Question

The oxidation states of sulphur in the anions SO32–, S2O42– and S2O62– follow the order-

• A. S2O42–< SO32–< S2O62–
• B. SO32–< S2O42–< S2O62–
• C. S2O42–< S2O62–< SO32–
• D. S2O62–< S2O42–< SO32–

Answer 1

Answer: (A)

S2O42–< SO32–< S2O62–

Answer 2

For SO₃²⁻ (sulfite ion), The oxidation state of sulfur can be calculated as follows:
Let the oxidation state of sulfur be x. So, you have:
x + 3(-2) = -2
x - 6 = -2
x = +4
For S₂O₄²⁻ (dithionite ion), The oxidation state of sulfur can be calculated as follows:
x + 4(-2) = -2
x - 8 = -2
x = +6
For S₂O₆²⁻ (tetrathionate ion), The oxidation state of sulfur can be calculated as follows:
x + 6(-2) = -2
x - 12 = -2
x = +10
Now, let's order these oxidation states:
+4 (SO₃²⁻) < +6 (S₂O₄²⁻) < +10 (S₂O₆²⁻)

So, the correct option is (A): S₂O₄²⁻ < SO₃²⁻ < S₂O₆²⁻