Among the following , the number of compounds that can react with `PCl_5` to give `POCl_3` is.
`O_2,CO_2,SO_2,H_2 O,H_2 SO_4,P_4 O_10`.

Explanation

To solve the problem of determining how many compounds can react with \( PCl_5 \) to give \( POCl_3 \), let's analyze each compound one by one.

### Step 1: List the compounds
The compounds given are:
1. \( O_2 \)
2. \( CO_2 \)
3. \( SO_2 \)
4. \( H_2O \)
5. \( H_2SO_4 \)
6. \( P_4O_{10} \)

### Step 2: Analyze each compound's reactivity with \( PCl_5 \)

1. **\( O_2 \)**:
- \( O_2 \) does not react with \( PCl_5 \) to form \( POCl_3 \).

2. **\( CO_2 \)**:
- \( CO_2 \) also does not react with \( PCl_5 \) to form \( POCl_3 \).

3. **\( SO_2 \)**:
- \( SO_2 \) reacts with \( PCl_5 \) to produce \( SOCl_2 \) and \( POCl_3 \).
- Reaction:
\[ SO_2 + PCl_5 \rightarrow SOCl_2 + POCl_3 \]
- This compound is counted.

4. **\( H_2O \)**:
- \( H_2O \) reacts with \( PCl_5 \) to produce \( POCl_3 \) and hydrochloric acid.
- Reaction:
\[ H_2O + PCl_5 \rightarrow POCl_3 + 2 HCl \]
- This compound is counted.

5. **\( H_2SO_4 \)**:
- \( H_2SO_4 \) reacts with \( PCl_5 \) to yield \( POCl_3 \), sulfur dioxide, and hydrochloric acid.
- Reaction:
\[ H_2SO_4 + PCl_5 \rightarrow 2 POCl_3 + SO_2 + 2 HCl \]
- This compound is counted.

6. **\( P_4O_{10} \)**:
- \( P_4O_{10} \) reacts with \( PCl_5 \) to produce \( POCl_3 \).
- Reaction:
\[ P_4O_{10} + 6 PCl_5 \rightarrow 10 POCl_3 \]
- This compound is counted.

### Step 3: Count the compounds that react
From the analysis:
- The compounds that react with \( PCl_5 \) to give \( POCl_3 \) are:
- \( SO_2 \)
- \( H_2O \)
- \( H_2SO_4 \)
- \( P_4O_{10} \)

Thus, there are **4 compounds** that can react with \( PCl_5 \) to give \( POCl_3 \).

### Final Answer:
The number of compounds that can react with \( PCl_5 \) to give \( POCl_3 \) is **4**.

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