What are the oxidation numbers of $\mathrm{S}$ atoms in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ ?

• A

  $6,-1,-1,6$
• Correct answer
  B

  $5,0,0,5$
• C

  $2.5,2.5,2.5,2.5$
• D

  $7,-2,-2,7$

Solution

When an atom is bonded to similar atoms, has zero oxidation state. $\therefore$ Two middle sulphur atoms have zero oxidation state. For other two sulphur atoms. $2 x+6 \times(-2)=-2$ $2 x-12=-2$ $\begin{aligned} 2 x & =-2+12 \\ 2 x & =+10 \\ x & =+5\end{aligned}$ $\therefore$ The oxidation number on $\mathrm{S}$-atoms in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ is $+5,0,0+5$

Hence, the correct answer is option: 'B'.