What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the fo

Exercises · 3.11

Q. What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.

(i)

F^-

(ii) Ar

(iii)

Mg^{2+}

(iv)

Rb^+

Chhattisgarh CgbseTextbookSubjective· 2mImportance★★★★★est

19% · 21/109 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 →

Isoelectronic species are atoms/ions with the same number of electrons. (i) $F^-$ is isoelectronic with Ne (or $O^{2-}$ , $Na^+$ , $Mg^{2+}$ , $Al^{3+}$ ).

(ii) Ar is isoelectronic with $Cl^-$ (or $K^+$ , $Ca^{2+}$ , $S^{2-}$ ).

(iii) $Mg^{2+}$ is isoelectronic with Ne (or $F^-$ , $O^{2-}$ , $Na^+$ , $Al^{3+}$ ).

(iv) $Rb^+$ is isoelectronic with Kr (or $Br^-$ , $Sr^{2+}$ , $Se^{2-}$ ).

The Core Idea: What “Isoelectronic” Really Means

The word “isoelectronic” breaks down beautifully: iso = same, electronic = electron arrangement. Two species (atoms or ions) are isoelectronic when they contain the exact same number of electrons. That’s it — no tricks about size, charge, or nuclear attraction. The number of electrons must match perfectly.

Why does this matter? Because isoelectronic species often show similar chemical behaviour in some contexts, and they form a neat series for comparing ionic radii (more protons → smaller radius, even with the same electron count). But for this question, you only need to count electrons.

The golden rule: For an atom, electrons = atomic number. For an ion, electrons = atomic number minus charge (remember: negative charge means extra electrons, positive charge means fewer).

Step-by-Step: Finding Isoelectronic Partners

1. (i) $F^-$ — Fluoride ion

Fluorine has atomic number 9. The $F^-$ ion has gained one electron, so:

\text{Electrons in } F^- = 9 + 1 = 10

A neutral atom with 10 electrons is neon (Ne), atomic number 10. Any ion that also has 10 electrons will work. Common examples: $O^{2-}$ (8 + 2 = 10), $Na^+$ (11 − 1 = 10), $Mg^{2+}$ (12 − 2 = 10), $Al^{3+}$ (13 − 3 = 10).

Tip

The 10-electron set is the neon configuration. Any species with this electron count is isoelectronic with $F^-$ . The most straightforward answer is usually the nearest noble gas — here, Ne.

2. (ii) Ar — Argon atom

Argon is a noble gas with atomic number 18. It is neutral, so:

\text{Electrons in Ar} = 18

A neutral atom with 18 electrons is argon itself — but you need a different species. Look for ions that also have 18 electrons. Common ones: $Cl^-$ (17 + 1 = 18), $K^+$ (19 − 1 = 18), $Ca^{2+}$ (20 − 2 = 18), $S^{2-}$ (16 + 2 = 18).

Watch out

A common mistake is to say “Ar is isoelectronic with itself.” The question asks for a different species, so pick an ion. The simplest is $Cl^-$ , the chloride ion.

3. (iii) $Mg^{2+}$ — Magnesium ion

Magnesium has atomic number 12. The $Mg^{2+}$ ion has lost two electrons:

\text{Electrons in } Mg^{2+} = 12 - 2 = 10

Again, 10 electrons — same as $F^-$ and Ne. So $Mg^{2+}$ is isoelectronic with Ne, $F^-$ , $O^{2-}$ , $Na^+$ , $Al^{3+}$ , etc.

Note

Notice that $F^-$ and $Mg^{2+}$ are both isoelectronic with Ne, even though one is an anion and the other a cation. The electron count is what matters, not the charge. …

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

←Previous question · exercises3.10 — How do atomic radius vary in a period and in a group? How do you expla…→Next question · exercises3.12 — Consider the following species: N³⁻, O²⁻, F^-, Na^+, Mg²⁺ and Al³⁺. (a…

Q.What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.

The Core Idea: What “Isoelectronic” Really Means

Step-by-Step: Finding Isoelectronic Partners

1. (i) F−F^-F− — Fluoride ion

2. (ii) Ar — Argon atom

3. (iii) Mg2+Mg^{2+}Mg2+ — Magnesium ion

🎓Unlock everything free for 14 days

1. (i) $F^-$ — Fluoride ion

3. (iii) $Mg^{2+}$ — Magnesium ion

Unlock everything free for 14 days