Calculate the oxidation number of each sulphur atom in the following compounds.
(a) `Na_(2)S_(2)O_(3)``" "`(b)`Na_(2)S_(4)O_(6)``" "`( c)`Na_(2)SO_(3)``" "`(d)`Na_(2)SO_(4)`

Explanation

The oxidation number of each sulphur atom in the following compounds are given below
(a) `N_(2)S_(2)O_(3)` Let us consider the structure of `Na_(2)S_(2)O_(3)`.
`Na^(+)O^(-)-underset(O)underset(||)overset(S)overset(uarr)(S)-O^(-)Na^(+)`
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.
`underset("For Na")(2(+1))+underset("For o-atoms")(3xx(-2))+x+underset("For coordinate S-atoms")(1(-2))=0`
`x=+6`
Therefore, the two sulphur atoms in `N_(2)S_(2)O_(3)` have -2 and +6 oxidation number.
(b) `Na_(2)S_(4)O_(6)` Let us consider the structure of `Na_(2)S_(4)O_(6)`.
`Na^(+)O^(-)-underset(O)underset(||)overset(O)overset(||)(S)-overset(0)(S)-overset(0)(S)-underset(O)underset(||)overset(O)overset(||)(S)-O^(-)Na^(+)`
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
`underset("For Na")(2(+1))+underset("For O")(6(-2))+2x+2(0)=0`
`2(+1)+x+4(-2)=0` or `x=+(10)/(2)=+5`
Therefore, the two central S-atoms have zero oxidation state and two terminal S-atoms have +5 Oxidation state each
( c) `Na_(2)SO_(3)` Let the oxidation number of S in `N_(2)SO_(3)` be x.
`2(+1)+x+3(-2)=0` or `x=+4`
(d) `Na_(2)SO_(4)` Let the oxidation number of S be x.
`2(+1)+x+4(-2)=0` or `x=+6`