$F e^{2 +}$ does not give prussian blue colour with $K_{4} [F e (C N)_{6}]$ , but on its reaction with (X) prussian blue colour appears with the reagent. (X) is :

M n O_{4}^{-} / H^{+}

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H_{2} S O_{4}

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N H_{3}

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all of above

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Solution

The correct option is B $M n O_{4}^{-} / H^{+}$
$F e^{2 +}$ gives prussian blue colour with $M n O_{4}^{-} / H^{+}$ . They acts as an oxidising agent, converts ferrous to ferric ion. This ferric ion reacts with $K_{4} [F e (C N)_{6}]$ to form prussian blue colour of ferric ferro cyanide.
$M n O_{4} / H^{+} \to [O]; F e^{2 +} \to F e^{3 +}$
$F e^{3 +} + K_{4} [F e (C N)_{6}] \to F e_{4} [F e (C N)_{6}]_{3} + 4 K^{+}$

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