Configurations like d⁵ and d¹⁰ have extra stability due to symmetry and exchange energy. Examples: Cr [Ar]3d⁵4s¹ and Cu [Ar]3d¹⁰4s¹ (anomalous configurations).
-- NCERT Class 11 Chemistry, Ch. 2, p. 64Half Filled Full Filled Stability
8 MCQs9-step worked example
Source: NCERT Structure of AtomOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Half Filled Full Filled Stability, explained for NEET
The trap: When asked for the ground-state electronic configuration of Cr (Z = 24) or Cu (Z = 29), most aspirants mechanically apply the Aufbau principle and write Cr as [Ar] 3d⁴ 4s² and Cu as [Ar] 3d⁹ 4s². Both are wrong. NEET exploits this confusion reliably.
The concept: Half-filled and completely filled d-subshells possess extra stability arising from two factors:
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Symmetrical distribution of electrons — electrons in half-filled or fully filled subshells are distributed symmetrically across all orbitals of that subshell, which lowers the overall energy.
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Exchange energy — the number of electrons with parallel spin that can exchange positions is maximized in half-filled (d⁵) and fully filled (d¹⁰) configurations. Greater exchange energy = greater stabilization.
Because of this extra stability, chromium adopts [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) and copper adopts [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²). One electron is promoted from 4s to 3d to achieve the more stable arrangement.
(Reference: NCERT Class 11 Chemistry Chapter 2, page 40.)
NEET connection: Questions test whether you recall the actual configuration or fall for the Aufbau-predicted one. The wrong option is almost always the "expected" Aufbau configuration — it looks correct to a student who hasn't internalized this exception.
Watch-out: This anomaly applies specifically to Cr and Cu in the 3d series. Do not overgeneralize — Mo and Ag show similar behaviour in the 4d series, but elements like Mn (3d⁵ 4s²) already have a half-filled d-subshell without needing promotion and are NOT anomalous.
Can you answer these Half Filled Full Filled Stability MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following ions has the extra stability of an exactly half-filled d subshell?
Show answer and why every option is right or wrong
Answer: B. B is correct. Mn is [Ar] 3d⁵ 4s²; losing the two 4s electrons gives Mn²⁺ = [Ar] 3d⁵, with one electron in each of the five d orbitals. Half-filled subshells are especially stable because of symmetrical distribution and maximum exchange energy, which is why Mn²⁺ resists further oxidation.
Why A is wrong: A is wrong because Fe²⁺ is [Ar] 3d⁶: one electron beyond half-filled. Losing that electron, to give Fe³⁺ (3d⁵), is exactly why iron(II) is readily oxidised.
Why C is wrong: C is wrong because Cr²⁺ is [Ar] 3d⁴: one electron short of half-filled. The neutral Cr atom is the famous half-filled case (3d⁵ 4s¹), but the ion is not.
Why D is wrong: D is wrong because Cu²⁺ is [Ar] 3d⁹: one short of a completely FILLED d subshell, not half-filled.
MCQ 2Easy RecallPractice
The ground-state electronic configuration of Cu (Z = 29) is:
Show answer and why every option is right or wrong
Answer: B. Cu adopts [Ar] 3d¹⁰ 4s¹ because the completely filled d¹⁰ configuration provides extra stability from maximal exchange energy. One 4s electron promotes to 3d (NCERT Class 11 Chemistry, Chapter 2, page 40).
Why A is wrong: A is the Aufbau-predicted configuration (3d⁹ 4s²) — it ignores the extra stability of the fully filled d¹⁰ subshell.
Why C is wrong: C has 12 electrons beyond Ar (10 + 2 = 12), giving Z = 30, which is Zn, not Cu (Z = 29).
Why D is wrong: D incorrectly places an electron in 4p and underfills the 3d subshell; this violates both energy ordering and the total electron count for Cu.
MCQ 3Easy RecallPractice
The extra stability of half-filled and completely filled orbitals is attributed to:
Show answer and why every option is right or wrong
Answer: D. NCERT (Class 11 Chemistry, Chapter 2, page 40) explicitly states two reasons: symmetrical electron distribution lowers repulsion, and exchange energy is maximized when electrons with parallel spin occupy all orbitals of a subshell equally.
Why A is wrong: A — nuclear charge affects overall energy levels but does not specifically explain why d⁵ and d¹⁰ are more stable than d⁴ or d⁹.
Why B is wrong: B — relativistic effects are relevant for very heavy elements (6th period onwards) and are not the NCERT explanation for Cr/Cu anomaly in the 3d series.
Why C is wrong: C — shielding explains why outer electrons feel a reduced nuclear charge, but it does not explain the special stability of half-filled/fully filled subshells relative to adjacent configurations.
MCQ 4Direct ApplicationPractice
Which of the following elements does NOT show an anomalous electronic configuration due to half-filled/fully filled d-orbital stability?
Show answer and why every option is right or wrong
Answer: D. Mn has the configuration [Ar] 3d⁵ 4s². It already has a half-filled d-subshell following normal Aufbau filling — no electron promotion is needed. There is no anomaly.
Why A is wrong: A — Cr is the textbook anomaly: expected [Ar] 3d⁴ 4s², actual [Ar] 3d⁵ 4s¹ due to d⁵ stability.
Why B is wrong: B — Cu is the other classic anomaly: expected [Ar] 3d⁹ 4s², actual [Ar] 3d¹⁰ 4s¹ due to d¹⁰ stability.
Why C is wrong: C — Mo (4d series) shows the same type of anomaly as Cr: expected [Kr] 4d⁴ 5s², actual [Kr] 4d⁵ 5s¹.
MCQ 5Direct ApplicationPractice
In the context of exchange energy, which configuration has more exchange pairs among d-electrons: 3d⁴ or 3d⁵ (all spins parallel)?
Show answer and why every option is right or wrong
Answer: A. Exchange pairs = n(n−1)/2 where n is the number of electrons with parallel spin. For d⁵: 5(4)/2 = 10 pairs. For d⁴: 4(3)/2 = 6 pairs. The d⁵ configuration has significantly more exchange energy, contributing to its extra stability.
Why B is wrong: B — 3d⁴ gives only 4(3)/2 = 6 exchange pairs, fewer than the 10 pairs of 3d⁵.
Why C is wrong: C — they are not equal: 6 ≠ 10. The jump from d⁴ to d⁵ adds 4 new exchange pairs, which is a large stabilization.
Why D is wrong: D — exchange energy is directly proportional to the number of parallel-spin electron pairs; this is the fundamental reason for half-filled stability.
MCQ 6Direct ApplicationPractice
The electronic configuration of Cu²⁺ (Z = 29) is:
Show answer and why every option is right or wrong
Answer: C. Cu neutral is [Ar] 3d¹⁰ 4s¹. When forming Cu²⁺, two electrons are removed — the 4s electron goes first, then one 3d electron. Result: [Ar] 3d⁹. Students who start from the wrong neutral config ([Ar] 3d⁹ 4s²) may still arrive at the correct ion accidentally, but the reasoning path through the anomalous neutral config is what NEET tests.
Why A is wrong: A — this configuration has only 27 electrons total (18 + 7 + 2), which would be Co, not Cu²⁺ (which should have 27 electrons as 29 − 2 = 27, but the correct distribution is [Ar] 3d⁹ not 3d⁷ 4s²). Electrons leave 4s before 3d.
Why B is wrong: B — electrons are removed from 4s before 3d during ionization. After removing 4s¹ from neutral Cu, the next electron comes from 3d¹⁰, giving 3d⁹, not 3d⁸ 4s¹.
Why D is wrong: D — 3d¹⁰ would mean only one electron was removed (from 4s¹), giving Cu⁺, not Cu²⁺.
MCQ 7CalculationPractice
Cr³⁺ has the electronic configuration [Ar] 3d³. This is derived by removing electrons from neutral Cr. Which electrons are removed and in what order?
Show answer and why every option is right or wrong
Answer: A. Neutral Cr is [Ar] 3d⁵ 4s¹ (anomalous). Ionization removes 4s first (higher principal quantum number), then 3d. Removing the one 4s electron gives Cr⁺ = [Ar] 3d⁵. Removing two more from 3d gives Cr³⁺ = [Ar] 3d³.
Why B is wrong: B — Cr has only one 4s electron (not two, because of the anomaly). A student using the wrong neutral config [Ar] 3d⁴ 4s² would pick this — that is the Cr/Cu anomaly trap.
Why C is wrong: C — 3d electrons are not removed first; 4s is removed before 3d during ionization despite filling in the opposite order during build-up.
Why D is wrong: D — 4s electrons are always removed before 3d during ionization, not after.
MCQ 8Concept TrapPractice
Among the following, which pair of elements shows anomalous electronic configurations due to the stability of half-filled and fully filled d-orbitals in the 3d series?
Show answer and why every option is right or wrong
Answer: C. Cr (3d⁵ 4s¹ instead of 3d⁴ 4s²) and Cu (3d¹⁰ 4s¹ instead of 3d⁹ 4s²) are the two 3d elements that show anomalous configurations. Mn and Zn already achieve d⁵ and d¹⁰ without any promotion (NCERT Class 11 Chemistry, Chapter 2, page 40).
Why A is wrong: A — Fe ([Ar] 3d⁶ 4s²) and Ni ([Ar] 3d⁸ 4s²) follow Aufbau normally; neither is half-filled nor fully filled in d-orbitals.
Why B is wrong: B — Mn ([Ar] 3d⁵ 4s²) and Zn ([Ar] 3d¹⁰ 4s²) DO have half-filled and fully filled d-subshells respectively, but they achieve this through normal Aufbau filling — there is no anomaly or electron promotion involved.
Why D is wrong: D — V ([Ar] 3d³ 4s²) and Co ([Ar] 3d⁷ 4s²) follow normal Aufbau; neither d-subshell is half-filled or fully filled.
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How do you solve a Half Filled Full Filled Stability question? A worked example
Pattern: Electronic configuration with Cr/Cu anomaly (anchored to trap T.CHE.U02.CR_CU_ANOMALY, mistake M.CHE.U02.CR_CU_CONFIG_REGULAR)
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Given
• Chromium, Z = 24• Required: ground-state electronic configuration and number of unpaired electrons
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Required
• Correct configuration accounting for anomaly• Count of unpaired electrons
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Concept
Half-filled d-subshells (d⁵) have extra stability due to maximized exchange energy and symmetrical electron distribution. Chromium achieves this by promoting one electron from 4s to 3d.
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Expected (Aufbau) vs. Actual
• Aufbau predicts: [Ar] 3d⁴ 4s² (fill 4s² first, then 3d⁴)• Actual: [Ar] 3d⁵ 4s¹ (one 4s electron promoted to achieve half-filled d⁵)
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Why the promotion occurs
Exchange pairs for d⁵ = 5(4)/2 = 10.
Exchange pairs for d⁴ = 4(3)/2 = 6.
Gain = 4 additional exchange pairs. This energy gain exceeds the cost of promoting one electron from 4s to 3d. - 6
Counting unpaired electrons
• 3d⁵: five orbitals, each with one electron (all parallel spin by Hund's rule) → 5 unpaired• 4s¹: one orbital with one electron → 1 unpaired• Total unpaired electrons = 6
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Final answer
Cr: [Ar] 3d⁵ 4s¹, 6 unpaired electrons.
Note: The numbers 5, 4, 6, 10 used in exchange-pair counting are exact integers and do not limit significant figures. - 8
Common trap
Writing [Ar] 3d⁴ 4s² gives only 4 unpaired electrons (4 in 3d, 0 in paired 4s²). This is the most common wrong answer in NEET for Cr-configuration questions. The distractor "4 unpaired electrons" directly exploits this mistake.
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Similar NEET-style question
"The number of unpaired electrons in the ground state of Cu (Z = 29) is ___."
(Answer: Cu = [Ar] 3d¹⁰ 4s¹ → 3d fully paired (0 unpaired) + 4s¹ (1 unpaired) = 1 unpaired electron. Trap answer: using [Ar] 3d⁹ 4s² gives 1 unpaired in 3d + 0 in 4s = 1 — same numerical answer but wrong reasoning path.)
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What to remember before solving Half Filled Full Filled Stability questions
Where do students lose marks on Half Filled Full Filled Stability?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student writes Cr as [Ar]3d⁴4s² (expected) instead of actual [Ar]3d⁵4s¹. Same for Cu: actual [Ar]3d¹⁰4s¹ (one e⁻ promoted from 4s to 3d).
When it triggers
Question asks for ground-state electronic configuration of Cr (Z=24) or Cu (Z=29).
How to avoid
Half-filled (d⁵) and fully filled (d¹⁰) configurations have extra stability from exchange energy and symmetry. Cr and Cu adopt these configurations by promoting one 4s electron.
Root cause: concept gap
Correction
Half-filled (d⁵) and full-filled (d¹⁰) configurations have extra exchange-energy stability. Cr and Cu adopt these by promoting one 4s electron.
More in Structure of Atom: 2 exam traps and mistakes · 5 formulas · 3 question patterns from its other lessons.
Half Filled Full Filled Stability questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Chemistry Chapter 2, p.40
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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