Question

Salt (A) gives brick red fumes (B) with conc.$H_{2}S{O}_4\text{and}{K}_{2}C{r}_2{O}_7$ which gives yellow solution (C) with $NaOH$ and it gives yellow precipitate (D) with acetic acid and lead acetate. What is the compound (C) ?

• A.
  $N{a}_{2}Cr{O}_4$
• B.
  $Cr{O}_{2}C{l}_2$
• C.
  $PbCr{O}_4$
• D.
  $NaCl$

Answer 1

The correct option is A

$N{a}_{2}Cr{O}_4$
$C{l}^{-}$ ion is confirmed by chromyl chloride test.
The given procedure is of chromyl chloride test.
Chromyl chloride (B) is produced when a chloride ion reacts with conc.$H_{2}S{O}_4\text{and}{K}_{2}C{r}_2{O}_7$.
Passing the chromyl chloride gas through $NaOH$ turns the solution yellow. This is due to the formation of $N{a}_{2}Cr{O}_4$ (C).
Treating $N{a}_{2}Cr{O}_4$ with with acetic acid and lead acetate forms yellow precipitate of lead chromate (D).

$\text{Chromyl chloride Test}:C{l}^{-}+conc.H_{2}S{O}_4+K_{2}C{r}_2{O}_7\to \underset{Brickred\left(B\right)}{Cr{O}_{2}C{l}_2}\uparrow Cr{O}_{2}C{l}_2+NaOH\to \underset{Yellowsolution\left(C\right)}{N{a}_{2}Cr{O}_4}+HClN{a}_{2}Cr{O}_4+Pb(C{H}_{3}COO{)}_2\to \underset{yellowprecipitate\left(D\right)}{PbCr{O}_4}$