In which of the following pairs, the ionisation energy of the first species is less than of the second

Options

A. O^(+) , O
B. S, P
C. N, P
D. Be^(+) , Be

Answer

To solve the question of which pair has the ionization energy of the first species less than that of the second, we will analyze each pair step by step.

### Step 1: Analyze the First Pair - Oxygen (O) and Oxygen Ion (O⁺)
- **Oxygen (O)** has the electronic configuration: 1s² 2s² 2p⁴.
- **Oxygen Ion (O⁺)** has lost one electron, so its configuration is: 1s² 2s² 2p³.

**Comparison**:
- O has a higher ionization energy than O⁺ because removing an electron from O⁺ means removing it from a more stable half-filled configuration (2p³), which requires more energy.
- Therefore, O⁺ has a higher ionization energy than O.

### Step 2: Analyze the Second Pair - Sulfur (S) and Phosphorus (P)
- **Sulfur (S)** has the electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁴.
- **Phosphorus (P)** has the electronic configuration: 1s² 2s² 2p⁶ 3s² 3p³.

**Comparison**:
- Phosphorus has a half-filled p subshell (3p³), which is more stable than sulfur's (3p⁴).
- Therefore, it is easier to remove an electron from sulfur than from phosphorus.
- Thus, the ionization energy of sulfur (S) is less than that of phosphorus (P).

### Step 3: Analyze the Third Pair - Nitrogen (N) and Phosphorus (P)
- **Nitrogen (N)** has the electronic configuration: 1s² 2s² 2p³.
- **Phosphorus (P)** has the electronic configuration: 1s² 2s² 2p⁶ 3s² 3p³.

**Comparison**:
- Nitrogen is in the second period and has a smaller atomic size and higher effective nuclear charge compared to phosphorus.
- Therefore, nitrogen has a higher ionization energy than phosphorus.

### Step 4: Analyze the Fourth Pair - Beryllium Ion (Be⁺) and Beryllium (Be)
- **Beryllium (Be)** has the electronic configuration: 1s² 2s².
- **Beryllium Ion (Be⁺)** has lost one electron, so its configuration is: 1s².

**Comparison**:
- The removal of an electron from Be results in a more stable configuration (1s²) for Be⁺.
- The ionization energy of Be⁺ is higher than that of Be because the effective nuclear charge increases after losing an electron, pulling the remaining electrons closer and making them harder to remove.

### Conclusion
From the analysis:
- The only pair where the ionization energy of the first species is less than that of the second is **Sulfur (S) and Phosphorus (P)**.

### Final Answer
**Option B: Sulfur (S) and Phosphorus (P)** is the correct answer.