Q.Which of the following species will have smallest size? Mg, Mg2+, Al, Al3+
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Ionic Radii Trends
Ionic Radii: What It Means and Why It Matters
Imagine an atom as a tiny, fuzzy sphere. When it loses an electron to become a positive ion (cation), or gains an electron to become a negative ion (anion), its size changes. That new size is the ionic radius — the distance from the nucleus to the outermost electron in the ion.
The key question: Why does the size change at all?
The Core Intuition: Two Forces at Play
Every electron in an atom is pulled toward the nucleus by electrostatic attraction. But electrons also repel each other. The balance between these two forces determines how "big" the electron cloud is.
When an atom loses an electron (becomes a cation), two things happen:
- The number of protons stays the same, but there are fewer electrons.
- The remaining electrons feel a stronger pull from the nucleus because there's less electron-electron repulsion to push them apart.
Result: The cation shrinks compared to the neutral atom.
When an atom gains an electron (becomes an anion):
- The number of protons stays the same, but there are more electrons.
- The extra electron adds more repulsion, pushing the electron cloud outward.
- The nucleus can't pull the extra electrons in as tightly.
Result: The anion expands compared to the neutral atom.
This is why, for the same element, the cation is always smaller than the neutral atom, and the anion is always larger. For example, a sodium atom (Na) has a radius of about 186 pm, but Na⁺ has a radius of only about 102 pm — nearly half the size.
The Precise Trend Across the Periodic Table
Now let's look at how ionic radii change as you move across a period and down a group.
Across a Period (Left to Right)
Consider the elements of Period 3: Na, Mg, Al, Si, P, S, Cl.
As you move right, the nuclear charge (number of protons) increases. Electrons are added to the same shell (n=3). The increasing positive charge pulls the electron cloud inward more strongly.
But here's the twist: cations and anions form at different places. The trend isn't smooth like atomic radii.
- On the left, elements form cations (Na⁺, Mg²⁺, Al³⁺). These are much smaller than their neutral atoms.
- On the right, elements form anions (P³⁻, S²⁻, Cl⁻). These are much larger than their neutral atoms.
So across a period, you see a sharp drop from the neutral atom to the cation, then a sharp rise to the anion, then a gradual decrease as you move further right among the anions.
A common mistake is to think ionic radii decrease smoothly across a period like atomic radii do. They don't — the change from cation to anion creates a huge jump. Always check whether you're comparing cations, anions, or neutral atoms.
Down a Group (Top to Bottom)
This is straightforward: ionic radii increase down a group.
Why? Each step down adds a new electron shell (n increases). The outermost electrons are farther from the nucleus, so the ion gets bigger.
For example:
- Li⁺: ~76 pm
- Na⁺: ~102 pm
- K⁺: ~138 pm
- Rb⁺: ~152 pm
- Cs⁺: ~167 pm
The same trend holds for anions: F⁻ < Cl⁻ < Br⁻ < I⁻.
The increase down a group is the most reliable trend for ionic radii. It holds for all ions — cations, anions, and even transition metal ions.
The Isoelectronic Series: A Special Case
Sometimes you compare ions that have the same number of electrons (isoelectronic). For example: O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺ all have 10 electrons (like neon).
Here, the trend is determined entirely by nuclear charge. More protons = stronger pull = smaller radius.
| Ion | Protons | Electrons | Radius (pm) |
|---|---|---|---|
| O²⁻ | 8 | 10 | 140 |
| F⁻ | 9 | 10 | 133 |
| Na⁺ | 11 | 10 | 102 |
| Mg²⁺ | 12 | 10 | 72 |
| Al³⁺ | 13 | 10 | 53.5 |
Removing electrons increases effective nuclear charge per remaining electron, shrinking a cation; among these four species, Al3+ has lost the most electrons relative to its nuclear charge. …
Al3+ is the smallest of the four species because it has the same 10-electron (neon-like) configuration as Mg2+ but a higher nuclear charge (+13 vs +12) pulling those electrons in more tightly.
Step 1 — Compare atoms to their cations: Losing electrons to form a cation always reduces size sharply, because the outer shell is removed and/or the remaining electrons feel a stronger effective nuclear charge. So Mg2+<Mg and Al3+<Al.
…
Showing the 12 most recent of 15 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The correct order of radii is(a) Na+ > F > Al+3 > Mg+2(b) N-3 < Na+ < Mg+2 < O-2(c) Al+3 < Mg+2 < O-2 < S-2(d) Mg+2 < N-3 < O-2 < F-
›Reveal solutionSolution
For a set of isoelectronic cations, higher positive charge means smaller radius (more protons pulling in the same electron cloud); for anions of increasing size (e.g. O2- vs S2-, which has an extra electron shell), radius increases going down a group.
Approximate ionic radii (pm): Al3+ ~ 53, Mg2+ ~ 72, O2- ~ 140, S2- ~ 184.
Check option (c): Al+3 < Mg+2 < O-2 < S-2
53 < 72 < 140 < 184 -- this is correctly increasing. Al3+ is smallest (highest positive charge pulling electrons in tightly, and it is a cation from period 3), Mg2+ is next (lower charge than Al3+), O2- is a fairly large anion (extra electron shell relative to a similarly-charged cation), and S2- is the largest here since it is a period-3 anion with one more electron shell than O2-.
Checking why the others are wrong: …
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is the correct order of size of the given species? I, I⁻, I⁺(a) I > I⁻ > I⁺(b) I⁺ > I⁻ > I(c) I > I⁺ > I⁻(d) I⁻ > I > I⁺
›Reveal solutionSolution
Anions are larger than the parent atom and cations are smaller, because the same nuclear charge acts on more or fewer electrons.
Reasoning: All three species have the same nuclear charge (53 protons for iodine), but different numbers of electrons.
- I− has gained an electron, so more electrons share the same nuclear pull, increasing electron-electron repulsion and expanding the electron cloud — largest.
- I (neutral) has its normal electron count. …
- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following ions has highest value of ionic radius?(a) Li+(b) B3+(c) O2-(d) F-
›Reveal solutionSolution
O2- has the largest ionic radius among the given isoelectronic-type ions.
Li+, B3+, O2- and F- are all small ions, and O2- and F- are isoelectronic with Ne (10 electrons), while Li+ and B3+ are isoelectronic with He (2 electrons). Among isoelectronic species, radius decreases as nuclear charge (and hence effective pull on the electron cloud) increases: for the 10-electron series, O2- (Z=8) is larger than F- (Z=9); for the 2-electron series, Li+ (Z=3 …
- CBSE 2025Set ANNUAL1 markQ.Arrange the following in increasing order of radius: Metallic radius, covalent radius, Vander waal's radius.
›Reveal solutionSolution
Increasing order: covalent radius < metallic radius < van der Waals radius.
These three radii measure atomic size under different bonding situations:
- Covalent radius: half the internuclear distance between two atoms joined by a single covalent bond — the tightest, shortest measure since the atoms' electron clouds overlap strongly.
- Metallic radius: half the internuclear distance between two adjacent atoms in a metallic crystal lattice — larger than covalent radius because metallic bonding is generally weaker/more diffuse than a covalent bond. …
- CBSE 2025Set ANNUAL1 markQ.Atomic size increases from left to right in a period.
›Reveal solutionSolution
The statement is False: atomic size decreases (not increases) from left to right across a period.
Across a period, electrons are added to the same principal shell while the nuclear charge (number of protons) increases with each element. The increasing effective nuclear charge pulls the electron cloud in more strongly, so atomic radius steadily decreases from …
- CBSE 2024Set ANNUAL1 markMCQQ.Which of the following cations has the smallest size?(a) Na^+(b) Mg^2+(c) Ca^2+(d) Al^3+
›Reveal solutionSolution
Al³⁺ is the smallest cation, being isoelectronic with Na⁺ and Mg²⁺ but with the highest nuclear charge.
Na⁺, Mg²⁺, and Al³⁺ are all isoelectronic species (10 electrons each, neon configuration). Among isoelectronic ions, size decreases as nuclear charge increases (more protons pull the same number of electrons in more tightly): Na+(Z=11)>Mg2+(Z=12)>Al3+(Z=13) in size order, …
- CBSE 2024Set ANNUAL1 markMCQQ.Out of the options given below, choose the correct order of atomic/ionic radii of sodium (Na) atom and sodium ion (Na+) in pm(a) A) 95, 186(b) B) 186, 95(c) C) 95, 95(d) D) 186, 186
›Reveal solutionSolution
[!TLDR]
B) 186, 95
Why
Na atom radius is about 186 pm; on losing its valence electron to form Na+, the radius shrinks to about 95 pm due to loss of a shell and …
- CBSE 2023Set ANNUAL1 markMCQQ.The ionic radii of N^3-, O^2-, F^-, Na^+ follow the order(a) N^3- > O^2- > F^- > Na^+(b) N^3- > Na^+ > O^2- > F^-(c) Na^+ > O^2- > N^3- > F^-(d) O^2- > F^- > Na^+ > N^3-
›Reveal solutionSolution
N3-, O2-, F-, and Na+ are all isoelectronic (10 electrons each, like neon), so their radii are governed purely by nuclear charge: more protons pull the same electron cloud in tighter, shrinking the radius as Z increases.
All four species have exactly 10 electrons (configuration of Ne: 1s²2s²2p⁶):
- N (Z=7) as N3-
- O (Z=8) as O2-
- F (Z=9) as F-
- Na (Z=11) as Na+ …
- CBSE 2023Set ANNUAL1 markMCQQ.The correct order of the atomic size of C, N, P, S follows the order.(a) N < C < P < S(b) C < N < S < P(c) C < N < P < S(d) N < C < S < P
›Reveal solutionSolution
Correct increasing size order: N < C < S < P.
Two periodic trends govern atomic size: it decreases across a period (nuclear charge rises) and increases down a group (a new shell is added). Approximate covalent radii (pm): C = 77, N = 75, P = 110, S = 104.
- Within period 2: C (77) > N (75), so N < C.
- Within period 3: P (110) > S (104), so S < P.
- Period-3 atoms are larger than period-2 atoms. …
- CBSE 2022Set ANNUAL1 markQ.Which of the following species will have largest size ? Mg, Mg2+, Al, Al3+
›Reveal solutionSolution
Cations are always smaller than their parent atoms, and Mg is bigger than Al (periodic trend); so the largest species of the four is neutral Mg.
Two trends combine here:
- Removing electrons to form a cation reduces electron-electron repulsion and increases effective nuclear charge per electron, shrinking the species. So Mg > Mg2+ and Al > Al3+. …
- CBSE 2022Set ANNUAL1 markQ.What will be the correct order of hydration energy of the following alkali metal ions: Na+, Rb+, K+, Li+?
›Reveal solutionSolution
Hydration energy decreases as ionic radius increases; among Li+, Na+, K+ and Rb+ (ionic size increasing in that order down Group 1), the correct order of hydration energy is Li+ > Na+ > K+ > Rb+.
Hydration energy is the energy released when gaseous ions are surrounded by water molecules (ion-dipole attraction). A smaller ion has a higher charge density (same +1 charge concentrated over a smaller volume), so it attracts the partially-negative oxygen ends of water molecules more strongly, releasing more energy on hydration.
…
- CBSE 2021Set ANNUAL1 markMCQQ.Which alkali metal gives hydrated salt?(a) Li(b) Na(c) K(d) Cs
›Reveal solutionSolution
Lithium, being the smallest alkali metal ion with the highest charge density, is strongly hydrated in solution and crystallises with water of hydration (e.g. LiCl⋅2H2O), unlike Na, K, or Cs salts which are typically anhydrous.
Step 1 — Why lithium is different: Lithium's ionic radius is the smallest among Group 1 elements. A small ion with a +1 charge has a very high charge density, which lets it polarise and attract surrounding water molecules strongly.
Step 2 — Consequence: This strong hydration means Li+ salts often crystallise out of aqueous solution carrying water molecules with them (e.g. LiCl⋅2H2O), whereas the larger, less charge-dense Na, K, and Cs ions form mostly anhydrous salts.
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.