Question

For given reaction

$$NaCl + {K_2}C{r_2}{O_7} + conc.{H_2}S{O_4}\xrightarrow{{heat}}{\text{Red coloured gas}}\xrightarrow[{Distillation}]{{NaOH}} \\\ {\text{Yellow solution}}\xrightarrow{{\left[ {Pb{{\left( {C{H_3}COO} \right)}_2}} \right]\left[ {C{H_3}COOH} \right]}}X \\\$$

The formula and colour of X is:
A.$Cr{O_2}C{l_2},red$
B.$C{r_2}{\left( {S{O_4}} \right)_3},green$
C.$PbO,yellow$
D.$PbCr{O_4},yellow$

Answer 1

We have to know that the reaction of sodium chloride and potassium dichromate with sulfuric acid in the presence of heat gives a red coloured gas, which on distillation with sodium hydroxide gives yellow solution. The red coloured gas should indicate compounds of chromium as the compounds of chromium are usually red in colour and the presence of yellow colour indicates the presence of lead compounds.

Complete answer:
The given reaction is,

$$NaCl + {K_2}C{r_2}{O_7} + conc.{H_2}S{O_4}\xrightarrow{{heat}}{\text{Red coloured gas}}\xrightarrow[{Distillation}]{{NaOH}} \\\ {\text{Yellow solution}}\xrightarrow{{\left[ {Pb{{\left( {C{H_3}COO} \right)}_2}} \right]\left[ {C{H_3}COOH} \right]}}X \\\$$

The given reaction is a chromyl chloride test. When we mix salt and the same quantity of solid potassium dichromate in the test tube and add a concentration solution of sulfuric acid. The mixture is heated in the test tube and the gas liberated is passed through the solution of sodium hydroxide. When we obtain a yellow solution, the solution is divided into two portions. The first portion is acidified with acetic acid and lead acetate solution is added. Yellow precipitate of lead chromate indicates the presence of chloride ions in the salt and so, X is lead chromate and its colour is yellow.
The completed chemical reaction is written as,
$4NaCl + {K_2}C{r_2}{O_7} + 6conc.{H_2}S{O_4}\xrightarrow{{heat}}2KHS{O_4} + 2Cr{O_2}C{l_2} + 4NaHS{O_4} + 3{H_2}O$
$Cr{O_2}C{l_2} + 4NaOH \to N{a_2}Cr{O_4} + 2NaCl + 2{H_2}O$
${\left( {C{H_3}COO} \right)_2}Pb + N{a_2}Cr{O_4} \to PbCr{O_4} + 2C{H_3}COONa$
The formula $PbCr{O_4}$ is lead chromate.
Therefore, the option (D) is correct.

Note:
We have to know that one of the qualitative analyses to show the presence of chloride ions is chromyl chloride test. Chlorides of mercury and silver are confirmed using chromyl chloride test. The reason is because the chlorides of silver and chlorides of mercury are covalent, and they do not generate chloride ions. This test is only helpful for compounds containing $C{l^ - }$ ionic bonds.