Question

Which of the following have been arrange in order of decreasing oxidation number of sulphur?

• A. $S{O}_2^{+}>S{O}_4^{2-}>S{O}_3^{2-}>HS{O}_4^{-}$
• B. $N{a}_2{S}_4{O}_6>H_2{S}_2{O}_7>N{a}_2{S}_2{O}_3>S_8$
• C. $H_{2}S{O}_4>S{O}_2>H_{2}S>H_2{S}_2{O}_8$
• D. $H_{2}S{O}_5>H_{2}S{O}_3>SC{l}_2>H_{2}S$

Answer 1

Answer: (B), (D)

The correct options are
A $N{a}_2{S}_4{O}_6>H_2{S}_2{O}_7>N{a}_2{S}_2{O}_3>S_8$
C $H_{2}S{O}_5>H_{2}S{O}_3>SC{l}_2>H_{2}S$

1.Oxidation number of $S$ in $H_2{S}_2{O}_7$ is +6,in $N{a}_2{S}_4{O}_6$ +2.5,in $N{a}_2{S}_2{O}_3$ is +2 and in $S_8$ is 0.

Oxidation number of sulphur in decreasing order:$H_2{S}_2{O}_7$>$N{a}_2{S}_4{O}_6$ >$N{a}_2{S}_2{O}_3$>$S_8$.

2.Oxidation number of $S$ in $S{O}_2^{+}$ is +5,in $S{O}_4^{2-}$ 6,in $S{O}_3^{2-}$ is +4 and in $HS{O}_4^{-}$ is +6.

Oxidation number of sulphur in decreasing order:$HS{O}_4^{-}$= $S{O}_4^{2-}$>$S{O}_2^{+}$>$S{O}_3^{2-}$.

3.Oxidation number of $S$ in $H_{2}S{O}_5$ is +8,in $H_{2}S{O}_3$ +4,in $SC{l}_2$ is +2 and in $H_{2}S$ is -2.

Oxidation number of sulphur in decreasing order:$H_{2}S{O}_5$>$H_{2}S{O}_3$ >$SC{l}_2$>$H_{2}S$.

4.Oxidation number of $S$ in $H_{2}S{O}_4$ is +6,in $S{O}_2$ +4,in $H_{2}S$ is -2 and in $H_2{S}_2{O}_8$ is +7.

Oxidation number of sulphur in decreasing order:$H_2{S}_2{O}_8$>$H_{2}S{O}_4$ >$S{O}_2$>$H_{2}S$.

So A and C are correct answer.