Which of the following sets contain only isoelectronic ions? (Note: more than one of the given options may be

NCERT Exemplar · Q23

Q. Which of the following sets contain only isoelectronic ions? (Note: more than one of the given options may be correct.)

(i) Zn^2+, Ca^2+, Ga^3+, Al^3+

(ii) K^+, Ca^2+, Sc^3+, Cl^-

(iii) P^3-, S^2-, Cl^-, K^+

(iv) Ti^4+, Ar, Cr^3+, V^5+

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Concept understanding — Isoelectronic Species

Isoelectronic Species – From Intuition to Precision

Imagine you are building atoms with LEGO blocks. Each block is a proton (positive charge) or an electron (negative charge). The number of protons decides which element you have — that's the atomic number $Z$ . The number of electrons decides the charge on the particle.

Now, here is the key idea: two different particles can have the same number of electrons. When that happens, their electron clouds are arranged in the same way. They become isoelectronic.

The Intuition

Think of a neutral neon atom. It has 10 protons and 10 electrons. Now take a sodium atom (11 protons, 11 electrons) and remove one electron. You get $\text{Na}^+$ , which has 11 protons but only 10 electrons. The electron count of $\text{Na}^+$ is exactly the same as that of neutral neon.

Even though $\text{Na}^+$ and $\text{Ne}$ are different elements with different nuclear charges, their electron configurations are identical: $1s^2 2s^2 2p^6$ . They are iso (same) electronic (electron arrangement).

Note

Isoelectronic species share the same number of electrons and therefore the same electronic configuration. They differ in nuclear charge ( $Z$ ).

The Precise Statement

Definition: Two or more atoms, ions, or molecules are said to be isoelectronic if they have the same number of electrons.

That is the entire definition. But the real power comes from what follows: because their electron clouds are identical in structure, their properties — like ionic radii, ionization energy, and chemical behaviour — show clear, predictable trends when you compare them.

How to Identify Isoelectronic Species

Count the electrons in each species.
- For a neutral atom: electrons = atomic number $Z$ .
- For a positive ion: electrons = $Z$ - (charge magnitude).
- For a negative ion: electrons = $Z$ + (charge magnitude).
Compare the counts. If they match, the species are isoelectronic.

Example: Which of these are isoelectronic? $\text{O}^{2-}$ , $\text{F}^-$ , $\text{Na}^+$ , $\text{Mg}^{2+}$ , $\text{Ne}$ .

$\text{O}^{2-}$ : $Z=8$ , electrons = $8+2=10$
$\text{F}^-$ : $Z=9$ , electrons = $9+1=10$
$\text{Na}^+$ : $Z=11$ , electrons = $11-1=10$
$\text{Mg}^{2+}$ : $Z=12$ , electrons = $12-2=10$
$\text{Ne}$ : $Z=10$ , electrons = $10$

All five have 10 electrons. They form an isoelectronic series.

\text{Electrons in an ion} = Z - (\text{charge})

The Critical Consequence: Size Trends

Here is where the concept becomes exam-relevant. In an isoelectronic series, as nuclear charge ( $Z$ ) increases, the ionic radius decreases.

Why? The same number of electrons is pulled more strongly by a larger positive nucleus. The electron cloud shrinks.

For the series above ( $\text{O}^{2-}$ , $\text{F}^-$ , $\text{Na}^+$ , $\text{Mg}^{2+}$ , $\text{Ne}$ ):

Species	$Z$	Electrons	Relative Radius
$\text{O}^{2-}$	8	10	Largest
$\text{F}^-$	9	10	↓
$\text{Ne}$	10	10	↓
$\text{Na}^+$	11	10	↓
$\text{Mg}^{2+}$	12	10	Smallest

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Concept: Isoelectronic Species

Isoelectronic species are atoms or ions that have the same number of electrons. To identify them, count the electrons in each species by adjusting the atomic number for the charge.

Step 1: Recall atomic numbers: $\text{Al} = 13$ , $\text{P} = 15$ , $\text{S} = 16$ , $\text{Cl} = 17$ , $\text{Ar} = 18$ , $\text{K} = 19$ , $\text{Ca} = 20$ , $\text{Sc} = 21$ , $\text{Ti} = 22$ , $\text{V} = 23$ , $\text{Cr} = 24$ , $\text{Ga} = 31$ , $\text{Zn} = 30$ .

Step 2: Count electrons for each option:

Option	Species	Electrons
(A)	$\text{Zn}^{2+}$	$30 - 2 = 28$
	$\text{Ca}^{2+}$	$20 - 2 = 18$
	$\text{Ga}^{3+}$	$31 - 3 = 28$
	$\text{Al}^{3+}$	$13 - 3 = 10$

Option (A) has 28, 18, 28, 10 electrons — not all equal.

| (B) | $\text{K}^{+}$ | $19 - 1 = 18$ |

| | $\text{Ca}^{2+}$ | $20 - 2 = 18$ |

| | $\text{Sc}^{3+}$ | $21 - 3 = 18$ |

| | $\text{Cl}^{-}$ | $17 + 1 = 18$ |

Option (B): all have 18 electrons ✓

| (C) | $\text{P}^{3-}$ | $15 + 3 = 18$ | …

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Isoelectronic ions have identical electron counts (atomic number $\pm$ charge). Every ion in sets (B) and (C) has $18$ electrons, so those are the isoelectronic sets.

(A) $\text{Zn}^{2+},\ \text{Ca}^{2+},\ \text{Ga}^{3+},\ \text{Al}^{3+}$ : $30-2=28$ , $20-2=18$ , $31-3=28$ , $13-3=10$ → counts $28,18,28,10$ differ → not isoelectronic.

(B) $\text{K}^+,\ \text{Ca}^{2+},\ \text{Sc}^{3+},\ \text{Cl}^-$ : $19-1=18$ , $20-2=18$ , $21-3=18$ , $17+1=18$ → all 18 electrons (argon core) → isoelectronic ✓ …

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

KCET 2025Set D-41 markMCQ
$_4$
›Reveal solutionSolution

$\ce{NH4Cl + NH3}$ is the Group III reagent: the common-ion effect keeps $[\ce{OH-}]$ deliberately low, so only the very-low- $K_{sp}$ hydroxides ( $\ce{Al^{3+}}$ , $\ce{Fe^{3+}}$ , $\ce{Cr^{3+}}$ ) precipitate — and $\ce{Al^{3+}}$ is the only such ion offered.

Step 1 — Identify the reagent

Solid $\ce{NH4Cl}$ together with aqueous $\ce{NH3}$ (ammonium hydroxide) is the classic Group III group reagent in the systematic qualitative analysis of cations. Group III comprises the ions that form highly insoluble hydroxides:
$\ce{Al^{3+}},\quad \ce{Fe^{3+}},\quad \ce{Cr^{3+}}$
Step 2 — Why $\ce{NH4Cl}$ is added — the common-ion effect

This is the concept the question is really testing. Aqueous ammonia is a weak base that ionises only slightly:
$\ce{NH4OH <=> NH4+ + OH^-}$
Adding solid $\ce{NH4Cl}$ floods the solution with $\ce{NH4+}$ , the common ion. By Le Chatelier's principle the equilibrium is pushed to the left, further suppressing the ionisation of $\ce{NH4OH}$ and keeping $[\ce{OH-}]$ deliberately low.

Why we want a low $[\ce{OH-}]$ : precipitation occurs only when the ionic product exceeds the solubility product:
$[\ce{M^{n+}}]\,[\ce{OH^-}]^{n} \;>\; K_{sp}$
The Group III hydroxides have extremely small $K_{sp}$ values, so even this small $[\ce{OH-}]$ is enough to precipitate them. But the hydroxides of the later groups (Group IV: $\ce{Ni^{2+}, Co^{2+}, Mn^{2+}, Zn^{2+}$ ; Group V: $\ce{Ba^{2+}, Ca^{2+}, Sr^{2+}}$ ) have much larger $K_{sp}$ values, so at this low $[\ce{OH-}]$ their ionic products stay below $K_{sp}$ and they stay in solution.

Without the $\ce{NH4Cl}$ , the higher $[\ce{OH-}]$ would drag down later-group hydroxides too, and the separation into groups would collapse. The common-ion effect is what makes the scheme selective.

Step 3 — Test each option

(A) $\ce{NH4+}$ ✗ — This is not even an analytical-group cation to be precipitated; it is supplied by the reagent itself. Ammonium salts are all soluble in water, so $\ce{NH4+}$ can never be precipitated here. (It is the Zero-group cation, detected instead by warming with NaOH and testing the $\ce{NH3}$ evolved.)

(B) $\ce{Al^{3+}}$ ✓ — A Group III cation. Aluminium hydroxide has an exceedingly small $K_{sp}$ ( $\approx 10^{-33}$ ), so it precipitates even at the low $[\ce{OH-}]$ maintained: …
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KCET 2024Set B-21 markMCQ
$\mathrm{N}_2$
›Reveal solutionSolution

All four options are isoelectronic pairs, so the discriminator is the bond order — only the 18-electron pair O $_2^{2-}$ /F $_2$ has bond order 1.

Step 1 — The MO bond-order formula.
$\text{Bond order} = \frac{N_b - N_a}{2}$
where $N_b$ and $N_a$ are the numbers of electrons in bonding and antibonding molecular orbitals.

Step 2 — Count the electrons in each species.

Species Electron count
N $_2$ $7+7 = 14$
CO $6+8 = 14$
NO $^+$ $7+8-1 = 14$
O $_2^{2-}$ $8+8+2 = 18$
F $_2$ $9+9 = 18$

So every option is an isoelectronic pair — the electron count alone cannot decide it. We must compute bond orders.

Step 3 — Bond order for the 14-electron species (N $_2$ , CO, NO $^+$ ).

MO configuration: $\sigma 1s^2\,\sigma^*1s^2\,\sigma 2s^2\,\sigma^*2s^2\,\pi 2p_x^2\,\pi 2p_y^2\,\sigma 2p_z^2$
$N_b = 2+2+2+2+2 = 10,\qquad N_a = 2+2 = 4$ $\text{B.O.} = \frac{10-4}{2} = 3 \quad (\text{triple bond}).$
This rules out (A) N $_2$ /CO, (B) N $_2$ /NO $^+$ and (D) CO/NO $^+$ — all bond order 3, not 1.

Step 4 — Bond order for the 18-electron species (O $_2^{2-}$ , F $_2$ ). …
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COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{CO}$
›Reveal solutionSolution
Isoelectronic species have the same number of electrons. $\mathrm{CO}$ has 14 electrons; so does $\mathrm{CN^-}$ .

Electron counts:

$\mathrm{CO}$ : $6\,(\mathrm{C}) + 8\,(\mathrm{O}) = 14$ electrons.

$\mathrm{O_2^+}$ : $2\times 8 - 1 = 15$ .

$\mathrm{CN^-}$ : $6\,(\mathrm{C}) + 7\,(\mathrm{N}) + 1\,(\text{extra}) = 14$ . ✓

$\mathrm{O_2^-}$ : $2\times 8 + 1 = 17$ .

$\mathrm{N_2^+}$ : $2\times 7 - 1 = 13$ . …
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KCET 2022Set B-31 markMCQ
$Q. Which noble gas has least tendency to form compounds? (A) Ar (B) Kr (C) He (D) Ne$
›Reveal solutionSolution

Reactivity of noble gases increases down the group as ionisation enthalpy falls, so the topmost gas listed — helium — is the least willing to form compounds.

Step 1 — Why noble gases are inert at all.

They have completely filled valence shells ( $ns^2np^6$ ; He is $1s^2$ ). There is no low-energy vacancy to accept an electron ( $\Delta_{eg}H$ is positive) and the filled shell is very stable, so removing an electron costs a huge amount of energy. With neither donation nor acceptance favourable, bonding is essentially shut off.

Step 2 — What lets some of them react anyway.

Compound formation becomes possible only when the atom is big and loosely held, so that its outer electrons can be pulled into bonds by a fiercely electronegative partner (F or O). Down the group the atom grows and the outer electrons are further from the nucleus and better shielded, so ionisation enthalpy falls:
$\mathrm{He\;(2372) > Ne\;(2080) > Ar\;(1520) > Kr\;(1351) > Xe\;(1170)\ \ kJ\,mol^{-1}}$
That is precisely why real chemistry starts only at the bottom: Xe gives a whole family ( $\mathrm{XeF_2, XeF_4, XeF_6, XeO_3}$ ), Kr manages a little ( $\mathrm{KrF_2}$ ), while Ar/Ne/He give essentially nothing under ordinary conditions.

Step 3 — Rank the four given gases.

Order of tendency to form compounds: $\mathrm{Kr > Ar > Ne > He}$ .

So the least tendency belongs to helium, which combines every unfavourable factor: …
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KCET 2022Set B-31 markMCQ
$Q. How many number of atoms are there in a cube based unit cell, having one atom on each corner and 2 atom on each body diagonal of cube? (A) 4 (B) 9 (C) 8 (D) 6$
›Reveal solutionSolution

Count corner atoms with the $\tfrac18$ sharing rule, then add the body-diagonal atoms, which lie entirely within the cell and count as $1$ each.

Concept — the contribution rules. An atom's contribution to a unit cell depends on how many cells share it:
$\text{corner} = \tfrac18, \qquad \text{edge} = \tfrac14, \qquad \text{face centre} = \tfrac12, \qquad \text{body (interior)} = 1$
Step 1 — Corner atoms.

A cube has $8$ corners, and each corner is shared between $8$ neighbouring unit cells:
$8 \times \frac{1}{8} = 1\ \text{atom}$
Step 2 — Body-diagonal atoms.

How many body diagonals does a cube have? A body diagonal joins two opposite vertices through the interior. With $8$ vertices paired off into opposite pairs:

$\frac{8}{2} = 4\ \text{body diagonals}$ …
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COMEDK 2022Set 20221 markMCQ
$\mathrm{O^{2-},F^-,Mg^{2+},Al^{3+},O_2,F_2}$
›Reveal solutionSolution

So four species (O^2-, F^-, Mg^2+, Al^3+) all have 10 electrons and are isoelectronic (all Ne-like).

Concept: Isoelectronic species have the SAME number of electrons.

O^2- : 8 + 2 = 10 electrons

F^- : 9 + 1 = 10 electrons

Mg^2+ : 12 - 2 = 10 electrons

Al^3+ : 13 - 3 = 10 electrons

O2 : 2 x 8 = 16 electrons

F2 : 2 x 9 = 18 electrons …
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COMEDK 2021Set 20211 markMCQ
$_3^{2-}$
›Reveal solutionSolution
Hence the pair CO3^2-, NO3^-.

Concept: Isoelectronic = same number of atoms and same total electrons; isostructural = same shape.

Count valence electrons / geometry:

CO3^2-: 4 atoms, total electrons 6 + 3(8) + 2 = 32; C is sp2, no lone pair -> TRIGONAL PLANAR

NO3^-: 4 atoms, total electrons 7 + 3(8) + 1 = 32; N is sp2 -> TRIGONAL PLANAR => same atom count, same electron count, same shape. Isoelectronic AND isostructural. …
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KCET 2019Set A-11 markMCQ
$10^{-5}$
›Reveal solutionSolution
Doping NaCl with $SrCl_2$ creates one cationic vacancy per $Sr^{2+}$ ion to maintain charge neutrality. For $10^{-5}$ mole of $SrCl_2$ , the number of vacancies is $6.022 \times 10^{18}$ .

Concept and intuition:

When $SrCl_2$ is added to NaCl, each $Sr^{2+}$ ion replaces two $Na^+$ ions in the crystal lattice. But the $Sr^{2+}$ ion has a +2 charge, while each $Na^+$ site normally holds a +1 charge. To keep the overall crystal electrically neutral, one $Na^+$ site is left empty for every $Sr^{2+}$ that enters. That empty site is a cationic vacancy. So the number of vacancies equals the number of $Sr^{2+}$ ions added.

Find the number of $Sr^{2+}$ ions from the given mole amount. We are told $10^{-5}$ mole of $SrCl_2$ is doped into 1 mole of NaCl. Each formula unit of $SrCl_2$ gives one $Sr^{2+}$ ion. So the number of $Sr^{2+}$ ions is:

$10^{-5} \times N_A$
where $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$ .

Calculate the number of vacancies. Since each $Sr^{2+}$ creates exactly one cationic vacancy: $\text{Number of vacancies} = 10^{-5} \times 6.022 \times 10^{23} = 6.022 \times 10^{18}$ …
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Species	Electron count
N $_2$	$7+7 = 14$
CO	$6+8 = 14$
NO $^+$	$7+8-1 = 14$
O $_2^{2-}$	$8+8+2 = 18$
F $_2$	$9+9 = 18$

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Q.Which of the following sets contain only isoelectronic ions? (Note: more than one of the given options may be correct.)

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