• # question_answer 63) Calculate the oxidation number of each sulphur atom in the following compounds :(a) $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$                          (b) $N{{a}_{2}}{{S}_{4}}{{O}_{6}}$                         (c)$\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}~~$                   (d)$\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$

(a) $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ $+2+2x-6=0$                 $x=+2$ (b) $\overset{+}{\mathop{N}}\,a\overset{\overset{-2}{\mathop{-}}\,}{\mathop{O}}\,-\underset{\underset{\underset{-2}{\mathop{O}}\,}{\mathop{||}}\,}{\overset{\overset{\overset{-2}{\mathop{O}}\,}{\mathop{||}}\,}{\mathop{{{S}^{x}}}}}\,-\overset{0}{\mathop{S}}\,-\overset{0}{\mathop{S}}\,-\underset{\underset{\underset{-2}{\mathop{O}}\,}{\mathop{||}}\,}{\overset{\overset{\overset{-2}{\mathop{O}}\,}{\mathop{||}}\,}{\mathop{S}}}\,-\overset{(-2)}{\mathop{\overset{-}{\mathop{O}}\,N\overset{+}{\mathop{a}}\,}}\,$                                 $2x+2-12=0$                                                 $x=+5$ Two sulphur atoms are in (+5) and remaining two are in zero oxidation state.