The ionisation energy is lowest for

Options

A. Nitrogen
B. Oxygen
C. Fluorine
D. Neon

Answer

To determine which element has the lowest ionization energy among nitrogen, oxygen, fluorine, and neon, we can follow these steps:

### Step 1: Write the Electronic Configurations
First, we need to write the electronic configurations of the four elements:

- **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
- **Neon (Ne)**: Atomic number 10 → Electronic configuration: 1s² 2s² 2p⁶

### Step 2: Analyze the Stability of the Configurations
Next, we analyze the stability of the electronic configurations:

- **Neon** has a complete outer shell (2p⁶), making it very stable and thus requiring a high amount of energy to remove an electron.
- **Fluorine** has one electron less than a full shell (2p⁵), which makes it relatively stable but less than neon.
- **Oxygen** has two electrons in the 2p subshell (2p⁴), which is less stable than fluorine.
- **Nitrogen** has a half-filled 2p subshell (2p³), which provides extra stability due to symmetry and exchange energy.

### Step 3: Compare Ionization Energies
Now, we compare the ionization energies based on the stability:

- **Neon** has the highest ionization energy due to its full outer shell.
- **Fluorine** has a high ionization energy but is less than neon.
- **Nitrogen** has a relatively high ionization energy due to its half-filled stability.
- **Oxygen** has the lowest ionization energy because it is less stable than the others.

### Conclusion
Thus, the element with the lowest ionization energy among nitrogen, oxygen, fluorine, and neon is **Oxygen (O)**.

### Final Answer
**Oxygen (O) has the lowest ionization energy.**

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