Which of the following order is correct for the first ionozation potential of `B, C`, and `N`?

Options

A. `B lt C lt N`
B. `N lt C lt B`
C. `N lt C lt B`
D. `N lt C lt B`

Answer

To determine the correct order of first ionization potential (IP) for Boron (B), Carbon (C), and Nitrogen (N), we can analyze their electronic configurations and the factors affecting ionization energy.

### Step 1: Write the electronic configurations
- **Boron (B)**: The electronic configuration is \(1s^2 2s^2 2p^1\).
- **Carbon (C)**: The electronic configuration is \(1s^2 2s^2 2p^2\).
- **Nitrogen (N)**: The electronic configuration is \(1s^2 2s^2 2p^3\).

### Step 2: Understand the concept of ionization energy
Ionization energy is the energy required to remove the outermost electron from an atom in the gaseous state. Factors affecting ionization energy include atomic size, nuclear charge, and electron configuration stability (such as half-filled or fully filled orbitals).

### Step 3: Analyze the stability of the orbitals
- **Nitrogen (N)** has a half-filled \(2p\) subshell (\(2p^3\)), which is more stable than the configurations of Boron and Carbon. This stability means that more energy is required to remove an electron from Nitrogen compared to Boron and Carbon.
- **Carbon (C)** has a \(2p^2\) configuration, while **Boron (B)** has a \(2p^1\) configuration. As we move from Boron to Carbon, the number of electrons in the \(2p\) subshell increases, and the atomic size decreases.

### Step 4: Compare the ionization energies
- **Nitrogen (N)** has the highest ionization energy due to its half-filled stability.
- **Carbon (C)** has a higher ionization energy than **Boron (B)** because it has a smaller atomic radius, leading to a stronger attraction between the nucleus and the outermost electron.
- Therefore, the order of ionization energy from lowest to highest is: Boron < Carbon < Nitrogen.

### Conclusion
The correct order for the first ionization potential of Boron (B), Carbon (C), and Nitrogen (N) is:
**B < C < N**