Which has the maximum `IE` ?

Options

A. `O^(o+)`
B. `N`
C. `O`
D. `Na`

Explanation

To determine which species has the maximum ionization enthalpy (IE), we will analyze the electronic configurations of the given options and understand the stability of these configurations. Here’s a step-by-step solution:

### Step 1: Identify the Options
The options provided are:
- A: O⁺
- B: N
- C: O
- D: Na

### Step 2: Write the Electronic Configurations
Next, we write the electronic configurations for each of the species:
- **A: O⁺**
The electronic configuration of O (Oxygen) is 1s² 2s² 2p⁴. When it loses one electron to become O⁺, it becomes 1s² 2s² 2p³.

- **B: N**
The electronic configuration of N (Nitrogen) is 1s² 2s² 2p³.

- **C: O**
The electronic configuration of O (Oxygen) is 1s² 2s² 2p⁴.

- **D: Na**
The electronic configuration of Na (Sodium) is 1s² 2s² 2p⁶ 3s¹.

### Step 3: Analyze Stability of Electronic Configurations
To determine which species has the highest ionization enthalpy, we consider the stability of their electronic configurations:
- Half-filled orbitals (like in N and O⁺) are more stable compared to incompletely filled orbitals (like in O and Na).
- O⁺ and N both have half-filled p orbitals, which are stable configurations.

### Step 4: Compare the Effective Nuclear Charge
To find the one with the maximum ionization enthalpy, we need to consider the effective nuclear charge (Z_eff):
- For **O⁺** (8 protons, 7 electrons): The effective nuclear charge is high because there are more protons than electrons.
- For **N** (7 protons, 7 electrons): The effective nuclear charge is lower than that of O⁺.

### Step 5: Conclusion
Since O⁺ has a higher effective nuclear charge compared to N, it will have a greater attraction between the nucleus and the remaining electrons. Therefore, O⁺ will have a higher ionization enthalpy than N.

### Final Answer
The species with the maximum ionization enthalpy is **A: O⁺**.

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