In Na2S2O3, the oxidation state of S can be determined as follows:

The oxidation state of Na is +1.

There are two Na atoms, so the total oxidation state contributed by Na is +2.

The overall charge of the compound is 0.

Let x be the oxidation state of S.

There are two S atoms, so the total oxidation state contributed by S is 2x.

To balance the charges, we must have:

+2 + 2x = 0

Solving for x, we get:

x = -1

Therefore, the oxidation state of S in Na2S2O3 is -1.