Question

The electronic configuration with the highest ionization enthalpy is :

• A. $[Ne] \, 3s^2 \, 3p^1$
• B. $[Ne] \, 3s^2 \, 3p^2$
• C. $[Ne] \, 3s^2 \, 3p^3$
• D. $[Ar] \, 3d^{10} \, 4s^2 \, 4p^3$

Answer 1

The Correct Option is C

The question asks about the electronic configuration with the highest ionization enthalpy among the given options. Ionization enthalpy refers to the energy required to remove the outermost electron from an atom in the gaseous state. It is influenced by factors such as the atomic size, nuclear charge, and electronic configuration.

1. $[Ne] \, 3s^2 \, 3p^1$: This configuration represents an atom having one electron in the 3p orbital.
2. $[Ne] \, 3s^2 \, 3p^2$: This represents an atom with two electrons in the 3p orbital.
3. $[Ne] \, 3s^2 \, 3p^3$: This configuration signifies a half-filled 3p subshell, which is a stable arrangement due to symmetrical distribution and exchange energy.
4. $[Ar] \, 3d^{10} \, 4s^2 \, 4p^3$: This is a post-transition element, having a higher principal quantum number level (4p), indicating less ionization enthalpy compared to lower energy levels.

Explanation:

The electronic configuration $[Ne] \, 3s^2 \, 3p^3$ corresponds to a half-filled p subshell, which inherently provides stability due to maximum parallel spin orientation and minimized electron-electron repulsions. Such stable configurations generally have higher ionization energies compared to configurations that are not half or completely filled. Due to these factors, configurations like $[Ne] \, 3s^2 \, 3p^3$ generally require more energy to remove the first electron.

Compared to other configurations, this results in $[Ne] \, 3s^2 \, 3p^3$ having the highest ionization enthalpy. Whereas, $[Ar] \, 3d^{10} \, 4s^2 \, 4p^3$ involves electrons at a higher energy level and larger radius, thereby reducing its ionization enthalpy.

Conclusion:

Therefore, the correct answer is the configuration with electrons in the most stable form: $[Ne] \, 3s^2 \, 3p^3$.