Choose the pair in which `IE_(1)` of first element is greater than `IE_(1)` of second element but in case of `IE_(2)` order is/are reversed

Options

A. N, O
B. P,S
C. Be,B
D. F,O

Answer

To solve the problem, we need to analyze the ionization energies (IE) of the given pairs of elements and determine which pairs satisfy the condition that the first ionization energy (IE1) of the first element is greater than that of the second element, while the second ionization energy (IE2) order is reversed.

### Step-by-Step Solution:

1. **Understanding Ionization Energy:**
- Ionization energy is the energy required to remove an electron from an atom. Generally, as you move across a period from left to right, the ionization energy increases due to increased nuclear charge.

2. **Analyzing Each Pair:**
- **Option 1: Nitrogen (N) and Oxygen (O)**
- Atomic Numbers: N = 7 (1s² 2s² 2p³), O = 8 (1s² 2s² 2p⁴)
- IE1: IE1(N) > IE1(O) because nitrogen has a half-filled p subshell which is more stable.
- IE2: After removing one electron from O (which becomes 2p³), the second ionization energy of N (removing from 2p²) is lower than that of O. Thus, IE2(O) > IE2(N).
- **Conclusion:** This pair satisfies the condition.

- **Option 2: Phosphorus (P) and Sulfur (S)**
- Atomic Numbers: P = 15 (1s² 2s² 2p⁶ 3s² 3p³), S = 16 (1s² 2s² 2p⁶ 3s² 3p⁴)
- IE1: IE1(P) > IE1(S) for the same reason as above (half-filled p subshell).
- IE2: After removing one electron from S (which becomes 3p³), the second ionization energy of P (removing from 3p²) is lower than that of S. Thus, IE2(S) > IE2(P).
- **Conclusion:** This pair also satisfies the condition.

- **Option 3: Beryllium (Be) and Boron (B)**
- Atomic Numbers: Be = 4 (1s² 2s²), B = 5 (1s² 2s² 2p¹)
- IE1: IE1(Be) > IE1(B) because Be has a filled s subshell which is stable.
- IE2: After removing one electron from B (which becomes 2s²), the second ionization energy of Be (removing from 2s¹) is higher than that of B. Thus, IE2(B) > IE2(Be).
- **Conclusion:** This pair satisfies the condition as well.

- **Option 4: Fluorine (F) and Oxygen (O)**
- Atomic Numbers: F = 9 (1s² 2s² 2p⁵), O = 8 (1s² 2s² 2p⁴)
- IE1: IE1(F) > IE1(O) because F is to the right of O in the same period.
- IE2: After removing one electron from F (which becomes 2p⁴), the second ionization energy of O (removing from 2p³) is lower than that of F. Thus, IE2(F) < IE2(O).
- **Conclusion:** This pair does not satisfy the condition.

3. **Final Answer:**
- The pairs that satisfy the condition are:
- Option 1: Nitrogen and Oxygen
- Option 2: Phosphorus and Sulfur
- Option 3: Beryllium and Boron