What is the oxidation number of sulfur in Na2S2O4?

Answer 1

To figure this out, we first need to figure out the charge on S2O4.

We know that Na has a +1 charge, so the Na2 portion of the salt will have an overall +2 charge (2 x Na).

Provided that Na2S2O4 is a neutral molecule, we know that the charge of S2O4 must balance Na2, which has a +2 charge. Therefore, the charge on S2O4 is -2.

If the charge on S2O4 is -2, that must mean that there is an excess of 2 electrons in the molecule. We know that Oxygen usually has a charge of -2, and given that there are 4 Oxygen atoms in S2O4, the charge contributed by Oxygen is 4 x (-2) = -8. Since the charge of S2O4 is -2, we know that the total charge contributed by sulfur is +6.

+6 divided by 2 Sulfur atoms gives you the oxidation state (or number) of S: +3

Answer 2

If we just take the anion S2O4^(2-)

Then using oxygen as the standard at '-2' , we notice that there are fpur oxygens so the oxygen moiety is 4 x -2 = -8

So we noiw create a little sum

2S + =8(the oxygen moiety) = -2(the charge on the anion)

2S - 8 = -2

Add '8' to both sides

2S = 6

Divide both sides by '2'

S = 3 (The oxudaton state of sulphur).

The Oxidation Number means that there are only '3' valance electrons out of '6', of sulphur involved in the bonding process.

NB Sulphur can exhibit several oxidation states. viz. -2, 0, 2, 3, 4, & 6.

Answer 3

+1 for each Na, -2 for each O, +4 for S

Answer 4

This is a ionic compound. S shows the +6 oxidation number.

Answer 5

Na is +1

O is -2

S is +4

Answer 6

Oxidation no of sulphur is +2.

Answer 7

+1 for each, -2 for each O, +6 for S

Answer 8

+2