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 based on the number of electrons in the outermost shell.

### 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 refers specifically to the energy needed to remove the first electron.

2. **Analyzing Electronic Configurations**:
- We will consider the following configurations:
- **NH2NP3**: This has 3 electrons in the p subshell (p³). This configuration is half-filled, which is relatively stable. Removing an electron will make it unstable, thus requiring a high amount of energy.
- **NH2NP4**: This configuration has 4 electrons in the p subshell (p⁴). Removing one electron will lead to a half-filled configuration (p³), which is stable. Therefore, less energy is required to remove this electron.
- **NH2NP5**: With 5 electrons in the p subshell (p⁵), removing one electron does not lead to a stable configuration. It remains unstable, and thus requires more energy than NH2NP4 but less than NH2NP6.
- **NH2NP6**: This is a noble gas configuration (p⁶), which is the most stable. Removing an electron from this configuration requires a very high amount of energy.

3. **Comparing Ionization Energies**:
- Based on the stability of the configurations:
- NH2NP3 has the highest ionization energy due to its stability.
- NH2NP4 has the lowest ionization energy because it becomes stable after losing one electron.
- NH2NP5 requires more energy than NH2NP4 but less than NH2NP6.
- NH2NP6 has the highest ionization energy due to its noble gas stability.

4. **Conclusion**:
- The configuration with the smallest first ionization energy is **NH2NP4**.