The correct sequence of decreasing number of `pi` - bonds in the structure of `H_2SO_3,H_2SO_4 and H_2S_2O_7` is :

Options

A. `H_2S_2O_7 gtH_2SO_3 gtH_2SO_4`
B. `H_2S_2O_7 gtH_2SO_4 gtH_2SO_3`
C. `H_2SO_4gtH_2S_2O_7 gtH_2SO_3`
D. `H_2SO_3gtH_2SO_4gtH_2S_2O_7`

Answer

To determine the correct sequence of decreasing number of π-bonds in the structures of H₂SO₃, H₂SO₄, and H₂S₂O₇, we will analyze the structures of each compound step by step.

### Step 1: Analyze H₂SO₃ (Sulfurous Acid)
- The structure of H₂SO₃ consists of one sulfur atom (S) bonded to three oxygen atoms (O).
- Two of the oxygen atoms are bonded to hydrogen (forming hydroxyl groups -OH), and one oxygen is double-bonded to sulfur.
- The structure can be represented as:
```
O
||
H-O-S-OH
```
- In this structure, there is **1 π-bond** (the double bond between sulfur and oxygen).

### Step 2: Analyze H₂SO₄ (Sulfuric Acid)
- The structure of H₂SO₄ consists of one sulfur atom bonded to four oxygen atoms.
- Two of the oxygen atoms are bonded to hydrogen (as hydroxyl groups), and two oxygen atoms are double-bonded to sulfur.
- The structure can be represented as:
```
O
||
H-O-S=O
||
O
```
- In this structure, there are **2 π-bonds** (two double bonds between sulfur and oxygen).

### Step 3: Analyze H₂S₂O₇ (Disulfuric Acid)
- The structure of H₂S₂O₇ consists of two sulfur atoms bonded to seven oxygen atoms.
- There are four double bonds between sulfur and oxygen, and two hydroxyl groups.
- The structure can be represented as:
```
O O
|| ||
H-O-S-O-S=O
|| |
O OH
```
- In this structure, there are **4 π-bonds** (four double bonds between sulfur and oxygen).

### Step 4: Summarize the Number of π-bonds
- H₂SO₃: 1 π-bond
- H₂SO₄: 2 π-bonds
- H₂S₂O₇: 4 π-bonds

### Step 5: Arrange in Decreasing Order
- The correct sequence of decreasing number of π-bonds is:
- H₂S₂O₇ (4 π-bonds)
- H₂SO₄ (2 π-bonds)
- H₂SO₃ (1 π-bond)

Thus, the final answer is:
**H₂S₂O₇ > H₂SO₄ > H₂SO₃**