Amongst the following elements (whose electronic configuration an given below) the one having highest ionization energy 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 among the given options has the highest ionization energy based on their electronic configurations, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- Write down the electronic configurations provided in the question.
- For example, let's assume the configurations are:
- Option A: 3s² 3p¹
- Option B: 3s² 3p³
- Option C: 4s² 3d¹⁰ 4p³
- Option D: 4s² 4p⁶

2. **Understand Ionization Energy**:
- Ionization energy is the energy required to remove the outermost electron from an atom in the gaseous state.
- Generally, elements with a full or half-full valence shell have higher ionization energies due to their stability.

3. **Analyze the Stability of Configurations**:
- Full configurations (like noble gases) and half-full configurations (like some transition metals) are more stable.
- For example:
- 3s² 3p¹ (Option A) is not stable.
- 3s² 3p³ (Option B) is half-filled and thus more stable.
- 4s² 3d¹⁰ 4p³ (Option C) has a filled d subshell, which provides additional stability.
- 4s² 4p⁶ (Option D) is a full outer shell (noble gas configuration).

4. **Consider Effective Nuclear Charge**:
- The effective nuclear charge (Z_eff) experienced by the outermost electrons affects ionization energy.
- In configurations with filled d orbitals, the inner d electrons shield the outer p electrons, making them easier to remove.
- In contrast, for half-filled configurations, the p electrons feel a stronger effective nuclear charge, making them harder to remove.

5. **Compare the Options**:
- Based on the above analysis:
- Option A (3s² 3p¹) has low ionization energy.
- Option B (3s² 3p³) has moderate ionization energy due to half-filled stability.
- Option C (4s² 3d¹⁰ 4p³) has lower ionization energy due to shielding from filled d orbitals.
- Option D (4s² 4p⁶) has the highest ionization energy due to full outer shell stability.

6. **Conclusion**:
- The element with the highest ionization energy among the given options is **Option D (4s² 4p⁶)**.