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

2 Answers
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 CN^-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

Oct 5, 2017

"Fe"^("III")["Fe"^("III")"Fe"^("II")("CN")_6]_3 (s)FeIII[FeIIIFeII(CN)6]3(s)


METHOD 1

In "Fe"_4["Fe"("CN")_6]_3Fe4[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)4Fe3+(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"^(-)CN, the oxidation state of the innermost inner-sphere iron is bb(+2)+2:

[bb"Fe"("CN")_6]^(4-)[Fe(CN)6]4

(+2) + 6(-1) = -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+)Fe3+ with {[bb("Fe")("Fe"("CN")_6)]_3}^(3-){[Fe(Fe(CN)6)]3}3

Each [bb("Fe")("Fe"("CN")_6)][Fe(Fe(CN)6)] is then a -11, so with each ["Fe"("CN")_6]^(4-)[Fe(CN)6]4 being a 4-4 charge, the bolded iron is a +3+3.

METHOD 2

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

"Fe"^(3+)Fe3+ and ["Fe"("CN")_6]^(4-)[Fe(CN)6]4

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


Therefore, the clearest way to write this is:

color(blue)("Fe"^("III")["Fe"^("III")"Fe"^("II")("CN")_6]_3 (s))FeIII[FeIIIFeII(CN)6]3(s)