When a mixture of \(NaCl\) and \(K_2Cr_2O_7\) is gently warmed with conc.\(H_2SO_4\)?
(A) Deep red vapours are evolved
(B) The vapours when passed through \(NaOH\) solution give yellow solution of \(Na_2CrO_4\)
(C) Chlorine gas is evolved
(D) Chromyl chloride is formed
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A, B and D only
This is the chromyl chloride test for chloride ions. When a chloride is warmed with \( \ce{K2Cr2O7} \) and concentrated \( \ce{H2SO4} \), chromyl chloride, CrO2Cl2, distils off as deep red (orange-red) vapours:
\( \ce{K2Cr2O7 + 4NaCl + 6H2SO4 -> 2CrO2Cl2 ^ + 2KHSO4 + 4NaHSO4 + 3H2O} \)
Chromium stays +6 (in \( \ce{Cr2O7^{2-}} \): 2x + 7(−2) = −2, x = +6; in \( \ce{CrO2Cl2} \): x + 2(−2) + 2(−1) = 0, x = +6). No redox occurs, so the chloride is not oxidised.
When the vapours are passed into NaOH solution, a yellow solution of sodium chromate is obtained:
\( \ce{CrO2Cl2 + 4NaOH -> Na2CrO4 + 2NaCl + 2H2O} \)
(A) Deep red vapours are evolved — Correct. These are the vapours of \( \ce{CrO2Cl2} \).
(B) The vapours when passed through NaOH solution give yellow solution of \( \ce{Na2CrO4} \) — Correct. Chromate ion is yellow.
(C) Chlorine gas is evolved — Incorrect. The chlorine leaves as part of \( \ce{CrO2Cl2} \); chromium is not reduced, so no \( \ce{Cl2} \) is set free.
(D) Chromyl chloride is formed — Correct. \( \ce{CrO2Cl2} \) is the red vapour formed in the first equation; chromium stays in the +6 state.
Note: The question is from beyond the NCERT textbook.