As always, the sum of the oxidation numbers EQUALS the charge on the ion, so we works out for Na+−[HSO4]. And typically oxygen has an oxidation number of −II and it does here, and sulfur bound to the more electronegative oxygen atom adopts a +VI oxidation state, which is the Group Number.........
And thus {HVISO4}−, and the sum of the oxidation numbers is: VI++4×II−+I=−1, the charge on bisulfate ion as required.........
We could look at the parent molecule H2SO4 and get precisely the same result...........
2×I++VI+−4×II−=0.
Another way of looking at H2SO4(l) is as (O=)2S(−O−H)2; the oxidation states are just the same, even the oxygen atoms are FORMALLY inequivalent; you pays your money and you takes your choice.
Can you now tell me the oxidation numbers for the individual elements in HNO3?
