Which one (atom`//`ion ) in the following pairs has highest electron gain enthalpy ?
(i)`O^(-)S (ii) O, S^(-) (iii) O^(-), S^(-) (iv) N^(-) P`

Explanation

To determine which atom or ion in the given pairs has the highest electron gain enthalpy, we will analyze each pair based on their electronic configurations and the stability of their resulting states after gaining an electron.

### Step-by-Step Solution:

1. **Understanding Electron Gain Enthalpy**:
- Electron gain enthalpy refers to the amount of energy released when an electron is added to a neutral atom in the gas phase. A more negative value indicates a greater tendency to gain an electron.

2. **Analyzing Pair (i): O^(-) vs S**:
- **O^(-)**: The electronic configuration is 1s² 2s² 2p⁶ (fully filled, stable).
- **S**: The electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.
- Since O^(-) is already stable and does not need to gain an electron, it will have a more negative electron gain enthalpy compared to S, which needs to gain electrons to achieve stability.

**Conclusion**: O^(-) has a higher electron gain enthalpy than S.

3. **Analyzing Pair (ii): O vs S^(-)**:
- **O**: The electronic configuration is 1s² 2s² 2p⁴.
- **S^(-)**: The electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ (fully filled, stable).
- S^(-) is more stable after gaining an electron compared to O, which is less stable and more likely to gain an electron.

**Conclusion**: S^(-) has a higher electron gain enthalpy than O.

4. **Analyzing Pair (iii): O^(-) vs S^(-)**:
- Both O^(-) and S^(-) are stable configurations (fully filled). However, O^(-) is smaller in size compared to S^(-), which means it can attract an additional electron more effectively.

**Conclusion**: O^(-) has a higher electron gain enthalpy than S^(-).

5. **Analyzing Pair (iv): N^(-) vs P**:
- **N^(-)**: The electronic configuration is 1s² 2s² 2p⁵.
- **P**: The electronic configuration is 1s² 2s² 2p⁶ 3s² 3p³.
- N^(-) has a higher electron-electron repulsion due to its smaller size, while P has half-filled orbitals which are stable. Thus, P is less likely to gain an electron compared to N^(-).

**Conclusion**: N^(-) has a higher electron gain enthalpy than P.

### Final Answers:
- (i) O^(-)
- (ii) S^(-)
- (iii) O^(-)
- (iv) N^(-)