On heating $NaCl + {K_2}C{r_2}{O_7} + conc.{H_2}S{O_4}$ ​, the gas comes out is
A. ${O_2}$
B. $C{l_2}$
C. $CrOC{l_2}$
D. $Cr{O_2}C{l_2}$

Hint: We must've been conscious that sodium chloride and potassium dichromate react with sulfuric acid when heated, producing a reddish gas that, when distilled with sodium hydroxide, yields a yellow solution. Given that chromium compounds are typically red in colour and lead compounds are typically yellow in colour, the reddish gas should be associated with chromium compounds.

Complete Step by Step Solution:
The test is a chromyl chloride reaction. When we combine salt with the same amount of solid potassium dichromate in a test tube, then we add a sulfuric acid concentration solution.

In a test tube, the combination is heated, and the gas that is released is passed through a sodium hydroxide solution. The solution is split into two parts once we get a yellow solution. Acetic acid is used to acidify the first portion, and then lead acetate solution is added. Lead chromate yellow precipitate is a sign that chloride ions are present in the salt.

The final chemical reaction is denoted by the following:
$4NaCl + {K_2}C{r_2}{O_7} + 6{H_2}S{O_4} \to 2Cr{O_2}C{l_2} + 2KHS{O_4} + 4NaHS{O_4} + 3{H_2}O$
Therefore, the option (D) is correct.

Note: We must be aware that the chromyl chloride test is one of the qualitative procedures to demonstrate the presence of chloride ions. The chromyl chloride test is used to establish the presence of mercury and silver chlorides. The covalent nature of silver and mercury chlorides, which prevents them from producing chloride ions, is the source of this. This test is only helpful for compounds containing $C{l^ - }$ ionic bonds.