Question

The order of increasing oxidation number ofS in $S_8,S_2{O}_8^{-2},S_2{O}_3^{-2},S_4{O}_6^{-2}$ is given below :

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

Answer 1

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

In S${}_8$ oxidation number of sulphur will be O in S${}_2$O${}_8^2$ oxidation number of sulphur will be

$2x16=2$

$2x14\Rightarrow x=7$ (which is not possible) sulphur 6 which is mix oxidation number of oxidation state will be + 6.

In $S_2{O}_3^2$

$2x6=2$

$x=+2$

In $S_4{O}_6-2$

$4x12=2$

$x=2.5$

$S_8<S_2{O}_3^{2-}<S_4{O}_6^{-2}<S_2{O}_8^{-2}$

So the order of increasing oxidation number of S is $S_8<S_2{O}_3^2<S_4{O}_6^2<S_2{O}_8^2$

Hence option D is correct.