A substance 'A' has green hydroxide but blue ferricyanide. Another substance 'B' has blue hydroxide but brown ferrocyanide. Substance 'A' and 'B' are, respectively :

F e S O_{4}

and

N i S O_{4}

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C u S O_{4}

and

N i S O_{4}

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C u S O_{4}

and

Z n S O_{3}

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F e S O_{4}

and

C u S O_{4}

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Solution

The correct option is B $F e S O_{4}$ and $C u S O_{4}$
Ferric hydroxide is green in colour which is obtained by the reaction of ferrous sulphate with caustic soda.

$F e S O_{4} + 2 N a O H \to F e (O H)_{3} (green colour) + N a_{2} S O_{4}$

When the solution of $F e S O_{4}$ is treated with potassium ferrocyanide solution, a blue coloured complex of ferric ferrocyanide is formed.

$F e S O_{4} + 3 K_{4} [F e (C N)_{6}] \to F e_{4} [F e (C N)_{6}] (blue colour) + K_{2} S O_{4}$

Copper hydroxide is blue in colour. When the solution of $C u S O_{4}$ is treated with potassium ferrocyanide solution, a brown coloured complex of copper ferrocyanide is formed.

$C u S O_{4} + 2 N a O H \to C u (O H)_{2} (blue colour) + N a_{2} S O_{4}$

$C u S O_{4} + K_{4} [F e (C N)_{6}] \to C u_{2} [F e (C N)_{6}] (brown colour) + K_{2} S O_{4}$

Hence, option D is correct.

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Similar questions

The colours of aqueous solutions of CuSO₄ and FeSO₄ as observed in the laboratory are:

K_{4} [F e (C N)_{6}]

F e C l_{3}

N H_{4} O H

C u S O_{4}

C u S O_{4}

N H_{4} O H + N i S O_{4}

C u {S O}_{4} \cdot 5 H_{2} O

F e {S O}_{4} \cdot 7 H_{2} O

In the following reaction, identify the substance, ‘X’ from the given options.
$F e + C u S O_{4} \to X + C u$

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