# How do you calculate the oxidation number of Na_2SO_4?

Aug 15, 2016

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

#### Explanation:

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

It means stable.

Aug 15, 2016

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 $S O {4}^{-} 2$ sulfur has a lower electro negativity than oxygen so will lose electrons to oxygen causing sulfur to be positive.

$S + 4 \times \left(- 2\right)$ == S + -8 = -2

S = +6

The $S O {4}^{-} 2$ sulfate ion is saturated containing the maximum number of oxygen so has the highest oxidation number for sulfur.

So $2 \times \left(+ 1\right) + 4 \times \left(- 2\right) + \left(+ 6\right)$ = 0