Exercises — Pauli exclusion principle

The rule for legality: $l$ must satisfy $0\le l\le n-1$, and $m_l$ must satisfy $-l\le m_l\le +l$.

• (a) $n=2\Rightarrow l$ can be $0$ or $1$; $l=1$ ✓, and $m_l=-1$ is in $[-1,1]$ ✓. Legal.
• (b) $n=3\Rightarrow l$ can be $0,1,2$. Here $l=3$ — too big. Forbidden.
• (c) $n=1\Rightarrow l=0$ only; $m_l=0$ ✓. Legal.
• (d) $n=2,l=0,m_l=0$ ✓. Legal.

Answer: (b). This is not a Pauli violation — it is an illegal address to begin with.

With $n,l,m_l$ fixed, the only freedom left is $m_s$, which has exactly 2 values ($+\frac{1}{2},-\frac{1}{2}$). Pauli forbids identical quartets, so the two electrons must take the two different spins. Answer: 2 electrons, with opposite spins, forced by the two-valuedness of $m_s$.

"$f$" means $l=3$. Number of $m_l$ orientations $=2l+1=2(3)+1=7$ orbitals. Each orbital holds 2 electrons, so max $=2\times 7=14$. Answer: 7 orbitals, 14 electrons.

Let us derive the shell capacity rather than recall it, so we see why the answer comes out. Step 1 — list the subshells. A shell $n$ contains all $l$ from $0$ to $n-1$. For $n=4$: $l=0,1,2,3$ (that is $4s,4p,4d,4f$). Step 2 — orbitals in each. Each $l$ has $2l+1$ orbitals: $1,3,5,7$. Step 3 — electrons in each. Double each for the two spins: $2,6,10,14$. Step 4 — add up. $2+6+10+14=32$. Now the general rule. The orbital counts $1,3,5,7,\dots$ are the odd numbers, and the picture below shows the beautiful fact that the first $n$ odd numbers add to $n^2$ (they tile a perfect square). So: $\text{orbitals in shell }n={\sum }_{l=0}^{n-1}(2l+1)=n^2,$ and doubling for spin gives $N=2n^2.$ For $n=4$: $2(4{)}^2=32$ ✓ — matching our brick-by-brick count. Answer: 32, and the formula $N=2n^2$ is earned, not assumed.

$2p$: $n=2,l=1$. Then $m_l\in \{-1,0,+1\}$ and $m_s\in \{+\frac{1}{2},-\frac{1}{2}\}$. That is $3\times 2=6$ sets: $(2,1,-1,+\frac{1}{2}),(2,1,-1,-\frac{1}{2}),(2,1,0,+\frac{1}{2}),(2,1,0,-\frac{1}{2}),(2,1,+1,+\frac{1}{2}),(2,1,+1,-\frac{1}{2})$. Answer: 6 distinct sets — exactly the $2p$ capacity.

Pauli-legal? Yes. Two electrons in one $2p$ orbital with opposite spins have sets $(2,1,m_l,+\frac{1}{2})$ and $(2,1,m_l,-\frac{1}{2})$ — they differ in $m_s$, so Pauli is satisfied. Actual ground state? No. Hund's rule says electrons occupy separate degenerate orbitals with parallel spins first, to minimise electron–electron repulsion. So carbon's real ground state puts the two $2p$ electrons in two different orbitals ($m_l=-1$ and $m_l=0$, say) with the same spin. The difference: Pauli sets the ceiling (what is forbidden). Hund picks the arrangement among allowed options. The paired-in-one-orbital version is a legal excited state, not the ground state.

With identical states $a=b$, the antisymmetric combination from the parent note is $\psi (1,2)=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_a(2)-{\psi }_a(2){\psi }_a(1)]=0.$ The two terms are identical and subtract to zero. Meaning: $|\psi {|}^2=0$ everywhere → zero probability → this configuration does not exist. That is why the two $1s$ electrons of helium are forced to opposite spins: $(1,0,0,+\frac{1}{2})$ and $(1,0,0,-\frac{1}{2})$.

The figure below makes this concrete. On the left (mint box) the two $1s$ electrons carry opposite spins (one up-arrow in lavender, one down-arrow in coral): the fourth number differs, the quartets are distinct, and $\psi \ne 0$ — an allowed atom. On the right (coral box) both electrons are drawn spin-up (two lavender arrows): all four numbers now match, the two terms of $\psi$ cancel exactly, and $\psi =0$ — the configuration is erased from existence. Read the two boxes as "allowed vs erased," and the arrows tell you which spins Pauli permits.

Shell $n=3$ holds $2n^2=18$ electrons total. Exactly half have each spin (every orbital's two electrons split into one $+\frac{1}{2}$ and one $-\frac{1}{2}$). $18/2=9$. Answer: 9 electrons with $m_s=-\frac{1}{2}$. Cross-check by subshell: $3s$ contributes 1, $3p$ (3 orbitals) contributes 3, $3d$ (5 orbitals) contributes 5 → $1+3+5=9$ ✓.

Pauli fixes subshell capacities: $s=2,p=6,d=10,f=14$. Period length equals the number of electrons added as you cross that row, and the order of filling is set by Aufbau principle (lowest energy first), not simply by $n$.

• Because energies overlap ($4s$ fills before $3d$, $5s$ before $4d$, etc.), the $d$ block "lags" one shell and the $f$ block lags two. The energy-ladder figure below shows this: the rungs are not grouped cleanly by $n$ — the $4s$ rung sits below the $3d$ rung, so nature climbs to $4s$ first, then drops back to fill $3d$.
• Filling order per period gives the added electrons:
  • Period 1: $1s$ → 2
  • Period 2: $2s2p$ → $2+6=8$
  • Period 3: $3s3p$ → $2+6=8$
  • Period 4: $4s3d4p$ → $2+10+6=18$
  • Period 5: $5s4d5p$ → $18$
  • Period 6: $6s4f5d6p$ → $2+14+10+6=32$
  • Period 7: $7s5f6d7p$ → $32$ Answer: $2n^2$ gives the maximum a shell can ever hold; period length is governed by the Aufbau filling order slicing those Pauli-capped subshells differently. Pauli sets each block's width, Aufbau sets when blocks appear.

Capacity per orbital = number of spin states = 3. Orbitals per shell (unchanged) $={\sum }_{l=0}^{n-1}(2l+1)=n^2$. Why does this sum equal $n^2$? The terms are the first $n$ odd numbers $1,3,5,\dots$, and (as the L2.2 figure shows) each new odd number wraps an "L-shaped" layer around a growing square, turning an $n\times n$ square into an $(n+1)\times (n+1)$ one — so their running total is always a perfect square, $n^2$. New shell max $=3\times n^2=3n^2$. For $n=2$: $3(2{)}^2=12$. Answer: $3n^2$; for $n=2$ that is 12. (Compare real world: $2n^2=8$.) This shows the "$2$" in $2n^2$ is literally the number of spin states, not something deeper about geometry.

Fermions require the total wavefunction to flip sign on exchange (antisymmetric), which forces $\psi \to 0$ when two states coincide → Pauli. Bosons require the wavefunction to stay the same on exchange (symmetric): $\psi (1,2)=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_b(2)+{\psi }_a(2){\psi }_b(1)].$ Set $a=b$: $\psi =\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_a(2)+{\psi }_a(1){\psi }_a(2)]=\sqrt{2}{\psi }_a(1){\psi }_a(2)\ne 0$. Answer: Two bosons in the same state give a perfectly nonzero wavefunction — bosons ignore Pauli. Real consequences (from Fermions and bosons): lasers (many photons in one mode) and Bose–Einstein condensates.

Not correctly. Pauli forbids two electrons from sharing the same single-particle state, and the state includes the spatial wavefunction, not just the $(n,l,m_l,m_s)$ quartet of one isolated atom. Two electrons bound to atoms a mile apart occupy states centred at different locations — their spatial parts barely overlap, so the full states are already distinct. There is no clash. The quartet $(n,l,m_l,m_s)$ is a bookkeeping shorthand valid within one atom's set of orbitals. Across separate atoms, the position degree of freedom keeps the states distinct automatically. Answer: The claim misapplies Pauli by ignoring the spatial part of the state; the true rule is "no two electrons in the same complete quantum state," which is satisfied trivially for well-separated atoms.

A full subshell of azimuthal number $l$ has every orbital $(m_l)$ doubly occupied. Net spin: each orbital carries one $+\frac{1}{2}$ and one $-\frac{1}{2}$, summing to $0$; over all $2l+1$ orbitals, total $m_s$-sum $=(2l+1)\cdot 0=0$. Net $\sum m_l$: the $m_l$ values are $-l,-l+1,\dots ,-1,0,1,\dots ,l$, and each is used twice (two electrons per orbital). Their sum: $2{\sum }_{m_l=-l}^l{m}_l=2\cdot 0=0,$ because the values are symmetric about $0$ and cancel in pairs $(+m_l)+(-m_l)=0$. Answer: Both totals are $0$ — a filled subshell is spherically bland (no net spin, no net $m_l$). This is why noble gases and filled-shell ions are so unreactive and (spin-) non-magnetic.