}}Cr{{O}_{4}}}}\,\xrightarrow{C{{H}_{3}}COOH}\underset{\text{Lead Chromate}}{\mathop{PbCr{{O}_{4}}}}\,\downarrow +2\underset{\text{Sodium Acetate}}{\mathop{C{{H}_{3}}COONa}}\,\]

- The second part also contains the chromate ions ($CrO_{4}^{-2}$) formed by the dissociation of sodium chromate ($N{{a}_{2}}Cr{{O}_{4}}$). We take the second part and acidify it. After that, we add small parts of alcohol in it and 1 ml of 10% hydrogen peroxide (${{H}_{2}}{{O}_{2}}$). The chromate ions ($CrO_{4}^{-2}$) reacts with hydrogen peroxide in the presence of the acid to form chromium pentoxide ($Cr{{O}_{5}}$). It dissolves in the alcohol added to form a blue coloured solution.
\[\begin{align}
  & \underset{\text{Chromate Ion}}{\mathop{CrO_{4}^{-2}}}\,+\underset{\text{Acid}}{\mathop{2{{H}^{+}}}}\,+\underset{\text{Hydrogen Peroxide}}{\mathop{2{{H}_{2}}{{O}_{2}}}}\,\to \underset{\text{Chromium Pentoxide}}{\mathop{Cr{{O}_{5}}}}\,+\underset{\text{Water}}{\mathop{3{{H}_{2}}O}}\, \\
 & Cr{{O}_{5}}\xrightarrow{\text{Alcohol}}\text{ Blue Colour} \\
\end{align}\]
This is the chromyl chloride test.

Note: Remember that for the test to come positive, there should be an ample amount of chloride ions in the solution. That means the chloride salt which will undergo this test must dissociate enough to produce the chloride ions that is necessary for this test. Chloride salts like mercuric ($HgC{{l}_{2}}$) and mercurous ($H{{g}_{2}}C{{l}_{2}}$) chloride will not give this test because they undergo very less or no dissociation in the reaction mixture. For these salts, the chromyl chloride test is carried out by preparing the soda extract first.