On strong heating `Pb(NO_(3))_(2)` gives:

Options

A. `PbO`
B. `NO_(2)`
C. `O_(2)`
D. `NO`

Answer

To solve the question regarding the decomposition of lead nitrate (Pb(NO₃)₂) upon strong heating, we can follow these steps:

### Step 1: Write the decomposition reaction
When lead nitrate is heated, it decomposes into lead oxide, nitrogen dioxide, and oxygen. The unbalanced equation for this reaction can be written as:

\[ \text{Pb(NO}_3\text{)}_2 \rightarrow \text{PbO} + \text{NO}_2 + \text{O}_2 \]

### Step 2: Balance the chemical equation
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides.

1. Start with the lead (Pb): There is 1 Pb on both sides.
2. Next, balance the nitrogen (N): There are 2 nitrogen atoms in Pb(NO₃)₂, so we need 2 NO₂ on the product side.
3. Now, balance the oxygen (O):
- On the left, we have 6 oxygen atoms from Pb(NO₃)₂.
- On the right, we have 2 from PbO, 4 from 2 NO₂, and 2 from O₂, totaling 8 oxygen atoms.

Thus, we need to adjust the coefficients to balance the reaction correctly:

\[ 2 \text{Pb(NO}_3\text{)}_2 \rightarrow 2 \text{PbO} + 4 \text{NO}_2 + \text{O}_2 \]

### Step 3: Identify the products formed
From the balanced equation, we can identify the products formed:
- **Lead oxide (PbO)**: This is a yellow solid.
- **Nitrogen dioxide (NO₂)**: This is a reddish-brown gas.
- **Oxygen (O₂)**: This is a colorless gas.

### Step 4: Conclusion
The products of the strong heating of lead nitrate (Pb(NO₃)₂) are:
- Lead oxide (PbO)
- Nitrogen dioxide (NO₂)
- Oxygen (O₂)

### Final Answer
On strong heating, Pb(NO₃)₂ gives:
1. Lead oxide (PbO)
2. Nitrogen dioxide (NO₂)
3. Oxygen (O₂)