Aufbau: orbitals filled in order of increasing energy (1s<2s<2p<3s<3p<4s<3d<...). Pauli: no two electrons in same atom have identical 4 quantum numbers. Hund: orbitals of same energy first filled singly with parallel spins.
-- NCERT Class 11 Chemistry, Ch. 2, p. 62Aufbau Pauli Hund
8 MCQs9-step worked example
Source: NCERT Structure of AtomOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Aufbau Pauli Hund, explained for NEET
The trap that costs marks here: You apply the Aufbau principle mechanically — fill 1s, 2s, 2p, 3s, 3p, 4s, 3d in sequence — and write Cr as [Ar] 3d⁴ 4s². The actual ground-state configuration is [Ar] 3d⁵ 4s¹. NEET exploits this directly.
The three rules governing electron filling:
Aufbau principle — Electrons occupy the lowest available energy orbital first. The filling order follows (n + l) rule: lower (n + l) fills first; for equal (n + l), lower n fills first. This gives the sequence: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → ...
Pauli exclusion principle — No two electrons in an atom can have the same set of four quantum numbers (n, l, mₗ, mₛ). Consequence: each orbital holds a maximum of 2 electrons with opposite spins.
Hund's rule of maximum multiplicity — Electrons occupy degenerate orbitals (same subshell) singly with parallel spins before pairing begins. In the 2p subshell (three orbitals), the fourth electron pairs — the first three go one each.
The Cr/Cu anomaly (NCERT Class 11 Chemistry Chapter 2, page 36):
Half-filled (d⁵) and fully filled (d¹⁰) d-subshells have extra stability due to exchange energy and symmetrical charge distribution. Chromium (Z = 24) adopts [Ar] 3d⁵ 4s¹ instead of [Ar] 3d⁴ 4s². Copper (Z = 29) adopts [Ar] 3d¹⁰ 4s¹ instead of [Ar] 3d⁹ 4s².
Watch-out: When writing configurations of ions (Cr³⁺, Cu²⁺), electrons are removed from 4s first (higher n), then 3d. Cr³⁺ = [Ar] 3d³. Cu²⁺ = [Ar] 3d⁹.
Can you answer these Aufbau Pauli Hund 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
According to the (n + l) rule, which orbital is filled immediately after 4s?
Show answer and why every option is right or wrong
Answer: B. B is correct. Orbitals fill in order of increasing (n + l); for equal (n + l), the one with lower n fills first. After 4s (n + l = 4), the candidates with n + l = 5 are 3d, 4p and 5s. Of these, 3d has the lowest n, so it fills next. That is why the first transition series starts right after calcium.
Why A is wrong: A is wrong because 4p also has n + l = 5, but its n (4) is higher than that of 3d (3). The tie is broken in favour of the lower n.
Why C is wrong: C is wrong because 4d has n + l = 6, higher than the n + l = 5 of 3d, so it fills later (after 5s).
Why D is wrong: D is wrong because 5s also has n + l = 5, but it has the highest n of the three candidates, so it fills last of them.
MCQ 2Direct ApplicationPractice
The number of unpaired electrons in a nitrogen atom (Z = 7) in its ground state is:
Show answer and why every option is right or wrong
Answer: C. C is correct. Nitrogen is 1s² 2s² 2p³. By Hund's rule, electrons entering the three degenerate 2p orbitals go in singly, with parallel spins, before any pairing, so the three 2p electrons occupy three different orbitals and all are unpaired.
Why A is wrong: A is wrong because 1 pairs two of the 2p electrons in one orbital. Hund's rule forbids pairing while an empty orbital of the same energy is available.
Why B is wrong: B is wrong because 5 counts all the valence electrons (2s² 2p³). The two 2s electrons are paired in a single orbital.
Why D is wrong: D is wrong because 0 would mean every electron is paired, which cannot be true for an atom with an odd number of electrons.
MCQ 3Easy RecallPractice
According to Hund's rule, the electronic configuration of nitrogen (Z = 7) in the ground state has the 2p electrons arranged as:
Show answer and why every option is right or wrong
Answer: A. A is correct. Nitrogen has three 2p electrons. Hund's rule requires them to occupy all three degenerate 2p orbitals singly with parallel spins before any pairing occurs.
Why B is wrong: B is wrong — pairing in one orbital before all three 2p orbitals are singly occupied violates Hund's rule of maximum multiplicity.
Why C is wrong: C is wrong — having two paired orbitals with only three electrons is impossible (that requires 5 electrons in 2p). This also violates Hund's rule.
Why D is wrong: D is wrong — although the electrons are singly placed, having anti-parallel spins (↑ ↓ ↑) violates the 'maximum multiplicity' requirement. All unpaired electrons must have parallel spins.
MCQ 4Direct ApplicationPractice
The Pauli exclusion principle states that:
Show answer and why every option is right or wrong
Answer: C. C is correct. Because the spin quantum number can take only two values, an orbital (fixed n, l and m) can hold at most two electrons, with opposite spins. This is what limits each subshell's capacity and so shapes the whole periodic table.
Why A is wrong: A is wrong because that is the Aufbau principle, which sets the ORDER in which orbitals are filled.
Why B is wrong: B is wrong because that is Hund's rule of maximum multiplicity, which says how electrons are distributed among orbitals of equal energy.
Why D is wrong: D is wrong because that is Heisenberg's uncertainty principle, which concerns measurement, not how electrons are arranged in orbitals.
MCQ 5Direct ApplicationPractice
The Aufbau principle predicts the filling order based on the (n + l) rule. Which of the following orbitals is filled BEFORE 4d?
Show answer and why every option is right or wrong
Answer: A. A is correct. 5s has (n + l) = 5 + 0 = 5. 4d has (n + l) = 4 + 2 = 6. Since 5s has lower (n + l), it fills before 4d.
Why B is wrong: B is wrong — 5p has (n + l) = 5 + 1 = 6, same as 4d. When (n + l) is equal, lower n fills first, so 4d fills before 5p.
Why C is wrong: C is wrong — 4f has (n + l) = 4 + 3 = 7, which is higher than 4d's (n + l) = 6. So 4f fills after 4d, not before.
Why D is wrong: D is wrong — 6s has (n + l) = 6 + 0 = 6, same as 4d. When (n + l) is equal, lower n fills first, so 4d fills before 6s.
MCQ 6Concept TrapPractice
An atom has the electronic configuration [Ar] 3d⁵ 4s¹. This configuration is adopted because:
Show answer and why every option is right or wrong
Answer: B. B is correct. The half-filled d⁵ configuration provides extra stability from symmetrical electron distribution and maximum exchange energy. This causes Cr to promote one 4s electron to achieve [Ar] 3d⁵ 4s¹ (NCERT Class 11 Chemistry Chapter 2, page 36).
Why A is wrong: A is wrong — 3d is not always lower in energy than 4s. For neutral atoms near Z = 20–30, the relative energies of 3d and 4s are close; the anomaly is specifically about the extra stabilisation of half-filled/full-filled configurations, not a general energy ordering reversal.
Why C is wrong: C is wrong — Pauli exclusion only forbids two electrons with identical quantum numbers in the same orbital. It does not forbid 4s². 4s can hold 2 electrons regardless of 3d occupancy.
Why D is wrong: D is wrong — Hund's rule applies to degenerate orbitals within the same subshell (e.g. the five 3d orbitals). It does not operate across different subshells (4s vs 3d).
MCQ 7CalculationPractice
The maximum number of electrons that can have the quantum numbers n = 3 and mₛ = +½ is:
Show answer and why every option is right or wrong
Answer: D. D is correct. For n = 3: l = 0, 1, 2 → total orbitals = 1 + 3 + 5 = 9. Each orbital can hold one electron with mₛ = +½. Therefore, maximum 9 electrons with n = 3 and mₛ = +½.
Why A is wrong: A is wrong — 3 would be correct for n = 3, l = 1, mₛ = +½ only (3p subshell spin-up). But the question doesn't restrict l, so all subshells contribute.
Why B is wrong: B is wrong — 18 is the total capacity of n = 3 shell (both spin-up and spin-down). The constraint mₛ = +½ halves this to 9.
Why C is wrong: C is wrong because 6 is the full capacity of the 3p subshell (both spins); with mₛ = +½ fixed, 3p gives only 3, and the whole n = 3 shell gives 9.
MCQ 8Easy RecallPractice
Which of the following electronic configurations violates the Pauli exclusion principle?
Show answer and why every option is right or wrong
Answer: D. D is correct (it violates Pauli). The 3d subshell has 5 orbitals, each holding at most 2 electrons, giving a maximum capacity of 10. 3d¹¹ exceeds this maximum, violating the Pauli exclusion principle.
Why A is wrong: A is valid — this is Mg (Z = 12). All subshell electron counts are within their maximum capacities (s: max 2, p: max 6).
Why B is wrong: B is valid — this is sodium (Z = 11). All counts are within subshell maxima.
Why C is wrong: C is valid — this is oxygen (Z = 8). 2p⁴ is within the 2p subshell maximum of 6.
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How do you solve a Aufbau Pauli Hund question? A worked example
Pattern: P.CHE.U02.AUFBAU_ELECTRONIC_CONFIG — Write electronic configuration handling the Cr/Cu anomaly.
- 1
Given
Element: Chromium, Z = 24.
- 2
Required
Ground-state electronic configuration of Cr.
- 3
Concept
Apply the Aufbau principle with (n + l) filling order. Recognize that half-filled (d⁵) configurations have extra stability — Cr is a known anomaly where one 4s electron promotes to 3d.
- 4
Formula/Rule
Aufbau filling order: 1s → 2s → 2p → 3s → 3p → 4s → 3d.
Anomaly rule: half-filled d⁵ → extra exchange-energy stability → one 4s electron promotes. - 5
Substitution / Setup
Expected (mechanical Aufbau): [Ar] 3d⁴ 4s² (24 − 18 = 6 electrons after Ar; 2 in 4s, 4 in 3d).
Apply anomaly: promote one 4s electron to 3d → [Ar] 3d⁵ 4s¹. - 6
Calculation / Verification
Electron count check: 18 (Ar) + 5 (3d) + 1 (4s) = 24 ✓
d⁵ = half-filled (5 electrons in 5 orbitals, one per orbital with parallel spins per Hund's rule) ✓ - 7
Final answer
Cr (Z = 24): [Ar] 3d⁵ 4s¹
- 8
Common trap
Writing [Ar] 3d⁴ 4s² by mechanically applying Aufbau without recognizing the half-filled stability exception. NEET 2024 tested this directly.
- 9
Similar NEET-style question
"The ground-state electronic configuration of Cu⁺ (Z = 29) is:"
Apply: Cu = [Ar] 3d¹⁰ 4s¹ (fully filled anomaly). Remove the 4s electron for Cu⁺ → [Ar] 3d¹⁰.
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What to remember before solving Aufbau Pauli Hund questions
More in Structure of Atom: 4 exam traps and mistakes · 5 formulas · 2 question patterns from its other lessons.
Aufbau Pauli Hund questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Aufbau Pauli Hund?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Write electronic configuration, especially handling Cr/Cu anomalies and ions.
InterpretationMedium
Common distractors
misses cr cu anomaly
Writes 3d⁴4s² instead of 3d⁵4s¹
Sources
NCERT refs: Class 11 Chemistry Chapter 2, p.36
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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