Answer the following question. Write condensed orbital notation of electronic configuration of the following e

Exercise · Q25

Q. Answer the following question. Write condensed orbital notation of electronic configuration of the following elements: a. Lithium (Z=3) b. Carbon (Z=6) c. Oxygen (Z=8) d. Silicon (Z=14) e. Chlorine (Z=17) f. Calcium (Z=20)

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Concept understanding — Electronic Configuration

Electronic Configuration & the Filling Rules

The big idea. Once you know the order in which orbitals fill and the three rules that govern that filling, you can write the electronic configuration of any atom or ion — and from it read off unpaired electrons, valence electrons, magnetic behaviour and stability. Three principles do all the work: Aufbau (which orbital fills first), Pauli (how many electrons fit), and Hund (how they arrange within a subshell).

The three rules

Aufbau principle (the (n + l) rule): electrons enter the lowest-energy orbital first. Order the subshells by n + l; a lower n + l fills first, and for a tie the one with the lower n fills first. This gives the familiar order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s. Note 4s fills before 3d.
Pauli exclusion principle: no two electrons share all four quantum numbers, so an orbital holds at most 2 electrons (opposite spins). A subshell holds 2(2l+1), a shell 2n².
Hund's rule of maximum multiplicity: within a set of degenerate orbitals, electrons occupy them singly with parallel spins before any pairing begins. This maximises the number of unpaired electrons.

Writing a configuration

Fill electrons in the Aufbau order until you reach the atom's electron count Z. Use a noble-gas core to shorten it: Fe (Z = 26) = [Ar] 3d⁶ 4s². Always check the superscripts add up to Z.

Ions — the 4s-out-first rule

For a cation, remove electrons from the orbital with the highest principal quantum number first — the 4s electrons leave before the 3d, even though 4s filled first:

Fe [Ar]3d⁶4s² → Fe²⁺ [Ar]3d⁶ → Fe³⁺ [Ar]3d⁵.

For an anion, simply add the electrons in Aufbau order (O 1s²2s²2p⁴ → O²⁻ 1s²2s²2p⁶). This is the single most common source of errors — never strip a 3d electron before the 4s.

The half-filled / fully-filled anomalies

Exactly half-filled (d⁵, p³, f⁷) and completely-filled (d¹⁰, p⁶, f¹⁴) subshells carry extra stability — their symmetrical distribution and larger exchange energy lower the energy. This overrides naive Aufbau for a few elements:

Cr = [Ar] 3d⁵ 4s¹ (not 3d⁴4s²), Cu = [Ar] 3d¹⁰ 4s¹ (not 3d⁹4s²),

and likewise Mo = [Kr]4d⁵5s¹, Ag = [Kr]4d¹⁰5s¹, Pd = [Kr]4d¹⁰.

Reading properties off the configuration

Unpaired electrons: apply Hund to the valence subshell (N 2p³ → 3 unpaired; O 2p⁴ → 2; Fe³⁺ 3d⁵ → 5).
Magnetism: an atom/ion with any unpaired electron is paramagnetic; all paired ⇒ diamagnetic. …

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Step 1. Lithium (Z=3): full $1s^2\,2s^1$ ; core = He ( $1s^2$ ); condensed $[He]\,2s^1$ .

Step 2. Carbon (Z=6): full $1s^2\,2s^2\,2p^2$ ; condensed $[He]\,2s^2\,2p^2$ .

Step 3. Oxygen (Z=8): full $1s^2\,2s^2\,2p^4$ ; condensed $[He]\,2s^2\,2p^4$ .

Step 4. Silicon (Z=14): full $1s^2\,2s^2\,2p^6\,3s^2\,3p^2$ ; core = Ne ( $1s^2\,2s^2\,2p^6$ ); condensed $[Ne]\,3s^2\,3p^2$ .

Step 5. Chlorine (Z=17): full $1s^2\,2s^2\,2p^6\,3s^2\,3p^5$ ; condensed $[Ne]\,3s^2\,3p^5$ . …

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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 17 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Write the electronic configuration of Mn (Z=25).$
›Reveal solutionSolution

The electronic configuration of Mn (Z = 25) is [Ar] 3d5 4s2 (1s2 2s2 2p6 3s2 3p6 3d5 4s2).

Manganese has atomic number 25, meaning 25 electrons to be filled according to the Aufbau principle (filling order of increasing n+l: 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...). Filling sequentially: 1s2 (2) then 2s2 (4) then 2p6 (10) then 3s2 (12) then 3p6 (18) then 4s2 (20) then 3d5 (25). Total = 25 electrons. Note that unlike chromium (which shows an anomalous configuration, 3d5 4s1, to gain extra stability from a half-filled 3d and half-filled 4s), manganese follows t …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The number of unpaired electrons in Chromium (z = 24) is :$
(a) 4
(b) 6
(c) 3
(d) 5
›Reveal solutionSolution
Chromium (Z = 24) has 6 unpaired electrons, due to its exceptional electron configuration.

By the normal Aufbau filling order, chromium (Z = 24) would be expected to have the configuration [Ar] 3d4 4s2. However, chromium shows an exception: one electron shifts from 4s to 3d to attain the extra stability of a half-filled d-subshell, giving the actual ground-state configuration [Ar] 3d5 4s1.

In this configuration:

The 3d5 subshell is exactly half-filled, so by Hund's rule all 5 electrons occupy the five d-orbitals singly, unpaired. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following has maximum number of unpaired electrons?$
(a) Zn
(b) Fe2+
(c) Ni3+
(d) Cu+
›Reveal solutionSolution
Fe2+ ([Ar]3d6) has 4 unpaired electrons, more than Zn, Ni3+, or Cu+.

Electron configurations:

Zn (Z=30): [Ar]3d10 4s2 — 3d10 is completely filled, so 0 unpaired electrons.

Fe2+ (Z=26, remove 2 electrons from 4s): [Ar]3d6. Filling 5 d-orbitals with 6 electrons by Hund's rule gives [↑↓][↑][↑][↑][↑] → 4 unpaired electrons. …
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CBSE 2025Set ANNUAL1 mark
$Q. Write the electronic configuration of Chromium [Given atomic number of Cr is 24]$
›Reveal solutionSolution

Chromium (Z=24) has the exceptional configuration [Ar] 3d5 4s1, not [Ar] 3d4 4s2, because a half-filled d subshell (3d5) together with a half-filled s subshell (4s1) is more stable (due to symmetrical distribution and extra exchange energy) than the configuration predicted by the simple Aufbau filling order.

By the normal Aufbau order, filling 24 electrons would give: 1s2 2s2 2p6 3s2 3p6 4s2 3d4.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The maximum number of unpaired electrons is present in$
(a) Fe2+
(b) Fe3+
(c) Fe4+
(d) Fe
›Reveal solutionSolution
Fe3+ = [Ar]3d5 (5 unpaired electrons), the maximum among Fe, Fe2+, Fe3+, Fe4+.

Fe (Z = 26) has configuration $[Ar]3d^6 4s^2$ . Removing electrons for the cations always removes the $4s$ electrons first, then $3d$ electrons:

$Fe$ : $3d^64s^2$ → in $3d^6$ , by Hund's rule 5 orbitals hold 5 unpaired electrons first, the 6th electron pairs up → 4 unpaired electrons.

$Fe^{2+}$ : $3d^6$ (loses both $4s$ electrons) → same as above, 4 unpaired electrons. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The correct electronic configuration of chromium (Z = 24) is$
(a) [Ar] 3d^5 4s^1
(b) [Ar] 3d^4 4s^2
(c) [Ar] 3d^6 4s^0
(d) [Ar] 3d^4 4s^1 4p^1
›Reveal solutionSolution

Cr's actual configuration is [Ar]3d⁵5 4s¹, not the 'expected' [Ar]3d⁴4s².

By the normal Aufbau order, Cr (Z=24) would be expected to have configuration [Ar]3d⁴4s². However, a half-filled d-subshell (3d⁵) is more symmetric and has extra exchange stabilization energy, so on …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. How many unpaired electrons are present in Nitrogen atom?$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

Nitrogen (Z = 7) has configuration 1s² 2s² 2p³, and by Hund's rule the three 2p electrons occupy the three separate p orbitals (pₓ, p_y, p_z) singly before any pairing, giving 3 unpaired electrons.

Electronic configuration of N (Z = 7): $1s^2\,2s^2\,2p^3$ .

…

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CBSE 2024Set sz1 markMCQ
$Q. Select the correct one: The correct ground state electronic configuration of chromium atom is:$
(a) [Ar] 3d5 4s1
(b) [Ar] 3d4 4s2
(c) [Ar] 3d6 4s0
(d) [Ar] 4d5 4s1
›Reveal solutionSolution

Chromium's ground-state configuration is [Ar] 3d5 4s1, not the Aufbau-predicted [Ar] 3d4 4s2, because a half-filled d subshell is extra stable.

By the normal Aufbau principle, chromium (Z = 24) would be expected to have configuration [Ar] 3d4 4s2. However, a completely half-filled (d5) subshell has extra stability due to symmetrical charge distribution and greater exchange energy compared to a d4 4s2 arrangement.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following angle corresponds to sp2 hybridisation ?$
(a) A) 90°
(b) B) 120°
(c) C) 180°
(d) D) 109°
›Reveal solutionSolution
[!TLDR]

B) 120°

Why

sp2 hybrid orbitals are arranged in a trigonal planar geometry wi …
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CBSE 2023Set ANNUAL1 mark
$Q. Write the electronic configuration of Cu. OR Write the electronic configuration of Cr.$
›Reveal solutionSolution

Copper's electronic configuration is [Ar]3d10 4s1, an exception to the straightforward aufbau filling order, driven by the extra stability of a completely filled d-subshell.

Copper has atomic number 29, so it has 29 electrons. Following the simple aufbau (building-up) principle, filling orbitals strictly in order of increasing energy, one would expect the configuration to be [Ar]3d9 4s2 (filling 4s fully before 3d, then placing the remaining electrons in 3d).

However, the actually observed (correct) electronic configuration of copper is:

[Ar] 3d10 4s1

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following configuration represent argon ?$
(a) ns^2
(b) ns^2 p^6
(c) ns^2 p^5
(d) ns^2 p^4
›Reveal solutionSolution

Argon's stable valence configuration is ns2 np6.

Argon has atomic number 18. Filling orbitals by the Aufbau order gives 1s2 2s2 2p6 3s2 3p6. The outermost shell (n = 3) therefore holds 3s2 3p6, i.e. the general noble-gas form ns2 np6 with a complete octet of 8 valence electrons — the reason argon is chemically i …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The orbital configuration of _24Cr is 3d^5 4s^1. The number of unpaired electrons in Cr^3+$
(g) is
(a) 3
(b) 2
(c) 1
(d) 4
›Reveal solutionSolution

Cr3+ has the 3d3 configuration with 3 unpaired electrons.

Ground-state chromium is [Ar] 3d5 4s1. When cations of transition metals form, electrons are removed first from the 4s orbital and then from 3d. For Cr3+ we remove 3 electrons: the single 4s electron and two of the 3d electrons, leaving [Ar] 3d3. In three degenerate 3d orbitals, three electrons occupy them singly (Hund's rule), gi …

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Q.Answer the following question. Write condensed orbital notation of electronic configuration of the following elements: a. Lithium (Z=3) b. Carbon (Z=6) c. Oxygen (Z=8) d. Silicon (Z=14) e. Chlorine (Z=17) f. Calcium (Z=20)

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Q.Answer the following question.
Write condensed orbital notation of electronic configuration of the following elements:
a. Lithium (Z=3) b. Carbon (Z=6) c. Oxygen (Z=8) d. Silicon (Z=14) e. Chlorine (Z=17) f. Calcium (Z=20)

Unlock everything free for 14 days