Oxidation state of sulphur in anions SO32-, S2O42- and S2O62- increases in the orders: (a) S2O62- < S2O42- < SO32- (b) SO32- < S2O42- < S2O62- (c) S2O42- < SO32- < S2O62- (d) S2O42- < S2O62- < SO32-

Correct answer: (c) S2O42- < SO32- < S2O62- Explanation: In SO32- x + 3(-2) = -2; x = +4 In S2O42- 2x + 4(-2) = -2 2x – 8 = -2 2x = 6; x = +3 In S2O62- 2x + 6(-2) = -2 2x = 10; x = +5 hence the correct order is S2O42- < SO32- < S2O62-

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Oxidation state of sulphur in anions SO3^2-, S2O4^2- and S2O6^2- increases in the orders:

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Last Post by Reyana09 3 years ago

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16/11/2021 11:32 am

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Palak23

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Oxidation state of sulphur in anions SO₃^2-, S₂O₄^2- and S₂O₆^2- increases in the orders:

(a) S₂O₆^2-< S₂O₄^2-< SO₃^2-

(b) SO₃^2-< S₂O₄^2-< S₂O₆^2-

(d) S₂O₄^2-< S₂O₆^2-< SO₃^2-

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16/11/2021 11:36 am

Reyana09

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Correct answer: (c) S₂O₄^2-< SO₃^2- < S₂O₆^2-

Explanation:

In SO₃^2-

x + 3(-2) = -2; x = +4

In S₂O₄^2-

2x + 4(-2) = -2

2x – 8 = -2

2x = 6; x = +3

In S₂O₆^2-

2x + 6(-2) = -2

2x = 10; x = +5

hence the correct order is

S₂O₄^2-< SO₃^2- < S₂O₆^2-

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Oxidation state of sulphur in anions SO3^2-, S2O4^2- and S2O6^2- increases in the orders:

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