How is charge used to assign oxidation numbers to the elements in a polyatomic ion?

1 Answer
Sep 18, 2016

Answer:

The sum of the oxidation numbers ALWAYS equals the charge on the ion.

Explanation:

Let's take 2 transition metal ions, dichromate, #Cr_2O_7^(2-)#, and permanganate, #MnO_4^-#.

Of course, oxygen normally takes a #-II# oxidation state, and it does so here. Given that the sum of the oxidation numbers equals the charge on the ion, then:

#2xxCr^("ON")+7xxO^("ON")=-2;#

#2xxCr^("ON")+7xx-2=-2;#

#2xxCr^("ON")=+12; Cr^("ON")=VI+#.

For permanganate,

#Mn^("ON")+4xxO^("ON")=-1;#

#Mn^("ON")+4xx(-2)=-1;#

#Mn^("ON")=+VII#.

The oxidation state of other charged metal oxides are calculated the same way: #CrO_4^(2-)# #-=# #Cr(VI+).#