Question

The correct sequence of decreasing number of $\pi$-bonds in the structures of H₂SO₃, H₂SO₄ and H₂S₂O₇ is:-

• A. H₂S₂O₇ > H₂SO₄ > H₂SO₃
• B. H₂SO₃ > H₂SO₄ > H₂S₂O₇
• C. H₂S₂O₇ > H₂SO₃ > H₂SO₄
• D. H₂SO₄ > H₂S₂O₇ > H₂SO₃

Answer 1

Answer: (A)

H₂S₂O₇ > H₂SO₄ > H₂SO₃

Answer 2

To determine the number of $\pi$-bonds in each compound, we need to draw their Lewis structures.

1. H₂SO₃ (Sulfurous acid)

   • Number of $\pi$-bonds = 1.

2. H₂SO₄ (Sulfuric acid)

   • Number of $\pi$-bonds = 2.

3. H₂S₂O₇ (Pyrosulfuric acid)

   • Number of $\pi$-bonds = 4.

Summary of $\pi$-bonds:

• H₂SO₃: 1 $\pi$-bond
• H₂SO₄: 2 $\pi$-bonds
• H₂S₂O₇: 4 $\pi$-bonds

Decreasing order of $\pi$-bonds:

H₂S₂O₇ (4) > H₂SO₄ (2) > H₂SO₃ (1)