The first loinsatisation in electron volts of nitrogen and oxygen atoms are respectively , given by

Options

A. `14.6,13.6`
B. `13.6,14.6`
C. `13.6,13.6`
D. `14.6,14.6`

Explanation

To determine the first ionization energies of nitrogen and oxygen, we need to analyze their electronic configurations and the stability of their outermost electrons.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers**:
- Nitrogen (N) has an atomic number of 7, meaning it has 7 electrons.
- Oxygen (O) has an atomic number of 8, meaning it has 8 electrons.

2. **Write the Electronic Configurations**:
- For Nitrogen: The electronic configuration is \(1s^2 2s^2 2p^3\).
- For Oxygen: The electronic configuration is \(1s^2 2s^2 2p^4\).

3. **Analyze the Outermost Electrons**:
- Nitrogen has 5 electrons in its outer shell (2 in 2s and 3 in 2p), while oxygen has 6 electrons in its outer shell (2 in 2s and 4 in 2p).
- The outermost electrons for nitrogen are in a half-filled p subshell (2p^3), which is more stable due to symmetry and exchange energy.

4. **Stability Consideration**:
- Half-filled subshells (like nitrogen's 2p^3) are more stable than those that are not half-filled (like oxygen's 2p^4). This means that it requires more energy to remove an electron from nitrogen than from oxygen.

5. **Conclusion on Ionization Energies**:
- Therefore, the first ionization energy of nitrogen is greater than that of oxygen:
\[ \text{First Ionization Energy of N} > \text{First Ionization Energy of O} \]

6. **Select the Correct Option**:
- Based on the analysis, we can conclude that option A (where the first ionization energy of nitrogen is greater than that of oxygen) is the correct answer.