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Valency and Oxidation State

The oxidation...

Question

The oxidation states of sulphur in the anions $S O_{3}^{2 -}, S_{2} O_{4}^{2 -}$ and $S_{2} O_{6}^{2 -}$ follow the order :

A

S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}

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B

S O_{3}^{2 -} < S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -}

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C

S_{2} O_{4}^{2 -} < S_{2} O_{6}^{2 -} < S O_{3}^{2 -}

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D

S_{2} O_{6}^{2 -} < S_{2} O_{4}^{2 -} < S O_{3}^{2 -}

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Solution

The correct option is C $S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$
Let the oxidation number of $S$ be $x$ .
The oxidation number of oxygen is $- 2$ .
$S_{2} O_{6}^{2 -} : 2 x - 12 = - 2 \Rightarrow x = 5$
$S O_{3}^{2 -} : x - 6 = - 2 \Rightarrow x = 4$
$S_{2} O_{4}^{2 -} : 2 x - 8 = - 2 \Rightarrow x = 3$
Hence, the correct order is $S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$ .

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Similar questions

Q. The oxidation states of sulphur in the anions

S O_{3}^{2 -}, S_{2} O_{4}^{2 -}

S_{2} O_{6}^{2 -}

follow the order

Q. In the reaction,

2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -}

, the eq. wt. of

{N a}_{2} S_{2} O_{3}

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{N a}_{2} S_{2} O_{3}

I_{2}

M_{1}

M_{2}

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{N a}_{2} S_{2} O_{3}

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2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}

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