Arrange the following in increasing order of size: O 2− ,F − ,Na + ,Mg 2+ .

📚 Chapter: Classification of Elements and Periodicity in Properties

📘 Detailed Solution

🔵 Step 1: Count the electrons

First, determine the number of electrons in each ion:

Ion	Atomic Number (Z)	Electrons
O2−	8	8 + 2 = 10
F−	9	9 + 1 = 10
Na+	11	11 − 1 = 10
Mg2+	12	12 − 2 = 10

All four ions have 10 electrons.

🟣 Step 2: Identify the isoelectronic species

Since all four ions contain 10 electrons, they are called isoelectronic species.

O²⁻ F⁻ Na⁺ Mg²⁺

For isoelectronic species, the number of electrons is the same. Therefore, the size depends mainly on the nuclear charge.

🟢 Step 3: Compare nuclear charge

Ion	Number of Protons	Relative Nuclear Attraction
O2−	8	Lowest
F−	9	Higher
Na+	11	Still higher
Mg2+	12	Highest

As the number of protons increases, the same 10 electrons are attracted more strongly towards the nucleus. Therefore, the ionic radius decreases.

Higher nuclear charge → Stronger attraction → Smaller size

🎬 Size Trend

Increasing Nuclear Charge

O²⁻ → F⁻ → Na⁺ → Mg²⁺

Size decreases →

✅ Final Answer

Mg²⁺ < Na⁺ < F⁻ < O²⁻

All four species are isoelectronic with 10 electrons. As nuclear charge increases from Mg (Z = 12) to O (Z = 8), the attraction on the electrons decreases and the ionic size increases.

📝 20 MCQs with Answers

1. Which of the following species are isoelectronic?

A) O²⁻, F⁻, Na⁺, Mg²⁺

B) O, F, Na, Mg

C) Na, Mg, Al, Si

D) H₂, He, Li

✅ Answer

A) All contain 10 electrons.

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2. The number of electrons in O²⁻ is:

A) 6

B) 8

C) 10

D) 12

✅ Answer

C) 10

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3. The number of electrons in Mg²⁺ is:

A) 10

B) 12

C) 14

D) 8

✅ Answer

A) 10

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4. Isoelectronic species have the same number of:

A) Protons

B) Neutrons

C) Electrons

D) Nucleons

✅ Answer

C) Electrons

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5. Which ion has the highest nuclear charge?

A) O²⁻

B) F⁻

C) Na⁺

D) Mg²⁺

✅ Answer

D) Mg²⁺

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6. Among isoelectronic ions, size decreases with increase in:

A) Electron number

B) Nuclear charge

C) Mass number only

D) Neutron number

✅ Answer

B) Nuclear charge

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7. Which is the largest species?

A) O²⁻

B) F⁻

C) Na⁺

D) Mg²⁺

✅ Answer

A) O²⁻

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8. Which is the smallest species?

A) O²⁻

B) F⁻

C) Na⁺

D) Mg²⁺

✅ Answer

D) Mg²⁺

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9. Atomic number of oxygen is:

A) 6

B) 7

C) 8

D) 9

✅ Answer

C) 8

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10. Atomic number of fluorine is:

A) 8

B) 9

C) 10

D) 11

✅ Answer

B) 9

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11. Atomic number of sodium is:

A) 9

B) 10

C) 11

D) 12

✅ Answer

C) 11

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12. Atomic number of magnesium is:

A) 10

B) 11

C) 12

D) 13

✅ Answer

C) 12

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13. The electronic configuration of all four ions is:

A) 1s²2s²2p⁶

B) 1s²2s²2p⁴

C) 1s²2s²2p⁵

D) 1s²2s²2p⁶3s²

✅ Answer

A) 1s²2s²2p⁶

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14. Which factor mainly determines the size order of these ions?

A) Nuclear charge

B) Colour

C) Physical state

D) Melting point

✅ Answer

A) Nuclear charge

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15. Increasing nuclear charge causes the electron cloud to be:

A) More strongly attracted

B) Less strongly attracted

C) Completely removed

D) Unaffected

✅ Answer

A) More strongly attracted

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16. The correct decreasing order of size is:

A) Mg²⁺ > Na⁺ > F⁻ > O²⁻

B) O²⁻ > F⁻ > Na⁺ > Mg²⁺

C) F⁻ > O²⁻ > Mg²⁺ > Na⁺

D) Na⁺ > Mg²⁺ > O²⁻ > F⁻

✅ Answer

B)

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17. In an isoelectronic series, the ion with more protons is generally:

A) Larger

B) Smaller

C) Equal in size

D) Unstable always

✅ Answer

B) Smaller

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18. O²⁻ has how many more electrons than neutral O?

A) 1

B) 2

C) 3

D) 4

✅ Answer

B) 2

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19. Na⁺ is formed when sodium:

A) Gains one electron

B) Loses one electron

C) Gains two electrons

D) Loses two electrons

✅ Answer

B) Loses one electron

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20. The correct increasing order is:

A) O²⁻ < F⁻ < Na⁺ < Mg²⁺

B) Mg²⁺ < Na⁺ < F⁻ < O²⁻

C) Na⁺ < Mg²⁺ < F⁻ < O²⁻

D) F⁻ < Mg²⁺ < Na⁺ < O²⁻

✅ Answer

B) Mg²⁺ < Na⁺ < F⁻ < O²⁻

Q. Arrange O²⁻, F⁻, Na⁺ and Mg²⁺ in increasing order of size.

Solution: All four ions are isoelectronic with 10 electrons. In an isoelectronic series, ionic size decreases as nuclear charge increases.

Mg²⁺ < Na⁺ < F⁻ < O²⁻

Q. Explain the size order of O²⁻, F⁻, Na⁺ and Mg²⁺.

Solution:

1. O²⁻, F⁻, Na⁺ and Mg²⁺ each contain 10 electrons.
2. Hence, they form an isoelectronic series.
3. The nuclear charge increases from O (Z=8) to Mg (Z=12), so the attraction on the same number of electrons increases and the ionic radius decreases.

Increasing size:
Mg²⁺ < Na⁺ < F⁻ < O²⁻

Q. What are isoelectronic species? Arrange O²⁻, F⁻, Na⁺ and Mg²⁺ in increasing order of ionic size.

Solution:

Species having the same number of electrons are called isoelectronic species.

Ion	Electrons	Z
O2−	10	8
F−	10	9
Na+	10	11
Mg2+	10	12

Since all have the same number of electrons, the species with greater nuclear charge has the smaller radius.

Mg²⁺ < Na⁺ < F⁻ < O²⁻

Q. Explain the variation of ionic size in an isoelectronic series with a suitable example.

Solution:

• Isoelectronic species have the same number of electrons.
• For example, O²⁻, F⁻, Na⁺ and Mg²⁺ all have 10 electrons.
• The number of protons, however, is different.
• As the number of protons increases, the attractive force between the nucleus and electrons increases.
• Therefore, the ionic radius decreases with increasing nuclear charge.

Size order:

Mg²⁺ < Na⁺ < F⁻ < O²⁻

Q. Explain in detail why O²⁻, F⁻, Na⁺ and Mg²⁺ have different ionic sizes even though they are isoelectronic. Arrange them in increasing order of size.

Detailed Solution:

The ions O²⁻, F⁻, Na⁺ and Mg²⁺ are all isoelectronic because each contains 10 electrons. Their electronic configuration is:

1s² 2s² 2p⁶

Ion	Atomic Number	Electrons	Protons
O2−	8	10	8
F−	9	10	9
Na+	11	10	11
Mg2+	12	10	12

Because the number of electrons is identical, the main factor controlling the size difference is the nuclear charge.

O²⁻ has only 8 protons, so its 10 electrons experience the weakest nuclear attraction and remain relatively far from the nucleus. Hence, it has the largest radius.

Mg²⁺ has 12 protons attracting the same 10 electrons. It therefore has the strongest nuclear attraction and the smallest radius.

Mg²⁺ < Na⁺ < F⁻ < O²⁻

Increasing ionic size:
Mg²⁺ < Na⁺ < F⁻ < O²⁻

🎯 Final Concept:

For an isoelectronic series:
Same number of electrons + Higher nuclear charge = Smaller ionic radius.

🎯 Quick Revision

O²⁻ = 10 e⁻
F⁻ = 10 e⁻
Na⁺ = 10 e⁻
Mg²⁺ = 10 e⁻

Isoelectronic Series → Higher Z → Smaller Size

Increasing Size:
Mg²⁺ < Na⁺ < F⁻ < O²⁻