The correct order of ionization energy in the following second-period elements is: B , Be , C ,N , O

Options

A. B lt Be lt C lt N lt O
B. B lt Be lt N lt C lt O
C. Be lt B lt C lt N gt O
D. B lt Be lt C lt O lt N

Answer

To determine the correct order of ionization energy for the second-period elements: B (Boron), Be (Beryllium), C (Carbon), N (Nitrogen), and O (Oxygen), we need to consider their electronic configurations and the general trends in ionization energy across the periodic table.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- Beryllium (Be): 1s² 2s²
- Boron (B): 1s² 2s² 2p¹
- Carbon (C): 1s² 2s² 2p²
- Nitrogen (N): 1s² 2s² 2p³
- Oxygen (O): 1s² 2s² 2p⁴

2. **Understand Ionization Energy**:
- Ionization energy is the energy required to remove the outermost electron from an atom. Generally, ionization energy increases across a period due to increasing nuclear charge and decreasing atomic size.

3. **Analyze Atomic Size**:
- As we move from left to right across the periodic table, the atomic size decreases. Smaller atoms have higher ionization energies because the outermost electrons are held more tightly by the nucleus.

4. **Consider Stability of Electron Configurations**:
- Nitrogen (N) has a half-filled p subshell (2p³), which is more stable and thus has a higher ionization energy compared to Oxygen (O), which has a partially filled p subshell (2p⁴).
- Beryllium (Be) has a fully filled s subshell (2s²), making it more stable than Boron (B), which has an unpaired electron in the p subshell (2p¹).

5. **Order the Elements Based on Ionization Energy**:
- From highest to lowest ionization energy:
- Nitrogen (N) > Oxygen (O) > Carbon (C) > Beryllium (Be) > Boron (B)

### Final Order:
The correct order of ionization energy for the elements B, Be, C, N, and O is:
**N > O > C > Be > B**