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^+

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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

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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. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$-$
›Reveal solutionSolution
The question asks which match is incorrect. By checking each property trend, we find that option (B) incorrectly orders metallic character; the correct order is K > Mg > Al > B, so (B) is the wrong match.

Concept and Intuition

This problem tests your understanding of periodic trends: ionization enthalpy, metallic character, ionic size, and electron gain enthalpy. Each option claims a specific order; we must verify each against known periodic behavior. The key is to recall that metallic character decreases across a period and increases down a group, while ionization enthalpy generally increases across a period and decreases down a group. Ionic size decreases with increasing positive charge for isoelectronic species. Electron gain enthalpy becomes more negative (more energy released) across a period, but fluorine is an exception due to its small size.

Step-by-step verification

Option (A): First Ionization Enthalpies – Na < Al?

Ionization enthalpy generally increases across a period (left to right).

Na (Group 1) has a low ionization enthalpy; Al (Group 13) is to the right, so Al’s first ionization enthalpy is higher than Na’s.

The given order “Na < Al” is correct.

Conclusion: (A) is a correct match.

Option (B): Metallic character – Al > Mg > B > K?

Metallic character decreases across a period and increases down a group.

K (Group 1, Period 4) is highly metallic, more so than Mg (Group 2, Period 3).

Al (Group 13, Period 3) is less metallic than Mg.

B (Group 13, Period 2) is a metalloid, least metallic here.

The correct decreasing order: K > Mg > Al > B.

The given order (Al > Mg > B > K) is wrong because K should be the most metallic, not the least.

Conclusion: (B) is an incorrect match.

Option (C): Ionic size – Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺?

These ions are isoelectronic (all have 10 electrons: Ne configuration).

For isoelectronic species, ionic radius decreases as nuclear charge increases.

Nuclear charges: Na⁺ (11), Mg²⁺ (12), Al³⁺ (13), Si⁴⁺ (14).

So size order: Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺ is correct.

Conclusion: (C) is a correct match.

Option (D): Electron gain enthalpy – F > Cl > Br > I?

Electron gain enthalpy becomes less negative down a group (Cl has the most negative, then Br, then I).

Fluorine is an exception: its small size causes high electron-electron repulsion, so its electron gain enthalpy is less negative than chlorine’s.

The correct order (most negative to least): Cl > F > Br > I.

The given order “F > Cl > Br > I” is incorrect because Cl should be before F.

Wait — the option says “F > Br > I” but omits Cl? Actually the option reads: “Electron Gain enthalpy – F > Br > I”. That is incomplete; it likely means F > Cl > Br > I? The printed text shows “F Br > I” which is ambiguous. However, typical exam questions list “F > Cl > Br > I” as incorrect because Cl has more negative electron gain enthalpy than F. So (D) is also incorrect. …
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KCET 2025Set D-41 markMCQ
$\mathrm{AlCl_3} + \mathrm{Cl^-} \rightarrow \mathrm{AlCl_4^-}$
›Reveal solutionSolution

Count the steric number (sigma bonds + lone pairs) on Al before and after: it goes from 3 ( $\mathrm{sp^2}$ ) to 4 ( $\mathrm{sp^3}$ ) when the chloride ion donates a lone pair.

Step 1 — Hybridisation in $\mathrm{AlCl_3}$ .

Aluminium: $Z = 13$ , configuration $[\mathrm{Ne}]\,3s^2 3p^1$ — three valence electrons. In $\mathrm{AlCl_3}$ it forms three $\sigma$ bonds to three Cl atoms and retains no lone pair.
$\text{Steric number} = (\text{3 } \sigma\text{ bonds}) + (\text{0 lone pairs}) = 3$
A steric number of 3 requires three equivalent hybrid orbitals, formed from one $s$ and two $p$ orbitals:
$\Rightarrow \mathbf{sp^2}\ \text{hybridisation, \textbf{trigonal planar}, bond angle } 120^\circ$
Crucially, this leaves Al with an empty, unhybridised $3p$ orbital perpendicular to the molecular plane. With only 6 electrons around it, Al is electron-deficient — an octet-incomplete Lewis acid.

Step 2 — What the reaction is.
$\mathrm{AlCl_3} + \mathrm{Cl^-} \rightarrow \mathrm{AlCl_4^-}$
This is a Lewis acid–base reaction. The chloride ion (a Lewis base, rich in lone pairs) donates a lone pair into aluminium's vacant $p$ orbital, forming a coordinate (dative) covalent bond. Aluminium thereby completes its octet.

Step 3 — Hybridisation in $\mathrm{AlCl_4^-}$ .

Al now has four $\sigma$ bonds (three normal covalent + one coordinate — they are indistinguishable once formed) and still no lone pair:
$\text{Steric number} = 4 + 0 = 4$
Four equivalent hybrid orbitals need one $s$ and all three $p$ orbitals:

$\Rightarrow \mathbf{sp^3}\ \text{hybridisation, \textbf{tetrahedral}, bond angle } 109.5^\circ$ …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-} \text { [Increasing Ionic Size] }$
›Reveal solutionSolution
Options (A), (B) and (C) list correct trends. Option (D) is wrong: N has a higher first ionisation enthalpy than O (stable half-filled 2p $^3$ ), so the correct order is B < C < O < N, making (D) 'not according to the property'.

Check each:

(A) $\mathrm{Mg^{2+}<Na^{+}<F^{-}<O^{2-}}$ (increasing ionic size): all isoelectronic (10 e $^-$ ); size increases as nuclear charge falls (Mg $^{2+}$ :12 → Na $^+$ :11 → F $^-$ :9 → O $^{2-}$ :8). Correct.

(B) $\mathrm{I<Br<F<Cl}$ (increasing $-\Delta_{eg}H$ ): magnitudes are Cl (349) > F (328) > Br (325) > I (295), so increasing order I < Br < F < Cl. Correct.

(C) $\mathrm{Na<K<Rb<Cs}$ (increasing metallic radii): radius increases down group 1. Correct. …
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KCET 2023Set D-21 markMCQ
$\mathrm{Al_2O_3, Li_2O}$
›Reveal solutionSolution

Test each oxide for reaction with both an acid and a base — only BeO and ZnO pass on both counts.

Step 1 — Definition

An amphoteric oxide reacts with an acid and with a base, forming a salt and water in each case. It sits at the boundary between metallic (basic) and non-metallic (acidic) character — typically the oxides of Be, Al, Zn, Sn, Pb.

Step 2 — Examine each candidate oxide

BeO — amphoteric ✓

Beryllium is the classic diagonal-relationship element (with Al); its small, highly polarising $\mathrm{Be^{2+}}$ gives its oxide covalent character.
$\mathrm{BeO + 2HCl \rightarrow BeCl_2 + H_2O}\quad(\text{acts as a base})$ $\mathrm{BeO + 2NaOH \rightarrow Na_2BeO_2 + H_2O}\quad(\text{sodium beryllate; acts as an acid})$
ZnO — amphoteric ✓
$\mathrm{ZnO + 2HCl \rightarrow ZnCl_2 + H_2O}$ $\mathrm{ZnO + 2NaOH \rightarrow Na_2ZnO_2 + H_2O}\quad(\text{sodium zincate})$
$\mathrm{Li_2O}$ — basic ✗

An alkali-metal oxide. It dissolves in water to give the strong base LiOH and reacts only with acids, never with alkali.

MgO — basic ✗

A typical alkaline-earth oxide, ionic and basic; it neutralises acids (hence its use as an antacid) but does not dissolve in NaOH. (Beryllium is the only group-2 element whose oxide is amphoteric — the anomalous first member.)

$\mathrm{BO_3}$ — not a valid species ✗ …
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COMEDK 2021Set 20211 markMCQ
$Q. The increasing order of atomic radii of the following group 13 elements is (A) Al < Ga < In < Tl (B) Ga < Al < In < Tl (C) Al < In < Tl < Ga (D) Al < Ga < Tl < In$
›Reveal solutionSolution

Hence increasing atomic radius: Ga < Al < In < Tl.

Concept: Atomic-radius trend in group 13 - the d-block (and later f-block) contraction breaks the simple 'increases down the group' rule.

Radii (pm): B 85, Al 143, Ga 135, In 167, Tl 170. …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. Which one of the following represent the correct increasing order of atomic radii of 4 elements, given below, which belong to group 13 of the Periodic Table? Ga, In, Al, Tl (A) Al < In < Ga < Tl (B) Ga < Al < In < Tl (C) Al < Ga < In < Tl (D) Ga < Al < Tl < In$
›Reveal solutionSolution
[!TLDR]

Because of poor 3d shielding, gallium is smaller than aluminium, so the increasing order of atomic radii is Ga < Al < In < Tl.

Concept

In group 13 (B, Al, Ga, In, Tl) atomic radius normally rises down the group as new shells are added. Gallium breaks this trend: after Al we cross the first transition series, and the poorly-shielding 3d electrons let the nucleus pull the valence electrons in more tightly, so Ga is slightly smaller than Al (a standard NCERT Class 11 p-block point).

Solution

Approximate NCERT atomic radii (pm): …
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KCET 2020Set A-11 markMCQ
$H_2$
›Reveal solutionSolution

Only an amphoteric metal liberates $H_2$ from both acid and alkali; among the options that metal is zinc.

Step 1 — The concept: amphoterism.

Most metals above hydrogen in the reactivity series displace $H_2$ from dilute acids. But dissolving in a strong base as well is a special property: it requires the metal (and its oxide/hydroxide) to be amphoteric — able to react with both acids and alkalis.

The common amphoteric metals are Zn, Al, Sn, Pb (and Be). Only these liberate hydrogen with NaOH, by forming a soluble complex/oxyanion (zincate, aluminate, stannate, plumbate).

Step 2 — Test zinc against the acid.

Zn lies above hydrogen in the reactivity series, so it displaces $H_2$ from dilute HCl:
$\mathrm{Zn} + 2\mathrm{HCl} \longrightarrow \mathrm{ZnCl_2} + \mathrm{H_2}\uparrow$
✓ Hydrogen evolved.

Step 3 — Test zinc against the alkali.

Because Zn is amphoteric, it also dissolves in hot aqueous NaOH, forming sodium zincate and liberating hydrogen:
$\mathrm{Zn} + 2\mathrm{NaOH} \longrightarrow \mathrm{Na_2ZnO_2} + \mathrm{H_2}\uparrow$
or, written with the hydroxo complex (the more modern representation):
$\mathrm{Zn} + 2\mathrm{NaOH} + 2\mathrm{H_2O} \longrightarrow \mathrm{Na_2[Zn(OH)_4]} + \mathrm{H_2}\uparrow$
✓ Hydrogen evolved. Zn satisfies both conditions.

Step 4 — Why the others fail the second test.

All of Fe, Mg and Ca are above hydrogen in the reactivity series, so each does give $H_2$ with dil. HCl: …
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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

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1. (i) $F^-$ — Fluoride ion

3. (iii) $Mg^{2+}$ — Magnesium ion

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