(i) Write the electronic configurations of the following ions: (a) H^- (b) Na^+ (c) O²⁻ (d) F^- (ii) What are

Exercises · 2.23

Q. (i) Write the electronic configurations of the following ions:

(a)

H^-

(b)

Na^+

(c)

O^{2-}

(d)

F^-

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a)

3s^1

(b)

2p^3

and

(c)

3p^5

(iii) Which atoms are indicated by the following configurations?

(a)

[He]\ 2s^1

(b)

[Ne]\ 3s^2\ 3p^3

(c)

[Ar]\ 4s^2\ 3d^1

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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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This problem covers three core ideas: writing electronic configurations for ions, deducing atomic numbers from outer-electron configurations, and identifying elements from noble-gas-core configurations. The key is to remember that ions gain or lose electrons from the neutral atom’s configuration, and that the outermost electrons determine the element’s group and period.

(i) Electronic configurations of the ions

The trick with ions is simple: cations lose electrons, anions gain electrons. Always start from the neutral atom’s configuration, then adjust.

1. (a) $H^-$

Neutral hydrogen has one electron: $1s^1$ . The $H^-$ ion has gained one extra electron, so it has two electrons total. Both go into the 1s orbital.

Configuration: $1s^2$

This is the same as the noble gas helium.

2. (b) $Na^+$

Neutral sodium (atomic number 11) has the configuration $1s^2\ 2s^2\ 2p^6\ 3s^1$ . To form $Na^+$ , it loses the single 3s electron.

Configuration: $1s^2\ 2s^2\ 2p^6$

This is the neon configuration.

3. (c) $O^{2-}$

Neutral oxygen (atomic number 8) is $1s^2\ 2s^2\ 2p^4$ . The $O^{2-}$ ion gains two electrons, which fill the 2p subshell completely.

Configuration: $1s^2\ 2s^2\ 2p^6$

Again, the neon configuration.

4. (d) $F^-$

Neutral fluorine (atomic number 9) is $1s^2\ 2s^2\ 2p^5$ . Gaining one electron completes the 2p subshell.

Configuration: $1s^2\ 2s^2\ 2p^6$

Also neon.

Note

All three ions $Na^+$ , $O^{2-}$ , and $F^-$ have the same electronic configuration — they are isoelectronic with neon. The $H^-$ ion is isoelectronic with helium.

(ii) Atomic numbers from outermost electron configurations

The outermost electrons tell you which shell and subshell the element’s last electron(s) occupy. The atomic number is the total number of electrons in the neutral atom.

1. (a) $3s^1$

This means the outermost electron is in the 3s subshell. The element is in period 3, group 1 (alkali metals). The only element with a single 3s electron is sodium.

Atomic number: 11

(Configuration: $1s^2\ 2s^2\ 2p^6\ 3s^1$ )

2. (b) $2p^3$

The outermost electrons are in the 2p subshell, with three electrons. This is period 2, group 15 (the nitrogen family). The element is nitrogen.

Atomic number: 7

(Configuration: $1s^2\ 2s^2\ 2p^3$ )

3. (c) $3p^5$

Outermost electrons in the 3p subshell, with five electrons. Period 3, group 17 (halogens). The element is chlorine.

Atomic number: 17

(Configuration: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^5$ ) …

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

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is a set of isoelectronic species?$
(a) CO, N2, CN-
(b) N2, O2+, NO
(c) CO, CN-, O2^-2
(d) NO2-, N2O, O3
›Reveal solutionSolution
Isoelectronic species have the same total number of electrons. Count total electrons for each species in every option and check whether all members of the set match.

Option (a): CO, N2, CN-

CO: C (6 e-) + O (8 e-) = 14 electrons.

N2: N (7 e-) + N (7 e-) = 14 electrons.

CN-: C (6 e-) + N (7 e-) + 1 extra electron (negative charge) = 14 electrons. All three have 14 electrons -- genuinely isoelectronic.

Option (b): N2 (14), O2+ (16 - 1 = 15), NO (7 + 8 = 15) -- not all equal (14, 15, 15).

…
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following ions has the largest size ?$
(a) Na^+
(b) Mg^2+
(c) O^2-
(d) F^-
›Reveal solutionSolution

Among the isoelectronic ions, O^2- (lowest nuclear charge) is the largest.

All four ions have 10 electrons (isoelectronic). For isoelectronic species, size increases as nuclear charge (Z) decreases, because fewer protons pull the same electrons less tightly. …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. O2-, F-, Na+ and Mg2+ are called as$
(a) Isoelectronic species
(b) Isoneutral species
(c) Isotopes
(d) Isobars
›Reveal solutionSolution
'Isoelectronic' means having the same number of electrons; all four given species have exactly 10 electrons, matching neon's electron count.

Check the electron count of each species by taking the parent element's atomic number and adjusting for charge:

O2- : Oxygen (Z=8) has 8 electrons; the 2- charge adds 2 electrons -> 8 + 2 = 10 electrons

F- : Fluorine (Z=9) has 9 electrons; the 1- charge adds 1 electron -> 9 + 1 = 10 electrons

Na+ : Sodium (Z=11) has 11 electrons; the 1+ charge removes 1 electron -> 11 - 1 = 10 electrons

Mg2+ : Magnesium (Z=12) has 12 electrons; the 2+ charge removes 2 electrons -> 12 - 2 = 10 electrons …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Out of isoelectronic ions O2-, F-, Na+ and Mg2+, which will have smallest radius?$
(a) O2-
(b) Na+
(c) Mg2+
(d) F-
›Reveal solutionSolution

All four species have 10 electrons, so the one with the highest nuclear charge pulls its electron cloud in the most, giving the smallest radius.

$O^{2-}$ , $F^-$ , $Na^+$ , and $Mg^{2+}$ are isoelectronic — each has $10$ electrons (same electronic configuration, $1s^2\,2s^2\,2p^6$ ).

For isoelectronic species, the radius is governed entirely by nuclear charge $Z$ (number of protons), since the number of electrons and their shielding is identical:

$Z(O)=8,\quad Z(F)=9,\quad Z(Na)=11,\quad Z(Mg)=12$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which one of the following sets of ions represents a collection of isoelectronic species? (At. Nos. F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 40, Sc = 21)$
(a) K+, Ca2+, Sc3+, Cl-
(b) Na+, Ca2+, Sc3+, F-
(c) K+, Cl-, Mg2+, Sc3+
(d) Na+, Mg2+, Al3+, Cl-
›Reveal solutionSolution

Isoelectronic species have the same number of electrons; checking each set shows only option (a) has all ions at 18 electrons.

Electron count of an ion = atomic number minus charge (for cations) or plus charge magnitude (for anions).

Option (a): K+ = 19-1 = 18; Ca2+ = 20-2 = 18; Sc3+ = 21-3 = 18; Cl- = 17+1 = 18. All equal to 18 -- isoelectronic.

Option (b): Na+ = 11-1 = 10; Ca2+ = 18; Sc3+ = 18; F- = 9+1 = 10. Mixed 10s and 18s -- not isoelectronic.

…

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CBSE 2023Set ANN1 mark
$Q. Which among the following is isoelectronic with O2- ?$
(a) Na
(b) Ca2+
(c) F-
(d) Mg
›Reveal solutionSolution
Isoelectronic species have the same number of electrons. O2⁻ has 10 electrons, so we need the option with 10 electrons.

O2⁻: Oxygen (Z = 8) has 8 electrons; O2⁻ gains 2 more electrons → 8 + 2 = 10 electrons.

Check each option:

Na (Z = 11): 11 electrons.

Ca2+ (Z = 20): 20 − 2 = 18 electrons. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following are iso-electronic ?$
(a) CO_2, NO_2
(b) NO_2^-, CO_2
(c) CN^-, CO
(d) SO_2, CO_2
›Reveal solutionSolution
CN- and CO are isoelectronic (14 electrons each).

Isoelectronic species contain the same number of electrons. Counting:

CN-: 6 (C) + 7 (N) + 1 (charge) = 14

CO: 6 (C) + 8 (O) = 14 …
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CBSE 2022Set TERM11 markMCQ
$Q. Which one of the following sets of ions represents a collection of isoelectronic species ? (At. No. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)$
(a) K+, Ca2+, Sc3+, Cl-
(b) Na+, Ca2+, Sc3+, Cl-
(c) K+, Cl-, Mg2+, Sc3+
(d) Na+, Mg2+, Al3+, Cl-
›Reveal solutionSolution

Isoelectronic species have the SAME number of electrons. Compute electrons for each ion in every option and find the set where all four match.

Atomic numbers given: F=9, Cl=17, Na=11, Mg=12, Al=13, K=19, Ca=20, Sc=21.

Option (a): K+ = 19-1 = 18; Ca2+ = 20-2 = 18; Sc3+ = 21-3 = 18; Cl- = 17+1 = 18. All four = 18 electrons -- isoelectronic (with Ar).

Option (b): Na+ = 11-1 = 10 (does not match the others at 18) -- not isoelectronic.

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. The anion O^- is isoelectronic with$
(a) N^2-
(b) F^-
(c) N^3-
(d) Ne
›Reveal solutionSolution
O^- has 9 electrons; among the options, N^2- also has 9 electrons, so they are isoelectronic.

Isoelectronic species are atoms/ions that have the identical number of electrons, regardless of their identity.

Count electrons for O^-: oxygen has atomic number Z = 8, so it has 8 electrons; gaining one extra electron to form O^- gives 8 + 1 = 9 electrons.

Now check each option:

(a) N^2-: nitrogen Z = 7 electrons; N^2- = 7 + 2 = 9 electrons. Matches.

(b) F^-: fluorine Z = 9 electrons; F^- = 9 + 1 = 10 electrons. Does not match. …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. Isoelectronic ions are:$
(a) Na+, O-
(b) F, O2-
(c) Al3+, P3+
(d) F-, O2-
›Reveal solutionSolution
Isoelectronic species contain the same number of electrons, even though their nuclear charges differ. Comparing electron counts for each pair identifies F⁻ and O²⁻ as the isoelectronic pair.

Electrons in an ion = atomic number − charge (for a cation, subtract the positive charge; for an anion, add the negative charge).

Na⁺, O⁻: Na (Z=11) → Na⁺ has $11-1=10$ electrons. O (Z=8) → O⁻ has $8+1=9$ electrons. $10 \\ne 9$ , not isoelectronic.

F, O²⁻: F (Z=9) as a neutral atom has 9 electrons. O²⁻ has $8+2=10$ electrons. $9 \\ne 10$ , not isoelectronic. …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. (c) Which of the following pairs represent isoelectronic$
(i) F-, Mg, Al3+
(ii) Mg2+, Na+ and Ne
(iii) Na, Mg2+ and F
(iv) Al, Ne and O2-
›Reveal solutionSolution

Species with the same number of electrons are isoelectronic; Mg2+ (10e-), Na+ (10e-) and Ne (10e-) match exactly.

Step 1: Count electrons in each species of every option.

Option (i): F- = 9+1=10, Mg = 12, Al3+ = 13-3=10 -> not all equal.

Option (ii): Mg2+ = 12-2=10, Na+ = 11-1=10, Ne = 10 -> all equal (10 each).

Option (iii): Na = 11, Mg2+ = 10, F = 9 -> not equal. …

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CBSE 2017Set ANNUAL1 mark
$Q. Arrange O^-2, Mg^+2, Na^+, F^-, Al^+3 in the increasing order of ionic radius.$
›Reveal solutionSolution
These five ions are isoelectronic (each has 10 electrons, the Ne configuration); among isoelectronic species, the ion with the higher nuclear charge (atomic number) is smaller.

All five species have 10 electrons:

O^-2: Z = 8, electrons = 8 + 2 = 10

F^-: Z = 9, electrons = 9 + 1 = 10

Na^+: Z = 11, electrons = 11 - 1 = 10

Mg^+2: Z = 12, electrons = 12 - 2 = 10

Al^+3: Z = 13, electrons = 13 - 3 = 10

Since all have the same number of electrons (same electron shells, i.e., the same shielding), the effective nuclear charge felt by each electron increases as the atomic number Z increases. A higher effective nuclear charge pulls the electron cloud in more tightly, making the ion smaller. So radius decreases steadily as Z increases from O (8) to Al (13).

…
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Q.(i) Write the electronic configurations of the following ions:

(i) Electronic configurations of the ions

(ii) Atomic numbers from outermost electron configurations

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