Oxidation Number of Sulphur in Na2S2O3

Oxidation number or state is defined as the charge that an atom appears to have when all other bonding electrons are divided equally between the atoms in a molecule. In Na2S2O3, the oxidation number of sulphur needs to be determined.

Steps to Determine Oxidation Number of Sulphur in Na2S2O3:

Step 1: Identify the Oxidation States of Other Elements

In Na2S2O3, sodium has a fixed oxidation state of +1, and oxygen has an oxidation state of -2.

Step 2: Determine the Total Charge of the Compound

Na2S2O3 is a neutral compound, so the total charge is 0.

Step 3: Calculate the Oxidation Number of Sulphur

Using the oxidation states of sodium (+1) and oxygen (-2), we can determine the oxidation number of sulfur as follows:

2(+1) + 3(-2) + 2x = 0

2 + (-6) + 2x = 0

-4 + 2x = 0

2x = +4

x = +2

Therefore, the oxidation number of sulfur in Na2S2O3 is +2.

Conclusion:

The oxidation number of sulphur in Na2S2O3 is +2.

The two sulphur atoms in Na2​S2​O3​ have -2 and +6 oxidation states.
The sulphur in the middle have made 6 bonds hence it has +6 oxidation state. While other sulphur shows -2 oxidation state.
Average oxidation number of S=−2+6​/2=+2