The energy required to remove an electron from the outermost shell of an isolate gaseous atom is known as `IE_(1)` of that atom. Similarly, the enrgy required for the removal of the electron from the unipositive ion, diapositive ion and tripositive ion are known as `IE_(2),IE_(3)` and `IE_(4)` respectively, and are called successive ionisation energies. The magnitude of the charge depends on the size of the orbital of electron. Electrons in smaller orbitals are on average close with each other and have more repulsion. Thus for `Be(2s^(2))`, the `IE_(1)` and `IE_(2)` are `9.3` and `18.2 eV "atom"^(1)`, whereas for `Ca(4s^(2))`, the values are `6.1` and `11.9 eV`.
The correct order of arrangement of the first ionisation energies of `C, N,O` and `F` ( in decreasing values) is

Options

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

Explanation

To determine the correct order of the first ionization energies of carbon (C), nitrogen (N), oxygen (O), and fluorine (F) in decreasing values, we can follow these steps:

### Step 1: Understand Ionization Energy
Ionization energy (IE) is the energy required to remove an electron from an atom in its gaseous state. The first ionization energy (IE₁) refers to the removal of the first electron.

### Step 2: Identify the Electronic Configurations
We need to write the electronic configurations of the elements in question:
- Carbon (C): 1s² 2s² 2p²
- Nitrogen (N): 1s² 2s² 2p³
- Oxygen (O): 1s² 2s² 2p⁴
- Fluorine (F): 1s² 2s² 2p⁵

### Step 3: Analyze the Trend in Ionization Energy Across a Period
As we move from left to right across a period in the periodic table, the ionization energy generally increases due to the increase in nuclear charge (more protons) which holds the electrons more tightly. However, there are exceptions due to electron configurations.

### Step 4: Consider the Stability of Half-Filled and Fully Filled Orbitals
- Nitrogen has a half-filled p subshell (2p³), which provides extra stability. This means it will have a higher ionization energy compared to its neighbors.
- Oxygen has one more electron than nitrogen, which leads to increased electron-electron repulsion in the 2p orbitals, making it easier to remove an electron compared to nitrogen.

### Step 5: Compare the Ionization Energies
Based on the above analysis:
- **Fluorine (F)** has the highest ionization energy because it has the most protons and a nearly filled p subshell (2p⁵).
- **Nitrogen (N)** comes next due to its half-filled stability.
- **Oxygen (O)** follows because of the increased repulsion in its p orbitals.
- **Carbon (C)** has the lowest ionization energy among these four.

### Conclusion
The correct order of the first ionization energies in decreasing values is:
**F > N > O > C**

### Final Answer
The correct option is **F, N, O, C**.

---