📘 Answer
The first ionization enthalpy of Beryllium (Be) is higher than that of Boron (B), even though ionization enthalpy generally increases from left to right across a period.
B: 1s² 2s² 2p¹
The reason is that the electron removed from Be is a 2s electron, whereas the electron removed from B is a 2p electron.
🔹 Why is the 2p electron of Boron easier to remove?
- The 2p electron in Boron has higher energy than the 2s electrons in Beryllium.
- The 2p electron is less penetrating toward the nucleus than a 2s electron.
- Therefore, the 2p electron experiences comparatively weaker attraction from the nucleus.
- Hence, less energy is required to remove the 2p electron from Boron.
⚛️ Electronic Configuration
| Element | Electronic Configuration | Electron Removed |
|---|---|---|
| Be | 1s² 2s² | 2s |
| B | 1s² 2s² 2p¹ | 2p |
The completely filled 2s² subshell of Be is relatively stable. In Boron, the first electron is present in the higher-energy 2p orbital and is therefore easier to remove.
📌 Important Periodic Trend Exception
Normally, ionization enthalpy increases from left to right across a period. However, there is an exception between Be and B.
Expected: IE ↑
Actual: IE ↓
This occurs because the electron removed from B is from the higher-energy 2p orbital, while the electron removed from Be is from the lower-energy 2s orbital.
✅ Final Answer
Beryllium has a higher first ionization enthalpy than Boron because Be has a stable, completely filled 2s² subshell, while Boron's outermost electron enters the higher-energy 2p orbital. The 2p electron is easier to remove.
20 MCQs with Answers
1. Which has higher first ionization enthalpy?
✅ Answer
B) Be
2. Electronic configuration of Be is:
✅ Answer
B) 1s² 2s²
3. Electronic configuration of B is:
✅ Answer
C
4. The electron removed from B during first ionization is from:
✅ Answer
C) 2p
5. The electron removed from Be during first ionization is from:
✅ Answer
B) 2s
6. Which orbital has higher energy in B?
✅ Answer
C) 2p
7. A 2p electron is generally:
✅ Answer
A
8. The 2s² configuration of Be is:
✅ Answer
B) Completely filled
9. Be and B belong to which period?
✅ Answer
B) 2nd
10. Atomic number of Be is:
✅ Answer
C) 4
11. Atomic number of B is:
✅ Answer
B) 5
12. The anomaly between Be and B is related to:
✅ Answer
B) Subshell energy
13. Which statement is correct?
✅ Answer
B
14. The 2p electron in B experiences comparatively:
✅ Answer
B) Weaker attraction
15. The 2s orbital is more penetrating than:
✅ Answer
B) 2p
16. Which element has a filled 2s subshell?
✅ Answer
B) Be
17. Ionization enthalpy is the energy required to:
✅ Answer
B
18. Which configuration is more stable between the two?
✅ Answer
A) Be: 2s²
19. The exception in IE trend from Be to B is:
✅ Answer
B) IE decreases
20. The correct order is:
✅ Answer
B) Be > B
Q. Why is the first ionization enthalpy of Be higher than that of B?
Solution: Be has the stable configuration 2s², whereas B has an electron in the higher-energy 2p orbital. Therefore, the 2p electron of B is easier to remove.
Q. Explain the electronic configurations of Be and B and their effect on ionization enthalpy.
B = 1s² 2s² 2p¹
In Be, the electron removed is a 2s electron from a completely filled 2s subshell. In B, the electron removed is a higher-energy 2p electron. Therefore, the electron in B is easier to remove.
Q. Explain the anomalous decrease in ionization enthalpy from Be to B.
Solution:
- Normally, IE increases from left to right across a period.
- Be has configuration 1s² 2s².
- B has configuration 1s² 2s² 2p¹.
- The electron removed from B is a higher-energy 2p electron.
- It is less penetrating and less strongly held than the 2s electron of Be.
Q. Give a detailed explanation of why Be has higher ionization enthalpy than B despite B having a higher atomic number.
Solution:
Although B has a higher nuclear charge than Be, the first electron removed from B belongs to the 2p orbital, while the electron removed from Be belongs to the 2s orbital.
- Be: 1s² 2s²
- B: 1s² 2s² 2p¹
- The 2p orbital has higher energy than the 2s orbital.
- The 2p electron is less penetrating.
- Therefore, the 2p electron is easier to remove.
Q. Explain in detail why the first ionization enthalpy of Beryllium is higher than that of Boron. Also explain this as an exception to the periodic trend.
Detailed Solution
Ionization enthalpy generally increases from left to right across a period because effective nuclear charge increases and atomic size decreases. However, an exception occurs between Be and B.
Electronic Configurations
B (Z = 5) = 1s² 2s² 2p¹
Reason 1 – Subshell Energy
The electron removed from Be is a 2s electron, whereas the electron removed from B is a 2p electron. The 2p electron has higher energy and is therefore easier to remove.
Reason 2 – Penetration
The 2s orbital penetrates closer to the nucleus than the 2p orbital. Consequently, the 2s electron is held more strongly by the nucleus.
Reason 3 – Stability of Be
Beryllium has a completely filled 2s² subshell, which is relatively stable. Removing an electron from this stable configuration requires more energy.
Conclusion
Thus, although the nuclear charge increases from Be to B, the first electron of B enters the higher-energy 2p subshell. This makes it easier to remove than the 2s electron of Be.
IE(Be) > IE(B)
Be has a higher IE because of its stable filled 2s² subshell, while B loses a higher-energy 2p electron more easily.
🎯 Quick Revision
B: 1s² 2s² 2p¹
Be electron removed: 2s
B electron removed: 2p
2p electron: Higher energy & easier to remove
Be: Stable filled 2s² subshell
IE(Be) > IE(B)
