Question

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

Answer 1

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 `CN^-` 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

Answer 2

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

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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 `"CN"^(-)`, the oxidation state of the innermost inner-sphere iron is `bb(+2)`:

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

`(+2) + 6(-1) = -4`

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

`"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:

`"Fe"^(3+)` and `["Fe"("CN")_6]^(4-)`

and similar logic leaves you with an `"Fe"^(+3)` and an `"Fe"^(+2)`.

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Therefore, the clearest way to write this is:

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