Define First Ionization Enthalpy. Why is the first IE of Nitrogen higher than Oxygen? Explain the trend of IE down a group.

📚 Chapter: Classification of Elements and Periodicity in Properties

📘 Definition

First Ionization Enthalpy is the minimum amount of energy required to remove the most loosely bound electron from one mole of isolated gaseous atoms in their ground state to form one mole of gaseous unipositive ions.

X(g) → X⁺(g) + e⁻

It is represented by Δ_iH₁ and is generally expressed in kJ mol⁻¹.

First Ionization Enthalpy = Energy required to remove the first electron from an isolated gaseous atom.

📘 Explanation

Although ionization enthalpy generally increases from left to right across a period, Nitrogen has a higher first ionization enthalpy than Oxygen. This is an important exception in the second period.

Element	Atomic Number	Electronic Configuration
N	7	1s² 2s² 2p³
O	8	1s² 2s² 2p⁴

🔹 Nitrogen

Nitrogen has a half-filled 2p³ configuration. Its three 2p orbitals contain one electron each.

2p³ → ↑   ↑   ↑

A half-filled subshell is relatively stable because of symmetrical distribution and exchange energy. Therefore, more energy is required to remove an electron from nitrogen.

🔹 Oxygen

Oxygen has the configuration 2p⁴. One of the 2p orbitals contains a pair of electrons.

2p⁴ → ↑↓   ↑   ↑

The two electrons in the same orbital experience greater electron-electron repulsion. Therefore, one electron can be removed comparatively easily from oxygen.

Therefore, IE₁(N) > IE₁(O)

Nitrogen has a stable half-filled 2p³ configuration, whereas oxygen has electron-electron repulsion in one paired 2p orbital.

📘 Trend Down a Group

The ionization enthalpy generally decreases as we move down a group in the periodic table.

Ionization Enthalpy ↓ down the group

🔹 Reasons

1. Increase in atomic size: A new electron shell is added at each step down the group.
2. Increase in shielding effect: The inner electrons shield the valence electron from the attraction of the nucleus.
3. Increase in distance: The outermost electron is farther from the nucleus.
4. Reduced effective nuclear attraction: The valence electron is held less strongly by the nucleus.
5. Therefore, less energy is required to remove the outermost electron.

Atomic size ↑   →   Shielding effect ↑   →   Ionization Enthalpy ↓

📊 Example: Group 1

Element	Trend in Atomic Size	Trend in IE
Li	Small	High
Na	↑	↓
K	↑	↓
Rb	↑	↓
Cs	Large	Low

Thus, ionization enthalpy generally decreases down a group.

📝 20 MCQs with Answers

1. First ionization enthalpy is the energy required to:

A) Add an electron to an atom

B) Remove the first electron from an isolated gaseous atom

C) Remove a proton

D) Add a proton

✅ Answer

B

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2. First ionization enthalpy is generally expressed in:

A) g mol⁻¹

B) L mol⁻¹

C) kJ mol⁻¹

D) mol L⁻¹

✅ Answer

C) kJ mol⁻¹

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3. The process X(g) → X⁺(g) + e⁻ represents:

A) Electron affinity

B) First ionization

C) Second ionization

D) Bond formation

✅ Answer

B) First ionization

---

4. Which has higher first ionization enthalpy?

A) N

B) O

C) Na

D) K

✅ Answer

A) N

---

5. Electronic configuration of nitrogen is:

A) 1s² 2s² 2p²

B) 1s² 2s² 2p³

C) 1s² 2s² 2p⁴

D) 1s² 2s² 2p⁵

✅ Answer

B

---

6. Nitrogen has high IE because of its:

A) Fully filled p-subshell

B) Half-filled p-subshell

C) Large size

D) Low nuclear charge

✅ Answer

B) Half-filled p-subshell

---

7. Oxygen has the configuration:

A) 2p²

B) 2p³

C) 2p⁴

D) 2p⁵

✅ Answer

C) 2p⁴

---

8. In oxygen, electron-electron repulsion occurs due to:

A) Unpaired electrons only

B) Paired electrons in one 2p orbital

C) Protons

D) Neutrons

✅ Answer

B

---

9. Ionization enthalpy generally decreases:

A) Across a period

B) Down a group

C) From right to left only

D) Randomly

✅ Answer

B) Down a group

---

10. Which factor increases down a group?

A) Atomic size

B) Ionization enthalpy

C) Effective nuclear attraction on valence electron

D) Electronegativity always

✅ Answer

A) Atomic size

---

11. Shielding effect down a group:

A) Decreases

B) Increases

C) Becomes zero

D) Remains unchanged

✅ Answer

B) Increases

---

12. Greater atomic size generally causes ionization enthalpy to:

A) Increase

B) Decrease

C) Become zero

D) Remain constant

✅ Answer

B) Decrease

---

13. Which is easier to ionize?

A) An atom with a tightly held electron

B) An atom with a loosely held valence electron

C) A nucleus

D) A proton

✅ Answer

B

---

14. The unusual IE trend between N and O is due to:

A) Half-filled stability and electron repulsion

B) Nuclear fission

C) Equal atomic radii

D) Different periods

✅ Answer

A

---

15. Which subshell is half-filled in nitrogen?

A) 2s

B) 2p

C) 3s

D) 3p

✅ Answer

B) 2p

---

16. Which element has the lowest IE among Li, Na and K?

A) Li

B) Na

C) K

D) All equal

✅ Answer

C) K

---

17. Down a group, the valence electron is:

A) Closer to nucleus

B) Farther from nucleus

C) Removed automatically

D) Converted into proton

✅ Answer

B

---

18. Which generally has higher IE?

A) Smaller atom

B) Larger atom

C) Larger ion always

D) None

✅ Answer

A) Smaller atom

---

19. Ionization enthalpy is a measure of:

A) Ease of removing an electron

B) Ease of adding a proton

C) Number of neutrons

D) Atomic mass only

✅ Answer

A

---

20. Correct statement is:

A) IE always decreases across a period

B) IE generally increases across a period and decreases down a group

C) IE always increases down a group

D) IE is independent of atomic size

✅ Answer

B

Q1. Define first ionization enthalpy.

It is the minimum energy required to remove the first electron from one mole of isolated gaseous atoms in their ground state to form one mole of gaseous unipositive ions.

X(g) → X⁺(g) + e⁻

---

Q2. Why is IE of nitrogen greater than oxygen?

Nitrogen has a stable half-filled 2p³ configuration, whereas oxygen has 2p⁴ configuration with one pair of electrons. Electron-electron repulsion in the paired orbital makes electron removal from oxygen easier.

IE₁(N) > IE₁(O)

Q1. Explain why ionization enthalpy decreases down a group.

1. Atomic size increases down the group.
2. Shielding effect increases due to addition of shells.
3. The valence electron is farther from the nucleus.
4. Nuclear attraction on the outer electron decreases.
5. Hence, less energy is required to remove it.

IE decreases down a group.

Q2. Why is nitrogen an exception in the IE trend between N and O?

Nitrogen has the particularly stable half-filled 2p³ configuration. Oxygen has 2p⁴ configuration, in which two electrons occupy the same 2p orbital and experience repulsion. Therefore, oxygen loses an electron more easily than nitrogen.

Q. Explain first ionization enthalpy and discuss its trend down a group.

First ionization enthalpy is the minimum energy required to remove the first electron from one mole of isolated gaseous atoms in their ground state.

X(g) → X⁺(g) + e⁻

Down a group:

• Number of shells increases.
• Atomic radius increases.
• Shielding effect increases.
• Valence electron becomes farther from the nucleus.
• Effective nuclear attraction decreases.

Therefore, first ionization enthalpy generally decreases down a group.

Q. Explain why the first ionization enthalpy of nitrogen is greater than that of oxygen.

The electronic configuration of nitrogen is 1s² 2s² 2p³. The three 2p electrons occupy the three 2p orbitals singly. This is a half-filled configuration, which possesses extra stability.

N: 2p³ → ↑   ↑   ↑

Oxygen has the configuration 1s² 2s² 2p⁴. One of its 2p orbitals contains a pair of electrons. The electrons in the same orbital experience repulsion.

O: 2p⁴ → ↑↓   ↑   ↑

Therefore, an electron can be removed more easily from oxygen than from nitrogen.

IE₁(N) > IE₁(O)

Q1. Define first ionization enthalpy and explain its periodic trend down a group.

Solution

First ionization enthalpy is the minimum energy required to remove the most loosely bound electron from one mole of isolated gaseous atoms in their ground state to form one mole of gaseous unipositive ions.

X(g) → X⁺(g) + e⁻

Trend Down a Group

Ionization enthalpy generally decreases down a group due to the following factors:

1. Increase in the number of electron shells.
2. Increase in atomic radius.
3. Increase in shielding effect.
4. Greater distance of the valence electron from the nucleus.
5. Decrease in effective nuclear attraction on the valence electron.
6. Consequently, less energy is needed to remove the electron.

Therefore, Ionization Enthalpy ↓ down a group.

Q2. Explain why the first ionization enthalpy of nitrogen is greater than that of oxygen.

Solution

Element	Configuration	Important Feature
N	1s² 2s² 2p³	Stable half-filled 2p subshell
O	1s² 2s² 2p⁴	Paired-electron repulsion

Nitrogen has a stable half-filled 2p³ configuration. In oxygen, one 2p orbital contains a pair of electrons. The mutual repulsion between these paired electrons makes removal of one electron easier.

Hence, the first ionization enthalpy of nitrogen is greater than that of oxygen.

IE₁(N) > IE₁(O)

🎯 Quick Revision

First IE: Energy required to remove first electron
General trend across period: IE ↑
General trend down group: IE ↓
Reason down group: Atomic size ↑ + Shielding ↑
Important exception: IE(N) > IE(O)
Reason: N has stable half-filled 2p³; O has paired-electron repulsion

Arrange the following in decreasing order of atomic size: Li,Na,K,Rb,Cs.

📚 Chapter: Classification of Elements and Periodicity in Properties

📘 Solution

🔵 Periodic Trend

All the given elements belong to Group 1 (Alkali Metals) of the periodic table.

As we move down a group, a new electron shell is added at each step. Therefore, the atomic radius generally increases because the outermost electron is farther from the nucleus and shielding effect also increases.

Atomic size increases ↓ down the group

📊 Given Elements

Element	Period	Shells	Relative Atomic Size
Li	2	2	Smallest
Na	3	3	↑
K	4	4	↑
Rb	5	5	↑
Cs	6	6	Largest

⬇️ Decreasing Order

Since atomic size increases from Li to Cs, the decreasing order is the reverse:

Cs > Rb > K > Na > Li

💡 Reason

• All elements belong to Group 1.
• Down the group, the number of electron shells increases.
• Shielding effect increases.
• The outermost electron becomes farther from the nucleus.
• Hence atomic radius increases from Li to Cs.

Li < Na < K < Rb < Cs
(Increasing atomic size)

✅ Final Answer

Decreasing order of atomic size:

Cs > Rb > K > Na > Li

📝 20 MCQs with Answers

1. Which is the largest atom among Li, Na, K, Rb and Cs?

A) Li

B) Na

C) K

D) Cs

✅ Answer

D) Cs

---

2. Which is the smallest atom among the given elements?

A) Li

B) Na

C) Rb

D) Cs

✅ Answer

A) Li

---

3. All the given elements belong to:

A) Group 1

B) Group 2

C) Group 17

D) Group 18

✅ Answer

A) Group 1

---

4. Atomic size generally ______ down a group.

A) Decreases

B) Increases

C) Remains constant

D) Becomes zero

✅ Answer

B) Increases

---

5. The correct decreasing order is:

A) Li > Na > K > Rb > Cs

B) Cs > Rb > K > Na > Li

C) K > Li > Na > Cs > Rb

D) Na > Li > Cs > K > Rb

✅ Answer

B) Cs > Rb > K > Na > Li

---

6. Which factor increases down Group 1?

A) Number of shells

B) Nuclear size only

C) Electronegativity

D) Ionisation energy

✅ Answer

A) Number of shells

---

7. Shielding effect down a group generally:

A) Decreases

B) Increases

C) Becomes zero

D) Remains exactly constant

✅ Answer

B) Increases

---

8. Which element has the highest principal quantum number among the given elements?

A) Li

B) Na

C) K

D) Cs

✅ Answer

D) Cs

---

9. Li belongs to which period?

A) 1

B) 2

C) 3

D) 4

✅ Answer

B) 2

---

10. Cs belongs to which period?

A) 4

B) 5

C) 6

D) 7

✅ Answer

C) 6

---

11. Which element has three electron shells?

A) Li

B) Na

C) K

D) Rb

✅ Answer

B) Na

---

12. Which has four electron shells?

A) Li

B) Na

C) K

D) Cs

✅ Answer

C) K

---

13. Atomic radius increases down Group 1 mainly because:

A) New shells are added

B) Electrons are removed

C) Nuclear charge decreases

D) Protons disappear

✅ Answer

A) New shells are added

---

14. Which has the greatest shielding effect?

A) Li

B) Na

C) Rb

D) Cs

✅ Answer

D) Cs

---

15. The outermost electron of Cs is located in:

A) 2nd shell

B) 3rd shell

C) 5th shell

D) 6th shell

✅ Answer

D) 6th shell

---

16. Which sequence represents increasing atomic size?

A) Cs < Rb < K < Na < Li

B) Li < Na < K < Rb < Cs

C) K < Na < Li < Cs < Rb

D) Rb < Cs < Li < Na < K

✅ Answer

B) Li < Na < K < Rb < Cs

---

17. Which pair has the larger atomic radius?

A) Li > Na

B) Na > K

C) K > Na

D) Cs < Rb

✅ Answer

C) K > Na

---

18. Rb is ______ in size compared with K.

A) Smaller

B) Larger

C) Equal

D) Unrelated

✅ Answer

B) Larger

---

19. The correct trend in atomic radius for Group 1 is:

A) Li > Na > K > Rb > Cs

B) Li < Na < K < Rb < Cs

C) Li = Na = K = Rb = Cs

D) Random

✅ Answer

B) Li < Na < K < Rb < Cs

---

20. Which statement is correct?

A) Atomic size decreases down Group 1

B) Atomic size increases down Group 1

C) Atomic size is independent of shells

D) Cs is the smallest among them

✅ Answer

B) Atomic size increases down Group 1

Q. Arrange Li, Na, K, Rb and Cs in decreasing order of atomic size. Give reason.

Solution:

All these elements belong to Group 1. Atomic size increases down the group because the number of electron shells and shielding effect increase.

Cs > Rb > K > Na > Li

Q. Explain the trend in atomic size of Li, Na, K, Rb and Cs.

Solution:

1. Li, Na, K, Rb and Cs belong to Group 1.
2. Moving down the group, a new electron shell is added.
3. Shielding effect increases and the outermost electron is farther from the nucleus.
4. Therefore, atomic size increases down the group.

Li < Na < K < Rb < Cs
Increasing atomic size

Q. Arrange Li, Na, K, Rb and Cs in decreasing order of atomic size and explain the trend.

Element	Period	Atomic Size Trend
Li	2	Smallest
Na	3	↑
K	4	↑
Rb	5	↑
Cs	6	Largest

The atomic size increases down the group because additional electron shells are added and shielding effect increases.

Decreasing order: Cs > Rb > K > Na > Li

Q. Discuss the periodic trend of atomic size down Group 1 using Li, Na, K, Rb and Cs.

Solution:

The alkali metals Li, Na, K, Rb and Cs are placed in Group 1. On moving from Li to Cs, the principal quantum number of the valence shell increases. Thus, each successive element contains one additional electron shell.

• Number of shells increases.
• Shielding effect increases.
• Distance of the valence electron from the nucleus increases.
• Effective attraction on the outermost electron decreases.
• Therefore, atomic radius increases down the group.

Increasing size:
Li < Na < K < Rb < Cs

Decreasing size:

Cs > Rb > K > Na > Li

Q. Arrange Li, Na, K, Rb and Cs in decreasing order of atomic size. Explain the periodic trend in detail.

Detailed Solution

Li, Na, K, Rb and Cs are alkali metals belonging to Group 1 of the periodic table. Their valence-shell configuration is generally ns¹.

Element	Electronic Configuration (Shell-wise)	Period
Li	2, 1	2
Na	2, 8, 1	3
K	2, 8, 8, 1	4
Rb	2, 8, 18, 8, 1	5
Cs	2, 8, 18, 18, 8, 1	6

Why does atomic size increase?

1. A new principal shell is added as we move down the group.
2. The valence electron is placed farther from the nucleus.
3. The number of inner-shell electrons increases.
4. Therefore, shielding effect increases.
5. The effective attraction of the nucleus on the valence electron decreases.
6. Hence, atomic radius increases from Li to Cs.

Down Group 1:
Number of shells ↑ → Shielding effect ↑ → Atomic radius ↑

Final Arrangement

Decreasing order of atomic size:

Cs > Rb > K > Na > Li

Increasing order:
Li < Na < K < Rb < Cs

🎯 Quick Revision

Group: 1
Elements: Li, Na, K, Rb, Cs
Trend down group: Atomic size ↑
Reason: New shells + increased shielding effect

Cs > Rb > K > Na > Li

Which has a larger radius: Cl or Cl − ? Why?

📚 Chapter: Classification of Elements and Periodicity in Properties

📘 Detailed Explanation

🔵 Answer

Cl⁻ has a larger radius than Cl.

A neutral chlorine atom Cl has 17 protons and 17 electrons. When it gains one electron, it forms the chloride ion Cl⁻, which has 17 protons but 18 electrons.

Cl < Cl⁻

⚛️ Why is Cl⁻ larger?

1. Cl gains one electron to form Cl⁻.
2. The nuclear charge remains the same because the number of protons remains 17.
3. The number of electrons increases from 17 to 18.
4. Electron-electron repulsion increases.
5. The increased repulsion causes the electron cloud to expand.
6. Therefore, the ionic radius of Cl⁻ is greater than the atomic radius of Cl.

Cl + e⁻ → Cl⁻

17 protons + 17 electrons → 17 protons + 18 electrons

📊 Comparison

Species	Protons	Electrons	Relative Radius
Cl	17	17	Smaller
Cl⁻	17	18	Larger

🔬 Electronic Configuration

Cl (Z = 17):

2, 8, 7

Cl⁻ (18 electrons):

2, 8, 8

Cl⁻ has a completely filled third shell with 8 electrons. The additional electron increases electron-electron repulsion, resulting in a larger electron cloud.

✅ Final Answer

Cl⁻ has a larger radius than Cl because chlorine gains one electron to form Cl⁻ while its nuclear charge remains unchanged. The extra electron increases electron-electron repulsion, causing the electron cloud to expand.

Cl < Cl⁻

📝 20 MCQs with Answers

1. Which has a larger radius?

A) Cl

B) Cl⁻

C) Both are equal

D) Cannot be determined

✅ Answer

B) Cl⁻

---

2. Cl⁻ is formed when Cl:

A) Loses one electron

B) Gains one electron

C) Loses one proton

D) Gains one proton

✅ Answer

B) Gains one electron

---

3. Atomic number of chlorine is:

A) 15

B) 16

C) 17

D) 18

✅ Answer

C) 17

---

4. Number of electrons in neutral Cl is:

A) 16

B) 17

C) 18

D) 19

✅ Answer

B) 17

---

5. Number of electrons in Cl⁻ is:

A) 16

B) 17

C) 18

D) 19

✅ Answer

C) 18

---

6. During formation of Cl⁻, the number of protons:

A) Increases

B) Decreases

C) Remains unchanged

D) Becomes zero

✅ Answer

C) Remains unchanged

---

7. The larger radius of Cl⁻ is mainly due to:

A) Increased nuclear charge

B) Increased electron-electron repulsion

C) Decreased number of electrons

D) Loss of a shell

✅ Answer

B) Increased electron-electron repulsion

---

8. Which species has the electronic configuration 2,8,8?

A) Cl

B) Cl⁻

C) Na

D) F

✅ Answer

B) Cl⁻

---

9. Cl⁻ is a:

A) Cation

B) Anion

C) Neutral atom

D) Molecule

✅ Answer

B) Anion

---

10. Which has more electrons?

A) Cl

B) Cl⁻

C) Both equal

D) Neither

✅ Answer

B) Cl⁻

---

11. An anion is generally ______ than its neutral atom.

A) Smaller

B) Larger

C) Equal

D) Zero

✅ Answer

B) Larger

---

12. Which statement is correct?

A) Cl has 18 electrons

B) Cl⁻ has 17 electrons

C) Cl⁻ has 18 electrons

D) Cl⁻ has 19 protons

✅ Answer

C) Cl⁻ has 18 electrons

---

13. Which factor remains unchanged when Cl becomes Cl⁻?

A) Number of electrons

B) Number of protons

C) Electron-electron repulsion

D) Ionic charge

✅ Answer

B) Number of protons

---

14. The charge on Cl⁻ is:

A) +1

B) −1

C) +2

D) 0

✅ Answer

B) −1

---

15. The valence shell of Cl contains:

A) 5 electrons

B) 6 electrons

C) 7 electrons

D) 8 electrons

✅ Answer

C) 7 electrons

---

16. Cl⁻ has a stable configuration similar to:

A) He

B) Ne

C) Ar

D) Kr

✅ Answer

C) Ar

---

17. Which order is correct?

A) Cl⁻ < Cl

B) Cl < Cl⁻

C) Cl = Cl⁻

D) Cl⁻ = nucleus

✅ Answer

B) Cl < Cl⁻

---

18. The formation of Cl⁻ involves:

A) Oxidation

B) Reduction

C) Sublimation

D) Neutralisation only

✅ Answer

B) Reduction

---

19. Which statement explains the larger radius of Cl⁻?

A) Same nuclear charge attracts more electrons

B) Nuclear charge doubles

C) Protons are lost

D) Electrons disappear

✅ Answer

A)

---

20. The best conclusion is:

A) Cations are always larger than atoms

B) Anions are generally larger than their parent atoms

C) Ion formation does not affect radius

D) Cl⁻ is smaller than Cl

✅ Answer

B)

Q. Which has a larger radius: Cl or Cl⁻? Give reason.

Solution:

Cl⁻ has a larger radius than Cl. Chlorine gains one electron to form Cl⁻. The nuclear charge remains the same, but electron-electron repulsion increases due to the additional electron, causing expansion of the electron cloud.

Cl < Cl⁻

Q. Explain why the radius of Cl⁻ is greater than that of Cl.

Solution:

1. Cl has 17 protons and 17 electrons.
2. Cl⁻ has 17 protons and 18 electrons.
3. The nuclear charge remains unchanged, while electron-electron repulsion increases.
4. This repulsion expands the electron cloud, making Cl⁻ larger.

Therefore, Cl⁻ > Cl in radius.

Q. Compare Cl and Cl⁻ in terms of their electronic configuration and radius.

Property	Cl	Cl⁻
Protons	17	17
Electrons	17	18
Electronic configuration	2, 8, 7	2, 8, 8
Radius	Smaller	Larger

The additional electron in Cl⁻ increases electron-electron repulsion while the nuclear charge remains unchanged. Therefore, the electron cloud expands.

Cl < Cl⁻

Q. Explain the difference between the atomic radius of Cl and the ionic radius of Cl⁻.

Solution:

Neutral chlorine has atomic number 17 and therefore contains 17 protons and 17 electrons. Its electronic configuration is 2,8,7.

When chlorine gains one electron, it forms Cl⁻. The ion now contains 17 protons and 18 electrons, with configuration 2,8,8.

The nuclear charge remains constant because no proton is added. However, the extra electron increases electron-electron repulsion. Consequently, the electron cloud expands and the ionic radius becomes larger.

Cl + e⁻ → Cl⁻
Electron count: 17 → 18

Hence, the radius of Cl⁻ is greater than the radius of Cl.

Q. Which has a larger radius, Cl or Cl⁻? Explain in detail using nuclear charge, number of electrons and electron-electron repulsion.

Detailed Solution

Chlorine has atomic number Z = 17. Therefore, a neutral chlorine atom contains 17 protons and 17 electrons.

Cl: 17 protons + 17 electrons → 2,8,7

When chlorine gains one electron, it forms the chloride ion:

Cl + e⁻ → Cl⁻
Cl⁻: 17 protons + 18 electrons → 2,8,8

Why does the size increase?

1. The number of protons remains 17.
2. The number of electrons increases from 17 to 18.
3. The same nuclear charge now attracts more electrons.
4. Electron-electron repulsion increases.
5. The electron cloud expands.
6. Therefore, Cl⁻ has a larger radius than Cl.

Species	Protons	Electrons	Radius
Cl	17	17	Smaller
Cl⁻	17	18	Larger

🎯 Final Answer:

Cl⁻ has a larger radius than Cl. When Cl gains an electron, the nuclear charge remains unchanged, but electron-electron repulsion increases. This expands the electron cloud and increases the ionic radius.

Cl < Cl⁻

🎯 Quick Revision

Cl: 17 e⁻ → 2,8,7
Cl⁻: 18 e⁻ → 2,8,8
Protons: Same = 17
Electron Repulsion: Increases in Cl⁻

Cl < Cl⁻