InterviewSolution
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InterviewSolution
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The oxidation states of sulphur in the anions `SO_(3)^(2-), S_(2)O_(4)^(2-)`, and `S_(2)O_(6)^(2-)` follow the orderA. ` S_2 O_6^(2-) lt S_2 O_4^(2-) lt SO_3^(2-)`B. ` S_2O_4^(2-) lt SO_3^(2-) lt S_2O_6^(2-)`C. `SO_3^(2-) lt S_(2)O_(6)^(2-) lt SO_(3)^(2-)`D. ` S_2O_4^(2-) lt S_2O_6^(2-) lt SO_3^(2-)` |
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Answer» Correct Answer - B We can work out the oxidation number of S by using the fact that in polyatomic ions, the algebraic sum of all the oxidation numbers must equal the charge on the ion. Let the oxidation number of S be x. (a) `SO_3^(2-) rArr + 3 (-2) =- 2` `x- 6=-2` ` x= +4` (b) `S_2O_4^(2-) rArr 2x +4 (-2) =-2` ` 2 x - 8 =- 2` ` x= +3` (c) ` S_2O_6^(2-) rArr 2x + 6(-2) =-2` ` 2x- 12 =-2` `x= +5` Thus, the order of oxidation number is `overset(+3)(S_(2))O_(4)^(2-) lt overset(+4)(S)O_(3)^(2-) lt overset(+5)(S_(2))O_(6)^(2-)` |
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