How to calculate oxidation number of Iron in ferriferrocyanide Fe4[Fe(CN)6]3 ?
2 Answers
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
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
"Fe"^("III")["Fe"^("III")"Fe"^("II")("CN")_6]_3 (s)FeIII[FeIIIFeII(CN)6]3(s)
METHOD 1
In
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
[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
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
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)