The first ionization energy is smallest for the atom with electronic configuration?

Options

A. `ns^2 np^3 `
B. `ns^2 np^4 `
C. `ns^2 np^5 `
D. `ns^2 np^6`

Answer

To determine which electronic configuration corresponds to the smallest first ionization energy, we need to analyze the stability of the configurations and how they relate to the energy required to remove an electron.

### Step-by-step Solution:

1. **Understanding Ionization Energy**:
- Ionization energy is the energy required to remove an electron from a neutral atom. The first ionization energy specifically refers to the removal of the first electron.

2. **Stability of Electronic Configurations**:
- Atoms with stable electronic configurations (like fully filled or half-filled subshells) require more energy to remove an electron. This is because they are energetically favorable and more stable.

3. **Identifying Electronic Configurations**:
- Let's consider the configurations mentioned in the transcript:
- \( \text{Ns}^2 \text{Np}^3 \) (half-filled)
- \( \text{Ns}^2 \text{Np}^4 \) (not stable)
- \( \text{Ns}^2 \text{Np}^5 \) (stable)
- \( \text{Ns}^2 \text{Np}^6 \) (fully filled)

4. **Analyzing the Configurations**:
- The configurations \( \text{Ns}^2 \text{Np}^3 \) and \( \text{Ns}^2 \text{Np}^6 \) are stable due to half-filled and fully filled subshells, respectively. Therefore, they will have higher ionization energies.
- The configuration \( \text{Ns}^2 \text{Np}^4 \) is less stable compared to the others, making it easier to remove an electron.

5. **Conclusion**:
- The configuration with the smallest first ionization energy is \( \text{Ns}^2 \text{Np}^4 \) because it is less stable and requires less energy to remove an electron.

### Final Answer:
- The first ionization energy is smallest for the atom with the electronic configuration \( \text{Ns}^2 \text{Np}^4 \).