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

Jun 12, 2017

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 $- I I$ and it does here, and sulfur bound to the more electronegative oxygen atom adopts a $+ V I$ oxidation state, which is the Group Number.........

And thus ${\left\{H \stackrel{V I}{S} {O}_{4}\right\}}^{-}$, and the sum of the oxidation numbers is: $V {I}^{+} + 4 \times I {I}^{-} + I = - 1$, the charge on bisulfate ion as required.........

We could look at the parent molecule ${H}_{2} S {O}_{4}$ and get precisely the same result...........

$2 \times {I}^{+} + V {I}^{+} - 4 \times I {I}^{-} = 0$.

Another way of looking at ${H}_{2} S {O}_{4} \left(l\right)$ is as ${\left(O =\right)}_{2} S {\left(- O - H\right)}_{2}$; 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 $H N {O}_{3}$?