What is the oxidation state of #Cr# in #K_2Cr_2O_7#?

1 Answer
Dec 22, 2016

Answer:

We have #Cr(VI+)#, the metal in its maximum oxidation state, its Group oxidation state.

Explanation:

#"Oxidation state/number"# is the charge left on the central atom, when all the bonding atoms are removed with the charge devolved to the more electronegative atom. The sum of the oxidation numbers of each species is the charge on the ion.

We have the #Cr_2O_7^(2-)# anion, and, as is typical, each oxygen atom has a formal #-II# oxidation number: thus #7xx(-II)+2xxCr_"oxidation number"=-2#. Clearly, #Cr_"oxidation number"=VI+#.

Because, chromium is in such a high oxidation state, it typically accepts electron density from other reagents, thereby oxidizing them, and reducing itself. Dichromate is thus a good oxidizing agent. Typically we see the colour change from red/orange, #Cr(VI+)#, to green #Cr(III+)#:

#Cr_2O_7^(2-) +14H^(+) + 6e^(-)rarr 2Cr^(3+) +7H_2O#

Are charge and mass balanced here? They should be.