Turnbull's blue and Prussian's blue respectively are ${K F e}^{I I} {[{F e}^{I I I} (C N)}_{6}], {K F e}^{I I I} {[{F e}^{I I} (C N)}_{6}]$

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Solution

Prussian blue and Turnbull's blue are identical in structure
Turnbull's blue and Prussian's blue respectively are ${K F e}^{I I} {[{F e}^{I I I} (C N)}_{6}], {K F e}^{I I I} {[{F e}^{I I} (C N)}_{6}]$
Fe(III) ions give deep blue colour with potassium ferrocyanide.
$3 K_{4} [F e (C N)_{6}] + 4 F e C l_{3} \to F e_{4} [F e (C N)_{6}]_{3} p u r s s i a n^{'} s b l u e + 12 K C l$
Turnbull's blue is KFe^{II}[Fe^{III}(CN)_6]
$F e S O_{4} + K_{3} [F e (C N)_{6}] ⟶ K F e^{I I} [F e^{I I I} (C N)_{6}] (H), T u r n b u l l^{'} s b l u e$

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