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

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

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

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

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

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Solution

The correct option is C $S_{2} O_{4}^{2 -} < S O_{3}^{2 -} < S_{2} O_{6}^{2 -}$
O.S. of S in $S O_{3}^{2 -},$
$x - 6 = - 2$
$x = + 4$

O.S. of S in $S_{2} O_{4}^{2 -},$
$2 x - 8 = - 2$
$x = 3$

O.S. of S in $S_{2} O_{6}^{2 -},$
$2 x - 12 = 2$
$x = + 5$

The correct order of increasing O.S. is;

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

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Similar questions

S O_{3}^{2 -}, S_{2} O_{4}^{2 -} a n d S_{2} O_{6}^{2 -}

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

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

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

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

S O_{3}^{2 -}

S_{2} O_{4}^{2 -}

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

{S O}_{3}^{2 -}, {S O}_{4}^{2 -}, S_{2} O_{4}^{2 -}, S_{2} O_{6}^{2 -}

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