The increasing order of ionisation energy among sulphur, phosphorous and oxygen is

Options

A. `SltPltO`
B. `PltSltO`
C. `OltSltP`
D. `OltPltS`

Answer

To determine the increasing order of ionization energy among sulfur (S), phosphorus (P), and oxygen (O), we need to consider a few key principles regarding ionization energy and the periodic table.

### Step-by-Step Solution:

1. **Understanding Ionization Energy**:
Ionization energy is the energy required to remove an electron from an atom in its gaseous state. Generally, ionization energy increases across a period (from left to right) and decreases down a group (from top to bottom) in the periodic table.

2. **Position of Elements in the Periodic Table**:
- Oxygen (O) is in Group 16 (or VI A) and Period 2.
- Sulfur (S) is in Group 16 (or VI A) and Period 3.
- Phosphorus (P) is in Group 15 (or V A) and Period 3.

3. **Comparing Ionization Energies**:
- **Oxygen vs. Sulfur**: As we move from sulfur to oxygen (down to up in the same group), the ionization energy increases. Therefore, O has a higher ionization energy than S.
- **Sulfur vs. Phosphorus**: Phosphorus is in Group 15 and has a half-filled p-orbital configuration (3 electrons in the p subshell), which makes it more stable and gives it a higher ionization energy compared to sulfur, which has a full p subshell configuration (4 electrons in the p subshell). Thus, P has a higher ionization energy than S.

4. **Final Order**:
Based on the above comparisons:
- Oxygen (O) has the highest ionization energy.
- Phosphorus (P) has a higher ionization energy than sulfur (S).
- Sulfur (S) has the lowest ionization energy.

Therefore, the increasing order of ionization energy is:
**Sulfur (S) < Phosphorus (P) < Oxygen (O)**.

### Conclusion:
The increasing order of ionization energy among sulfur, phosphorus, and oxygen is:
**S < P < O**.