Atomic and Ionic Radii, Isoelectronic Species (ions) | NCERT Solved Examples | Conceptual Important Question Answers

Anand Classes presents a comprehensive collection of important questions and answers on atomic and ionic radii, specially designed for students preparing for JEE, NEET, and CBSE exams. This compilation focuses on understanding the size variations in atoms and ions, helping learners grasp key concepts such as nuclear charge, electron configuration, and periodic trends. Each question is paired with a detailed explanation, making complex topics easier to remember and apply during exams. Whether you are revising or practicing, these carefully crafted Q&As will strengthen your foundation in chemistry and boost your confidence.

Q: What are Isoelectronic Species?

Answer : Isoelectronic species are atoms, ions, or molecules that have the same number of electrons but may differ in:

• The number of protons (nuclear charge)
• The element they belong to
• The ionic charge

Even though they have the same number of electrons, their sizes vary because of differences in nuclear charge. Greater nuclear charge pulls electrons more strongly, reducing the size.

Q: Find one species that is isoelectronic with each : (i) F⁻ (ii) Ar (iii) Mg²⁺ (iv) Rb⁺ [NCERT]

Answer : (i) F⁻ :

• Atomic number of F = 9 → neutral fluorine has 9 electrons
• F⁻ ion has 1 extra electron, so total = 10 electrons
• Any species with 10 electrons will be isoelectronic. Example: Ne (Neon)
  Answer: Ne

(ii) Ar :

• Atomic number of Ar = 18 → has 18 electrons
• Any species with 18 electrons is isoelectronic. Example: K⁺ (19 protons – 1 electron removed = 18 electrons)
• Answer: K⁺

(iii) Mg²⁺ :

• Atomic number of Mg = 12 → neutral magnesium has 12 electrons
• Mg²⁺ means 2 electrons removed → total = 10 electrons
• Any species with 10 electrons is isoelectronic. Example: Na⁺ (11 protons – 1 electron removed = 10 electrons)
  Answer: Na⁺

(iv) Rb⁺ :

• Atomic number of Rb = 37 → neutral rubidium has 37 electrons
• Rb⁺ means 1 electron removed → total = 36 electrons
• Any species with 36 electrons is isoelectronic. Example: Kr (Krypton)
  Answer: Kr

✅ Final Answers Table

Q: Which of the following species are isoelectronic? (i) O²⁻ (ii) Mg²⁺ (iii) Na (iv) F (v) Cl⁻ (vi) Al³⁺ (vii) Ne (viii) S²⁻ (ix) Ca²⁺ (x) K [NCERT]

1. Meaning of Isoelectronic Species

• Isoelectronic species are atoms or ions that have the same number of electrons.
• They can be from different elements but must have identical electron counts.
• Example: O²⁻ (10 electrons) and Ne (10 electrons) are isoelectronic.

2. How to Identify Them

• Neutral atoms → Number of electrons = Atomic number (Z).
• Anions (negative ions) → Electrons = Z + (number of extra electrons gained).
• Cations (positive ions) → Electrons = Z − (number of electrons lost).

3. Step-by-Step for the Given List

4. Isoelectronic Groups

• Group 1: 10 electrons → O²⁻, Mg²⁺, Al³⁺, Ne
• Group 2: 18 electrons → Cl⁻, S²⁻, Ca²⁺
• Not Isoelectronic with others here → Na (11 e⁻), F (9 e⁻), K (19 e⁻)

5. Size Trend in Isoelectronic Series

Even though species in a group have the same electrons, their size differs due to nuclear charge (Z).

• Rule: Greater nuclear charge → Electrons pulled closer → Smaller size.
• Example (10-electron group, increasing size):
  Al³⁺ < Mg²⁺ < Ne < O²⁻
• Example (18-electron group, increasing size):
  Ca²⁺ < Cl⁻ < S²⁻

Q: Which of the following species will have the largest and the smallest size? [NCERT] : Mg, Mg²⁺, Al, Al³⁺

Explanation:

1. Atomic size trend in a period:
   • Atomic radii decrease from left to right across a period in the periodic table.
   • Therefore, Mg (in Group 2) will be larger than Al (in Group 13).
2. Effect of cation formation:
   • When an atom loses electrons to form a cation, the number of protons remains the same but electrons decrease.
   • This increases the effective nuclear charge, pulling the remaining electrons closer to the nucleus.
   • Hence: Mg²⁺ is smaller than Mg, and Al³⁺ is smaller than Al.\text{Mg²⁺ is smaller than Mg, and Al³⁺ is smaller than Al.}
3. Isoelectronic comparison:
   • Al³⁺, Mg²⁺, and Na⁺ are examples of isoelectronic species (same number of electrons).
   • Among isoelectronic species, the one with the higher positive nuclear charge (Z) will have a smaller radius.
   • Between Mg²⁺ and Al³⁺: $$Z_{\text{Mg}} = 12, \quad Z_{\text{Al}} = 13$$ So, Al³⁺ is smaller than Mg²⁺.
4. Final size order: $$\text{Al³⁺} < \text{Mg²⁺} < \text{Al} < \text{Mg}$$

Answer:

• Largest species: Mg
• Smallest species: Al³⁺

Q : Which of the following atoms and ions are isoelectronic? (i) Al³⁺ (ii) F (iii) Cl⁻ (iv) O²⁻ (v) Na (vi) Mg²⁺. Arrange the isoelectronic ions in the decreasing order of their size.

Answer :

Step 1: Find the number of electrons in each species

We compare the number of electrons, because “isoelectronic” means having the same number of electrons.

Step 2: Identify Isoelectronic Species

From the table:

• Al³⁺ → 10 electrons
• O²⁻ → 10 electrons
• Mg²⁺ → 10 electrons

✅ Isoelectronic species: Al³⁺, O²⁻, Mg²⁺

Step 3: Understand size trend in isoelectronic series

In an isoelectronic series:

• Same electrons but different nuclear charge (Z).
• Higher Z → Greater attraction between nucleus and electrons → Smaller atomic radius.

Step 4: Compare nuclear charges

• O²⁻ → Z = 8 (smallest nuclear charge → weakest pull → largest size)
• Mg²⁺ → Z = 12 (stronger pull → smaller size)
• Al³⁺ → Z = 13 (strongest pull → smallest size)

Step 5: Final Size Order

$$O^{2-} > Mg^{2+} > Al^{3+}$$

💡 Key idea:
For isoelectronic species — more positive charge → smaller size, more negative charge → larger size.

Q: Arrange the following ions in the order of increasing size : Be²⁺, Cl⁻, S²⁻, Na⁺, Mg²⁺, Br⁻ and Why ?

Answer :

Step 1: Understand the concept

• Cations (positive ions) are smaller than their neutral atoms because they lose electrons → fewer electron–electron repulsions + stronger pull of nucleus.
• Anions (negative ions) are larger than their neutral atoms because they gain electrons → more electron–electron repulsions + weaker effective nuclear pull.
• Across a period (left to right), atomic size decreases due to increasing nuclear charge.
• Down a group, atomic size increases because new electron shells are added.

Step 2: Classify the ions

• Cations: Be²⁺, Mg²⁺, Na⁺ (smallest sizes)
• Anions: Cl⁻, S²⁻, Br⁻ (largest sizes)

Step 3: Arrange within each type

Cations (fewest electrons → smallest): $$\text{Be}^{2+} < \text{Mg}^{2+} < \text{Na}^+$$

(Be²⁺ is smallest because it has only 2 protons pulling on 0 extra electrons after losing both valence electrons; Mg²⁺ has more shells than Be²⁺ but still smaller than Na⁺; Na⁺ has the most shells among these cations.)

Anions (most electrons → largest): $$\text{Cl}^- < \text{S}^{2-} < \text{Br}^-$$

(Cl⁻ is smaller than S²⁻ because S²⁻ has more electrons with same shells → more repulsion; Br⁻ is largest due to being in a lower period with more shells.)

Step 4: Combine cations + anions in increasing order

$$\boxed{\text{Be}^{2+} < \text{Mg}^{2+} < \text{Na}^+ < \text{Cl}^- < \text{S}^{2-} < \text{Br}^-}$$

Answer

(i) Br or Br⁻ → Larger: Br⁻

• Reason: Br⁻ is formed by gaining one electron, increasing electron-electron repulsion and expanding the electron cloud. This makes Br⁻ larger than neutral Br.

(ii) O²⁻ or F⁻ → Larger: O²⁻

• Reason: O²⁻ gains two electrons, resulting in a much greater repulsion among electrons and a significant increase in size compared to F⁻, which gains only one electron.

(iii) K or K⁺ → Larger: K

• Reason: K⁺ is formed by losing one electron, which removes an entire outer shell, making it much smaller than neutral K.

(iv) Li⁺ or Na⁺ → Larger: Na⁺

• Reason: Both are cations, but Na⁺ has more electron shells (3 shells) compared to Li⁺ (2 shells). More shells mean a larger atomic radius.

(v) P or As → Larger: As

• Reason: As lies below P in the periodic table, so it has more electron shells, making it larger in size.

(vi) Na⁺ or Mg²⁺ → Larger: Na⁺

• Reason: Both are isoelectronic (10 electrons), but Mg²⁺ has a higher nuclear charge (+12) than Na⁺ (+11), pulling electrons closer and making Mg²⁺ smaller.

Q: Arrange the following in order of increasing radii with reason :
(i) I, I⁺, I⁻
(ii) N, O, P
(iii) F, Cl, Br

Answer

(i) I, I⁺, I⁻ → Order: I⁺ < I < I⁻

• Reason:
  • Cation (I⁺): Smaller because loss of an electron increases effective nuclear charge, pulling electrons inward.
  • Neutral I: Intermediate size.
  • Anion (I⁻): Largest because gaining an electron increases repulsion and expands the electron cloud.

(ii) N, O, P → Order: O < N < P

• Reason:
  • Across a period (N → O): Size decreases as nuclear charge increases.
  • Down a group (N → P): Size increases due to the addition of electron shells.
  • Combining effects: P (down the group) is largest, O (right in the period) is smallest.

(iii) F, Cl, Br → Order: F < Cl < Br

• Reason: All are halogens in the same group. Atomic radius increases down the group due to more shells being added, even though nuclear charge increases.

Q: For each of the following pairs, state which one is larger in size and why :
(a) Li, F
(b) O, Se
(c) Fe²⁺, Fe³⁺
(d) Br, Br⁻
(e) Na⁺, F⁻
(f) K, K⁺

Answer

(a) Li, F → Larger: Li

• Across the period (Li → F), size decreases due to increasing nuclear charge.
• Li is on the left of the period → larger radius than F.

(b) O, Se → Larger: Se

• Both are in Group 16.
• Down the group, size increases due to additional electron shells.
• Se is below O → larger radius.

(c) Fe²⁺, Fe³⁺ → Larger: Fe²⁺

• Both are cations, but Fe³⁺ has more positive charge → stronger nuclear pull → smaller size.

(d) Br, Br⁻ → Larger: Br⁻

• Anion (Br⁻) gains an electron → increased repulsion → larger size than neutral atom.

(e) Na⁺, F⁻ → Larger: F⁻

• Na⁺ is a cation → much smaller due to loss of a shell.
• F⁻ is an anion → larger due to extra electron repulsion.

(f) K, K⁺ → Larger: K

• Neutral atom is larger.
• K⁺ loses an entire shell → much smaller.

Q : Account for the difference in size of Na⁺ (0.095 nm) and Mg²⁺ (0.065 nm), both of which have the same noble gas configuration.

Answer:

• Same noble gas configuration:
  Both Na⁺ and Mg²⁺ have the electron configuration of neon (1s² 2s² 2p⁶).
• Reason for size difference:
  • Nuclear charge: Mg²⁺ has +12 protons, while Na⁺ has +11 protons.
  • Higher nuclear charge in Mg²⁺ → electrons are pulled more strongly towards the nucleus.
  • This stronger electrostatic attraction shrinks the electron cloud.
• Result: Mg²⁺ has a smaller radius (0.065 nm) compared to Na⁺ (0.095 nm), despite having the same number of electrons.

Q: A boy has reported the radii of Cu, Cu⁺, and Cu²⁺ as 0.096 nm, 0.122 nm, and 0.072 nm respectively. However, he interchanged the values by mistake. Assign the correct values to the different species.

Answer :

Step 1 — Recall the size trend for neutral atom and cations

For the same element:
Neutral atom > monocation (Cu⁺) > dication (Cu²⁺)

Reason:

• Removal of electrons increases effective nuclear charge (Zeff).
• Higher Zeff pulls the remaining electrons closer, reducing size.

Step 2 — Arrange given values in decreasing order

Given values:
0.122 nm > 0.096 nm > 0.072 nm

Step 3 — Assign correct radii

• Largest size → neutral Cu → 0.122 nm
• Intermediate size → Cu⁺ → 0.096 nm
• Smallest size → Cu²⁺ → 0.072 nm

✅ Final Correct Assignment:

• Cu → 0.122 nm
• Cu⁺ → 0.096 nm
• Cu²⁺ → 0.072 nm

Q : Arrange the following ions in the order of increasing size: Be²⁺, Cl⁻, S²⁻, Na⁺, Mg²⁺, Br⁻ and why ? [NCERT]

Answer :

Step 1 — Size trend

1. Cations (positive ions) are smaller than their parent atoms due to loss of electrons and higher effective nuclear charge.
2. Anions (negative ions) are larger than their parent atoms due to gain of electrons and increased electron–electron repulsion.
3. For ions with the same charge, size increases down the group in the periodic table.
4. For isoelectronic species (same number of electrons), higher nuclear charge = smaller size.

Step 2 — Compare each ion

• Be²⁺ → smallest (very high positive charge, only 2 protons’ worth of shielding electrons left in 1s²).
• Mg²⁺ → larger than Be²⁺ but still a small cation.
• Na⁺ → larger than Mg²⁺ (one less positive charge, lower Zeff).
• Cl⁻ → large anion.
• S²⁻ → larger than Cl⁻ (same period, but S²⁻ has more electron–electron repulsion).
• Br⁻ → largest (same charge as Cl⁻ but lower in the group → more shells).

Step 3 — Final order of increasing size

Be²⁺ < Mg²⁺ < Na⁺ < Cl⁻ < S²⁻ < Br⁻

Q : Select from each group the species which has the smallest radius:
(a) O, O⁻, O²⁻
(b) K⁺, Sr²⁺, Ar
(c) Si, P, Cl [NCERT]

Answer :

Step-by-step reasoning

(a) O, O⁻, O²⁻

• Neutral O: baseline size.
• O⁻: one extra electron → increased repulsion → larger.
• O²⁻: two extra electrons → even more repulsion → largest.
• Smallest = O.

(b) K⁺, Sr²⁺, Ar

• K⁺: loses 1 electron → smaller than neutral K.
• Sr²⁺: loses 2 electrons, more positive charge → much smaller than neutral Sr.
• Ar: neutral atom, size in between cations and large neutral atoms.
• Smallest = Sr²⁺ (highest positive charge, smallest radius).

(c) Si, P, Cl

• All are neutral atoms in Period 3.
• Across a period, nuclear charge increases → size decreases.
• Si (Z=14) > P (Z=15) > Cl (Z=17) in nuclear charge, so Cl is the smallest.

✅ Final Answers:
(a) O
(b) Sr²⁺
(c) Cl

📘 Important Practice Questions on Atomic & Ionic Radii

1. Between Al³⁺ and Al, which has a smaller radius? Explain.

Answer: Al³⁺
Reasoning:

• Al is neutral with electron configuration: 1s² 2s² 2p⁶ 3s² 3p¹.
• Al³⁺ loses its 3 valence electrons → electron configuration becomes 1s² 2s² 2p⁶ (same as Ne).
• Loss of electrons → stronger pull of the nucleus on remaining electrons → radius decreases drastically.
• Hence, Al³⁺ is smaller.

2. Arrange in increasing order of radii: F⁻, F, F⁺

Answer: F⁺ < F < F⁻
Reasoning:

• F⁺: Loss of an electron → stronger attraction → smallest.
• F: Neutral size, in between.
• F⁻: Gain of an electron → more repulsion → largest.

3. Which is larger: O²⁻ or S²⁻?

Answer: S²⁻
Reasoning:

• Both have the same charge, but S is in Period 3 and O is in Period 2.
• Down a group, number of shells increases → larger atomic radius.
• Hence S²⁻ > O²⁻.

4. Arrange in decreasing radius: Ca²⁺, K⁺, Cl⁻

Answer: Cl⁻ > K⁺ > Ca²⁺
Reasoning:

• Cl⁻: Anion → larger than neutral Cl.
• K⁺: Cation → smaller than neutral K, but has more shells than Ca²⁺.
• Ca²⁺: Loses 2 electrons → smallest due to high positive charge.

5. Select the smallest radius: Li⁺, Be²⁺, B³⁺

Answer: B³⁺
Reasoning:

• All have lost electrons, but higher positive charge → smaller size.
• B³⁺ has the highest charge (+3) and smallest radius.

6. Arrange in increasing order: Na, Na⁺, Mg²⁺

Answer: Mg²⁺ < Na⁺ < Na
Reasoning:

• Mg²⁺: Highest positive charge → smallest.
• Na⁺: Smaller than neutral Na but larger than Mg²⁺.
• Na: Neutral → largest.

7. Between Cl and Ar, which is smaller?

Answer: Ar
Reasoning:

• Both are in the same period, but Ar has more protons → stronger pull → smaller radius.

8. Arrange in increasing order: N³⁻, O²⁻, F⁻

Answer: F⁻ < O²⁻ < N³⁻
Reasoning:

• More negative charge → greater electron repulsion → larger size.
• N³⁻ has the most extra electrons → largest radius.

9. Which is larger: Zn²⁺ or Cu²⁺?

Answer: Zn²⁺
Reasoning:

• Zn is to the right of Cu? No — actually, Zn comes after Cu in the periodic table (Period 4, d-block), but effective nuclear charge for Cu²⁺ is slightly higher due to fewer protons in Zn.
• In fact, Zn²⁺ is slightly larger than Cu²⁺ because Cu²⁺ has a greater nuclear charge and pulls electrons in more strongly.

10. Arrange in decreasing size: Rb, K, Na

Answer: Rb > K > Na
Reasoning:

• All are alkali metals.
• Down the group → more shells → larger size.

Great! Here’s a compiled list of 25 important questions and answers on atomic and ionic radii, complete with clear explanations. Perfect for JEE, NEET, and CBSE revision.

25 Important Questions & Answers on Atomic and Ionic Radii

📚 Buy Study Material & Join Our Coaching

For premium study materials specially designed for JEE, NEET, NDA, and CBSE/ICSE Classes, visit our official study material portal:
👉 https://anandclasses.net.in/

To enroll in our offline or online coaching programs, visit our coaching center website:
👉 https://anandclasses.co.in/

📞 Call us directly at: +91-94631-38669

💬 WhatsApp Us Instantly

Need quick assistance or want to inquire about classes and materials?

📲 Click below to chat instantly on WhatsApp:
👉 Chat on WhatsApp

🎥 Watch Video Lectures

Get access to high-quality video lessons, concept explainers, and revision tips by subscribing to our official YouTube channel:
👉 Neeraj Anand Classes – YouTube Channel