On heating `Pb(NO_3)_2` the products formed ate :

Options

A. ` PbO, N_2,O_2`
B. `Pb (NO_2)_2, O_2`
C. `Pb , NO_2, O_2`
D. `Pb, N_2,O_2`

Explanation

To determine the products formed when lead(II) nitrate, \( Pb(NO_3)_2 \), is heated, we can follow these steps:

### Step 1: Write the decomposition reaction
When lead(II) nitrate is heated, it undergoes thermal decomposition. The general reaction can be written as:
\[ Pb(NO_3)_2 \rightarrow PbO + NO_2 + O_2 \]

### Step 2: Identify the products
From the decomposition of lead(II) nitrate, we can identify the products formed:
1. **Lead(II) oxide (PbO)**: This is a yellow solid that is often used in paints.
2. **Nitrogen dioxide (NO2)**: This is a brown gas that is paramagnetic due to the presence of an unpaired electron in its structure.
3. **Oxygen (O2)**: This is a colorless gas.

### Step 3: Balance the reaction
To ensure that the reaction is balanced, we need to check the number of atoms of each element on both sides of the equation:
- **Left-hand side (LHS)**: 1 lead (Pb), 2 nitrogen (N), and 6 oxygen (O).
- **Right-hand side (RHS)**:
- From \( PbO \): 1 lead (Pb) and 1 oxygen (O).
- From \( NO_2 \): 2 nitrogen (N) and 4 oxygen (O).
- From \( O_2 \): 2 oxygen (O).

Now, let's balance the equation:
\[ 2 Pb(NO_3)_2 \rightarrow 2 PbO + 4 NO_2 + O_2 \]
This gives us:
- LHS: 2 Pb, 4 N, 12 O
- RHS: 2 Pb, 4 N, 12 O

### Conclusion
The balanced equation shows that when lead(II) nitrate is heated, the products formed are:
- Lead(II) oxide (PbO)
- Nitrogen dioxide (NO2)
- Oxygen (O2)

Thus, the final answer is:
**The products formed are PbO, NO2, and O2.**

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