Consider the following representation of oxy-acid, $$H_{n_1}S_{2}O_{n_2},$$ (where S is central sulphur atom and $$n_1$$ 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 :
For any oxyacid of sulphur number of H-atom $$= 2; \,H_{n_1} S_2O_{n_2} H_2S_2O_3, H_2S_2O_4, H_2S_2O_5, H_2S_2O_6, H_2S_2O_7, H_2S_2O_8.$$
For $$H_2S_2O_4; \dfrac{n_2}{n_1} = 2; \Rightarrow$$ Oxidation state of sulphur atom $$= +3$$
For $$H_2S_2O_6; \dfrac{n_2}{n_1} = 4; \Rightarrow$$ Oxidation state of sulphur atom $$= +6$$
Sum of oxidation state $$= 3 + 6 = 9$$