With which of the following electronic configuration of an atom has the lowest ionization enthalpy:

Options

A. `1s^(2) 2s^(2)2p^(6)`
B. ` 1s^(2) 2s^(2) 2p^(5)`
C. `1s^(2) 2s^(2) 2p^(3) `
D. `1s^(2) 2s^(2) 2p^(6) 3s^(1)`

Answer

To determine which electronic configuration has the lowest ionization enthalpy, we need to analyze each configuration based on the stability of the electron arrangement and the energy required to remove an electron.

### Step-by-Step Solution:

1. **Understanding Ionization Enthalpy**:
- Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom. Atoms with stable electron configurations (like noble gases) have high ionization enthalpy, while those with less stable configurations (like alkali metals) have lower ionization enthalpy.

2. **Analyzing Each Configuration**:
- **Configuration A: 1s² 2s² 2p⁶**
- This configuration corresponds to a noble gas (Neon). It has a complete octet and is very stable, requiring a large amount of energy to remove an electron. Thus, it has high ionization enthalpy.

- **Configuration B: 1s² 2s² 2p⁵**
- This configuration corresponds to a halogen (Fluorine). It is one electron short of a complete octet, making it relatively stable but still requires a significant amount of energy to remove an electron. Hence, it has high ionization enthalpy.

- **Configuration C: 1s² 2s² 2p³**
- This configuration corresponds to an element with a half-filled p subshell (Nitrogen). Half-filled configurations are also stable, but they still require considerable energy to remove an electron, resulting in high ionization enthalpy.

- **Configuration D: 1s² 2s² 2p⁶ 3s¹**
- This configuration corresponds to an alkali metal (like Sodium). It has a complete octet in the first two shells and one electron in the third shell. The outermost electron is relatively easy to remove, leading to low ionization enthalpy.

3. **Conclusion**:
- Among the given configurations, **Configuration D (1s² 2s² 2p⁶ 3s¹)** has the lowest ionization enthalpy because it corresponds to an alkali metal, which readily loses its outermost electron.

### Final Answer:
**Configuration D: 1s² 2s² 2p⁶ 3s¹ has the lowest ionization enthalpy.**

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