Consider the following species: $N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}$ and $Al^{3+}$. (a) What is common in them? (b) Arrange them in the order of increasing ionic radii.

Consider the following species: N³⁻, O²⁻, F^-, Na^+, Mg²⁺ and Al³⁺. (a) What is common in them? (b) Arrange th

Exercises · 3.12

N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}

(a) What is common in them?

(b) Arrange them in the order of increasing ionic radii.

Goa GbshseTextbookSubjective· 2mImportance★★★★★est

20% · 22/109 Questions

🔒 Locked · start free trial →

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.

Note

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.

Watch out

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⁻.

Important

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

Start your 14-day free trial to unlock the full solution →

Concept: Isoelectronic Species and Ionic Radii Trends

All six species have 10 electrons (the neon configuration: $1s^2 2s^2 2p^6$ ). Species with the same number of electrons are called isoelectronic.

For isoelectronic species, ionic radius depends solely on nuclear charge $Z$ . A higher $Z$ pulls the electron cloud more tightly inward, shrinking the radius. The nuclear charges are:

Ion	$N^{3-}$	$O^{2-}$	$F^-$	$Na^+$	$Mg^{2+}$	$Al^{3+}$
$Z$	7	8	9	11	12	13

Start your 14-day free trial to unlock the full solution →

All six species are isoelectronic (10 electrons each, neon configuration). Ionic radius decreases as nuclear charge increases: $Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$ .

(a) What is common in them?

The unifying feature is their electron count. Let's verify:

$N^{3-}$ : nitrogen has 7 protons, gains 3 electrons → $7 + 3 = 10$ electrons
$O^{2-}$ : oxygen has 8 protons, gains 2 electrons → $8 + 2 = 10$ electrons
$F^-$ : fluorine has 9 protons, gains 1 electron → $9 + 1 = 10$ electrons
$Na^+$ : sodium has 11 protons, loses 1 electron → $11 - 1 = 10$ electrons
$Mg^{2+}$ : magnesium has 12 protons, loses 2 electrons → $12 - 2 = 10$ electrons
$Al^{3+}$ : aluminium has 13 protons, loses 3 electrons → $13 - 3 = 10$ electrons

All six species have the same electronic configuration: $1s^2 \, 2s^2 \, 2p^6$ , which is the stable neon configuration. Species with identical electron counts are called isoelectronic.

(b) Arrange them in order of increasing ionic radii

When ions are isoelectronic, the electron cloud experiences different nuclear attractions depending on the number of protons. The key principle: more protons pull the same electron cloud tighter.

Think of it this way. All six ions have 10 electrons arranged in the same orbitals, but the nuclear charge varies from $Z = 7$ (nitrogen) to $Z = 13$ (aluminium). A stronger positive charge in the nucleus exerts a greater pull on the electron cloud, contracting it inward. Conversely, a weaker nuclear charge allows the electrons to spread out more.

Let's organize by nuclear charge:

Species	Protons ( $Z$ )	Electrons	Net charge
$N^{3-}$	7	10	$-3$
$O^{2-}$	8	10	$-2$
$F^-$	9	10	$-1$
$Na^+$	11	10	$+1$
$Mg^{2+}$	12	10	$+2$
$Al^{3+}$	13	10	$+3$

Now we can rank them:

$N^{3-}$ has the largest radius. With only 7 protons holding 10 electrons, the electron–electron repulsion dominates and the ion is very diffuse.
$O^{2-}$ is next. One more proton than nitrogen pulls the cloud slightly tighter.
$F^-$ follows. With 9 protons, the attraction is stronger still, shrinking the radius further. …

Start your 14-day free trial to unlock the full solution →

←Previous question · exercises3.11 — What do you understand by isoelectronic species? Name a species that w…→Next question · exercises3.13 — Explain why cations are smaller and anions larger in radii than their …

Q.Consider the following species: N3−,O2−,F−,Na+,Mg2+N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}N3−,O2−,F−,Na+,Mg2+ and Al3+Al^{3+}Al3+.

(a) What is common in them?

(b) Arrange them in order of increasing ionic radii

🎓Unlock everything free for 14 days

Q.Consider the following species: $N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}$ and $Al^{3+}$ .

Unlock everything free for 14 days