How do you calculate the oxidation number of #Na_2SO_4#?
One Sodium (+1) and one sulfate (-2)
Na2(+1) = 21 = 2
SO4(-2) = 1-2 =-2.
It means stable.
Add up all of oxidation numbers of atoms and ions in the compound the result ( zero) is the oxidation number of the compound.
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 = +6