Question

How do you calculate the oxidation number of `Na_2SO_4`?

Answer 1

One Sodium (+1) and one sulfate (-2)

Explanation:

Na2(+1) = 21 = 2
SO4(-2) = 1-2 =-2.

It means stable.

Answer 2

Add up all of oxidation numbers of atoms and ions in the compound the result ( zero) is the oxidation number of the compound.

Explanation:

Na number 11 has an oxidation of +1
Sodium wants to lose one electron creating the electron structure of Neon which is stable.

Oxygen number 8 has an oxidation number of -2 Oxygen has an electro negativity of 3.5 the second highest of all elements. So Oxygen always gains ( takes) two electrons creating the electron structure of Neon which is stable.

Sulfur number 16 has oxidation numbers of -2, +2 +4 and +6.
When combined in the `SO4^-2` sulfur has a lower electro negativity than oxygen so will lose electrons to oxygen causing sulfur to be positive.

`S + 4 xx (-2)` == S + -8 = -2

S = +6

The `SO4^-2` sulfate ion is saturated containing the maximum number of oxygen so has the highest oxidation number for sulfur.

So `2 xx(+1) + 4 xx(-2) + (+6)` = 0