What oxidation state does #"Cr"# have in #"CrO"_5#?

1 Answer
Truong-Son N.
Jan 22, 2018

Well, shouldn't it be #+6#? #"Cr"# only has six electrons in its valence orbitals.

As such, one must know ahead of time that the structure is going to contain some peroxides, i.e. #"O"_2^(2-)# ligands. There are, in fact, two of those, and one #"O"^(2-)#, an oxide.

https://upload.wikimedia.org/

Thus, we have an oxidation state based on having one #"O"^(2-)# and two #"O"_2^(2-)# ligands, making #"Cr"# have a #+6# oxidation state.

#stackrel(color(blue)(+6))("Cr")stackrel(color(blue)(-2))(("=O"))stackrel(color(blue)(" "2xx-2))(("O"_2)_2)#

This is, of course, an unstable oxide, and it decomposes readily.

#"CrO"_5(s) -> "CrO"_3(s) + "O"_2(g)#

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