What is the oxidation number of hydrogen in #HSO_4^(-)#, #"bisulfate ion"#?

1 Answer
Jun 12, 2017

Answer:

The oxidation number of hydrogen here is #+I#.

Explanation:

As always, the sum of the oxidation numbers EQUALS the charge on the ion, so we works out for #Na^(+)""^(-)[HSO_4]#. 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 #{Hstackrel(VI)SO_4}^-#, and the sum of the oxidation numbers is: #VI^++4xxII^(-)+I=-1#, the charge on bisulfate ion as required.........

We could look at the parent molecule #H_2SO_4# and get precisely the same result...........

#2xxI^(+)+VI^(+)-4xxII^(-)=0#.

Another way of looking at #H_2SO_4(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 #HNO_3#?