Question

calculate the oxidation number of Sulphur in NaSo4​

Answer 1

Answer:

+7

Explanation:

Na has +1 oxidation number.

O has -2.

Let S oxidation number be x.

+1 + x + 4(-2) = 0(The are 4 atoms of O, so 4*-2, also 0 is taken as the compound has an overall charge of zero.)

1 + x -8 = 0

x = 7.

Sulphur has an oxidation number 7 in NaSO4.

Answer 2

Answer:

The oxidation number of sulfur in $NaSO_{4}$ is +7.

Explanation:

We know that the $NaSO_{4}$ is a neutral molecule.

Therefore the total charge of the molecule is zero.

The oxidation number of sodium = +1

The oxidation number of the oxygen = -2

Assume that the value of the oxidation number of sulfur is x.

That is, (+1)+ x +(4×-2) = 0

            x - 7 = 0

            x = +7

Therefore the oxidation number of sulfur in $NaSO_{4}$ is +7.