In an ionic or molecular species, the sum of the oxidation numbers is EQUAL to the charge on the ion. Formal rules of assignment, are given below...
Here we gots #Cu(NO_3)_2#, a NEUTRAL salt... Now we break this up into its constituent ions, we get #Cu^(2+)=Cu(+II)#, and #2xxNO_3^(-)#. Given the rules, we gots #O(-II)#, and #N(+V)#...The sum of the oxidation numbers, here, give ZERO, as is required.
#1.# #"The oxidation number of a free element is always 0."#
#2.# #"The oxidation number of a mono-atomic ion is equal"# #"to the charge of the ion."#
#3.# #"For a given bond, X-Y, the bond is split to give "X^+# #"and"# #Y^-#, #"where Y is more electronegative than X."#
#4.# #"The oxidation number of H is +1, but it is -1 in when"# #"combined with less electronegative elements."#
#5.# #"The oxidation number of O in its"# compounds #"is usually -2, but it is -1 in peroxides."#
#6.# #"The oxidation number of a Group 1 element"# #"in a compound is +1."#
#7.# #"The oxidation number of a Group 2 element in"# #"a compound is +2."#
#8.# #"The oxidation number of a Group 17 element in a binary compound is -1."#
#9.# #"The sum of the oxidation numbers of all of the atoms"# #"in a neutral compound is 0."#
#10.# #"The sum of the oxidation numbers in a polyatomic ion"# #"is equal to the charge of the ion."#
