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JEE Advance - Chemistry (2009 - Paper 1 Offline - No. 15)

p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue colouration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the formation of Z.
p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue colouration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the formation of Z.
p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue colouration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the formation of Z.
The compound Z is
$$M{g_2}[Fe{(CN)_6}]$$
$$Fe[Fe{(CN)_6}]$$
$$F{e_4}{[Fe{(CN)_6}]_3}$$
$${K_2}Z{n_3}{[Fe{(CN)_6}]_2}$$

Explanation

The compound X is H$$_2$$S:

$$\mathrm{\mathop {N{a_2}S}\limits_{(X)} + 2{H^ + } \to {H_2}S + 2N{a^ + }}$$

IIT-JEE 2009 Paper 1 Offline Chemistry - Salt Analysis Question 4 English Explanation 1

Thus, the compound Y is FeCl$$_3$$.

Compound Y on reaction with potassium hyxacyanoferrate(II) forms intense blue precipitate which dissolves on addition of reagent

$$\mathrm{4FeC{l_3} + 3{K_4}\left[ {Fe(II){{(CN)}_6}} \right] \to \mathop {F{e_4}{{[Fe{{(CN)}_6}]}_3}}\limits_{Intense\,blue\,ppt.} + 12KCl}$$

Compound Y on reaction with potassium hexacyanoferrate (III) forms brown colouration due to

$$\mathrm{Fe[Fe(CN)_6]}$$

IIT-JEE 2009 Paper 1 Offline Chemistry - Salt Analysis Question 4 English Explanation 2

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