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

To determine the oxidation state of sulfur in the given anions, we can apply the formula method for calculating oxidation states. The formula is based on the oxidation state rules, considering the usual oxidation states of the other elements in the compound.

1. SO₃^2– Anion:
   • Let the oxidation state of sulfur (S) be \( x \).
   • For oxygen (O), the oxidation state is usually -2.
   • In SO₃^2–, we have one sulfur and three oxygens: \( x + 3(-2) = -2 \).
   • This simplifies to: \( x - 6 = -2 \).
   • Solving for \( x \), we get \( x = +4 \).
2. S₂O₄^2– Anion:
   • Let the oxidation state of sulfur (S) be \( x \).
   • For oxygen (O), the oxidation state is -2.
   • In S₂O₄^2–, we have two sulfurs and four oxygens: \( 2x + 4(-2) = -2 \).
   • This simplifies to: \( 2x - 8 = -2 \).
   • Solving for \( x \), we get \( 2x = +6 \), so \( x = +3 \).
3. S₂O₆^2– Anion:
   • Let the oxidation state of sulfur (S) be \( x \).
   • For oxygen (O), the oxidation state is -2.
   • In S₂O₆^2–, we have two sulfurs and six oxygens: \( 2x + 6(-2) = -2 \).
   • This simplifies to: \( 2x - 12 = -2 \).
   • Solving for \( x \), we get \( 2x = +10 \), so \( x = +5 \).

Comparing the oxidation states obtained:

• Oxidation state of sulfur in S₂O₄^2– is \( +3 \).
• Oxidation state of sulfur in SO₃^2– is \( +4 \).
• Oxidation state of sulfur in S₂O₆^2– is \( +5 \).

Thus, the order of increasing oxidation states for sulfur is: S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}.

This matches the given correct option: S₂O₄^2– < SO₃^2– < S₂O₆^2–