Electronic configuration of chromium is : (A) $[Ar]\,3d^4\,4s^1$ (B) $[Ar]\,3d^4\,4s^2$ (C) $[Ar]\,3d^5\,4s^1$ (D) $[Ar]\,3d^5\,4s^2$

> [!TLDR] > Chromium has an anomalous electronic configuration due to the extra stability of a half-filled d-subshell. The correct configuration is $[Ar]\,3d^5\,4s^1$, which corresponds to option **(C)**. The question tests a classic exception in electronic configuration — one that every student must know for exams like JEE, NEET, or board exams. The key is understanding **why** chromium doesn't follow the expected pattern. ## The Concept: Stability of Half-Filled and Fully-Filled Subshells When we write electronic configurations using the Aufbau principle, electrons fill orbitals in order of increasing energy: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, ...$ For chromium (atomic number 24), the expected configuration would be $[Ar]\,3d^4\,4s^2$ — filling the 4s before the 3d, then putting 4 electrons in the 3d subshell. But nature prefers stability. A **half-filled** subshell (each orbital has one electron, all parallel spins) has extra stability due to: - Symmetrical distribution of electron density - Maximum exchange energy (Hund's rule of maximum multiplicity) - Reduced electron-electron repulsion For the 3d subshell (which has 5 orbitals), a half-filled configuration means $3d^5$. Chromium can achieve this by promoting one electron from the 4s subshell to the 3d subshell. > [!WARNING] > A common mistake is to think that the 4s subshell is always filled before the 3d. While 4s is lower in energy than 3d for neutral atoms, once electrons start filling the 3d, the energy levels shift. For chromium, the stability gain from $3d^5$ outweighs the small energy cost of moving an electron from 4s. ## Step-by-Step Solution **1. Determine the atomic number and base configuration** Chromium has atomic number 24. The noble gas core is argon ($[Ar]$), which accounts for 18 electrons. So we need to place the remaining $24 - 18 = 6$ electrons. **2. Apply the Aufbau principle (expected order)** The order of filling after argon is: $4s$ then $3d$. So the expected configuration w

Electronic configuration of chromium is : (A) [Ar]3d⁴4s¹ (B) [Ar]3d⁴4s² (C) [Ar]3d⁵4s¹ (D) [Ar]3d⁵4s²

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[Ar]\,3d^4\,4s^1

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

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Concept understanding — Electron Configuration

Electron Configuration: Where Do Electrons Actually Live?

Imagine a school building. Students don't just wander randomly — they sit in specific classrooms, on specific floors, in specific rows. Electrons in an atom behave similarly. They don't buzz around the nucleus chaotically. They occupy specific energy levels (floors), sublevels (classrooms), and orbitals (seats).

The electron configuration is simply the address system that tells you exactly which "seats" every electron in an atom is sitting in.

The Intuition: Why Can't Electrons Sit Anywhere?

Two big rules force electrons into this orderly arrangement:

Energy matters. Electrons want to be as close to the nucleus as possible (lowest energy). The first "floor" (n=1) is the most comfortable. Higher floors cost more energy.
No crowding. A famous rule called the Pauli Exclusion Principle says: no two electrons in the same atom can have the exact same set of four quantum numbers. In plain language: each orbital (seat) can hold at most two electrons, and they must spin in opposite directions.

So electrons fill up from the bottom floor upward, like students filling a theatre from the front row back.

The Precise Statement

n

For example, the configuration of carbon (6 electrons) is:

1s^2 \; 2s^2 \; 2p^2

This reads: "Two electrons in the 1s orbital, two in the 2s orbital, and two in the 2p orbitals."

The Filling Order: The Aufbau Principle

Electrons don't fill levels in simple numerical order. Here's the actual sequence (memorise this — it's exam gold):

1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p \rightarrow 5s \rightarrow 4d \rightarrow 5p \rightarrow 6s \rightarrow 4f \rightarrow 5d \rightarrow 6p \rightarrow 7s \rightarrow 5f \rightarrow 6d \rightarrow 7p

Notice: 4s fills before 3d. This is because the 4s orbital is actually lower in energy than 3d. This trips up many students.

Tip

Use the diagonal rule (Madelung's rule) to remember the order: draw arrows diagonally across the $(n+\ell)$ chart. Or just remember the mnemonic: "Silly People Don't Fail" for the sublevel order within each shell.

How to Write Any Configuration (Step-by-Step)

Let's do iron (Fe, atomic number 26).

Step 1: Know the total electrons = 26.

Step 2: Follow the filling order, counting electrons as you go:

$1s^2$ (2 used, 24 left)
$2s^2$ (4 used, 22 left)
$2p^6$ (10 used, 16 left)
$3s^2$ (12 used, 14 left)
$3p^6$ (18 used, 8 left)
$4s^2$ (20 used, 6 left)
$3d^6$ (26 used, 0 left)

Step 3: Write it in order of increasing $n$ (standard notation):

1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 3d^6 \; 4s^2

Watch out

Many textbooks write configurations in order of filling (4s before 3d), but IUPAC standard lists them by principal quantum number $n$ (3d before 4s). Check which convention your exam uses. For CBSE/ICSE, write in order of increasing $n$ : $1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p...$

The Three Golden Rules (Memorise These)

Rule	What it says	Why it matters
Aufbau Principle	Electrons fill lowest energy orbitals first	Determines the order of filling

Why this formula?

Electron Configuration: Why the Rules Work

Let's build this from the ground up — not just what the rules are, but why they exist.

The Core Question

Why do electrons arrange themselves in specific shells, subshells, and orbitals — and not just pile up anywhere?

The answer lies in three fundamental principles, each rooted in physics and quantum mechanics.

1. The Aufbau Principle: Why "Build Up" in Order?

What it says: Electrons fill orbitals from lowest to highest energy.

Why it holds: Nature seeks the lowest possible energy state (ground state). An atom is most stable when its electrons occupy the lowest available energy levels.

Think of it like water flowing downhill — electrons "fall" into the lowest energy orbitals first.

The energy ordering (for multi-electron atoms) is:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s...

Why this order? It comes from the $(n + \ell)$ rule:

$n$ = principal quantum number (shell)
$\ell$ = azimuthal quantum number (subshell: $s=0$ , $p=1$ , $d=2$ , $f=3$ )

Orbitals fill in order of increasing $(n + \ell)$ . If two have the same $(n + \ell)$ , the one with lower $n$ fills first.

Example: $4s$ has $(4+0)=4$ , $3d$ has $(3+2)=5$ . So $4s$ fills before $3d$ — even though $4s$ is a higher shell number.

2. Pauli Exclusion Principle: Why Only Two Per Orbital?

What it says: No two electrons in an atom can have the same set of all four quantum numbers.

Why it holds: This is a fundamental law of quantum mechanics — electrons are fermions (spin-1/2 particles). Fermions obey the Pauli exclusion principle, which arises from the antisymmetry of the wavefunction.

The four quantum numbers:

$n$ (shell)
$\ell$ (subshell shape)
$m_\ell$ (orbital orientation)
$m_s$ (spin: $+\frac{1}{2}$ or $-\frac{1}{2}$ )

Since only $m_s$ can differ for electrons in the same orbital, maximum 2 electrons per orbital — one spin-up ( $\uparrow$ ) and one spin-down ( $\downarrow$ ).

Key result: The $s$ subshell ( $\ell=0$ , one orbital) holds 2 electrons. The $p$ subshell ( $\ell=1$ , three orbitals) holds 6 electrons. The $d$ subshell ( $\ell=2$ , five orbitals) holds 10 electrons.

3. Hund's Rule: Why Spread Out First?

What it says: Within a subshell, electrons occupy empty orbitals singly before pairing up — and all unpaired electrons have parallel spins.

Why it holds: Electrons repel each other (Coulomb repulsion). By occupying different orbitals, they stay farther apart, reducing repulsion energy.

The spin alignment (all parallel) comes from exchange energy — a quantum mechanical effect where parallel spins have a slightly lower energy state due to wavefunction symmetry.

Example for carbon ( $1s^2 2s^2 2p^2$ ):

Correct: $\uparrow$ | $\uparrow$ | (two unpaired, parallel)
Wrong: $\uparrow\downarrow$ | | (paired in one orbital — higher repulsion)

The Big Picture: Why These Three Rules Together? …

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Chromium has an anomalous electronic configuration due to the extra stability of a half-filled d-subshell. The correct configuration is $[Ar]\,3d^5\,4s^1$ , which corresponds to option (C).

The question tests a classic exception in electronic configuration — one that every student must know for exams like JEE, NEET, or board exams. The key is understanding why chromium doesn't follow the expected pattern.

The Concept: Stability of Half-Filled and Fully-Filled Subshells

When we write electronic configurations using the Aufbau principle, electrons fill orbitals in order of increasing energy: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, ...$

For chromium (atomic number 24), the expected configuration would be $[Ar]\,3d^4\,4s^2$ — filling the 4s before the 3d, then putting 4 electrons in the 3d subshell.

But nature prefers stability. A half-filled subshell (each orbital has one electron, all parallel spins) has extra stability due to:

Symmetrical distribution of electron density
Maximum exchange energy (Hund's rule of maximum multiplicity)
Reduced electron-electron repulsion

For the 3d subshell (which has 5 orbitals), a half-filled configuration means $3d^5$ . Chromium can achieve this by promoting one electron from the 4s subshell to the 3d subshell.

Watch out

A common mistake is to think that the 4s subshell is always filled before the 3d. While 4s is lower in energy than 3d for neutral atoms, once electrons start filling the 3d, the energy levels shift. For chromium, the stability gain from $3d^5$ outweighs the small energy cost of moving an electron from 4s.

Step-by-Step Solution

1. Determine the atomic number and base configuration

Chromium has atomic number 24. The noble gas core is argon ( $[Ar]$ ), which accounts for 18 electrons. So we need to place the remaining $24 - 18 = 6$ electrons.

2. Apply the Aufbau principle (expected order)

The order of filling after argon is: $4s$ then $3d$ . So the expected configuration would be:

[Ar]\,4s^2\,3d^4

This matches option (B).

3. Check for stability-driven exceptions

For chromium, the $3d$ subshell is just one electron short of half-filled ( $3d^5$ ). By promoting one electron from the $4s$ to the $3d$ , we get:

[Ar]\,4s^1\,3d^5

This gives:

A half-filled $3d$ subshell ( $3d^5$ ) — extra stable …

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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 35 on this concept.

CBSE 2026Set 56/2/11 markMCQ
$[Ar]\,3d^4\,4s^1$
›Reveal solutionSolution
Chromium has an anomalous electronic configuration due to the extra stability of a half-filled d-subshell. The correct configuration is $[Ar]\,3d^5\,4s^1$ , which corresponds to option (C).

The question tests a classic exception in electronic configuration — one that every student must know for exams like JEE, NEET, or board exams. The key is understanding why chromium doesn't follow the expected pattern.

The Concept: Stability of Half-Filled and Fully-Filled Subshells

When we write electronic configurations using the Aufbau principle, electrons fill orbitals in order of increasing energy: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, ...$

For chromium (atomic number 24), the expected configuration would be $[Ar]\,3d^4\,4s^2$ — filling the 4s before the 3d, then putting 4 electrons in the 3d subshell.

But nature prefers stability. A half-filled subshell (each orbital has one electron, all parallel spins) has extra stability due to:

Symmetrical distribution of electron density

Maximum exchange energy (Hund's rule of maximum multiplicity)

Reduced electron-electron repulsion

For the 3d subshell (which has 5 orbitals), a half-filled configuration means $3d^5$ . Chromium can achieve this by promoting one electron from the 4s subshell to the 3d subshell.

Watch out

A common mistake is to think that the 4s subshell is always filled before the 3d. While 4s is lower in energy than 3d for neutral atoms, once electrons start filling the 3d, the energy levels shift. For chromium, the stability gain from $3d^5$ outweighs the small energy cost of moving an electron from 4s.

Step-by-Step Solution

1. Determine the atomic number and base configuration

Chromium has atomic number 24. The noble gas core is argon ( $[Ar]$ ), which accounts for 18 electrons. So we need to place the remaining $24 - 18 = 6$ electrons.

2. Apply the Aufbau principle (expected order)

The order of filling after argon is: $4s$ then $3d$ . So the expected configuration would be:
$[Ar]\,4s^2\,3d^4$
This matches option (B).

3. Check for stability-driven exceptions

For chromium, the $3d$ subshell is just one electron short of half-filled ( $3d^5$ ). By promoting one electron from the $4s$ to the $3d$ , we get:
$[Ar]\,4s^1\,3d^5$
This gives:

A half-filled $3d$ subshell ( $3d^5$ ) — extra stable …
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CBSE 2026Set A1 markMCQ
$Q. If the electronic configuration of a transition element in its +3 oxidation state is [Ar]3d^4, then its atomic number would be$
(a) 25
(b) 26
(c) 22
(d) 19
›Reveal solutionSolution

[Ar]3d4 for the M3+ ion means 22 electrons in the ion; the neutral atom therefore has 25 electrons -> atomic number 25 (Mn).

Count the electrons in the +3 ion. [Ar] contributes 18 electrons and 3d^4 adds 4 more:

electrons in M3+ = 18 + 4 = 22

Since the ion carries a +3 charge, the neutral atom has 3 more electrons:

Z = 22 + 3 = 25

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Anomalous electronic configurations in 3d-series are of:$
(a) Cr and Fe
(b) Cr and Cu
(c) Cu and Zn
(d) Fe and Cu
›Reveal solutionSolution
Cr ([Ar]3d^5 4s^1) and Cu ([Ar]3d^10 4s^1) show anomalous configurations because half-filled and fully-filled d-orbitals are extra stable.

By the normal Aufbau filling order, chromium (Z=24) would be expected to have the configuration [Ar]3d^4 4s^2, and copper (Z=29) would be expected to have [Ar]3d^9 4s^2. Experimentally, however, one electron from the 4s orbital shifts into the 3d subshell:

Cr: [Ar]3d^5 4s^1 (giving a half-filled d^5 configuration)

Cu: [Ar]3d^10 4s^1 (giving a completely filled d^10 configuration) …
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CBSE 2026Set ANNUAL1 mark
$Q. Write the general electronic configuration of d-block elements.$
›Reveal solutionSolution

d-block elements have the general outer configuration (n−1)d¹⁻¹⁰ ns⁰⁻².

The d-block (transition) elements are those in which the last electron enters the d-orbital of the penultimate (n−1) shell, while the outermost ns orbital is normally filled first. Across a d-block series, the (n−1)d orbitals are progressively filled from d¹ to d¹⁰ while the ns orbital holds 0, 1 or 2 electrons (0 or 1 in a few excepti …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Electronic configuration of a transition element X in +3 oxidation state is [Ar] 3d^5. What is its atomic number ?$
(a) 25
(b) 26
(c) 27
(d) 24
›Reveal solutionSolution

If X³⁺ has configuration [Ar]3d⁵, the neutral atom X has configuration [Ar]3d⁶4s², which is iron (Z = 26).

For transition metals, when forming a cation the electrons are removed first from the 4s subshell and then from 3d (the reverse order of filling).

Given: X³⁺ = [Ar]3d⁵ (i.e., 3 electrons have been removed relative to the neutral atom).

To get the neutral atom X, add back 3 electrons in the correct order — first fill 4s (2 electrons) then 3d (1 electron):

X (neutral) = [Ar]3d⁶4s²

Counting electrons: Ar core = 18, plus 3d⁶ (6) plus 4s² (2) = 18 + 6 + 2 = 26 electrons ⇒ atomic number = 26.

…

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CBSE 2025Set D1 markMCQ
$Q. The electronic configuration of copper (Z = 29) is$
(a) [Ar] 3d9 4s2
(b) [Ar] 3d10 4s1
(c) [Ar] 3d8 4s2
(d) [Ar] 3d10 4s2
›Reveal solutionSolution

Cu (Z = 29) is [Ar] 3d10 4s1 because a completely filled 3d subshell gives extra stability.

The expected configuration would be [Ar] 3d9 4s2, but a fully filled d subshell (d10) is more stable than d9. So one 4s electron shifts into 3d, giving the actual ground-state configuration:

…

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CBSE 2025Set D1 markMCQ
$Q. Which of the following is the electronic configuration of Lanthanide elements?$
(a) (n-2)f^1-14 (n-1)s2 p6 d^0,1 ns2
(b) (n-2)f^0-14 (n-1)d10 ns2
(c) (n-2)f^0-14 (n-1)d^0,1 ns2
(d) (n-2)d^0,1 (n-1)f^0-14 ns1
›Reveal solutionSolution

Lanthanides fill the (n-2)f subshell: general form (n-2)f^1-14 (n-1)d^0,1 ns2.

In the lanthanoids the electrons progressively enter the antepenultimate (third from outermost) 4f subshell, i.e. the (n-2)f level with n = 6. The (n-1) = 5 shell is completed as s2 p6 with d either 0 or 1 electron, and the outermost ns2 = 6s2. The general valence configuration is ther …

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CBSE 2025Set ANNUAL1 mark
$Q. Write the electronic configuration of Cr3+ (Z = 24).$
›Reveal solutionSolution

Chromium's ground-state configuration is an exception ([Ar] 3d5 4s1); to form Cr3+, the 4s electron and two 3d electrons are removed, leaving [Ar] 3d3.

Neutral chromium (Z = 24) has the anomalous ground-state configuration:

$\text{Cr}: [\text{Ar}]\,3d^5\,4s^1$

(This exception occurs because a half-filled 3d subshell, combined with a filled 4s, gives extra stability.)

To form Cr3+, three electrons are removed. Electrons are always removed first from the outermost occupied subshell (4s), then from 3d: …

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CBSE 2024Set B1 mark
$Q. Answer in one word/sentence: Write the general electronic configuration of f-block elements.$
›Reveal solutionSolution

f-block elements fill electrons progressively into the (n-2)f subshell, with the outer ns^2 (and sometimes one (n-1)d electron) — general configuration (n-2)f^(1-14)(n-1)d^(0-1)ns^2.

The f-block (inner transition elements: lanthanoids and actinoids) is characterised by the filling of the antepenultimate (n-2)f orbitals, while the outermost ns^2 subshell remains filled, and the (n-1)d subshell may hold 0 or 1 electron depending on the element (due to the very close energies of (n-2)f and (n-1)d orbitals). The ge …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The elements with Anomalous electronic configuration in the 3d series are of$
(a) Cr and Fe
(b) Cu and Zn
(c) Fe and Cu
(d) Cr and Cu
›Reveal solutionSolution

Cr and Cu deviate from the expected (n-1)d^x ns^2 filling pattern because a half-filled or fully-filled d subshell is extra stable, so one 4s electron shifts into the 3d subshell.

Expected filling (Aufbau) for Cr would be [Ar]3d4 4s2, but the ACTUAL configuration is [Ar]3d5 4s1 - the half-filled 3d5 subshell (symmetric electron distribution, exchange energy stabilisation) is more stable.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following has equal number of Electrons with Chromium?$
(a) Fe2+
(b) Fe3+
(c) Cu+
(d) Zn2+
›Reveal solutionSolution
Chromium (Z=24) has 24 electrons as a neutral atom; checking each ion's electron count against 24 picks out Fe²⁺.

Chromium, atomic number 24, has 24 electrons in its neutral (ground) state ([Ar] 3d⁵ 4s¹). Now count electrons in each ion (electrons = atomic number − ionic charge):

Fe (Z=26): Fe²⁺ = 26 − 2 = 24 electrons ✓ matches Cr

Fe³⁺ = 26 − 3 = 23 electrons …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The number of d-electrons retained in Fe2+ (At. No. of Fe = 26) is$
(a) 6
(b) 3
(c) 4
(d) 5
›Reveal solutionSolution

Transition metal cations lose electrons from the ns sub-shell before the (n-1)d sub-shell, so Fe2+ keeps all 6 of its d-electrons.

Neutral Fe (Z = 26) electronic configuration: [Ar] 3d6 4s2.

When a transition metal atom ionises, electrons are removed from the outermost ns orbital first (it is higher in energy than (n-1)d once the atom is formed), not from the d sub-shell. …

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Q.Electronic configuration of chromium is : (A) [Ar] 3d4 4s1[Ar]\,3d^4\,4s^1[Ar]3d44s1 (B) [Ar] 3d4 4s2[Ar]\,3d^4\,4s^2[Ar]3d44s2 (C) [Ar] 3d5 4s1[Ar]\,3d^5\,4s^1[Ar]3d54s1 (D) [Ar] 3d5 4s2[Ar]\,3d^5\,4s^2[Ar]3d54s2

The Concept: Stability of Half-Filled and Fully-Filled Subshells

Step-by-Step Solution

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Q.Electronic configuration of chromium is : (A) $[Ar]\,3d^4\,4s^1$ (B) $[Ar]\,3d^4\,4s^2$ (C) $[Ar]\,3d^5\,4s^1$ (D) $[Ar]\,3d^5\,4s^2$

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