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Coordination Polyhedra

Turnbull's bl...

Question

Turnbull's blue and Prussian's blue respectively are :
(I). ${F e}^{I I} [{F e}^{I I} ({C N)}_{6}]^{2 -}$
(II). ${F e}^{I I I} [{F e}^{I I I} ({C N)}_{6}]$
(III). ${F e}^{I I} [{F e}^{I I I} ({C N)}_{6}]^{-}$
(IV). ${F e}^{I I I} [{F e}^{I I} ({C N)}_{6}]^{-}$

I, II

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I, III

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III, IV

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IV, III

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Solution

The correct option is C III, IV
Turnbull's blue and Prussian's blue are ${F e}^{I I} [{F e}^{I I I} ({C N)}_{6}]^{-}$ and ${F e}^{I I I} [{F e}^{I I} ({C N)}_{6}]^{-}$ respectively.
$3 F e C l_{2} + 2 K_{3} [F e (C N)_{6}] ⟶ F e_{3}^{I I} [F e^{I I I} (C N)_{6}]_{2} T u r n b u l l^{'} s b l u e + 6 K C l$
$3 K_{4} [F e (C N)_{6}] + 4 F e C l_{3} \to F e_{4} [F e (C N)_{6}]_{3} p u r s s i a n^{'} s b l u e + 12 K C l$

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