Amongst the following elements whose electronic configuration are given below, the one having the highest enthalpy is

Options

A. `[Ne]3s^(2)3p^(1)`
B. `[Ne]3s^(2)3p^(3)`
C. `[Ne]3s^(-2)3p^(2)`
D. `[Ar]3d^(10)4s^(2)4p^(3)`

Explanation

To determine which element has the highest enthalpy among the given electronic configurations, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
Let's denote the electronic configurations provided in the question. For example:
- Element A: \( 1s^2 2s^2 2p^1 \) (P1)
- Element B: \( 1s^2 2s^2 2p^3 \) (P3)
- Element C: \( 1s^2 2s^2 2p^2 \) (P2)
- Element D: \( 1s^2 2s^2 3p^3 \) (P3)

2. **Determine the Valence Electron Configuration**:
Focus on the valence electrons, particularly the p-orbitals:
- P1: 1 valence electron in p-orbital
- P2: 2 valence electrons in p-orbital
- P3: 3 valence electrons in p-orbital (two occurrences)

3. **Stability of Half-Filled Configuration**:
The half-filled p-orbital configuration (P3) is more stable than P1 and P2 configurations. Therefore, both configurations with P3 are more stable.

4. **Compare the Energy Levels**:
- The 3p orbital (from Element D) is at a higher principal quantum number (n=3) compared to the 2p orbital (from Element B), which means it is further from the nucleus.
- The effective nuclear charge felt by electrons in the 3p orbital is less than that felt by electrons in the 2p orbital.

5. **Determine the Highest Enthalpy**:
Since the 3p orbital (from Element D) is less stable due to being further from the nucleus and experiencing less effective nuclear charge, it will have a higher enthalpy than the 2p orbital (from Element B).

6. **Conclusion**:
Therefore, the element with the highest enthalpy is the one with the 3p3 configuration, which corresponds to Element D.

### Final Answer:
The element with the highest enthalpy is the one with the electronic configuration \( 3p^3 \).

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