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

2 Answers
Oct 5, 2017

Answer:

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

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 #"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)#.


Therefore, the clearest way to write this is:

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