Which of the following has lowest ionisation energy

Options

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

Explanation

To determine which of the given elements has the lowest ionization energy, we will analyze the elements: Sulfur (S), Oxygen (O), Nitrogen (N), and Fluorine (F).

### Step-by-Step Solution:

1. **Understand Ionization Energy**:
- Ionization energy is the energy required to remove an electron from an isolated atom in its gaseous state.

2. **Identify the Elements**:
- The elements in question are:
- 1. Oxygen (O)
- 2. Nitrogen (N)
- 3. Fluorine (F)
- 4. Sulfur (S)

3. **Consider the Periodic Trends**:
- Ionization energy generally increases across a period (left to right) due to increasing nuclear charge.
- Ionization energy generally decreases down a group (top to bottom) due to increased distance from the nucleus and increased electron shielding.

4. **Locate the Elements on the Periodic Table**:
- Oxygen (O) and Nitrogen (N) are in the second period, while Sulfur (S) is in the third period, and Fluorine (F) is also in the second period.

5. **Compare Ionization Energies**:
- The experimental values for the ionization energies are:
- Sulfur (S): 999.4 kJ/mol
- Oxygen (O): 1313.7 kJ/mol
- Nitrogen (N): 1402.1 kJ/mol
- Fluorine (F): 1680 kJ/mol

6. **Identify the Lowest Ionization Energy**:
- From the values, it is clear that Sulfur (S) has the lowest ionization energy at 999.4 kJ/mol.

7. **Conclusion**:
- Therefore, among the given options, Sulfur (S) has the lowest ionization energy.

### Final Answer:
**Sulfur (S) has the lowest ionization energy.**

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