Which of the following compound is possible :

Options

A. `PbI_(4)`
B. `AlO_(2)`
C. `NaF_(2)`
D. `PbO_(2)`

Answer

To determine which of the given compounds is possible, we will analyze each option by calculating the oxidation states of the elements involved.

### Step-by-Step Solution:

1. **Option A: PbI4**
- Let the oxidation state of Pb be \( x \).
- The oxidation state of iodine (I) is \(-1\).
- The equation for the compound is:
\[ x + 4(-1) = 0 \]
- Simplifying gives:
\[ x - 4 = 0 \implies x = +4 \]
- Lead (Pb) in the +4 oxidation state can act as an oxidizing agent. However, I- is a weak reducing agent and cannot reduce Pb from +4 to 0. Therefore, PbI4 is not stable and the compound is not possible.

2. **Option B: AlO2**
- Let the oxidation state of Al be \( x \).
- The oxidation state of oxygen (O) is \(-2\).
- The equation for the compound is:
\[ x + 2(-2) = 0 \]
- Simplifying gives:
\[ x - 4 = 0 \implies x = +4 \]
- The maximum oxidation state of aluminum (Al) is +3. Therefore, AlO2 is not possible.

3. **Option C: NaF2**
- Let the oxidation state of Na be \( x \).
- The oxidation state of fluorine (F) is \(-1\).
- The equation for the compound is:
\[ x + 2(-1) = 0 \]
- Simplifying gives:
\[ x - 2 = 0 \implies x = +2 \]
- Sodium (Na) typically has an oxidation state of +1. Therefore, NaF2 is not possible.

4. **Option D: PbO2**
- Let the oxidation state of Pb be \( x \).
- The oxidation state of oxygen (O) is \(-2\).
- The equation for the compound is:
\[ x + 2(-2) = 0 \]
- Simplifying gives:
\[ x - 4 = 0 \implies x = +4 \]
- Lead (Pb) can exist in the +4 oxidation state, making PbO2 (lead dioxide) a valid compound. This compound is known to be used in batteries as an electrode.

### Conclusion:
The only possible compound among the options given is **PbO2**.