
The incorrect statement in respect of chromyl chloride test is:
A: Formation of red vapours
B: Formation of lead chromate
C: Formation of chromyl chloride
D: Liberation of chlorine
Answer
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Hint: The chromyl chloride test refers to a chemical test for the detection of chloride ions qualitatively. Any chloride salt, for instance, NaCl, when undergoes a heat treatment in the presence of acidified potassium dichromate as well as concentrated sulphuric acid, the orange red coloured fumes of chromyl chloride evolves out, thus confirming the existence of chloride ions. If no chloride is present in the salt, no red fumes will be produced and no corresponding compounds will be formed with fluorides, cyanides, bromides, and iodides.
Complete answer:
> Step 1: Take a small amount of soluble chloride salt with an equal amount of powdered \[{K_2}C{r_2}{O_7}\].
> Step 2: Heat the mixture with conc. \[{H_2}S{O_4}\].
\[{K_2}C{r_{2}}{O_7} + {\text{ }}4KCl{\text{ }} + {\text{ }}6{H_2}S{O_4} \to {\text{ }}2Cr{O_2}C{l_2}{\text{ }} + {\text{ }}6KHS{O_4}{\text{ }} + {\text{ }}3{H_2}O\]
> Step 3: The orange red vapours of formed chromyl chloride (\[Cr{O_2}C{l_2}\]) is passed through the NaOH solution taken in another test tube.
\[Cr{O_2}C{l_2}{\text{ }} + {\text{ }}4NaOH \to 2N{a_2}Cr{O_4} + {\text{ }}2NaCl{\text{ }} + {\text{ }}2{H_2}O\]
> Step 4: The solution turns yellow.
> Step 5: Add acetic acid to the solution and then lead acetate solution.
\[N{a_2}Cr{O_4} + {\text{ }}{\left( {C{H_3}COO} \right)_2}Pb \to PbCr{O_4} + {\text{ }}2C{H_3}COONa\]
> Step 6: Yellow PPT (\[PbCr{O_4}\]) is formed.
As you can clearly see that the option A is the correct statement because red vapours are formed during step 3. Similarly option B and C are also correct statements because lead chromate is formed in step 5 while chromyl chloride is formed in step 2.
As a result, the incorrect statement in respect of chromyl chloride test is (D) i.e. Liberation of chlorine.
Note: Exposure to chromyl chloride vapour causes irritation in the respiratory system and also severely irritates the eyes. Ingestion can also cause severe internal damage.
Complete answer:
> Step 1: Take a small amount of soluble chloride salt with an equal amount of powdered \[{K_2}C{r_2}{O_7}\].
> Step 2: Heat the mixture with conc. \[{H_2}S{O_4}\].
\[{K_2}C{r_{2}}{O_7} + {\text{ }}4KCl{\text{ }} + {\text{ }}6{H_2}S{O_4} \to {\text{ }}2Cr{O_2}C{l_2}{\text{ }} + {\text{ }}6KHS{O_4}{\text{ }} + {\text{ }}3{H_2}O\]
> Step 3: The orange red vapours of formed chromyl chloride (\[Cr{O_2}C{l_2}\]) is passed through the NaOH solution taken in another test tube.
\[Cr{O_2}C{l_2}{\text{ }} + {\text{ }}4NaOH \to 2N{a_2}Cr{O_4} + {\text{ }}2NaCl{\text{ }} + {\text{ }}2{H_2}O\]
> Step 4: The solution turns yellow.
> Step 5: Add acetic acid to the solution and then lead acetate solution.
\[N{a_2}Cr{O_4} + {\text{ }}{\left( {C{H_3}COO} \right)_2}Pb \to PbCr{O_4} + {\text{ }}2C{H_3}COONa\]
> Step 6: Yellow PPT (\[PbCr{O_4}\]) is formed.
As you can clearly see that the option A is the correct statement because red vapours are formed during step 3. Similarly option B and C are also correct statements because lead chromate is formed in step 5 while chromyl chloride is formed in step 2.
As a result, the incorrect statement in respect of chromyl chloride test is (D) i.e. Liberation of chlorine.
Note: Exposure to chromyl chloride vapour causes irritation in the respiratory system and also severely irritates the eyes. Ingestion can also cause severe internal damage.
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