When a mixture of \(K_2Cr_2O_7\) and \(NaCl\) is warmed with concentrated \(H_2SO_4\) red vapours of \((A)\) are evolved, which turn \(NaOH\) yellowish due to formation of compound \((B)\). \(A\) and \(B\) respectively are :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A: CrO2Cl2 and B: Na2CrO4
This is the chromyl chloride test for chloride ions. Warming a chloride with \( \ce{K2Cr2O7} \) and concentrated \( \ce{H2SO4} \) gives orange-red vapours of chromyl chloride:
\( \ce{K2Cr2O7 + 4NaCl + 6H2SO4 -> 2CrO2Cl2 ^ + 2KHSO4 + 4NaHSO4 + 3H2O} \)
When the vapours are passed into NaOH solution, the yellow sodium chromate is formed:
\( \ce{CrO2Cl2 + 4NaOH -> Na2CrO4 + 2NaCl + 2H2O} \)
Why the other options are incorrect
Option (1) CrOCl2 — Incorrect. The red vapours are \( \ce{CrO2Cl2} \) (chromium in the +6 state); \( \ce{CrOCl2} \) is not formed in this test.
Option (2) CrO5 and Na2Cr2O7 — Incorrect. \( \ce{CrO5} \) (blue chromium peroxide) is formed with \( \ce{H2O2} \), not with NaCl; and in alkaline NaOH the dichromate changes to yellow chromate, not the reverse.
Option (4) CrO5 and Na2Cr2O4 — Incorrect. \( \ce{CrO5} \) forms only with \( \ce{H2O2} \). \( \ce{Na2Cr2O4} \) is not a real chromate: sodium chromate is \( \ce{Na2CrO4} \) and sodium dichromate is \( \ce{Na2Cr2O7} \).