# oxidation no. of S in S2O3 2-

In AL level, the two S atoms in S2O32- have different oxidation number.

The two different oxidation numbers are 0 and +4.

*** How do we know? How to calculate it?

Update:

Isn't that the two different oxidation numbers are 0 and +4?

Rating

yes, thiosulphate is like sulphate with one of oxygen replaced by sulphur. terminal sulphur attacted to central sulphur with double bond, while central sulphur has =O, -O(-) and -O(-).

however, we look at the oxidation number in some other way.

when considering oxidation number, we look at the number of bonds as well as difference in electronegativity among them.

look at terminal sulphur -- two bonds with another sulphur. however, they have same electronegativity, so neither one draws electrons from another (the bonds are non-polar). thus it doesn't "gain" or "lose" any electrons, and O.N. remains zero.

for central sulphur, it has 2 bonds to terminal sulphur and totally 4 bonds to oxygen. as oxygen is more electronegative than sulphur, in each bond oxygen will draw the bonding electrons to them.

(remember that a covalent bond is formed by sharing an electron from each bonding atom -- in this case, one from oxygen and one from sulphur.)

as there're four S-O bonds, central sulphur "loses" 4 electrons, so its O.N. is +4.

Sally just missed considering the electronegativity difference between S and O, and incorrectly think that the terminal-S retains the O.N. of original oxygen.