# How to calculate oxidation number of Iron in ferriferrocyanide Fe4[Fe(CN)6]3 ?

Oct 5, 2017

see below

#### Explanation:

It is Berlin blue ( Ferric ferrocyanide. Ferric hexacyanoferrate (II)) . Iron can have two oxydation number : +3 and +2.
You can understand that the first iron has the number +three because it needs 3 groups ferrocyanide to make the compound. the groups ferrocyanide have the charge -4, given from 6 ions $C {N}^{-}$ with charge -1 and a atom of iron with charge +2.
So you have some iron with oxydation number + 3 and some with number +2

mathematically you have combining atoms and oxidation numbers:
4 (+3) + 3 [+2 + 6 (-1)] = 0

Oct 5, 2017

"Fe"^("III")["Fe"^("III")"Fe"^("II")("CN")_6]_3 (s)

METHOD 1

In "Fe"_4["Fe"("CN")_6]_3, or Prussian blue, we can consider how it was synthesized:

4"Fe"^(3+)(aq) + 3["Fe"("CN")_6]^(4-)(aq) -> "Fe"["Fe"("Fe"("CN")_6)]_3 (s)

From knowing from the left-hand side what the charge of the ferrocyanide is, and knowing that cyanide is ${\text{CN}}^{-}$, the oxidation state of the innermost inner-sphere iron is $\boldsymbol{+ 2}$:

[bb"Fe"("CN")_6]^(4-)

$\left(+ 2\right) + 6 \left(- 1\right) = - 4$

And similarly, the second-innermost inner-sphere irons must balance out the remaining charge with the outer-sphere iron as follows:

${\text{Fe}}^{3 +}$ with {[bb("Fe")("Fe"("CN")_6)]_3}^(3-)

Each [bb("Fe")("Fe"("CN")_6)] is then a $- 1$, so with each ["Fe"("CN")_6]^(4-) being a $4 -$ charge, the bolded iron is a $+ 3$.

METHOD 2

You can also simply un-criss-cross to find that we have:

${\text{Fe}}^{3 +}$ and ["Fe"("CN")_6]^(4-)

and similar logic leaves you with an ${\text{Fe}}^{+ 3}$ and an ${\text{Fe}}^{+ 2}$.

Therefore, the clearest way to write this is:

color(blue)("Fe"^("III")["Fe"^("III")"Fe"^("II")("CN")_6]_3 (s))