In which of the following compounds the sulphur exhibit + 6 oxidation number     

In the given compounds $({\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{8}},{\text{K}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{8}},{\text{S}}_{\text{2}}{\text{O}}_8^{2-})$   2 oxygen atoms shows peroxide linkage. 

In above compounds there is one peroxy linkage hence two of the eight Oxygen atoms have -1  oxidation state.  So, for ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}$and

${\text{K}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}$ ,the oxidation number of Sulphur is,

$\begin{array}{l}2\times (+1)+2x+6(-2)+2(-1)=0\\ +2+2x-12-2=0\\ 2x-12=0\\ 2x=+12\\ x=\frac{+12}{2}=+6\end{array}$

 For  ${\text{S}}_{\text{2}}{\text{O}}_8^{2-}$  compound 

$\begin{array}{l}2x+6(-2)+2(-1)=-2\\ 2x-14=-2\\ 2x=+12\\ x=\frac{+12}{2}=+6\end{array}$

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{5}}\\ 2\times (+1)+x+3\times (-2)+2(-1)=0\\ \left(\text{for\hspace{0.17em}H}\right)\left(\text{for\hspace{0.17em}S}\right)\left(\text{for\hspace{0.17em}O}\right)(\text{for\hspace{0.17em}O}-\text{O})\\ 2+x-6-2=0\\ x=+6\end{array}$