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The oxidation states of sulphur in the anions $SO_3^{2-}$, $S_2O_4^{2-}$ and $S_2O_6^{2-}$ follow the order -
A
$S_2O_4^{2-} \lt SO_3^{2-} \lt S_2O_6^{2-}$
B
$SO_3^{2-} \lt S_2O_4^{2-} \lt S_2O_6^{2-}$
C
$S_2O_4^{2-} \lt S_2O_6^{2-} \lt SO_3^{2-}$
D
$S_2O_6^{2-} \lt S_2O_4^{2-} \lt SO_3^{2-}$
Detailed Solution
Take O = –2 and let x be the oxidation state of S.
$SO_3^{2-}$: $x + 3(-2) = -2 \Rightarrow x = +4$
$S_2O_4^{2-}$ (dithionite): $2x + 4(-2) = -2 \Rightarrow 2x = 6 \Rightarrow x = +3$
$S_2O_6^{2-}$ (dithionate): $2x + 6(-2) = -2 \Rightarrow 2x = 10 \Rightarrow x = +5$
Order: +3 $\lt$ +4 $\lt$ +5
So $S_2O_4^{2-} \lt SO_3^{2-} \lt S_2O_6^{2-}$.
$SO_3^{2-}$: $x + 3(-2) = -2 \Rightarrow x = +4$
$S_2O_4^{2-}$ (dithionite): $2x + 4(-2) = -2 \Rightarrow 2x = 6 \Rightarrow x = +3$
$S_2O_6^{2-}$ (dithionate): $2x + 6(-2) = -2 \Rightarrow 2x = 10 \Rightarrow x = +5$
Order: +3 $\lt$ +4 $\lt$ +5
So $S_2O_4^{2-} \lt SO_3^{2-} \lt S_2O_6^{2-}$.