Question

The order of increasing Oxidation number of $S$ in $S_8,S_2{O}_8^{2-},S_2{O}_3^{2-},S_4{O}_6^{2-}$ is:

• A. $S_8<S_2{O}_3^{2-}<S_4{O}_6^{2-}<S_2{O}_8^{2-}$
• B. $S_2{O}_8^{2-}<S_2{O}_3^{2-}<S_4{O}_6^{2-}<S_8$
• C. $S_8<S_2{O}_8^{2-}<S_2{O}_3^{2-}<S_4{O}_6^{2-}$
• D. $S_2{O}_8^{2-}<S_8<S_4{O}_6^{2-}<S_2{O}_3^{2-}$

Answer 1

The correct option is A $S_8<S_2{O}_3^{2-}<S_4{O}_6^{2-}<S_2{O}_8^{2-}$

For $S_8$, as sulfur is present in its elemental state. So, oxidation number (O.N.) of $S=0.$

For $S_2{O}_8^{2-}$
Here 2 oxygen have peroxide linkage so they will be having oxidation number -1, while other 6 will be having -2.
let oxidation number of S is $x$,
$\Rightarrow 2x+6\times (-2)+2\times (-1)=-2$
$\Rightarrow x=6$.


For $S_2{O}_3^{2-}$
Here all oxygen have oxidation number -2.
let oxidation number of S is $y$,
$\Rightarrow 2y+3\times (-2)=-2$
$\Rightarrow y=2$.

For $S_4{O}_6^{2-}$
Here all oxygen have oxidation number -2.
let oxidation number of S is $z$,
$\Rightarrow 4z+6\times (-2)=-2$
$\Rightarrow z=2.5$.

The oxidation numbers of $S$ are shown below along with the compounds:

$\underset{0}{S_8},\underset{+6}{S_2{O}_8^{2-}},\underset{+2}{S_2{O}_3^{2-}},\underset{+2.5}{S_4{O}_6^{2-}}$.

Hence the order of increasing O.N. of $S$ is $S_8<S_2{O}_3^{2-}<S_4{O}_6^{2-}<S_2{O}_8^{2-}$.