23. Calculate the oxidation number of each sulphur atom in the following compounds.

(a) Na2S2O3

(b) Na2S4O6

(c) Na2SO3

(d) Na2SO4


The oxidation number of each sulphur atom in the following compounds are given below
(a) Na2S2O3 Let us consider the structure of Na2S2O3.
There is a coordinate bond between two sulphur atoms. The oxidation number of acceptor S-atom is -2. Let, the oxidation number of other S-atom be x.
2(+1)For Na+3×(-2)For O-atoms+x+1(-2)=0For coordinate S-atomx=+6
Therefore, the two sulphur atoms in Na2S2O3 have -2 and +6 oxidation number.
(b) Na2S4O6 Let us consider the structure of Na2S4O6.
In this structure, two central sulphur atoms have zero oxidation number because electron pair forming the S-S bond remain in the centre. Let, the oxidation number of (remaining S-atoms) S-atom be x.
2(+1)For Na+6(-2)For O+2x+2(0)=02-12+2x=0 or x=+102=+5
Therefore, the two central S-atoms have zero oxidation state and two terminal S-atoms have +5 oxidation state each.
(c) Na2SO3 Let the oxidation number of S in Na2SO3 be x.
2(+1)+x+3(-2)=0 or x=+4
(d) Na2SO4 Let the oxidation number of S be x.
2(+1)+x+4(-2)=0 or x=+6