Worked examples — Pauli exclusion principle

This page is the drill hall for Pauli exclusion principle. The parent note derived the rule and counted occupancy. Here we hunt down every distinct kind of question the exam or reality can ask, put them in a table, then solve one example per row. If you meet a Pauli problem in the wild that this page did not cover, tell me — but I don't think you will.

Everything rests on four labels, the Quantum numbers $(n,l,m_l,m_s)$. If any of those four symbols feels shaky, read that note first — I use them from line one below.

Recall The four addresses (peek if unsure)

• $n$ = which shell. A whole number $1,2,3,\dots$
• $l$ = which subshell shape. Runs $0$ to $n-1$. $l=0$ is "s", $1$ is "p", $2$ is "d", $3$ is "f".
• $m_l$ = which orientation of that shape. Runs $-l,\dots ,0,\dots ,+l$ — that is $2l+1$ values.
• $m_s$ = which spin. Only two values, $+\frac{1}{2}$ (call it ↑) or $-\frac{1}{2}$ (call it ↓).

Pauli's one law: no two electrons in one atom may carry the same complete quartet $(n,l,m_l,m_s)$.

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The scenario matrix

Every Pauli question I have ever seen falls into one of these cells. Each cell gets a fully worked example below, labelled [Cell X].

Cell	Case class	The question it asks	Example
A	Minimal case — a single orbital	Can these two exact quartets coexist?	Ex 1
B	Boundary of a subshell — filling exactly to the ceiling	How many electrons max here, and why?	Ex 2
C	Over-fill (illegal) input	What breaks if you add one too many?	Ex 3
D	Degenerate / partial fill — fewer electrons than slots	Which arrangement is legal (Pauli) vs preferred (Hund)?	Ex 4
E	Whole-shell counting — sum over subshells	Prove the $2n^2$ ceiling by brute force	Ex 5
F	Limiting / large-$n$ behaviour	How fast does capacity grow? Any cap?	Ex 6
G	Real-world word problem	Why does a white-dwarf star not collapse?	Ex 7
H	Exam twist — spot the illegal quartet in a list	Which listed electron cannot exist?	Ex 8
I	Zero / edge input — $l$ near its extremes	What are the allowed $m_l$ at the smallest and largest $l$?	Ex 9

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Example 1 — [Cell A] The smallest possible clash

Forecast: Look at the four slots of each address. How many match? Guess whether they can coexist before reading on.

1. Compare the quartets slot by slot. $n$: $2=2$. $l$: $1=1$. $m_l$: $0=0$. $m_s$: $+\frac{1}{2}=+\frac{1}{2}$. All four identical. Why this step? Pauli acts on the complete set. You cannot judge legality until you have laid all four numbers side by side.

2. Apply the law. Identical quartets → forbidden. Using the takeaway from the definition box above: identical addresses force the wavefunction $\psi =0$, so the state describes nothing. Why this step? This is the raw rule — no arithmetic, just a "same/different" check.

3. Minimal fix. The first three numbers ($n,l,m_l$) define one orbital, and the orbital happily holds 2 electrons. So change only $m_s$: make Q be $(2,1,0,-\frac{1}{2})$. Why this step? We change the fewest labels. Since $m_s$ has a spare value, flipping spin costs no extra energy structure — both electrons stay in the same orbital.

Verify: New quartets $(2,1,0,+\frac{1}{2})$ and $(2,1,0,-\frac{1}{2})$ differ in exactly one slot → distinct → legal. Slots that differ $=1$, which is the smallest possible non-zero difference. ✓

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Example 2 — [Cell B] Filling a subshell to its ceiling

Forecast: $d$ means $l=2$. How many orbitals? Multiply by 2. Write your number down first.

1. Read off $l$. "d" ⟹ $l=2$. Why this step? The letter is $l$; nothing else determines subshell shape.

2. Count orientations $m_l$. They run $-l$ to $+l$: $-2,-1,0,+1,+2$ → that's $2l+1=5$ orbitals. Why this step? Each distinct $m_l$ is a separate orbital (separate $(n,l,m_l)$ triple), and Pauli treats each independently.

3. Double for spin. Each orbital holds 2 (↑ and ↓). Total $=5\times 2=10$. Why this step? $m_s$ contributes the factor 2 — the only reason an orbital isn't just 1 electron.

Figure (Cell B): five black boxes labelled $m_l=-2,-1,0,+1,+2$ — the five 3d orbitals. Inside every box sits one red up-arrow (spin $+\frac{1}{2}$) and one black down-arrow (spin $-\frac{1}{2}$). Counting all arrows gives $5\times 2=10$ electrons — the full 3d subshell.

Verify: Look at the figure — five black boxes, one per $m_l$ value, and inside each a red up-arrow (spin $+\frac{1}{2}$) beside a black down-arrow (spin $-\frac{1}{2}$): count them and you get $5\times 2=10$ arrows. Formula $2(2l+1)=2(5)=10$. ✓ Every quartet has $n=3,l=2$, one of the five $m_l$, and one of two $m_s$ — $5\times 2=10$ distinct addresses, none repeated. ✓

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Example 3 — [Cell C] The illegal over-fill

Forecast: How many distinct quartets can share $n=1,l=0$? Count them before reading.

1. Fix the shared triple. $1s$ means $n=1,l=0$. Then $m_l$ must be $0$ (only value when $l=0$). So all three electrons share the triple $(1,0,0)$. Why this step? We nail down everything except spin, so the only freedom left is $m_s$.

2. Enumerate the free label. $m_s\in \{+\frac{1}{2},-\frac{1}{2}\}$ — exactly two options. Why this step? This is the whole game: with 3 electrons and 2 spin values, by the pigeonhole idea two electrons must share the same $m_s$.

3. Force the collision. Electrons 1 and 2 take $+\frac{1}{2}$ and $-\frac{1}{2}$. Electron 3 must reuse one of those → its quartet equals an existing one → $\psi =0$ → forbidden. Why this step? This is Pauli biting: the third address does not exist inside $1s$.

Verify: Legal count in $1s$ $=2(2\cdot 0+1)=2$. Trying to place 3 exceeds 2, so exactly $3-2=1$ electron is impossible. The real fix: the third electron starts a new shell ($n=2$), which is why lithium opens a new row of the periodic table. ✓

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Example 4 — [Cell D] Partial fill: Pauli's ceiling vs Hund's choice

Forecast: Three electrons, three orbitals. Guess: do they pile up or spread out?

1. Count the slots. $2p$: $l=1$ → $m_l=-1,0,+1$ → 3 orbitals, ceiling $=6$ electrons. Why this step? Establishes the Pauli ceiling — the hard wall we may not cross.

2. Test the "all in one orbital" idea. One orbital holds only 2 (one ↑, one ↓). A third in that same orbital repeats a quartet → forbidden. In the figure below, this is the left group: three arrows crammed in one box, struck through with a red ✕ because the third electron repeats a quartet. Why this step? Pauli already rules out stacking all three in one $m_l$: (a) answer is no.

3. List legal arrangements. With 3 electrons and 3 boxes, legal options include one-per-box (all ↑) or two-in-one-plus-one. All obey Pauli. Pauli does not pick between them. Why this step? Shows Pauli sets the ceiling, not the seating.

4. Bring in Hund. Hund's rule says spread out singly with parallel spins first → $\uparrow \uparrow \uparrow$, one per orbital. In the figure, this is the right group: three separate boxes, each with one red up-arrow. Why this step? Pauli = law of the maximum; Hund = law of the arrangement. Both must be quoted.

Figure (Cell D): LEFT group — one box crammed with three arrows, crossed out by a bold red ✕ because the third electron would repeat a quartet (Pauli-forbidden). RIGHT group — three separate boxes, each holding a single red up-arrow, showing the true $2p^3$ ground state where electrons spread out with parallel spins (Hund's rule).

Verify: Compare the two groups in the figure — the red-crossed left group is Pauli-forbidden; the right group is the true ground state. Ground state $2p^3=\uparrow \uparrow \uparrow$ uses quartets $(2,1,-1,+\frac{1}{2}),(2,1,0,+\frac{1}{2}),(2,1,+1,+\frac{1}{2})$ — three distinct addresses (different $m_l$) → Pauli satisfied, and all spins parallel → Hund satisfied. ✓

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Example 5 — [Cell E] Prove a shell holds $2n^2$ by brute counting

Forecast: Which subshells live in $n=4$? Add their capacities in your head first.

1. List allowed $l$. $l$ runs $0$ to $n-1=3$: so $l=0,1,2,3$ → subshells $4s,4p,4d,4f$. Why this step? The range of $l$ is fixed by $n$; this enumerates every subshell present.

2. Capacity of each. $2(2l+1)$:

   • $4s$: $2(1)=2$
   • $4p$: $2(3)=6$
   • $4d$: $2(5)=10$
   • $4f$: $2(7)=14$ Why this step? Each subshell's ceiling is orbitals-times-two, from Ex 2's logic.

3. Sum. $2+6+10+14=32$. Why this step? The shell ceiling is the sum of its subshell ceilings — Pauli slots never overlap between subshells.

Verify: Formula $2n^2=2(4{)}^2=2\cdot 16=32$. ✓ Matches brute count exactly. (The odd numbers $1+3+5+7=16=4^2$ — sum of first $n$ odds is $n^2$.) ✓

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Example 6 — [Cell F] Limiting behaviour: how fast does capacity grow?

Forecast: Guess the ratio of the $n=7$ shell to the $n=1$ shell.

1. Evaluate endpoints. $2(1{)}^2=2$ and $2(7{)}^2=98$. Why this step? $n=1$ (tiniest, the He shell) and $n=7$ (deepest occupied in real atoms) frame the range.

2. Ratio. $98/2=49=7^2$. Why this step? Capacity scales as $n^2$, so the ratio is the square of the $n$-ratio — the "grows quadratically" fact made concrete.

3. Any cap on $n$? Mathematically no — $n$ can be any positive integer, so $2n^2\to \infty$. Physically, electrons that far out are barely bound and real atoms stop around $n=7$. Why this step? Separates the rule's limit (none) from physics' practical limit.

4. Gap between shells. $2(n+1{)}^2-2n^2=2(2n+1)=4n+2$. Why this step? Shows the increments grow linearly ($6,10,14,18,\dots$) even though totals grow quadratically.

Verify: $2(1{)}^2=2$, $2(7{)}^2=98$, ratio $49$. Gap check: $n=1\to 2$ gives $4(1)+2=6$, and indeed $8-2=6$. ✓

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Example 7 — [Cell G] Real-world: why a white dwarf doesn't collapse

Forecast: What stops gravity, if not heat and not (only) charge? Guess before reading.

1. Identify the fermions. The star's matter is a dense sea of electrons — spin-$\frac{1}{2}$ fermions. Each must occupy a distinct quantum state. Why this step? Pauli's antisymmetry argument (parent note) applies to any fermions in shared space, not just atomic orbitals.

2. Gravity tries to co-locate them. Crushing electrons into the same tiny volume forces them toward the same momentum/spatial state. Why this step? Two electrons squeezed to the same state → identical quartets → $\psi =0$ → not allowed.

3. The resulting push. To stay distinct, electrons must occupy higher-momentum states — this fast motion produces degeneracy pressure, resisting collapse. No temperature required. Why this step? This is Pauli as a mechanical effect: exclusion → forced high momentum → outward pressure.

4. Rule out "it's just charge." The parent note's Mistake #2: degeneracy pressure survives even for hypothetical neutral fermions. It is statistical, not Coulombic. Why this step? Confirms the effect is Pauli, not electrostatics — the whole point of the example.

Verify (sanity, not numeric): Neutron stars (uncharged neutrons, also spin-$\frac{1}{2}$) also resist collapse by the same degeneracy pressure — proving the mechanism is Pauli/antisymmetry, independent of charge. ✓

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Example 8 — [Cell H] Exam twist: spot the impossible electron

Forecast: Guess how many of the four are legal.

1. Check (i): is $l<n$? $n=2$ allows $l=0,1$ only. Here $l=2$. Illegal — $l$ must be $\le n-1$. Why this step? $l$'s range is capped by $n$; violating it means the subshell doesn't exist in that shell.

2. Check (ii): is $|m_l|\le l$? $l=1$ allows $m_l=-1,0,+1$. Here $m_l=-2$. Illegal — orientation out of range. Why this step? $m_l$ can never exceed $l$ in magnitude; there is no such orbital.

3. Check (iii): is $m_s$ a valid spin? From the definition box, spin is only $\pm \frac{1}{2}$. Here $m_s=+1$. Illegal — no such spin. Why this step? $m_s$ has exactly two allowed values; anything else is not an electron spin state — this is the trap the examiner set with the "$+1$".

4. Check (iv): run all four range tests. $n=4\ge 1$ ✓; $l=2$ with $2\le n-1=3$ ✓; $m_l=-1$ with $|-1|\le 2$ ✓; $m_s=-\frac{1}{2}\in \{\pm \frac{1}{2}\}$ ✓. Every constraint passes → legal. Why this step? All four range rules hold simultaneously, so this address is genuinely occupiable — the one survivor.

Verify: Legal count $=1$ (only iv). Illegal $=3$ (i, ii, iii). ✓ Note: none of these were rejected by Pauli itself — they fail the range rules of Quantum numbers that Pauli presupposes. Pauli only forbids duplicates; these fail before we even get to duplication. ✓

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Example 9 — [Cell I] Zero / edge inputs of $l$

Forecast: Which $l$ is largest for $n=3$, and how many $m_l$ does it have?

1. Largest $l$ in $n=3$. $l_{\text{max}}=n-1=2$ (the $3d$ subshell). Why this step? $l$ tops out at $n-1$; that's the widest subshell present.

2. Edge $m_l$ values. They run from the low edge $-l=-2$, through the middle $0$, to the high edge $+l=+2$: the full list is $\{-2,-1,0,+1,+2\}$, which is $2l+1=5$ values. Why this step? $m_l$'s extremes are exactly $\pm l$; the middle is always $0$. Naming both edges and the centre proves no orientation is skipped between them.

3. Electron count for $3d$. Five orbitals, each doubled by spin: $2(2l+1)=2(5)=10$ electrons. Why this step? Same doubling rule as Ex 2; confirms the widest subshell of $n=3$ holds 10.

4. Degenerate edge $l=0$. Now the range "runs from $-0$ to $+0$": it collapses to a single value, $m_l=0$. So $2l+1=1$ orbital → $2(1)=2$ electrons. Why this step? This is the smallest, degenerate case — showing the formula $2l+1$ still returns $1$ orbital (never $0$), so an $s$ subshell always has exactly one orbital and holds 2 electrons.

Verify: $3d$: $m_l$ list has 5 entries, $2(2\cdot 2+1)=10$ ✓. $s$ subshell ($l=0$): one $m_l$ value, $2(2\cdot 0+1)=2$ ✓. Both the largest and smallest $l$ behave exactly as the range rule predicts. ✓

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Wrapping up

Recall The whole matrix in one breath (test yourself before revealing)

• A minimal clash → change only $m_s$. B subshell ceiling $=2(2l+1)$. C over-fill → 3rd electron impossible, new shell opens. D Pauli sets the ceiling, Hund seats them. E shell $=\sum 2(2l+1)=2n^2$. F grows as $n^2$, no math cap, gap $=4n+2$. G degeneracy pressure holds up a white dwarf. H range rules kill bad quartets before Pauli acts. I edges of $l$: $l=n-1$ widest, $l=0$ degenerate single orbital.

Every cell of the scenario matrix now has a fully worked example, from the single-orbital clash all the way to a collapsing star and the degenerate $l=0$ edge. The one idea threaded through all nine: an electron's four-number address must be unique, and $m_s$ has only two values — everything else (the ceilings, the $2n^2$, the exam traps) is bookkeeping on top of that.

Where to go next:

• Return to the derivation and counting in Pauli exclusion principle (the parent note).
• Prefer Hindi–English? Try the Hinglish version.
• See how these ceilings get filled in order via the Aufbau principle, and arranged within a subshell via Hund's rule.
• Trace the $2n^2$ result into the shape of the table at Periodic table periodicity.

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Connections

• Quantum numbers — every cell above is a legality check on the four labels
• Aufbau principle — fills lowest energy first; each example respects the ceilings Pauli sets
• Hund's rule — chooses seating in the partial-fill Cell D
• Fermions and bosons — Cell G's white-dwarf pressure comes from fermion antisymmetry
• Electron spin — the $m_s=\pm \frac{1}{2}$ that saves Cells A and C
• Periodic table periodicity — the $2n^2$ of Cells E and F sets period lengths