Question

The oxidation states of Sulphur in the anions SO32– , S2O42– and S2O62– follow the order :

• A. S2O62– < S2O42 < SO32–

• B. S2O42– < SO32– < S2O62–

• C. SO32– < S2O42– < S2O62–
• D. S2O42 < S2O62– < SO32–

Answer 1

Answer: (B)

S2O42– < SO32– < S2O62–

Answer 2

(2)

SO32– 1$\Rightarrow$ (x) + 3(–2) = – 2 $\therefore$x = + 4

S2O42– 2 $\Rightarrow$ (x)+ 4 (–2) = – 2 $\therefore$x = + 3

S2O62– 2 $\Rightarrow$ (x)+ 6 (–2) = – 2 $\therefore$x = + 5