Question

\[Cr{O_2}C{l_2}\] is formed while testing:
A) $N{O_3}^ - $
B) $C{l^ - }$
C) $C{r^{3 + }}$
D) $F{e^{3 + }}$

Answer 1

Hint: Chromyl chloride is a chemical compound and its chemical formula is given as \[Cr{O_2}C{l_2}\]. They are dark red blood color liquid where the molecules of chromyl chloride are tetrahedral in shape.

Complete step by step answer:
Chromyl Chloride is formed during Chromyl Chloride test. Chromyl chloride test is a qualitative analysis test used for the conformation for $C{l^ - }$ ions. Let’s take an example to further get a clearer understanding of the test. A sample of chlorine-containing salt is heated with potassium chromate ${K_2}C{r_2}{O_7}$ and concentrated sulfuric acid (${H_2}S{O_4}$ ). If chloride is present, chromyl chloride is formed and red fumes are given out. Chromyl chloride test reaction is given as:
${K_2}C{r_2}{O_7} + 4NaCl + 6{H_2}S{O_4} \to 2Cr{O_2}C{l_2} + 2KHS{O_4} + 4NaHS{O_4} + 3{H_2}O$
Chromyl Chloride Test Mechanism:
The mechanism of chromyl chloride test is simple. On reacting potassium dichromate with sulphuric acid, chromate trioxide (oxidation state = +6) is formed. The color of chromate trioxide ($Cr{O_3}$) varies from dark red to brown color.
${K_2}C{r_2}{O_7} + {H_2}S{O_4} \to Cr{O_3} + {H_2}O$
Now salt-containing chloride ($NaCl$) is reacted with sulphuric acid that gives sodium bisulphate $NaHS{O_4}$ and hydrochloric acid ($HCl$ ) is also formed.
$NaCl + {H_2}S{O_4} \to NaHS{O_4} + HCl$
In the next step, chromate trioxide is reacted with hydrochloric acid, which produces chromyl chloride (\[Cr{O_2}C{l_2}\]) which gives out red fumes.
$Cr{O_3} + HCl \to Cr{O_2}C{l_2}$

So the correct option is (B).

Note: For salts such as chlorides of mercury and silver chromyl chloride test is not applicable. This is because the chlorides of mercury and silver are covalent and they do not generate $C{l^ - }$ ions. The chromyl chloride test is applicable only for compounds having $C{l^ - }$ ionic bonds.