Nitrogen form `N_2`, but when phosphorous form `P_2`, it's at a time convert in `P_4`, reason is : -

Options

A. (a) Triple bond present between phosphorous atom
B. (b) `ppi-ppi` bonding is weak
C. (c) `ppi-ppi` bonding is strong
D. (d) Multiple bond form easily

Answer

To solve the question regarding the formation of nitrogen and phosphorus molecules, we need to analyze the bonding characteristics of both elements.

### Step-by-Step Solution:

1. **Understanding Nitrogen Bonding**:
- Nitrogen exists primarily as a diatomic molecule, \(N_2\), which features a triple bond between the two nitrogen atoms. This triple bond consists of one sigma bond and two pi bonds.
- The small atomic size of nitrogen allows for effective overlap of p-orbitals, resulting in strong p-pi-p-pi bonding.

2. **Understanding Phosphorus Bonding**:
- Phosphorus can exist in several allotropes, but the most common forms are \(P_2\) and \(P_4\). When phosphorus attempts to form \(P_2\), it quickly converts to \(P_4\).
- In \(P_4\), each phosphorus atom forms three sigma bonds with three other phosphorus atoms, creating a tetrahedral structure.

3. **Comparison of Atomic Sizes**:
- Nitrogen has a smaller atomic radius compared to phosphorus. This smaller size facilitates the formation of multiple bonds (like the triple bond in \(N_2\)).
- Conversely, phosphorus, being larger, struggles to form multiple bonds due to the increased distance between the nuclei, which weakens the bonding interactions.

4. **Bonding Strength**:
- The p-pi-p-pi bonding in nitrogen is strong due to effective orbital overlap.
- In phosphorus, however, the p-pi-p-pi bonding is weak because of the larger atomic size, which leads to less effective overlap of p-orbitals.

5. **Conclusion**:
- Therefore, phosphorus prefers to form \(P_4\) with single bonds rather than existing as \(P_2\) with multiple bonds. The weak p-pi-p-pi bonding in phosphorus is the reason for this preference.

### Final Answer:
The correct reason for why nitrogen forms \(N_2\) while phosphorus forms \(P_4\) is that **p-pi-p-pi bonding is weak in phosphorus**.