In which of the following there is maximum `p pi -p pi` bonding ?

Options

A. `NF_(3)`
B. `NI_(3)`
C. `BI_(3)`
D. `BF_(3)`

Explanation

To determine which compound exhibits the maximum `p pi - p pi` bonding, we need to analyze the bonding characteristics of the given compounds. The key concept here is back bonding, which occurs under specific conditions.

### Step-by-Step Solution:

1. **Understanding Back Bonding**:
Back bonding occurs when there is a sigma bond already formed between two atoms, and one of the atoms has a lone pair while the other has a vacant orbital. Additionally, one of the atoms must belong to the second period of the periodic table.

2. **Identifying the Compounds**:
We need to consider the compounds that could potentially exhibit `p pi - p pi` bonding. Common examples include BF3, SiF4, and others that may involve elements from the second period.

3. **Analyzing BF3**:
- **Sigma Bond Formation**: In BF3, boron (B) forms three sigma bonds with three fluorine (F) atoms.
- **Lone Pairs and Vacant Orbitals**: Each fluorine atom has three lone pairs and boron has an empty p orbital. This satisfies the condition for back bonding.
- **Periodicity**: Both boron and fluorine are in the second period.

4. **Conclusion on BF3**:
Since all conditions for back bonding are satisfied in BF3, we can conclude that there is significant `p pi - p pi` bonding due to the overlap of the p orbitals of fluorine with the vacant p orbital of boron.

5. **Comparing with Other Compounds**:
Other compounds like SiF4 do not exhibit the same level of `p pi - p pi` bonding because silicon does not have vacant p orbitals that can participate in back bonding in the same manner as boron in BF3.

### Final Answer:
The compound with the maximum `p pi - p pi` bonding is **BF3**.