The formation of `PH_(4)^(+)` is diffficult compaired to `NH_(4)^(+)` because:

Options

A. lone pair of phosphorus is optically inert
B. lone pair of phosphorus resides in almost pure p-orbital
C. lone pair of phosphorus resides at `sp^(3)` orbital
D. lone pair of phosphorus resides in almost pure s-orbitals

Explanation

To understand why the formation of \( \text{PH}_4^+ \) is more difficult compared to \( \text{NH}_4^+ \), we can analyze the electronic configurations and the nature of the orbitals involved in bonding.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- For Nitrogen (\( \text{N} \)): Atomic number = 7
- Electronic configuration: \( 1s^2 2s^2 2p^3 \)
- For Phosphorus (\( \text{P} \)): Atomic number = 15
- Electronic configuration: \( 1s^2 2s^2 2p^6 3s^2 3p^3 \)

2. **Analyze the Valence Orbitals**:
- The valence orbitals for nitrogen are \( 2s \) and \( 2p \).
- The valence orbitals for phosphorus are \( 3s \) and \( 3p \).

3. **Consider the Size and Effective Nuclear Charge**:
- Nitrogen is smaller in size compared to phosphorus. This means that the valence electrons in nitrogen experience a stronger effective nuclear charge due to being closer to the nucleus.
- Phosphorus, being larger, has its valence electrons further from the nucleus, resulting in a weaker effective nuclear charge.

4. **Hybridization of Orbitals**:
- In \( \text{NH}_4^+ \), the lone pair of electrons on nitrogen can hybridize effectively (sp³ hybridization), which allows for directional bonding.
- In \( \text{PH}_4^+ \), the lone pair of electrons on phosphorus resides mainly in a pure \( s \) orbital, which is non-directional. This means it does not hybridize well with \( p \) orbitals to form directional bonds.

5. **Directional vs Non-Directional Properties**:
- The hybridized orbitals in nitrogen allow for strong directional bonding, facilitating the formation of \( \text{NH}_4^+ \).
- The non-directional nature of the pure \( s \) orbital in phosphorus leads to a decreased tendency to form \( \text{PH}_4^+ \) because it cannot effectively overlap with other orbitals to form stable bonds.

6. **Conclusion**:
- The combination of these factors (size, effective nuclear charge, and orbital hybridization) leads to the conclusion that the formation of \( \text{PH}_4^+ \) is more difficult compared to \( \text{NH}_4^+ \).

### Final Answer:
The formation of \( \text{PH}_4^+ \) is difficult compared to \( \text{NH}_4^+ \) because the lone pair of phosphorus resides in a pure \( s \) orbital, which is non-directional, while the lone pair of nitrogen is in hybridized orbitals that are directional.