Question

The oxidation states of sulphur in the anions SO₃^2–, S₂O₄^2– and S₂O₆^2– follow the order- 

• A. S₂O₄^2–< SO₃^2–< S₂O₆^2–
• B. SO₃^2–< S₂O₄^2–< S₂O₆^2–
• C. S₂O₄^2–< S₂O₆^2–< SO₃^2– 
• D. S₂O₆^2–< S₂O₄^2–< SO₃^2–

Answer 1

The Correct Option is A

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⁻