Q.Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
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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? …
Concept: Electron Configuration – filling order follows the Aufbau principle (n+l rule) and the (n-1)d, (n-2)f blocks fill after the ns orbital.
Reasoning steps:
-
Atomic number 61 (Promethium, Pm):
After Xe (54 electrons), the next 7 electrons go into 6s² and then 4f.
Configuration: [Xe]6s24f5 (since 4f fills before 5d).
-
Atomic number 91 (Protactinium, Pa):
After Rn (86 electrons), the next 5 electrons fill 7s² and then 5f², with one electron going into 6d¹ (anomaly due to stability).
Configuration: [Rn]7s25f26d1.
-
Atomic number 101 (Mendelevium, Md):
After Rn (86), the next 15 electrons fill 7s², 5f¹³, and then 6d⁰ (no 6d electron).
Configuration: [Rn]7s25f13.
-
Atomic number 109 (Meitnerium, Mt): …
The key is to follow the Aufbau principle (n+l rule) and account for the special stability of half-filled and fully-filled orbitals. The configurations are: 61: [Xe]4f56s2; 91: [Rn]5f26d17s2; 101: [Rn]5f137s2; 109: [Rn]5f146d77s2.
Why Electron Configuration Works This Way
Electrons fill orbitals in order of increasing energy, not just increasing principal quantum number n. The rule is: an orbital with lower (n+l) fills first; if two have the same (n+l), the one with lower n fills first. This is the Aufbau principle, and it explains why the 4f subshell fills after 6s, and 5f after 7s.
For elements beyond lanthanum (atomic number 57), the 4f orbitals begin to fill. Similarly, beyond actinium (89), the 5f orbitals fill. But there are exceptions — half-filled (f⁷) and fully-filled (f¹⁴) subshells are extra stable, so sometimes an electron from the s-orbital moves into the f-orbital to achieve that stability.
Let’s work through each element.
1. Atomic number 61 — Promethium (Pm)
The nearest noble gas is xenon (Xe, Z=54). That gives us a core of [Xe].
Remaining electrons: 61−54=7 electrons.
The filling order after Xe is: 6s (2 electrons), then 4f (up to 14 electrons), then 5d, then 6p.
So we put 2 electrons into 6s: 6s2.
That leaves 7−2=5 electrons. These go into the 4f subshell: 4f5.
No special stability is reached here (f⁷ would be half-filled, but we only have 5), so no exception occurs.
Configuration: [Xe]4f56s2
For lanthanides (Z=58 to 71), the 4f subshell fills after 6s. The 5d orbital is usually empty or has at most 1 electron in this series — only exceptions are La, Ce, Gd, and Lu.
2. Atomic number 91 — Protactinium (Pa)
Nearest noble gas: radon (Rn, Z=86). Core: [Rn].
Remaining electrons: 91−86=5 electrons.
After Rn, the filling order is: 7s (2), then 5f (14), then 6d (10), then 7p.
First, 2 electrons go into 7s: 7s2.
That leaves 5−2=3 electrons. According to the Aufbau order, the next orbital is 5f. So we would expect 5f3.
But here’s the catch: for protactinium, the 5f and 6d orbitals are very close in energy. Experimentally, the configuration is [Rn]5f26d17s2, not [Rn]5f37s2. Why? Because having one electron in the 6d orbital (which is slightly lower in energy for Pa) is more stable than putting all three into 5f.
A common mistake is to blindly follow the Aufbau order for actinides. The 5f and 6d orbitals are very close in energy, and for elements like Pa, U, Np, and Cm, you get 6d electrons. Always check the actual configuration — don’t assume the simple filling order holds.
Configuration: [Rn]5f26d17s2
3. Atomic number 101 — Mendelevium (Md)
Core: [Rn] (Z=86).
Remaining electrons: 101−86=15 electrons. …
Method: Aufbau Principle with (n + ℓ) Rule
This is the standard method for writing ground-state electron configurations. It uses the order of increasing orbital energy determined by the sum (n+ℓ) — and for equal sums, by lower n first.
Steps
- Identify the atomic number (Z) — this equals the total number of electrons in a neutral atom.
- Follow the Aufbau order (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p...).
- Fill each subshell to its maximum capacity:
- s: 2 electrons
- p: 6 electrons
- d: 10 electrons
- f: 14 electrons
- Stop when the total electrons equal Z.
- Write in order of increasing principal quantum number n (standard notation), not in filling order.
Solutions
Atomic number 61 — Promethium (Pm)
- Z = 61
- Fill: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f⁵
- Final configuration (by n): 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f5 5s2 5p6 6s2
Atomic number 91 — Protactinium (Pa)
- Z = 91
- Fill: ... up to 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f² 6d¹
- Final configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 5f2 6s2 6p6 6d1 7s2
Note: Pa is an anomaly — the 5f and 6d are very close in energy. The above is the accepted ground state.
Atomic number 101 — Mendelevium (Md)
- Z = 101
- Fill: ... up to 7s² 5f¹³
- Final configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 5f13 6s2 6p6 7s2
Atomic number 109 — Meitnerium (Mt)
- Z = 109
- Fill: ... up to 7s² 5f¹⁴ 6d⁷
- Final configuration: …
🧠 The Core Idea First
Electronic configuration follows the Aufbau principle (fill lowest energy orbitals first), Hund’s rule (maximize unpaired spins), and the Pauli exclusion principle. But for elements beyond atomic number 57 (La), the energy ordering of orbitals changes due to nuclear charge and shielding effects.
The correct filling order is:
1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
✗ Common Mistake #1: Forgetting the f-block (lanthanides & actinides)
The error:
Students write configurations for atomic numbers 61, 91, 101, 109 as if they are normal d-block elements, skipping the f-subshell entirely.
Example of wrong answer for Z=61:
[Xe]6s24f1 ✗ (a 57-electron count — and not even real La, whose actual configuration is [Xe]5d16s2)
Why it happens:
They memorise the order but forget that after La (Z=57), the 4f subshell starts filling — not 5d.
How to avoid:
- Remember: Lanthanides (Z=58 to 71) fill the 4f subshell.
- Actinides (Z=90 to 103) fill the 5f subshell.
- Use the n + ℓ rule to confirm: 4f has n+ℓ = 4+3 = 7, 5d has 5+2 = 7, but 4f is lower in energy because of lower n.
Correct for Z=61 (Promethium, Pm):
[Xe]6s24f5 ✓
✗ Common Mistake #2: Misplacing the 5f and 6d orbitals for Z=91 and 101
The error:
For Z=91 (Protactinium), students write [Rn]7s25f3 ✗ — but the actual configuration has a 5f² 6d¹ arrangement.
Why it happens:
They assume the 5f subshell fills strictly after 7s, ignoring that 6d can be slightly lower in energy for early actinides.
How to avoid:
- For early actinides (Th, Pa, U, Np), the 6d orbital may get one electron before 5f fills completely.
- Memorise the exceptions for Pa (Z=91): [Rn]7s25f26d1 ✓
- For later actinides (Am onwards), 5f fills normally.
Correct for Z=91 (Protactinium):
[Rn]7s25f26d1 ✓
Correct for Z=101 (Mendelevium, Md):
[Rn]7s25f13 ✓ (no 6d electron here)
✗ Common Mistake #3: Forgetting the d-block exception for Z=109
The error:
For Z=109 (Meitnerium, Mt), students write [Rn]7s25f146d7 — the same electrons, but listed in filling order. This is a presentation slip rather than wrong chemistry: the convention is to list subshells in order of increasing n, so present it as [Rn]5f146d77s2.
Why it happens:
They write orbitals in filling order (7s before 6d) but forget that in the periodic table, we write by increasing n (principal quantum number), not filling order.
How to avoid: …
Showing the 12 most recent of 34 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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