📘 Solution
The elements C, N, O and F belong to the 2nd period. Generally, first ionization enthalpy increases from left to right across a period because effective nuclear charge increases and atomic size decreases.
However, there is an important exception between N and O. Nitrogen has a stable half-filled 2p³ configuration, whereas oxygen has 2p⁴ configuration with one pair of electrons in a 2p orbital.
⚛️ Electronic Configurations
| Element | Electronic Configuration | Special Feature |
|---|---|---|
| C | 1s² 2s² 2p² | 2p² |
| N | 1s² 2s² 2p³ | Half-filled 2p³ |
| O | 1s² 2s² 2p⁴ | Paired 2p electron |
| F | 1s² 2s² 2p⁵ | High effective nuclear charge |
🔹 Why is IE of N higher than O?
Nitrogen has the configuration 2p³, in which the three 2p orbitals are singly occupied. This is a relatively stable half-filled subshell.
Oxygen has 2p⁴. In one of its 2p orbitals, two electrons are paired. The electron-electron repulsion within this paired orbital makes one electron comparatively easier to remove.
O (2p⁴) → Electron Pairing Repulsion → Lower IE
📊 Increasing Order
Considering the general periodic trend and the N–O exception:
Therefore, the increasing order of first ionization enthalpy is:
✅ Final Answer
Increasing order:
The unusual position of O and N is due to the greater stability of the half-filled 2p³ configuration of nitrogen.
20 MCQs with Answers
1. Which has the highest first IE among C, N, O and F?
✅ Answer
D) F
2. Which has the lowest first IE among C, N, O and F?
✅ Answer
A) C
3. Correct increasing order is:
✅ Answer
A) C < O < N < F
4. C, N, O and F belong to which period?
✅ Answer
B) 2nd
5. Nitrogen has which valence configuration?
✅ Answer
C) 2s²2p³
6. Oxygen has which valence configuration?
✅ Answer
C) 2s²2p⁴
7. Why is IE of N higher than O?
✅ Answer
B
8. In oxygen, lower IE is partly due to:
✅ Answer
A
9. Which element has a half-filled 2p subshell?
✅ Answer
B) N
10. Across a period, effective nuclear charge generally:
✅ Answer
A) Increases
11. Atomic size across a period generally:
✅ Answer
B) Decreases
12. Which has configuration 1s²2s²2p⁵?
✅ Answer
D) F
13. Which has configuration 1s²2s²2p²?
✅ Answer
A) C
14. The first IE of N is higher than O because:
✅ Answer
A
15. Which order follows the general increase in IE across the period?
✅ Answer
A
16. The exception in the C–N–O–F sequence occurs between:
✅ Answer
B) N and O
17. The atomic number of nitrogen is:
✅ Answer
C) 7
18. The atomic number of oxygen is:
✅ Answer
C) 8
19. The correct order of first IE is:
✅ Answer
A
20. Which principle explains the higher IE of N compared with O?
✅ Answer
A
Q. Arrange C, N, O and F in increasing order of first ionization enthalpy.
Solution: The general IE trend increases across the period, but N has a stable half-filled 2p³ configuration and therefore has higher IE than O.
Q. Why is the first ionization enthalpy of nitrogen greater than that of oxygen?
Solution:
- N has configuration 2p³, which is half-filled and relatively stable.
- O has configuration 2p⁴, containing a paired electron.
- Electron-electron repulsion in the paired orbital makes removal of an electron from O easier.
Q. Explain the increasing order of first ionization enthalpy of C, N, O and F.
Solution:
All four elements belong to the second period. Effective nuclear charge increases from C to F, so IE generally increases. However, nitrogen has a stable half-filled 2p³ configuration, while oxygen has 2p⁴ with electron pairing repulsion.
Thus, the expected C < N < O < F sequence is modified because IE(N) > IE(O).
Q. Explain the periodic trend in first ionization enthalpy and arrange C, N, O and F in increasing order.
Solution:
- First ionization enthalpy generally increases from left to right across a period.
- This is mainly due to increasing effective nuclear charge and decreasing atomic size.
- C, N, O and F are elements of the second period.
- N has a stable half-filled 2p³ configuration.
- O has 2p⁴ configuration and electron pairing causes additional repulsion.
Q. Discuss the first ionization enthalpy trend of C, N, O and F and give the correct increasing order with reasons.
Detailed Solution
Carbon, nitrogen, oxygen and fluorine belong to the second period. As we move from left to right, nuclear charge increases while electrons are added to the same principal shell. Therefore, effective nuclear charge generally increases and atomic radius decreases. Consequently, first ionization enthalpy generally increases.
Electronic Configurations
N = 1s² 2s² 2p³
O = 1s² 2s² 2p⁴
F = 1s² 2s² 2p⁵
Important Exception: N and O
Nitrogen has a half-filled 2p³ subshell. This configuration is relatively stable because the three 2p orbitals contain one electron each.
Oxygen has 2p⁴ configuration. One of its 2p orbitals contains a pair of electrons. Repulsion between the paired electrons makes one electron easier to remove.
Final Order
Therefore, the increasing order of first ionization enthalpy is C < O < N < F.
🎯 Quick Revision
N: 2p³ → Half-filled → Higher IE
O: 2p⁴ → Pairing repulsion → Lower than N
F: 2p⁵ → Highest among these four
C < O < N < F
