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Question

The oxidation states of $S$ atom in $S_{4} O_{6}^{2 -}$ from left to right respectively are

$+ 6, 0, 0, + 6$
$+ 3, + 1, + 1, + 3$
$+ 5, 0, 0, + 5$
$+ 4, + 1, + 1, + 4$

A

+ 4, + 1, + 1, + 4

B

+ 5, 0, 0, + 5

C

+ 6, 0, 0, + 6

D

+ 3, + 1, + 1, + 3

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Solution

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The two middle sulphur in structure have 0 oxidation state because any atom bonded with similar atoms has an oxidation state of 0. The total oxidation state of sulphur in the compound is 10. The oxidation state of leftmost and rightmost sulphur is +5, as oxygen is more electronegative.

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