Explain why IE of Beryllium is higher than Boron.

📚 Chapter: Classification of Elements and Periodicity in Properties

📘 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.

Be: 1s² 2s²

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?

1. The 2p electron in Boron has higher energy than the 2s electrons in Beryllium.
2. The 2p electron is less penetrating toward the nucleus than a 2s electron.
3. Therefore, the 2p electron experiences comparatively weaker attraction from the nucleus.
4. Hence, less energy is required to remove the 2p electron from Boron.

2p electron → Higher Energy → Easier Removal → Lower IE

⚛️ 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.

Therefore, IE(Be) > IE(B)

📌 Important Periodic Trend Exception

Normally, ionization enthalpy increases from left to right across a period. However, there is an exception between Be and B.

Be → 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.

IE(Be) > IE(B)

20 MCQs with Answers

1. Which has higher first ionization enthalpy?

A) B

B) Be

C) C

D) Li

✅ Answer

B) Be

---

2. Electronic configuration of Be is:

A) 1s² 2s¹

B) 1s² 2s²

C) 1s² 2p²

D) 1s² 2s² 2p¹

✅ Answer

B) 1s² 2s²

---

3. Electronic configuration of B is:

A) 1s² 2s²

B) 1s² 2s¹

C) 1s² 2s² 2p¹

D) 1s² 2p³

✅ Answer

C

---

4. The electron removed from B during first ionization is from:

A) 1s

B) 2s

C) 2p

D) 3s

✅ Answer

C) 2p

---

5. The electron removed from Be during first ionization is from:

A) 1s

B) 2s

C) 2p

D) 3s

✅ Answer

B) 2s

---

6. Which orbital has higher energy in B?

A) 1s

B) 2s

C) 2p

D) None

✅ Answer

C) 2p

---

7. A 2p electron is generally:

A) Easier to remove than a 2s electron

B) Harder to remove than a 2s electron

C) A proton

D) A neutron

✅ Answer

A

---

8. The 2s² configuration of Be is:

A) Half-filled

B) Completely filled

C) Empty

D) Unstable

✅ Answer

B) Completely filled

---

9. Be and B belong to which period?

A) 1st

B) 2nd

C) 3rd

D) 4th

✅ Answer

B) 2nd

---

10. Atomic number of Be is:

A) 2

B) 3

C) 4

D) 5

✅ Answer

C) 4

---

11. Atomic number of B is:

A) 4

B) 5

C) 6

D) 7

✅ Answer

B) 5

---

12. The anomaly between Be and B is related to:

A) Nuclear size

B) Subshell energy

C) Neutron mass

D) Isotopes

✅ Answer

B) Subshell energy

---

13. Which statement is correct?

A) IE(B) > IE(Be)

B) IE(Be) > IE(B)

C) IE(Be) = IE(B)

D) Both have zero IE

✅ Answer

B

---

14. The 2p electron in B experiences comparatively:

A) Stronger attraction

B) Weaker attraction

C) No attraction

D) Infinite attraction

✅ Answer

B) Weaker attraction

---

15. The 2s orbital is more penetrating than:

A) 1s

B) 2p

C) Both equally

D) 3s only

✅ Answer

B) 2p

---

16. Which element has a filled 2s subshell?

A) Li

B) Be

C) B

D) C

✅ Answer

B) Be

---

17. Ionization enthalpy is the energy required to:

A) Add an electron

B) Remove an electron from a gaseous atom

C) Remove a proton

D) Add a neutron

✅ Answer

B

---

18. Which configuration is more stable between the two?

A) Be: 2s²

B) B: 2p¹

C) Both are identical

D) Neither

✅ Answer

A) Be: 2s²

---

19. The exception in IE trend from Be to B is:

A) IE increases

B) IE decreases

C) IE becomes zero

D) No change

✅ Answer

B) IE decreases

---

20. The correct order is:

A) B > Be

B) Be > B

C) B = Be

D) Cannot be compared

✅ 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.

IE(Be) > IE(B)

Q. Explain the electronic configurations of Be and B and their effect on ionization enthalpy.

Be = 1s² 2s²
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.

Hence, IE(Be) > IE(B).

Q. Explain the anomalous decrease in ionization enthalpy from Be to B.

Solution:

1. Normally, IE increases from left to right across a period.
2. Be has configuration 1s² 2s².
3. B has configuration 1s² 2s² 2p¹.
4. The electron removed from B is a higher-energy 2p electron.
5. It is less penetrating and less strongly held than the 2s electron of Be.

Therefore, IE of Be is higher than IE of B.

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.

Thus, IE(Be) > IE(B), which is an important exception to the general periodic trend.

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

Be (Z = 4) = 1s² 2s²

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.

Be: 2s² → Stable Filled Subshell

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.

Final Answer:

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

Be: 1s² 2s²
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)

Why do s-block elements have very low IE?

📚 Chapter: Classification of Elements and Periodicity in Properties

📘 Answer

s-block elements generally have low first ionization enthalpy because their outermost electron is present in an s-orbital and is relatively easy to remove.

s-block → ns¹ or ns²

The valence electrons are comparatively far from the nucleus and experience effective nuclear attraction that is relatively low. Hence, only a small amount of energy is required to remove them.

🔹 Main Reasons for Low Ionization Enthalpy

1. Large atomic size: s-block elements generally have a relatively large atomic radius, so the valence electron is farther from the nucleus.
2. Low effective nuclear attraction: Inner-shell electrons shield the valence electron from the nucleus.
3. High shielding effect: The inner electrons reduce the effective attraction experienced by the outermost electron.
4. Single or two valence electrons: Group 1 elements have ns¹ and Group 2 elements have ns² configurations. The outer electron can be removed comparatively easily.
5. Formation of stable ions: Removal of the valence electron often leads to a stable noble-gas configuration.

⚛️ Example: Sodium

Sodium has atomic number 11 and electronic configuration:

Na = 1s² 2s² 2p⁶ 3s¹

Its outermost electron is present in the 3s orbital. It is relatively far from the nucleus and is shielded by the inner 10 electrons. Therefore, it can be removed easily.

Na → Na⁺ + e⁻

After losing one electron, sodium forms Na⁺ with the stable noble-gas configuration of neon:

Na⁺ = 1s² 2s² 2p⁶

📊 Group 1 vs Group 2

Group	General Configuration	IE
Group 1	ns¹	Very Low
Group 2	ns²	Low, but higher than Group 1

Group 1 elements generally have lower first ionization enthalpy than Group 2 elements in the same period because removal of one electron from ns¹ gives a stable noble-gas configuration.

✅ Final Answer

s-block elements have very low ionization enthalpy because their valence electrons are relatively far from the nucleus and are strongly shielded by inner-shell electrons. Therefore, the effective nuclear attraction on the valence electron is low and the electron can be removed easily.

Large Size + High Shielding → Weak Attraction → Low IE

20 MCQs with Answers

1. s-block elements generally have:

A) Very high IE

B) Low IE

C) Zero IE

D) Negative IE

✅ Answer

B) Low IE

---

2. The valence electron of Group 1 elements is present in:

A) p-orbital

B) d-orbital

C) s-orbital

D) f-orbital

✅ Answer

C) s-orbital

---

3. General valence configuration of Group 1 elements is:

A) ns²

B) ns¹

C) np¹

D) nd¹

✅ Answer

B) ns¹

---

4. General valence configuration of Group 2 elements is:

A) ns¹

B) ns²

C) np²

D) nd²

✅ Answer

B) ns²

---

5. Which element belongs to the s-block?

A) Na

B) Cl

C) Fe

D) Br

✅ Answer

A) Na

---

6. Which factor reduces nuclear attraction on outer electrons?

A) Shielding effect

B) Atomic number only

C) Proton mass

D) Neutron number

✅ Answer

A) Shielding effect

---

7. Higher shielding generally causes ionization enthalpy to:

A) Increase

B) Decrease

C) Become infinite

D) Remain zero

✅ Answer

B) Decrease

---

8. Sodium has which valence configuration?

A) 3s¹

B) 3s²

C) 3p¹

D) 2p⁶

✅ Answer

A) 3s¹

---

9. Which has lower first IE in the same period?

A) Na

B) Mg

C) Al

D) Si

✅ Answer

A) Na

---

10. Removal of one electron from Na produces:

A) Na⁻

B) Na⁺

C) Na²⁺

D) Na²⁻

✅ Answer

B) Na⁺

---

11. Which configuration is obtained by Na⁺?

A) 2,8

B) 2,8,1

C) 2,7

D) 2,8,8

✅ Answer

A) 2,8

---

12. Which group contains alkali metals?

A) Group 1

B) Group 2

C) Group 17

D) Group 18

✅ Answer

A) Group 1

---

13. Which group contains alkaline earth metals?

A) Group 1

B) Group 2

C) Group 16

D) Group 18

✅ Answer

B) Group 2

---

14. In general, atomic size down Group 1:

A) Decreases

B) Increases

C) Remains unchanged

D) Becomes zero

✅ Answer

B) Increases

---

15. As atomic size increases, ionization enthalpy generally:

A) Increases

B) Decreases

C) Becomes zero always

D) Doubles

✅ Answer

B) Decreases

---

16. Which is generally more easily removed?

A) Inner-shell electron

B) Valence electron of an alkali metal

C) Proton

D) Neutron

✅ Answer

B

---

17. s-block elements are mainly found in:

A) Groups 1 and 2

B) Groups 13–18

C) Groups 3–12

D) Only Group 18

✅ Answer

A) Groups 1 and 2

---

18. Which generally has lower first IE?

A) K

B) Na

C) Li

D) Mg

✅ Answer

A) K

---

19. The outermost electron in s-block elements is generally:

A) Easy to remove

B) Impossible to remove

C) In the nucleus

D) A proton

✅ Answer

A) Easy to remove

---

20. Correct relationship is:

A) Shielding ↑ → IE ↑

B) Atomic size ↑ → IE ↓

C) Atomic size ↑ → IE ↑ always

D) Shielding has no effect

✅ Answer

B) Atomic size ↑ → IE ↓

Q. Why do s-block elements have low ionization enthalpy?

Solution:

s-block elements have relatively large atomic size and significant shielding of their valence electrons by inner-shell electrons. Therefore, the outermost electron experiences weaker effective nuclear attraction and can be removed easily.

Hence, s-block elements generally have low ionization enthalpy.

Q. Give three reasons for the low ionization enthalpy of s-block elements.

Solution:

1. They generally have relatively large atomic size.
2. Their valence electrons experience considerable shielding by inner electrons.
3. The effective nuclear attraction on the outer electron is relatively low.

Q. Explain why Group 1 elements have very low first ionization enthalpy.

Solution:

1. Group 1 elements have the general configuration ns¹.
2. The single valence electron is relatively far from the nucleus.
3. It is shielded by the inner-shell electrons.
4. Removal of this electron gives a stable noble-gas configuration.

Therefore, Group 1 elements have very low first ionization enthalpy.

Q. Explain the factors responsible for low ionization enthalpy of s-block elements with an example.

Solution:

The low ionization enthalpy of s-block elements is mainly due to their relatively large atomic size, high shielding effect and weak effective nuclear attraction on the valence electron.

For example, sodium has configuration:

Na = 2,8,1

Its single 3s electron is far from the nucleus and is shielded by the inner 10 electrons. Therefore, it can be removed easily:

Na → Na⁺ + e⁻

Thus, s-block elements generally have low ionization enthalpy.

Q. Explain in detail why s-block elements have low ionization enthalpy. Discuss the effect of atomic size, shielding effect and electronic configuration.

Detailed Solution

s-block elements are those elements in which the last electron enters an s-orbital. They mainly include Groups 1 and 2. Their general outer electronic configurations are ns¹ and ns².

1. Large Atomic Size

The valence electron is relatively far from the nucleus. Greater distance reduces the attractive force between the nucleus and the valence electron, making its removal easier.

2. Shielding Effect

The inner-shell electrons shield the valence electron from the attraction of the nucleus. Therefore, the effective nuclear attraction experienced by the outer electron is reduced.

Shielding ↑ → Effective Attraction ↓ → IE ↓

3. Electronic Configuration

Group 1 elements have ns¹ configuration. Removal of this single electron gives a stable noble-gas configuration. Therefore, the first electron can be removed comparatively easily.

Example: Sodium

Na = 1s² 2s² 2p⁶ 3s¹

The 3s electron is the valence electron. It is relatively far from the nucleus and is shielded by the inner electrons. On removal:

Na → Na⁺ + e⁻

Na⁺ obtains the stable configuration of neon: 1s² 2s² 2p⁶.

Final Conclusion:

s-block elements have low ionization enthalpy because their valence electrons are relatively far from the nucleus, strongly shielded by inner electrons, and are held comparatively weakly.

Large Atomic Size + Shielding + ns¹/ns² → Low IE

🎯 Quick Revision

s-block: Groups 1 & 2
Outer Configuration: ns¹ / ns²
Atomic Size: Relatively large
Shielding: High
Valence Electron: Relatively easy to remove
Ionization Enthalpy: Low

Shielding ↑ + Size ↑ → IE ↓

Why do s-block elements have very low IE?

📚 Chapter: Classification of Elements and Periodicity in Properties

📘 Answer

s-block elements generally have low first ionization enthalpy because their outermost electron is present in an s-orbital and is relatively easy to remove.

s-block → ns¹ or ns²

The valence electrons are comparatively far from the nucleus and experience effective nuclear attraction that is relatively low. Hence, only a small amount of energy is required to remove them.

🔹 Main Reasons for Low Ionization Enthalpy

1. Large atomic size: s-block elements generally have a relatively large atomic radius, so the valence electron is farther from the nucleus.
2. Low effective nuclear attraction: Inner-shell electrons shield the valence electron from the nucleus.
3. High shielding effect: The inner electrons reduce the effective attraction experienced by the outermost electron.
4. Single or two valence electrons: Group 1 elements have ns¹ and Group 2 elements have ns² configurations. The outer electron can be removed comparatively easily.
5. Formation of stable ions: Removal of the valence electron often leads to a stable noble-gas configuration.

⚛️ Example: Sodium

Sodium has atomic number 11 and electronic configuration:

Na = 1s² 2s² 2p⁶ 3s¹

Its outermost electron is present in the 3s orbital. It is relatively far from the nucleus and is shielded by the inner 10 electrons. Therefore, it can be removed easily.

Na → Na⁺ + e⁻

After losing one electron, sodium forms Na⁺ with the stable noble-gas configuration of neon:

Na⁺ = 1s² 2s² 2p⁶

📊 Group 1 vs Group 2

Group	General Configuration	IE
Group 1	ns¹	Very Low
Group 2	ns²	Low, but higher than Group 1

Group 1 elements generally have lower first ionization enthalpy than Group 2 elements in the same period because removal of one electron from ns¹ gives a stable noble-gas configuration.

✅ Final Answer

s-block elements have very low ionization enthalpy because their valence electrons are relatively far from the nucleus and are strongly shielded by inner-shell electrons. Therefore, the effective nuclear attraction on the valence electron is low and the electron can be removed easily.

Large Size + High Shielding → Weak Attraction → Low IE

20 MCQs with Answers

1. s-block elements generally have:

A) Very high IE

B) Low IE

C) Zero IE

D) Negative IE

✅ Answer

B) Low IE

---

2. The valence electron of Group 1 elements is present in:

A) p-orbital

B) d-orbital

C) s-orbital

D) f-orbital

✅ Answer

C) s-orbital

---

3. General valence configuration of Group 1 elements is:

A) ns²

B) ns¹

C) np¹

D) nd¹

✅ Answer

B) ns¹

---

4. General valence configuration of Group 2 elements is:

A) ns¹

B) ns²

C) np²

D) nd²

✅ Answer

B) ns²

---

5. Which element belongs to the s-block?

A) Na

B) Cl

C) Fe

D) Br

✅ Answer

A) Na

---

6. Which factor reduces nuclear attraction on outer electrons?

A) Shielding effect

B) Atomic number only

C) Proton mass

D) Neutron number

✅ Answer

A) Shielding effect

---

7. Higher shielding generally causes ionization enthalpy to:

A) Increase

B) Decrease

C) Become infinite

D) Remain zero

✅ Answer

B) Decrease

---

8. Sodium has which valence configuration?

A) 3s¹

B) 3s²

C) 3p¹

D) 2p⁶

✅ Answer

A) 3s¹

---

9. Which has lower first IE in the same period?

A) Na

B) Mg

C) Al

D) Si

✅ Answer

A) Na

---

10. Removal of one electron from Na produces:

A) Na⁻

B) Na⁺

C) Na²⁺

D) Na²⁻

✅ Answer

B) Na⁺

---

11. Which configuration is obtained by Na⁺?

A) 2,8

B) 2,8,1

C) 2,7

D) 2,8,8

✅ Answer

A) 2,8

---

12. Which group contains alkali metals?

A) Group 1

B) Group 2

C) Group 17

D) Group 18

✅ Answer

A) Group 1

---

13. Which group contains alkaline earth metals?

A) Group 1

B) Group 2

C) Group 16

D) Group 18

✅ Answer

B) Group 2

---

14. In general, atomic size down Group 1:

A) Decreases

B) Increases

C) Remains unchanged

D) Becomes zero

✅ Answer

B) Increases

---

15. As atomic size increases, ionization enthalpy generally:

A) Increases

B) Decreases

C) Becomes zero always

D) Doubles

✅ Answer

B) Decreases

---

16. Which is generally more easily removed?

A) Inner-shell electron

B) Valence electron of an alkali metal

C) Proton

D) Neutron

✅ Answer

B

---

17. s-block elements are mainly found in:

A) Groups 1 and 2

B) Groups 13–18

C) Groups 3–12

D) Only Group 18

✅ Answer

A) Groups 1 and 2

---

18. Which generally has lower first IE?

A) K

B) Na

C) Li

D) Mg

✅ Answer

A) K

---

19. The outermost electron in s-block elements is generally:

A) Easy to remove

B) Impossible to remove

C) In the nucleus

D) A proton

✅ Answer

A) Easy to remove

---

20. Correct relationship is:

A) Shielding ↑ → IE ↑

B) Atomic size ↑ → IE ↓

C) Atomic size ↑ → IE ↑ always

D) Shielding has no effect

✅ Answer

B) Atomic size ↑ → IE ↓

Q. Why do s-block elements have low ionization enthalpy?

Solution:

s-block elements have relatively large atomic size and significant shielding of their valence electrons by inner-shell electrons. Therefore, the outermost electron experiences weaker effective nuclear attraction and can be removed easily.

Hence, s-block elements generally have low ionization enthalpy.

Q. Give three reasons for the low ionization enthalpy of s-block elements.

Solution:

1. They generally have relatively large atomic size.
2. Their valence electrons experience considerable shielding by inner electrons.
3. The effective nuclear attraction on the outer electron is relatively low.

Q. Explain why Group 1 elements have very low first ionization enthalpy.

Solution:

1. Group 1 elements have the general configuration ns¹.
2. The single valence electron is relatively far from the nucleus.
3. It is shielded by the inner-shell electrons.
4. Removal of this electron gives a stable noble-gas configuration.

Therefore, Group 1 elements have very low first ionization enthalpy.

Q. Explain the factors responsible for low ionization enthalpy of s-block elements with an example.

Solution:

The low ionization enthalpy of s-block elements is mainly due to their relatively large atomic size, high shielding effect and weak effective nuclear attraction on the valence electron.

For example, sodium has configuration:

Na = 2,8,1

Its single 3s electron is far from the nucleus and is shielded by the inner 10 electrons. Therefore, it can be removed easily:

Na → Na⁺ + e⁻

Thus, s-block elements generally have low ionization enthalpy.

Q. Explain in detail why s-block elements have low ionization enthalpy. Discuss the effect of atomic size, shielding effect and electronic configuration.

Detailed Solution

s-block elements are those elements in which the last electron enters an s-orbital. They mainly include Groups 1 and 2. Their general outer electronic configurations are ns¹ and ns².

1. Large Atomic Size

The valence electron is relatively far from the nucleus. Greater distance reduces the attractive force between the nucleus and the valence electron, making its removal easier.

2. Shielding Effect

The inner-shell electrons shield the valence electron from the attraction of the nucleus. Therefore, the effective nuclear attraction experienced by the outer electron is reduced.

Shielding ↑ → Effective Attraction ↓ → IE ↓

3. Electronic Configuration

Group 1 elements have ns¹ configuration. Removal of this single electron gives a stable noble-gas configuration. Therefore, the first electron can be removed comparatively easily.

Example: Sodium

Na = 1s² 2s² 2p⁶ 3s¹

The 3s electron is the valence electron. It is relatively far from the nucleus and is shielded by the inner electrons. On removal:

Na → Na⁺ + e⁻

Na⁺ obtains the stable configuration of neon: 1s² 2s² 2p⁶.

Final Conclusion:

s-block elements have low ionization enthalpy because their valence electrons are relatively far from the nucleus, strongly shielded by inner electrons, and are held comparatively weakly.

Large Atomic Size + Shielding + ns¹/ns² → Low IE

🎯 Quick Revision

s-block: Groups 1 & 2
Outer Configuration: ns¹ / ns²
Atomic Size: Relatively large
Shielding: High
Valence Electron: Relatively easy to remove
Ionization Enthalpy: Low

Shielding ↑ + Size ↑ → IE ↓