The difference in the oxidation numbers of two types of sulphul atoms in `Na_(2)S_(4)O_(6)` is…..

Answer

5

Explanation

`{:(" O O"),(" || ||"),(overset(+)(Na)-overset(-)(O)-S-overset(0)(S)-overset(0)(S)-S^(+5)-overset(-)(O).overset(+)(Na)),(" +5|| ||"),(" O O"):}`
Since each of the two terminal sulphur atoms is connected to two oxygen atoms by a double bond and one oxygen by a single bonds, therefore, the oxidation state of each of these terminal atoms is +5. Since two central atoms are linked to each other by a single bond and each sulphur is further attached to similar species on either side, the attached to similar species on either side, the electron pair forming the s-s bond remains in the centre. Hence, each of the two sulphur atoms has an oxidation state of zero. However, the average oxidation state of the four sulphur atoms `= (2xx5+2xx0)/(4)=2.5`.