Chemistry - Complete d-block Question with Solution | TestHub

Problem: In which reaction reddish brown vapour is evolved?

Options:

A. ${\mathrm{CrO}}_4^{2-}+{\mathrm{H}}_2{\mathrm{SO}}_{4\text{ (conc.) }}+{\mathrm{H}}_2{\mathrm{O}}_2\to$

B. ${\mathrm{Mn}}^{2+}+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{S}}_2{\mathrm{O}}_8+{\mathrm{H}}^{+}\to$

C. ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7(\mathrm{s})+\text{ conc. }{\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{Cl}}^{-}\stackrel{\Delta }{\to }$

D. ${\mathrm{MnO}}_2+\mathrm{HCl}⟶$

Correct Answer: C

Solution:

Option C involves the chromyl chloride test. When solid potassium dichromate (${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$) is heated with concentrated sulfuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$) and a chloride salt (${\mathrm{Cl}}^{-}$), reddish-brown vapors of chromyl chloride (${\mathrm{CrO}}_2{\mathrm{Cl}}_2$) are evolved.

The reaction is:

$${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7(\mathrm{s})+4\mathrm{NaCl}(\mathrm{s})+6{\mathrm{H}}_2{\mathrm{SO}}_4(\text{ conc. })\stackrel{\Delta }{\to }2{\mathrm{CrO}}_2{\mathrm{Cl}}_2(\mathrm{g})+2{\mathrm{KHSO}}_4(\mathrm{s})+4{\mathrm{NaHSO}}_4(\mathrm{s})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$

Chromyl chloride (${\mathrm{CrO}}_2{\mathrm{Cl}}_2$) is a volatile, reddish-brown liquid that produces reddish-brown vapors.

Option A produces blue peroxochromate (${\mathrm{CrO}}_5$).

Option B produces pink ${\mathrm{MnO}}_4^{-}$ (permanganate ion).

Option D produces greenish-yellow chlorine gas (${\mathrm{Cl}}_2$).

Therefore, the reaction in option C evolves reddish-brown vapor.