The correct order of `IE_(2)` of `C, N, O` and `F` is

Options

A. `O gt F gt N gt C`
B. `F gt O gt N gt C`
C. `C gt N gt O gt F`
D. `O gt N gt F gt C`

Explanation

To determine the correct order of second ionization energy (IE₂) for carbon (C), nitrogen (N), oxygen (O), and fluorine (F), we will follow these steps:

### Step 1: Write the electronic configurations
- **Carbon (C)**: Atomic number = 6
Electronic configuration: 1s² 2s² 2p²

- **Nitrogen (N)**: Atomic number = 7
Electronic configuration: 1s² 2s² 2p³

- **Oxygen (O)**: Atomic number = 8
Electronic configuration: 1s² 2s² 2p⁴

- **Fluorine (F)**: Atomic number = 9
Electronic configuration: 1s² 2s² 2p⁵

### Step 2: Analyze the first ionization energies (IE₁)
- For **C**: After removing one electron, it becomes C⁺: 1s² 2s² 2p¹
- For **N**: After removing one electron, it becomes N⁺: 1s² 2s² 2p²
- For **O**: After removing one electron, it becomes O⁺: 1s² 2s² 2p³
- For **F**: After removing one electron, it becomes F⁺: 1s² 2s² 2p⁴

### Step 3: Determine the stability of the resulting ions
- **O⁺** has a half-filled 2p subshell (2p³), which is more stable and requires more energy to remove another electron.
- **F⁺** also has a stable configuration (2p⁴), but it is less stable than O⁺ because it is one electron away from a stable noble gas configuration.
- **N⁺** has a half-filled 2p subshell (2p²), which is stable but less than O⁺.
- **C⁺** is the least stable among these because it has a 2p¹ configuration.

### Step 4: Compare the second ionization energies (IE₂)
- **IE₂ for O**: Highest due to stability of half-filled subshell.
- **IE₂ for F**: Second highest, as it is one electron away from a stable configuration.
- **IE₂ for N**: Third, as it has a half-filled subshell but is less stable than O.
- **IE₂ for C**: Lowest, as it is the least stable configuration.

### Conclusion: Correct order of IE₂
The correct order of second ionization energy (IE₂) for C, N, O, and F is:
**O > F > N > C**