For `d^(4) ` ions the fourth electron enters one of the ` e_(g ) ` orbitals giving rise to the configuration ` t_(2g)^(3) e_(g) ^(1)` , when

Options

A. `Delta _(0) gt P`
B. `Delta _(0) lt P`
C. ` Delta_(0)=P`
D. `Delta _(0) gt P`

Explanation

To solve the question regarding the electron configuration of `d^4` ions and the behavior of the fourth electron in an octahedral field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electron Configuration**:
- For a `d^4` ion, there are four electrons to be placed in the d-orbitals. In an octahedral field, the d-orbitals split into two sets: the lower energy `t_(2g)` orbitals and the higher energy `e_(g)` orbitals.

2. **Filling the Orbitals**:
- The first three electrons will fill the `t_(2g)` orbitals first, as they are lower in energy. This gives us the configuration `t_(2g)^(3)`.
- The fourth electron will then enter one of the `e_(g)` orbitals, resulting in the configuration `t_(2g)^(3) e_(g)^(1)`.

3. **Considering Pairing Energy vs. Splitting Energy**:
- The energy required to pair electrons in the same orbital is known as pairing energy (P), while the energy difference between the `t_(2g)` and `e_(g)` orbitals is known as the octahedral splitting energy (Δo).
- In this case, if the pairing energy is greater than the splitting energy (P > Δo), electrons will prefer to occupy the higher energy `e_(g)` orbital rather than pair up in the `t_(2g)` orbitals.

4. **Conclusion**:
- Since the fourth electron goes into the `e_(g)` orbital without pairing, we conclude that the pairing energy is higher than the splitting energy. Thus, the configuration `t_(2g)^(3) e_(g)^(1)` is stable and does not involve pairing.

5. **Final Answer**:
- The answer to the question is that for `d^4` ions, the fourth electron enters one of the `e_(g)` orbitals, resulting in the configuration `t_(2g)^(3) e_(g)^(1)`.