When a mixture of NaCl and $K_{2} C r_{2} O_{7}$ is gently warmed with $conc . H_{2} {SO}_{4}$
(i) Deep –red vapour is evolved
(ii) The vapour formed when passed into NaOH solution gives yellow precipitate of $N a_{2} C r O_{4}$
(iii) Chlorine gas is not evolved at all
(iv) Chromyl chloride is formed
(v) $S n C l_{4}$ used instead of NaCl also gives same observations
Then identify the correct option?

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All statements are correct

three statements are correct

two statements are correct

None of the statements are correct

answer is B.

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Detailed Solution

i) $4 C l^{-} + C r_{2} O_{7}^{2 -} + 6 H^{+} \to 2 C r_{2} O_{2} C l_{2} ↑ + 3 H_{2} O$

$C r O_{2} C l_{2} + 4 (O H^{-}) \to C r O_{4}^{2 -} + 2 C l^{-} + 2 H_{2} O$

$6 C l^{-} + C r_{2} O_{7}^{2 -} + 14 H^{+} \to 3 C l_{2} ↑ 2 C r^{3 +} + 7 H_{2} O$

ii) $S n C l_{4}$ -do not respond to chromyl chloride test

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