Question

The oxidation states of sulphur in the anions SO₃²⁻, S₂O₄²⁻
and S₂O₆²⁻ follow the order [2003]
(a) S₂O₆²⁻ < S₂O₄²⁻ < SO₃²⁻
(b) S₂O₄²⁻ < SO₃²⁻ < S₂O₆²⁻
(c) SO₃²⁻ < S₂O₄²⁻ < S₂O₆²⁻
(d) S₂O₄²⁻ < S₂O₆²⁻ < SO₃²⁻

Answer 1

Answer:

option d Should be correct

Answer 2

S₂O₄²⁻ < SO₃²⁻ < S₂O₆²⁻ is the correct order.

Explanation:

• The oxides of sulphur are several in number.

• Among those given here in the question, there are S₂O₄²⁻, SO₃²⁻ and S₂O₆²⁻.

• The oxidation number of sulphur in S₂O₄²⁻ is
• $2(x)+4(-2)=-2$
• $2x=8 - 2$
• $2x=6$
• $x=3$.
• The oxidation number of sulphur in SO₃²⁻ is
• $x+3(-2)=-2$
• $x=6-2$
• $x=4$
• The oxidation number of sulphur in S₂O₆²⁻ is
• $2(x)+6(-2)=-2$
• $2x=12-2$
• $2x=10$
• $x=5$
• So the trend of the oxidation number is S₂O₄²⁻ < SO₃²⁻ < S₂O₆²⁻.

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