What is the oxidation number of chromium in #K_2Cr_2O_7#?

1 Answer
May 24, 2017

Answer:

We has #Cr(VI+)#........

Explanation:

The ion is #Cr_2O_7^(2-)#. The sum of the oxidation numbers of the individual elements must equal the charge on the ion. And thus #2xxCr_"oxidation number"+7xxO_"oxidation number"+=-2#.

The oxidation number of oxygen in its compounds is usually #-II#, and it is here..........

So #Cr_"oxidation number"=(-2+14)/2=+6#, i.e. #Cr^(VI+)#.