Q.The correct electron configuration of Cu2+ is –
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🔒 Start your 14-day free trial to unlock the full solution →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
Electron configuration is the distribution of electrons of an atom or molecule in atomic orbitals. It is written as a sequence of:
- Principal quantum number n (the energy level: 1, 2, 3...)
- Sublevel letter (s, p, d, f) — tells you the shape of the orbital
- Superscript — the number of electrons in that sublevel
For example, the configuration of carbon (6 electrons) is:
1s22s22p2
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→2s→2p→3s→3p→4s→3d→4p→5s→4d→5p→6s→4f→5d→6p→7s→5f→6d→7p
Notice: 4s fills before 3d. This is because the 4s orbital is actually lower in energy than 3d. This trips up many students.
Use the diagonal rule (Madelung's rule) to remember the order: draw arrows diagonally across the (n+ℓ) 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:
- 1s2 (2 used, 24 left)
- 2s2 (4 used, 22 left)
- 2p6 (10 used, 16 left)
- 3s2 (12 used, 14 left)
- 3p6 (18 used, 8 left)
- 4s2 (20 used, 6 left)
- 3d6 (26 used, 0 left)
Step 3: Write it in order of increasing n (standard notation):
1s22s22p63s23p63d64s2
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+ℓ) rule:
- n = principal quantum number (shell)
- ℓ = azimuthal quantum number (subshell: s=0, p=1, d=2, f=3)
Orbitals fill in order of increasing (n+ℓ). If two have the same (n+ℓ), 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)
- ℓ (subshell shape)
- mℓ (orbital orientation)
- ms (spin: +21 or −21)
Since only ms can differ for electrons in the same orbital, maximum 2 electrons per orbital — one spin-up (↑) and one spin-down (↓).
Key result: The s subshell (ℓ=0, one orbital) holds 2 electrons. The p subshell (ℓ=1, three orbitals) holds 6 electrons. The d subshell (ℓ=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 (1s22s22p2):
- Correct: ↑ | ↑ | (two unpaired, parallel)
- Wrong: ↑↓ | | (paired in one orbital — higher repulsion)
The Big Picture: Why These Three Rules Together? …
Neutral copper is an exception, with a filled 3d subshell and a single 4s electron, and forming an ion removes the 4s electron before any d-electron. Taking away two electrons from copper therefor …
Neutral Cu is an exception, [Ar]3d10 4s1; removing 2 electrons to form Cu2+ removes the single 4s electron first, then one 3d electron, giving [Ar]3d9 4s0.
Copper (Z=29) has the anomalous ground-state configuration [Ar]3d10 4s1 (a full 3d subshell is more stable than a half-filled one with an extra 4s electron, so one 4s electron is "borrowed" to complete the 3d subshell).
When Cu loses electrons to form Cu2+, the RULE for transition metal ions is that (n-1)d and ns electrons - once the ion is formed - are removed from the ns orbital first (4s empties before 3d, because for ions the (n-1)d orbital lies below the ns orbital in energy).
…
Showing the 12 most recent of 35 on this concept.
- CBSE 2026Set 56/2/11 markMCQQ.Electronic configuration of chromium is : (A) [Ar]3d44s1 (B) [Ar]3d44s2 (C) [Ar]3d54s1 (D) [Ar]3d54s2
›Reveal solutionSolution
Chromium has an anomalous electronic configuration due to the extra stability of a half-filled d-subshell. The correct configuration is [Ar]3d54s1, 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]3d44s2 — 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 3d5. Chromium can achieve this by promoting one electron from the 4s subshell to the 3d subshell.
Watch outA 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 3d5 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]4s23d4
This matches option (B).
3. Check for stability-driven exceptions
For chromium, the 3d subshell is just one electron short of half-filled (3d5). By promoting one electron from the 4s to the 3d, we get:
[Ar]4s13d5
This gives:
- A half-filled 3d subshell (3d5) — extra stable …
- CBSE 2026Set A1 markMCQQ.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
…
- CBSE 2026Set ANNUAL1 markMCQQ.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) …
- CBSE 2026Set ANNUAL1 markQ.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 …
- CBSE 2026Set ANNUAL1 markMCQQ.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.
…
- CBSE 2025Set D1 markMCQQ.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:
…
- CBSE 2025Set D1 markMCQQ.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 …
- CBSE 2025Set ANNUAL1 markQ.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:
Cr:[Ar]3d54s1
(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: …
- CBSE 2024Set B1 markQ.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 …
- CBSE 2024Set ANNUAL1 markMCQQ.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.
…
- CBSE 2024Set ANNUAL1 markMCQQ.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 …
- CBSE 2024Set ANNUAL1 markMCQQ.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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