I came across a compound $\ce{[FeO4]^{2-}}$ but $\ce{Fe}$ has +6 charge according to my calculations. How this is possible ? Or it is possible but my periodic table is kind of non-detailed one. (On the table I got $d$-metals possible charges).
$\begingroup$
$\endgroup$
asked Jun 1, 2013 at 20:09
ordinary chemistry studentordinary chemistry student
8901 gold badge10 silver badges21 bronze badges
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
If you write out the electron configuration of Fe, you will find it has $3\text{d}^6 4\text{s}^2$ in its valence level, meaning it can theoretically take on any oxidation state from +1 to +8 (and even some negative states, for that matter).
In practice it is extremely uncommon to find any oxidation states other than +2 or +3 for iron - which explains why a simple periodic table containing only the most common naturally occurring states will omit them - but they are possible, just not very stable or long-lived. The polyatomic ion you have observed is called ferrate.
-
1$\begingroup$ I've always enjoyed citing catalase as an example of a ubiquitous compound where iron takes on uncommon oxidation values. $\endgroup$ Commented Jun 1, 2013 at 20:40
-
$\begingroup$ Actually iron(VI) is not all that exotic, being in use commercially in water treatment. If we regard each oxide ligand as donating a pi electron pair as well as a sigma pair (a reasonable model for oxide ligands to electron-deficient metal-ion centers), then ferrate(VI) meets the 18-electron rule. $\endgroup$ Commented May 28, 2023 at 21:11