# What is the oxidation number of sulfur in "NaHSO"_4?

Nov 13, 2015

$+ 6$

#### Explanation:

Sodium hydrogen sulfate, ${\text{NaHSO}}_{4}$, is an ionic compound made up of sodium cations, ${\text{Na}}^{+}$, and hydrogen sulfate anions, ${\text{HSO}}_{4}^{-}$.

The oxidation state of sulfur can thus be determined by looking at the oxidation states of the two other elements that make up the hydrogen sulfate anion, hydrogen and oxygen.

Since you're dealing with a polyatomic ion, you know that the sum of the oxidation states of each individual atom that makes up the anion must add up to give the charge, which in this case is $\left(1 -\right)$.

Oxygen will almost always have an oxidation state of $- 2$, and this case is no exception. Likewise, hydrogen will almost always have an oxidation number equal to $+ 1$, as is the case here.

This means that you have

${\overbrace{+ 1 \times 1}}^{\textcolor{b l u e}{\text{one H atom")) + overbrace(ON_"S")^(color(blue)("one S atom")) + overbrace(4 xx (-2))^(color(blue)("four O atoms}}} = - 1$

$O {N}_{\text{S}} = - 1 - 1 + 8 = \textcolor{g r e e n}{+ 6}$

The oxidation state of sulfur in sodium hydrogen sulfate is $+ 6$.