Consider the following representation of oxy-acid, `H_(n_(1))S_(2)O_(n_(2))`, (where S is central sulphur atom annd `n_(1) and n_(2)` are natural numbers.) if there are two possible oxy-acid of sulphur A and B contains ratio of `n_(2):n_(1)` are 2 and 4 respectively, then sum of oxidation state of 'S' atom in both oxy-acid will be:

Answer

9

Explanation

for any oxyacid of sulphur number of H-atoms=2, `H_(n_(1))S_(2)O_(n_(2)),H_(2)S_(2)O_(3),H_(2)S_(2)O_(4),H_(2)S_(2)O_(5),H_(2)S_(2)O_(6),H_(2)S_(2)O_(7),H_(2)S_(2)O_(8)`.
For `H_(2)S_(2)O_(4),(n_(2))/(n_(1))=2,implies` Oxidation state of sulphur atom=+3
For `H_(2)S_(2)O_(6),(n_(2))/(n_(1))=4,implies`Oxidation state of sulphur atom=+6
sum of oxidation state=3+6=9.