Electronic Configuration

8 MCQs9-step worked example

Source: NCERT Structure of AtomPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The configuration of Cr is not [Ar] 3d⁴ 4s². It is [Ar] 3d⁵ 4s¹. If you wrote the former in a NEET paper, you lost a mark and gained a negative — that is the trap this lesson exists to fix.

Electronic configuration is the distribution of electrons among available orbitals following three rules: Aufbau principle (fill lowest energy first), Pauli exclusion (no two electrons share all four quantum numbers), and Hund's rule (maximise spin multiplicity within a subshell). For most elements up to Z = 30, straightforward application of (n + l) ordering gives the correct ground-state configuration.

The exceptions that NEET tests repeatedly are chromium (Z = 24) and copper (Z = 29). The expected filling gives Cr: [Ar] 3d⁴ 4s² and Cu: [Ar] 3d⁹ 4s². The actual configurations are:

Cr: [Ar] 3d⁵ 4s¹ — half-filled 3d achieves extra exchange-energy stability.
Cu: [Ar] 3d¹⁰ 4s¹ — fully filled 3d achieves extra stability.

The underlying reason: exchange energy increases with the number of parallel-spin electron pairs. A half-filled or fully filled d-subshell maximises these pairs, and the energy gain from exchange exceeds the small 4s–3d gap for these specific elements.

NCERT Class 11 Chemistry Chapter 2, page 36, explicitly states this principle as part of the electronic configuration rules for transition elements.

Watch-out for ions: When forming Cr³⁺ or Cu²⁺, electrons are removed from 4s first (higher principal quantum number), then from 3d. Cr³⁺ is [Ar] 3d³, not [Ar] 3d¹ 4s². Cu²⁺ is [Ar] 3d⁹.

Practice 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

The ground-state electronic configuration of chromium (Z = 24) is:

MCQ 2Easy RecallPractice

The ground-state electronic configuration of Cu (Z = 29) is:

MCQ 3Easy RecallPractice

Which of the following explains why Cr and Cu deviate from expected Aufbau filling?

MCQ 4Direct ApplicationPractice

The electronic configuration of Cr³⁺ ion is:

MCQ 5Direct ApplicationPractice

The electronic configuration of Cu²⁺ ion is:

MCQ 6Direct ApplicationPractice

Among Fe (Z = 26), Co (Z = 27), Ni (Z = 28), and Cu (Z = 29), which has a ground-state configuration that deviates from simple Aufbau prediction?

MCQ 7Concept TrapPractice

The number of unpaired electrons in Cr (Z = 24) in its ground state is:

MCQ 8Easy RecallPractice

Which pair of elements in the first transition series (Z = 21 to Z = 30) shows anomalous electronic configuration due to d-subshell stability?

Worked Example

Pattern: Electronic configuration writing with Cr/Cu anomaly (P.CHE.U02.AUFBAU_ELECTRONIC_CONFIG, observed NEET 2024)

1
Given
- Cr: atomic number 24 - Cu⁺: atomic number 29, charge +1 (28 electrons)
2
Required
Ground-state electronic configuration and number of unpaired electrons for each species.
3
Concept
Electronic configuration follows Aufbau principle with exceptions for Cr and Cu due to extra exchange-energy stability of half-filled (d⁵) and fully filled (d¹⁰) subshells. For ions, electrons are removed from the highest principal quantum number (4s before 3d).
4
Formula/Rule
- Aufbau order: 1s < 2s < 2p < 3s < 3p < 4s < 3d... - Exception: Cr adopts 3d⁵ 4s¹; Cu adopts 3d¹⁰ 4s¹ - Ion formation: remove from highest n first (4s before 3d)
5
Substitution
**(a) Cr (24 electrons):** - Expected: [Ar] 3d⁴ 4s² (18 + 4 + 2 = 24) ✗ - Actual: [Ar] 3d⁵ 4s¹ (18 + 5 + 1 = 24) ✓ (half-filled d⁵ stability) **(b) Cu⁺ (28 electrons):** - Neutral Cu: [Ar] 3d¹⁰ 4s¹ (29 electrons) - Remove 1 electron from 4s (highest n): [Ar] 3d¹⁰ (28 electrons)
6
Calculation
**(a) Cr unpaired electrons:** - 3d⁵: 5 orbitals, each singly occupied → 5 unpaired - 4s¹: 1 orbital, singly occupied → 1 unpaired - Total: 6 unpaired electrons **(b) Cu⁺ unpaired electrons:** - 3d¹⁰: 5 orbitals, each doubly occupied → 0 unpaired - Total: 0 unpaired electrons
7
Final answer
| Species | Configuration | Unpaired e⁻ | |---------|--------------|-------------| | Cr | [Ar] 3d⁵ 4s¹ | 6 | | Cu⁺ | [Ar] 3d¹⁰ | 0 | Note: Z values (24, 29) are counting integers (exact) and do not affect any significant-figure consideration in this problem.
8
Common trap
Writing Cr as [Ar] 3d⁴ 4s² (forgetting the anomaly) loses the mark AND earns a negative marking penalty. For Cu⁺, the trap is removing the 3d electron instead of 4s — always remove from highest n first in ions.
9
Similar NEET-style question
"The number of unpaired electrons in Cu²⁺ (Z = 29) is ___." [Answer: 1. Configuration: [Ar] 3d⁹. Nine d-electrons fill four orbitals doubly and one singly.]

Before solving, remember these

Law

Aufbau, Pauli, Hund's rules

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. 36

Formulas

Bohr energy (hydrogen-like)

E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV

Energy of nth orbit. Negative (bound). Ground state H: -13.6 eV.

Symbol	Quantity	SI Unit
E_n	orbit energy	eV
Z	nuclear charge	-
n	principal	-

Valid when

Hydrogen-like atom
Non-relativistic

Bohr radius (hydrogen-like)

r_{n} = (0.529 \overset{˚}{A}) \frac{n ^{2}}{Z}

Radius of nth Bohr orbit for hydrogen-like atom of nuclear charge Z.

Symbol	Quantity	SI Unit
n	principal quantum number	-
Z	nuclear charge	-
r_n	orbit radius	Å

Valid when

Hydrogen-like (one-electron) atom
Non-relativistic

de Broglie wavelength

λ = h / (m v)

Wavelength associated with moving particle of momentum mv.

Symbol	Quantity	SI Unit
h	Planck 6.626e-34	J*s
m	mass	kg
v	velocity	m/s

Valid when

Non-relativistic

Heisenberg uncertainty

Δ x Δ p \geq \frac{h}{4 π}

Position and momentum cannot both be known with arbitrary precision.

Symbol	Quantity	SI Unit
Δx	position uncertainty	m
Δp	momentum uncertainty	kg*m/s

Valid when

Quantum scale; meaningful only when Δx, Δp comparable to atomic dimensions

Rydberg formula (H spectrum)

\frac{1}{λ} = R_{H} Z^{2} (1/ n_{1}^{2} - 1/ n_{2}^{2})

Spectral wavelengths of hydrogen-like atoms. Lyman (n1=1, UV), Balmer (n1=2, visible), Paschen (n1=3, IR).

Symbol	Quantity	SI Unit
lambda	wavelength	m
R_H	Rydberg 1.097e7	1/m
Z	nuclear charge	-
n1, n2	integers, n2>n1	-

Valid when

One-electron atom

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Cr / Cu electronic configuration anomaly

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.

Trap

Hydrogen-like Z² factor

Category: Similar Terms

Student forgets Z² scaling when applying Bohr formulas to He⁺ (Z=2) or Li²⁺ (Z=3).

When it triggers

Question involves hydrogen-like ion (He+, Li2+, etc.).

How to avoid

E_n = -13.6 × Z²/n² eV. r_n = (0.529/Z) × n² Å. He+: 4× more bound than H. Li²⁺: 9× more bound. Always include Z².

Mistake

Writes Cr as [Ar]3d⁴4s² (expected) instead of actual [Ar]3d⁵4s¹. Same for Cu.

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.

Mistake

Drops Z² factor when applying Bohr formulas to He⁺ or Li²⁺.

Root cause: formula misuse

Correction

Always include Z². E_n = -13.6 × Z²/n². For He+: 4× more energetic than H. For Li²⁺: 9×.

Past Year Questions

9 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Higher yield of NO in - 40 - N 2 (g ) + O 2 (g ) 2 N O (g ) can be obtained at [H of the reaction = +180.7 kJ mol–1] A. Higher temperature B. Lower temperature C. Higher concentration of N 2 D. Higher concentration of O 2 Choose the correct answer from the options given below :

1A, C, D only

2A, D only

3B, C only

4B, C, D only

NTA Answer: Option 1(final)

NEET 2024Revised key

Match List I with List II. List I List II (Molecule) (Number and types of bond/s between two carbon atoms) A. ethane I. one -bond and two -bonds B. ethene II. two -bonds C. carbon molecule, C III. one -bonds 2 D. ethyne IV. one -bond and one -bond Choose the correct answer from the options given below:

1A-I, B-IV, C-II, D-III

2A-IV, B-III, C-II, D-I

3A-III, B-IV, C-II, D-I

4A-III, B-IV, C-I, D-II

NTA Answer: Option 3(revised_final)

NEET 2024Revised key

Match List I with List II List I List II (Quantum Number) (Information provided) A. m I. Shape of orbital l B. m II. Size of orbital s C. l III. Orientation of orbital D. n IV. Orientation of spin of electron Choose the correct answer from the options given below :

1A-I, B-III, C-II, D-IV

2A-III, B-IV, C-I, D-II

3A-III, B-IV, C-II, D-I

4A-II, B-I, C-IV, D-III

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

Given below are two statements : Statement I : [Co(NH ) ]3+ is a homoleptic complex whereas [Co(NH ) Cl ]+ is a heteroleptic complex. 3 6 3 4 2 Statement II : Complex [Co(NH ) ]3+ has only one kind of ligands but [Co(NH ) Cl ]+ has more than one kind 3 6 3 4 2 of ligands. In the light of the above statements, choose the correct answer from the options given below.

1Both Statement I and Statement II are true

2Both Statement I and Statement II are false

3Statement I is true but Statement II is false

4Statement I is false but Statement II is true

NTA Answer: Option 1(revised_final)

NEET 2023

Select the correct statements from the following A. Atoms of all elements are composed of two fundamental particles. B. The mass of the electron is 9.10939 × 10–31 kg. C. All the isotopes of a given element show same chemical properties: D. Protons and electrons are collectively known as nucleons. E. Dalton’s atomic theory, regarded the atom as an ultimate particles of matter Choose the correct answer from the options given below

1C, D and E only

2A and E only

3B, C and E only

4A, B and C only

NTA Answer: Option 3(final)

NEET 2023

On balancing the given redox reaction, c aCrO2 bSO2(aq)cH(aq) 2aCr3(aq)bSO2(aq) H O(l) 2 7 3 4 2 2 the coefficients a, b and c are found to be, respectively-

13, 8, 1

21, 8, 3

38, 1, 3

41, 3, 8

NTA Answer: Option 4(final)

NEET 2022

Match List-I with List-II List-I List-II (a) Li (i) absorbent for carbon dioxide (b) Na (ii) electrochemical cells (c) KOH (iii) coolant in fast breeder reactors (d) Cs (iv) photoelectric cell Choose the correct answer from the options given below :

1(a) - (ii), (b) – (iii), (c) – (i), (d) – (iv)

2(a) - (iv), (b) – (i), (c) – (iii), (d) – (ii)

3(a) - (iii), (b) – (iv), (c) – (ii), (d) – (i)

4(a) - (i), (b) – (iii), (c) – (iv), (d) – (ii)

NTA Answer: Option 1(final)

NEET 2021

Statement I : Acid strength increases in the order given as HF << HCl << HBr << HI. Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases. In the light of the above statements, choose the correct answer from the options given below.

1Statement I is incorrect but Statement II is true

2Both statement I and Statement II are true

3Both Statement I and Statement II are false

4Statement I : correct but statement II is false

NTA Answer: Option 2(final)

NEET 2021

From the following pairs of ions which one is not an iso-electronic pair? Fe2+, Mn2+

1O2–, F–

2Na+, Mg2+

3Mn2+, Fe3+

4Answer

NTA Answer: Option 1(final)

How NEET usually asks this

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

Bohr energy difference between two levels; compute photon energy/wavelength using Rydberg or E_n formulas.

Multi StepMedium

Common distractors

forgets z squared

Drops Z² for hydrogen-like

Apply Δx·Δp ≥ h/(4π) to find minimum uncertainty given the other.

Direct ApplicationEasy

Common distractors

uses h instead of h over 4pi

Drops 4π factor

Given orbital (e.g. 3p_z) or electron, identify n, l, m_l, m_s. Or check forbidden quantum number combinations.

InterpretationEasy

Common distractors

uses wrong l range

Uses l ≤ n instead of l < n

Sources

NCERT refs: Class 11 Chemistry Chapter 2, p.36

Test yourself on this topic with real past-paper questions:

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