Quantum Mechanical Model – Rules for filling electrons in orbitals – Aufbau principle: NEET MCQs with Answers

15 free questions on Quantum Mechanical Model – Rules for filling electrons in orbitals – Aufbau principle for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Rules for filling electrons in orbitals – Aufbau principle. Pauli's exclusion principle and Hund's rule, electronic configuration of elements, extra stability of half-filled and completely filled orbitals.

Practise all 48 Quantum Mechanical Model – Rules for filling electrons in orbitals – Aufbau principle questions free Take a free timed test

Question 1

Previous-year question
An ion has 18 electrons in the outermost shell. It is [AIPMT 1990]
  1. Cu⁺
  2. Th⁴⁺
  3. Cs⁺
  4. K⁺
Show answer
Answer: (A) Cu⁺
Explanation: Cu has Z = 29; Cu⁺ therefore has 28 electrons. Its configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰. The outermost shell is n = 3, containing 3s² + 3p⁶ + 3d¹⁰ = 2 + 6 + 10 = 18 electrons. This is the well-known ‘pseudo-noble-gas’ or 18-electron outer configuration. By comparison, K⁺ and Cs⁺ both attain 8 electrons in the outermost shell (a true noble-gas configuration), so they do not qualify.

Question 2

Previous-year question
In a given atom no two electrons can have the same values of all the four quantum numbers. This is called [AIPMT 1991]
  1. Hund’s rule
  2. Aufbau principle
  3. Uncertainty principle
  4. Pauli’s exclusion principle
Show answer
Answer: (D) Pauli’s exclusion principle
Explanation: This is the statement of PAULI’S EXCLUSION PRINCIPLE (1925). Its immediate consequence is that an orbital — already defined by fixed n, l and mₗ — can hold at most two electrons, and only if their spins are opposite (+½ and −½). For contrast: Hund’s rule governs how electrons distribute themselves among DEGENERATE orbitals; the Aufbau principle governs the ORDER in which subshells are filled; the uncertainty principle concerns the simultaneous measurement of position and momentum.

Question 3

Previous-year question
The order of filling of electrons in the orbitals of an atom will be [AIPMT 1991]
  1. 3d, 4s, 4p, 4d, 5s
  2. 4s, 3d, 4p, 5s, 4d
  3. 5s, 4p, 3d, 4d, 5s
  4. 3d, 4p, 4s, 4d, 5s
Show answer
Answer: (B) 4s, 3d, 4p, 5s, 4d
Explanation: The (n + l) values are 4s = 4, 3d = 5, 4p = 5, 5s = 5 and 4d = 6. When n + l is equal, the orbital with lower n fills first, giving 4s → 3d → 4p → 5s → 4d.

Question 4

Previous-year question
The electronic configuration of Cu (atomic number = 29) is [AIPMT 1991]
  1. 1s², 2s² 2p⁶, 3s² 3p⁶, 4s², 3d⁹
  2. 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s¹
  3. 1s², 2s² 2p⁶, 3s² 3p⁶, 4s² 4p⁶, 5s² 5p¹
  4. 1s², 2s² 2p⁶, 3s² 3p⁶, 4s² 4p⁶, 3d³
Show answer
Answer: (B) 1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰, 4s¹
Explanation: Cu (Z = 29) is an exception: instead of 3d⁹4s² it adopts 3d¹⁰4s¹, because a completely filled d subshell has extra stability (greater exchange energy and symmetry).

Question 5

Previous-year question
The electronic configuration of a calcium atom can be written as [AIPMT 1992]
  1. [Ne] 4p²
  2. [Ar] 4s²
  3. [Ne] 4s²
  4. [Kr] 4p²
Show answer
Answer: (B) [Ar] 4s²
Explanation: Calcium (Z = 20) has 18 electrons in the argon core and the remaining 2 electrons in 4s, giving [Ar]4s².

Question 6

Previous-year question
For which one of the following sets of four quantum numbers will an electron have the highest energy? [AIPMT 1994]
  1. n = 3, l = 2, m = 1, s = +½
  2. n = 4, l = 2, m = −1, s = +½
  3. n = 4, l = 1, m = 0, s = −½
  4. n = 5, l = 0, m = 0, s = −½
Show answer
Answer: (B) n = 4, l = 2, m = −1, s = +½
Explanation: By the (n + l) rule: 3d = 5, 4d = 6, 4p = 5 and 5s = 5. The 4d electron (n = 4, l = 2) has the largest n + l and therefore the highest energy.

Question 7

Previous-year question
Which of the following configurations is correct for iron? [AIPMT 1999]
  1. 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵
  2. 1s², 2s² 2p⁶, 3s² 3p⁶, 4s², 3d⁵
  3. 1s², 2s² 2p⁶, 3s² 3p⁶, 4s², 3d⁷
  4. 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁶, 4s²
Show answer
Answer: (D) 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁶, 4s²
Explanation: Iron (Z = 26) is [Ar]3d⁶4s²: 2 + 2 + 6 + 2 + 6 + 6 + 2 = 26 electrons. Unlike Cr and Cu, iron is not an exception to the Aufbau order.
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

Previous-year question
If n = 6, the correct sequence for the filling of electrons will be [AIPMT 2011]
  1. ns ⟶ (n−1)d ⟶ (n−2)f ⟶ np
  2. ns ⟶ (n−2)f ⟶ np ⟶ (n−1)d
  3. ns ⟶ np ⟶ (n−1)d ⟶ (n−2)f
  4. ns ⟶ (n−2)f ⟶ (n−1)d ⟶ np
Show answer
Answer: (D) ns ⟶ (n−2)f ⟶ (n−1)d ⟶ np
Explanation: Substitute n = 6 into each expression: ns = 6s, (n−1)d = 5d, (n−2)f = 4f, np = 6p. Now order these by the (n + l) rule: 6s: 6 + 0 = 6 4f: 4 + 3 = 7 5d: 5 + 2 = 7 (same as 4f, so the lower n — 4f — fills first) 6p: 6 + 1 = 7 (also 7, but with the highest n, so it fills last) Filling order is therefore 6s → 4f → 5d → 6p, i.e. ns ⟶ (n−2)f ⟶ (n−1)d ⟶ np. This is exactly the sequence followed across period 6 of the periodic table: Cs/Ba (6s), the lanthanides (4f), the 5d transition metals, then Tl to Rn (6p).

Question 9

Previous-year question
The outer electronic configuration of Gd (atomic number 64) is [NEET 2013]
  1. 4f⁵ 5d⁴ 6s¹
  2. 4f⁷ 5d¹ 6s²
  3. 4f³ 5d⁵ 6s²
  4. 4f⁴ 5d⁵ 6s¹
Show answer
Answer: (B) 4f⁷ 5d¹ 6s²
Explanation: Gadolinium has Z = 64, so its configuration is [Xe] 4f⁷ 5d¹ 6s². Check the electron count: Xe = 54, plus 7 + 1 + 2 = 10, giving 64. ✓ Gd is one of the well-known ANOMALIES of the lanthanide series. Instead of filling 4f⁸, it adopts 4f⁷ 5d¹ because the exactly half-filled 4f⁷ subshell carries extra stability from maximum exchange energy and spherical symmetry. This parallels the familiar Cr and Cu anomalies in the 3d series.

Question 10

Previous-year question
Which is the correct order of increasing energy of the listed orbitals in the atom of titanium? [AIPMT 2015]
  1. 3s 4s 3p 3d
  2. 4s 3s 3p 3d
  3. 3s 3p 3d 4s
  4. 3s 3p 4s 3d
Show answer
Answer: (D) 3s 3p 4s 3d
Explanation: Apply the (n + l) rule (Aufbau principle): 3s: 3 + 0 = 3 3p: 3 + 1 = 4 4s: 4 + 0 = 4 3d: 3 + 2 = 5 3s has the lowest value. For the tie between 3p and 4s (both 4), the lower n wins, so 3p comes before 4s. 3d has the highest value. Increasing energy: 3s < 3p < 4s < 3d. This is exactly the ordering that explains why the 4s orbital fills before 3d in the first transition series, to which titanium belongs.

Question 11

Previous-year question
The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of electrons in which one of the following? [NEET 2015 (Cancelled) | AIPMT 2015]
  1. s-electrons in Mg (Z = 12)
  2. p-electrons in Cl (Z = 17)
  3. d-electrons in Fe (Z = 26)
  4. p-electrons in Ne (Z = 10)
Show answer
Answer: (B) p-electrons in Cl (Z = 17)
Explanation: Fe²⁺ is [Ar]3d⁶, so it has 6 d-electrons. Mg has 6 s-electrons (1s²2s²3s²), Fe has 6 d-electrons and Ne has 6 p-electrons, but Cl (1s²2s²2p⁶3s²3p⁵) has 11 p-electrons.

Question 12

Previous-year question
The 4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is [NEET (National) 2019]
  1. 6p > 5p > 5f > 4d
  2. 5p > 5f > 4d > 5p
  3. 5f > 6p > 4d > 5p
  4. 5f > 6p > 5p > 4d
Show answer
Answer: (D) 5f > 6p > 5p > 4d
Explanation: Apply the (n + l) rule: the orbital with the higher (n + l) value has the higher energy; if two orbitals have the same (n + l), the one with the higher n has the higher energy. 4d: n + l = 4 + 2 = 6 5p: n + l = 5 + 1 = 6 5f: n + l = 5 + 3 = 8 6p: n + l = 6 + 1 = 7 Highest first: 5f (8) > 6p (7), then the tie between 5p and 4d (both 6) is broken by n — 5p has n = 5 and 4d has n = 4, so 5p > 4d. Decreasing order: 5f > 6p > 5p > 4d.

Question 13

Match each species in List I with its outer electronic configuration in List II. List I: (A) Fe³⁺ (B) Cr (C) Fe⁺ (D) Co³⁺. List II: (I) 3d⁶ (II) 3d⁵4s¹ (III) 3d⁵ (IV) 3d⁶4s¹
  1. A-III, B-II, C-I, D-IV
  2. A-III, B-II, C-IV, D-I
  3. A-I, B-IV, C-III, D-II
  4. A-II, B-III, C-I, D-IV
Show answer
Answer: (B) A-III, B-II, C-IV, D-I
Explanation: Iron is [Ar]3d⁶4s²; Fe⁺ loses one 4s electron to give 3d⁶4s¹, and Fe³⁺ loses both 4s electrons and one 3d electron to give 3d⁵. Chromium is 3d⁵4s¹ because a half-filled d subshell is extra stable. Cobalt is 3d⁷4s², so Co³⁺ is 3d⁶.

Question 14

An atom's electrons are placed as follows: one electron (↑) in 2s; two electrons with the same spin (↑↑) in the first 2p orbital; one electron (↑) in the second 2p orbital; the third 2p orbital is empty. Which rule or rules does this arrangement break?
  1. Only the Pauli exclusion principle
  2. The Pauli exclusion principle, Hund's rule and the Aufbau principle
  3. Only Hund's rule
  4. Only the Aufbau principle
Show answer
Answer: (B) The Pauli exclusion principle, Hund's rule and the Aufbau principle
Explanation: Two electrons with the same spin in one orbital break the Pauli exclusion principle. Putting two electrons in one 2p orbital while another 2p orbital is empty breaks Hund's rule. Placing electrons in 2p while 2s is only half-filled breaks the Aufbau principle.

Question 15

In the following arrangements of electrons in the 2s and three 2p orbitals, ↑↓ means a pair of electrons with opposite spins. Which arrangement violates the Aufbau principle?
  1. 2s: ↑↓; 2p: ↑↓, ↑↓, empty
  2. 2s: ↑↓; 2p: ↑, ↑, ↑
  3. 2s: ↑↓; 2p: ↑↓, ↑, empty
  4. 2s: ↑; 2p: ↑↓, ↑↓, ↑
Show answer
Answer: (D) 2s: ↑; 2p: ↑↓, ↑↓, ↑
Explanation: The Aufbau principle requires the lower-energy 2s orbital to be filled completely before electrons enter 2p. Leaving 2s half-filled while 2p holds five electrons breaks this rule. Pairing electrons in a 2p orbital while another 2p orbital is empty breaks Hund's rule, not the Aufbau principle, and three unpaired parallel 2p electrons after a filled 2s is correct.

Want more Quantum Mechanical Model – Rules for filling electrons in orbitals – Aufbau principle questions?

E-GYAN NIDHI has 48 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in ATOMIC STRUCTURE

↑