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Fe 2+ gives b...

Question

${F e}^{2 +}$ gives blue colour, called Turnbull's blue with $K_{3} [{F e (C N)}_{6}]$ While ${F e}^{3 +}$ gives blue colour, called Prussian blue with $K_{4} [{F e (C N)}_{6}]$

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Solution

${F e}^{2 +}$ gives blue colour, called.Turnbull's blue. with $K_{3} [{F e (C N)}_{6}]$ . While ${F e}^{3 +}$ gives blue colour, called.Prussian blue with. $K_{4} [{F e (C N)}_{6}]$
$F e S O_{4} (C) + K_{3} [F e (C N)_{6}] ⟶ K F e^{I I} [F e^{I I I} (C N)_{6}] (H), T u r n b u l l^{'} s b l u e$
$3 K_{4} [F e (C N)_{6}] + 4 F e C l_{3} \to F e_{4} [F e (C N)_{6}]_{3} p u r s s i a n^{'} s b l u e + 12 K C l$

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