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 36 on this concept.
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.Consider the elements with atomic number (Z) from 11 to 18. The ratio of number of s-electrons to p-electrons in element X is 2:3 and in element Y is 3:5. What are X and Y? (A) Mg,Al (B) Si,P (C) P,S (D) S,Cl
›Reveal solutionSolution
This tests electron configuration bookkeeping across period 3 — count s- and p-electrons for each element and match the given ratios.
Concept and Intuition
Every element from Na (Z=11) to Ar (Z=18) fills the 3s and 3p subshells after the fixed 1s22s22p6 core. The core always contributes 2 s-electrons (1s2) + 2 s-electrons (2s2) = 4 s-electrons and 6 p-electrons (2p6), fixed for all of them. On top of that, each element adds its own 3s and 3p electrons.
Step-by-Step Solution
- Write full configurations for Z = 11 to 18 and tally total s vs p electrons:
- Na (11): 1s22s22p63s1 → s = 2+2+1 = 5, p = 6 → ratio 5:6
- Mg (12): 3s2 → s = 6, p = 6 → ratio 1:1
- Al (13): 3s23p1 → s = 6, p = 7 → ratio 6:7
- Si (14): 3s23p2 → s = 6, p = 8 → ratio 3:4
- P (15): 3s23p3 → s = 6, p = 9 → ratio 2:3
- S (16): 3s23p4 → s = 6, p = 10 → ratio 3:5 …
- Write full configurations for Z = 11 to 18 and tally total s vs p electrons:
- AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQQ.The ground state electronic configuration of Gadolinium (Z = 64) is (A) [Xe]4f75d16s2 (B) [Xe]4f85d06s2 (C) [Xe]4f95d06s1 (D) [Xe]4f75d26s1
›Reveal solutionSolution
Gadolinium is a well-known lanthanide exception: it configures as [Xe]4f75d16s2, not [Xe]4f85d06s2, because a half-filled 4f subshell is extra stable.
Concept and Intuition
Across the lanthanide series, electrons are added to the 4f subshell while 5d and 6s stay essentially fixed. But nature strongly favours half-filled (f7) and fully-filled (f14) subshells because of the extra exchange stabilisation energy among electrons with parallel spins in degenerate orbitals. When adding the electron that would make 4f8, it is instead energetically cheaper to keep 4f at the stable half-filled 4f7 and place the extra electron in 5d. This is exactly analogous to why Cr is [Ar]3d54s1 instead of 3d44s2, and Cu is [Ar]3d104s1 instead of 3d94s2.
Step-by-Step Solution
- Gadolinium, Z = 64, is in the lanthanide series (element after Xe core, filling 4f).
- The "expected" regular filling (following Aufbau strictly across lanthanides) would be [Xe]4f85d06s2 (since 64 − 54 (Xe) − 2 (6s^2) = 8 electrons distributed to 4f).
- However, 4f7 (half-filled) is a specially stable configuration (maximum number of unpaired parallel-spin electrons, maximum exchange energy). …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.The ion with 4f7 configuration is (A) Pr3+ (B) Lu3+ (C) Eu2+ (D) Ce4+
›Reveal solutionSolution
Tests recall of lanthanide electronic configurations and which ion retains a 4f7 (half-filled, extra-stable) shell. Answer: Eu2+.
Concept and Intuition
Lanthanides fill the 4f subshell after [Xe]6s2. A half-filled (f7) or fully-filled (f14) 4f subshell is extra stable (exchange energy is maximised), so ions that can attain these configurations by losing only their outer 6s electrons (instead of also digging into 4f) are especially favoured. Eu is the classic case: its +2 oxidation state is unusually accessible (compared to the "normal" lanthanide +3 state) precisely because Eu2+ retains the stable half-filled 4f7 core.
Step-by-Step Solution
- Write ground-state atomic configurations (using the [Xe] core, Z=54):
- Pr (Z=59): [Xe]4f36s2
- Eu (Z=63): [Xe]4f76s2
- Lu (Z=71): [Xe]4f145d16s2
- Ce (Z=58): [Xe]4f15d16s2
- Remove electrons to form the ions in the options, starting with the outermost (6s, then 5d, then 4f if needed):
- Pr3+: remove 6s2 and one 4f electron ⇒[Xe]4f2
- Eu2+: remove just 6s2⇒[Xe]4f7 …
- Write ground-state atomic configurations (using the [Xe] core, Z=54):
- AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQQ.In Am3+ (Z=95), the number of f-electrons with (n+l) value equal to 8 is (where n, l represent principal and azimuthal quantum numbers) (A) 7 (B) 6 (C) 5 (D) 4
›Reveal solutionSolution
(n+l)=8 singles out the 5f subshell; since Am³⁺ has the configuration [Rn]5f6, the number of such f-electrons is 6.
Concept and Intuition
The quantum number rule n+l pins down a specific subshell for a given l. For f-electrons, l=3, so n+l=8 forces n=5 — i.e. these are exactly the 5f electrons. So the question reduces to: how many 5f electrons does Am3+ have?
Step-by-Step Solution
- Americium (Z=95) ground-state configuration: [Rn]5f77s2 (half-filled 5f, like the lanthanide analogue Eu, [Xe]4f76s2).
- Forming Am3+ requires removing 3 electrons. For actinide/lanthanide ions, the outer ns electrons are removed first, then (if needed) one from the f subshell: remove both 7s2 electrons, then one 5f electron.
- Resulting configuration: Am3+=[Rn]5f6. …
- AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQQ.The number of d-electrons in Fe2+(Z=26) is not equal to (A) the number of p-electrons in Ne (Z = 10) (B) the number of s-electrons in Mg (Z = 12) (C) the number of d-electrons in Fe (Z = 26) (D) the number of p-electrons in Cl (Z = 17)
›Reveal solutionSolution
Fe²⁺ has 6 d-electrons; comparing this against each option's electron count shows only Cl's p-electron count (11) fails to match.
Concept and Intuition
When a transition metal ion forms, electrons are removed from the outermost (4s) orbital first, not from the (n-1)d orbital, even though 4s fills before 3d in the neutral atom. This is a classic conceptual trap in electron configuration problems.
Step-by-Step Solution
- Fe (Z=26): [Ar]3d64s2. Removing 2 electrons to form Fe²⁺ removes both 4s electrons first: Fe²⁺ = [Ar]3d6 → 6 d-electrons.
- (A) Ne (Z=10): 1s22s22p6 → p-electrons = 6. Equal.
- (B) Mg (Z=12): 1s22s22p63s2 → s-electrons = 2+2+2=6. Equal.
- (C) Fe neutral (Z=26): 3d6 → d-electrons = 6. Equal. …
- AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQQ.The number of electrons with magnetic quantum number, ml=0 in the elements with atomic numbers Z = 24 and Z = 29 are respectively (A) 12, 13 (B) 12, 12 (C) 13, 12 (D) 14, 15
›Reveal solutionSolution
Counting electrons occupying the ml=0 orbital in each subshell of Cr (Z=24) and Cu (Z=29) — using their actual (exception) configurations — gives 12 and 13 respectively.
Concept and Intuition
Each subshell has (2l+1) orbitals with ml values from −l to +l; exactly one of these has ml=0 (the 's'-orbital itself for l=0, or the middle orbital for p/d). To find how many electrons in an atom have ml=0, go subshell by subshell: for a completely filled subshell, the ml=0 orbital is full (2 electrons); for a partially filled subshell, use Hund's rule (electrons singly occupy every orbital, including ml=0, before any pairing).
Step-by-Step Solution
- Chromium (Z=24) has the well-known exception configuration 1s22s22p63s23p63d54s1 (half-filled 3d and 4s for extra stability).
- Go through each subshell's ml=0 orbital: 1s (2), 2s (2), 2p full so its ml=0 orbital has 2, 3s (2), 3p full so 2, 3d5 is exactly half-filled — by Hund's rule each of the 5 d-orbitals (including ml=0) holds exactly 1 electron, so 1; 4s1 contributes 1.
- Total for Cr: 2+2+2+2+2+1+1=12. …
- AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQQ.In an element with atomic number (Z) 25, the number of electrons with (n+l) value equal to 3 and 4 are x and y respectively. The value of (x+y) is (A) 21 (B) 12 (C) 14 (D) 16
›Reveal solutionSolution
Writing out Mn's electron configuration and grouping subshells by their (n+l) value shows 2p+3s give x=8 electrons at (n+l)=3, and 3p+4s give y=8 electrons at (n+l)=4, so x+y=16.
Concept and Intuition
The Aufbau principle fills orbitals in order of increasing (n+l), and for equal (n+l), in order of increasing n (this is exactly why 4s fills before 3d: 4s has n+l=4+0=4 while 3d has n+l=3+2=5). To find "how many electrons have a given (n+l) value", we just need the atom's full electron configuration and then tag each occupied subshell with its (n+l).
Step-by-Step Solution
- Z=25 is manganese; ground-state configuration: 1s22s22p63s23p63d54s2 (total 2+2+6+2+6+5+2=25 electrons, confirming the configuration is right).
- Tag each subshell with (n+l): 1s→1, 2s→2, 2p→3, 3s→3, 3p→4, 3d→5, 4s→4.
- Subshells with (n+l)=3: 2p (6 electrons) and 3s (2 electrons) ⇒x=6+2=8. …
- AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQQ.Correct set of four quantum numbers for the valence electron of strontium (Z = 38) is (A) 5,0,0,+21 (B) 5,1,0,+21 (C) 5,1,1,+21 (D) 6,0,0,+21
›Reveal solutionSolution
Strontium's valence electrons occupy the 5s orbital, giving quantum numbers n=5,l=0,ml=0,ms=+21.
Concept and Intuition
Strontium (Z=38) sits in Group 2, Period 5 — its electron configuration is [Kr]5s2, so its valence electrons are in a filled 5s subshell (an s-orbital always has l=0, ml=0).
Step-by-Step Solution
- Electron configuration of Sr: 1s22s22p63s23p63d104s24p65s2=[Kr]5s2. …
- AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQQ.Identify the pair of elements in which number of electrons in (n−1) shell is same (A) Fe, Mn (B) Zn, Fe (C) K, Sc (D) Mn, Cr
›Reveal solutionSolution
Counting electrons in the second-outermost (n−1) shell for each candidate pair shows Mn and Cr
both have 13 electrons there (both being 3d5 configurations), making them the matching pair.
Concept and Intuition
For 4th-period transition elements, the outermost shell is n=4 (the 4s electrons), so the
"(n−1) shell" is the 3rd shell, containing the 3s, 3p, and 3d electrons. Counting these for each
candidate element reveals which pairs coincide — this is a good test of knowing the exceptional
configurations of Cr and Cu (half-filled/fully-filled 3d stability).
Step-by-Step Solution
- Fe (Z=26): [Ar]3d64s2 → (n−1)=3rd shell electrons: 2+6+6=14.
- Mn (Z=25): [Ar]3d54s2 → 3rd shell: 2+6+5=13.
- Zn (Z=30): [Ar]3d104s2 → 3rd shell: 2+6+10=18.
- K (Z=19): [Ar]4s1 → 3rd shell: 2+6+0=8.
- Sc (Z=21): [Ar]3d14s2 → 3rd shell: 2+6+1=9. …
- AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQQ.Which of the following lanthanoids have [Xe] 4fx5d16s2 configuration in their ground state? (x = 1-14) (A) Pr, Tb, Yb (B) Ce, Yb, Lu (C) Ce, Gd, Lu (D) Gd, Tb, Lu
›Reveal solutionSolution
Only La, Ce, Gd, and Lu among the lanthanoids adopt the 4fx5d16s2 pattern (extra stability of empty/half-filled/fully-filled 4f); restricting to x=1–14 excludes La, leaving Ce, Gd, Lu.
Concept and Intuition
Most lanthanoids follow the expected [Xe]4fn6s2 filling pattern (no 5d electron), because 4f orbitals are lower in energy and get filled preferentially. However, four lanthanoids break this pattern and instead have one electron in 5d, because that gives their 4f subshell a specially stable configuration — empty (4f0, La), exactly half-filled (4f7, Gd), or completely filled (4f14, Lu) — plus Ce, whose 4f15d16s2 arises from near-degenerate 4f/5d energies at the start of the series.
Step-by-Step Solution
- Recall the standard exceptions to lanthanoid configuration: La: [Xe]5d16s2 (4f0); Ce: [Xe]4f15d16s2; Gd: [Xe]4f75d16s2; Lu: [Xe]4f145d16s2.
- The question specifies x=1–14, which excludes La (whose x=0). …
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.The total number of electrons with magnetic quantum number ml=0 in Fe3+ and Cr+ is respectively (A) 14, 13 (B) 13, 14 (C) 11, 11 (D) 13, 13
›Reveal solutionSolution
Fe3+ and Cr+ both end up with the identical configuration [Ar]3d5, so both have exactly 11 electrons with magnetic quantum number ml=0.
Concept and Intuition
For each subshell, only the orbital with ml=0 contributes to this count (e.g. the pz orbital in a p subshell, or dz2 in a d subshell); an s orbital is always ml=0. In a half-filled d5 configuration, Hund's rule places one electron in each of the five d orbitals (ml=−2,−1,0,1,2), so the ml=0 orbital there holds exactly one electron (not two).
Step-by-Step Solution
- Fe (Z=26): [Ar]3d64s2. Removing 3 electrons to form Fe3+ removes both 4s electrons first, then one 3d electron: Fe3+ = [Ar]3d5.
- Cr (Z=24): [Ar]3d54s1 (anomalous half-filled stability). Removing 1 electron to form Cr+ removes the single 4s electron: Cr+ = [Ar]3d5.
- Both ions share the exact configuration 1s22s22p63s23p63d5. …
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.Which of the following ions has [Xe] 4f76s0 as outer electronic configuration? (A) Yb2+ (B) Eu2+ (C) Sm2+ (D) Tm2+
›Reveal solutionSolution
Europium's ground state is [Xe]4f⁷6s²; losing the 6s electrons gives Eu²⁺ = [Xe]4f⁷6s⁰, favoured by the extra stability of a half-filled f-subshell.
Concept and Intuition
Most lanthanides prefer the +3 oxidation state, but a few show unusually stable +2 (or +4) states when doing so gives a half-filled (f7) or fully-filled (f14) or empty (f0) subshell — extra-stable electronic configurations analogous to the familiar half-filled/fully-filled stability seen in d-block chemistry. Eu²⁺ (f⁷), Yb²⁺ (f¹⁴), and Sm²⁺ (f⁶, less stable) are the classic examples.
Step-by-Step Solution
- Europium, Z=63: ground-state electron configuration is [Xe]4f76s2 (note Eu breaks the "normal" filling trend seen in La–Lu to preserve the extra-stable half-filled 4f⁷).
- Removing the two outer 6s electrons (as always happens first upon ionisation) gives Eu2+:[Xe]4f76s0 — exactly matching the configuration asked for. …
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