Q. The chemical formula of Prussian blue is formed by the reaction of ferric ion and ferrocyanide is

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A

$K_{4}\left[\right.Fe\left(\right.CN\left(\left.\right)_{6}\left]\right.$

B

$Na_{4}\left[\right.Fe\left(\right.CN\left(\left.\right)_{6}\left]\right.$

C

$Fe_{4}\left[\right.Fe\left(\right.CN\left(\left.\right)_{6}\left(\left]\right.\right)_{3}$

D

None of these

Solution:

The chemical formula of Prussian blue is $Fe_{4}\left[\right.Fe\left(\right.CN\left(\left.\right)_{6}\left(\left]\right.\right)_{3}$ .
$Fe^{3 +}$ reacts with sodium ferrocyanide to give Ferric hexacyanoferrate(II) (Prussian blue) as follows
$4 \left(\text{Fe}\right)^{3 +} + 3 \left(\text{Na}\right)_{4} \left[\right. \text{Fe} \left(\text{CN} \left(\left.\right)_{6}\right] \rightarrow \underset{\text{Prussian blue}}{\left(\text{Fe}\right)_{4} \left(\left[\right. \text{Fe} \left(\right. \text{CN} \left.\right)_{6} \left]\right.\right)_{3}} + 12 \left(\text{Na}\right)^{+}$

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