Quantum numbers — n (principal), l (azimuthal), mₗ (magnetic), mₛ (spin): Worked examples

The scenario matrix

Cell	Badge	Case class	What makes it tricky	Cleared by
A	🟢	Fully valid set	just confirm each rule passes	Ex 1
B	🟢⚪	Zero / degenerate input ( $l = 0$ , smallest $n$ )	edge of every range	Ex 2
C	🔴	Invalid: $l$ too big ( $l \geq n$ )	the most common trap	Ex 3
D	🔴	Invalid: $m_{l}$ out of range ( $∣ m_{l} ∣ > l$ )	projection exceeds magnitude	Ex 3
E	🔴	Invalid: wrong $m_{s}$ / non-integer $n$	forbidden values	Ex 4
F	🔢	Counting states (orbitals & electrons)	sum over $l$ , factor of 2	Ex 5
G	📐	Angular-momentum geometry ( $∣ L ∣$ , $L_{z}$ , tilt)	needs a picture	Ex 6
H	🔢➡️	Limiting / large- $n$ behaviour	how ranges grow	Ex 7
I	🌍	Real-world word problem (Zeeman splitting)	words → numbers	Ex 8
J	🔄	Exam twist (given a count, find $n$ or $l$ )	work the rules backwards	Ex 9

We will now clear every cell.

Example 1 — Case A: a fully valid address

Worked example Is

$(n, l, m_{l}, m_{s}) = (4, 2, - 1, + \frac{1}{2})$ allowed? Forecast: Run down the four rules in order. Before reading, guess: does it survive all four checks? What orbital is this?

Step 1 — Check $n$ . $n = 4$ is a positive whole number. Why this step? $n$ is the outermost box; if it fails, nothing inside is defined. $n = 4 \geq 1$ ✓.

Step 2 — Check $l$ . Rule: $0 \leq l \leq n - 1 = 3$ . Here $l = 2$ , and $0 \leq 2 \leq 3$ ✓. Why this step? $l$ lives inside the $n$ -box, so its ceiling is $n - 1$ . Using the letter code above, $l = 2$ is the letter d.

Step 3 — Check $m_{l}$ . Rule: $- l \leq m_{l} \leq + l$ , i.e. $- 2 \leq m_{l} \leq + 2$ . Here $m_{l} = - 1$ ✓. Why this step? $m_{l}$ is the projection of angular momentum; it can never point "more sideways" than the total allows.

Step 4 — Check $m_{s}$ . Must be $+ \frac{1}{2}$ or $- \frac{1}{2}$ . Here $+ \frac{1}{2}$ ✓.

Verify: All four boxes nest correctly, so this is a real 4d electron with spin up. Sanity check: since $l = 2 \leq 3$ , the 4d subshell genuinely exists (it does — it fills after 5s). ✓

Example 2 — Case B: the smallest, most degenerate case ( $n = 1$ , $l = 0$ )

Worked example List

every complete address available in the $n = 1$ shell. Forecast: How many electrons can the first shell hold, and how many different addresses is that? Guess before reading.

Step 1 — Find allowed $l$ . $l$ runs $0 \to n - 1 = 0$ . So only $l = 0$ (an s subshell). Why this step? When $n = 1$ , the ceiling $n - 1$ equals the floor $0$ , so the range collapses to a single value — this is the degenerate edge.

Step 2 — Find allowed $m_{l}$ . With $l = 0$ : $m_{l}$ runs $- 0 \to + 0$ , so only $m_{l} = 0$ . That is $2 l + 1 = 1$ orientation. Why this step? An s-orbital is a sphere — a sphere looks the same from every direction, so there is nothing to "orient". One orbital only.

Step 3 — Attach spin. Each orbital takes two spins: $m_{s} = + \frac{1}{2}$ and $m_{s} = - \frac{1}{2}$ . Why this step? An "orbital" is fixed by the first three numbers $(n, l, m_{l})$ ; the fourth number $m_{s}$ still has two legal values, so each orbital splits into exactly two distinct four-number addresses. This factor of 2 is where the "2" in every electron count comes from (see Pauli exclusion principle).

Step 4 — Enumerate. The complete list of addresses: $(1, 0, 0, + \frac{1}{2}), (1, 0, 0, - \frac{1}{2})$ Why this step? Listing them explicitly proves there are no hidden extra states — we have walked every branch of the tree ( $1$ value of $l$ × $1$ value of $m_{l}$ × $2$ values of $m_{s}$ ), so the count is complete, not a guess.

Verify: Count $= 2$ electrons. Direct multiplication: $1 \times 1 \times 2 = 2$ ✓. This is exactly helium's full 1s² shell. (The general shortcut $2 n^{2}$ is derived in Example 5; here we simply counted by hand.) ✓

Example 3 — Cases C & D: two ways to break the range rules

Worked example Which of these are invalid, and

why? (a) $(2, 2, 0, + \frac{1}{2})$ (b) $(3, 1, + 2, - \frac{1}{2})$ Forecast: One breaks the $l$ -rule, one breaks the $m_{l}$ -rule. Guess which is which before reading.

Part (a) — Case C ( $l$ too big). Step 1: Check $l \leq n - 1$ . Here $n = 2$ so $n - 1 = 1$ , but $l = 2$ . Since $2 > 1$ , invalid. Why this step? The $n = 2$ shell only contains s ( $l = 0$ ) and p ( $l = 1$ ). A "2d" orbital does not exist — the radial equation gives no finite, normalizable solution for $l \geq n$ .

Part (b) — Case D ( $m_{l}$ out of range). Step 1: Check $l$ first: $l = 1 \leq n - 1 = 2$ ✓ — this part is fine, it's a 3p subshell. Step 2: Check $m_{l}$ : rule is $- l \leq m_{l} \leq + l$ , i.e. $- 1 \leq m_{l} \leq + 1$ . But $m_{l} = + 2$ . Since $2 > 1$ , invalid. Why this step? $m_{l}$ is the shadow ( $L_{z}$ ) of the angular-momentum arrow on the $z$ -axis. A shadow can never be longer than the arrow, so $∣ m_{l} ∣$ can never exceed $l$ .

Verify: (a) fails at rule 2 (the $l$ -box), (b) fails at rule 3 (the $m_{l}$ -box). Both correctly rejected. A p-subshell only ever offers $m_{l} \in {- 1, 0, + 1}$ — three values, and $+ 2$ is not among them. ✓

Example 4 — Case E: forbidden values that aren't even in the range game

Worked example Reject these and name the broken rule: (a)

$(0, 0, 0, + \frac{1}{2})$ (b) $(3, 1, 0, + 1)$ (c) $(2.5, 1, 0, - \frac{1}{2})$ Forecast: These don't fail by being "too big" — they use values that are simply not permitted at all.

Part (a): $n = 0$ . Rule: $n = 1, 2, 3, \dots$ , never zero. Why? Put $n = 0$ into $E_{n} = - \frac{13.6 Z ^{2}}{n ^{2}} eV$ (with $Z$ the atomic number, defined above) and you divide by zero — the energy blows up, and there is no allowed $l$ (the range $0 \leq l \leq - 1$ is empty). Invalid.

Part (b): $m_{s} = + 1$ . Rule: $m_{s} \in {+ \frac{1}{2}, - \frac{1}{2}}$ only. Why? An electron's spin quantum number is fixed at magnitude $\frac{1}{2}$ ; its projection can only be $\pm \frac{1}{2}$ . There is no " $+ 1$ " spin state for a single electron. Invalid.

Part (c): $n = 2.5$ . Rule: $n$ must be a whole number. Why? $n$ counts radial nodes plus one; a fractional count has no physical meaning and gives no normalizable wavefunction. Invalid.

Verify: Each set violates a definition, not a range bound — the difference from Example 3. All three rejected. ✓

Example 5 — Case F: counting orbitals and electrons in $n = 4$

Worked example How many orbitals and how many electrons does the

$n = 4$ shell hold? Forecast: Guess the electron total, then prove it the long way by adding up subshells — and watch the $2 n^{2}$ formula emerge.

Step 1 — List the subshells. $l = 0, 1, 2, 3$ → s, p, d, f. Why this step? $l$ ceiling is $n - 1 = 3$ , and this shell reaches all the way to f.

Step 2 — Count orbitals per subshell using $2 l + 1$ : $s : 1, p : 3, d : 5, f : 7.$ Why this step? $2 l + 1$ counts the distinct $m_{l}$ orientations.

Step 3 — Total orbitals. $1 + 3 + 5 + 7 = 16 = n^{2} = 4^{2}$ . Why this step? The sum of the first $n$ odd numbers is always $n^{2}$ — a neat identity, so we don't even have to add.

Step 4 — Total electrons. Multiply by 2 for the two spins: $2 \times 16 = 32$ . Why this step? Each orbital (one $(n, l, m_{l})$ triple) holds two electrons — one $m_{s} = + \frac{1}{2}$ and one $m_{s} = - \frac{1}{2}$ — exactly the spin doubling we saw by hand in Example 2. Multiplying the orbital count by 2 converts "orbitals" into "electron slots", and doing this to $n^{2}$ orbitals is precisely how the general capacity $2 n^{2}$ is born.

Verify: Formula $2 n^{2} = 2 (4)^{2} = 32$ ✓. Cross-check subshell widths: $2 + 6 + 10 + 14 = 32$ ✓. Both routes agree. ✓

Example 6 — Case G: the geometry of a 3d electron's angular momentum

Worked example For a 3d electron, find

$∣ L ∣$ , list the allowed $L_{z}$ values, and describe the picture. Forecast: Will the arrow ever point straight up along $z$ ? Guess yes/no before reading.

Figure — alt text / standalone caption. On a warm cream background with a dotted grid, a vertical black line marks the $z$ -axis. The plum dot at the centre is the nucleus (the origin). Five teal solid arrows all start at the origin and all have the same length $6 ℏ$ — these are the angular-momentum vector $L$ in its five allowed tilts; a fainter mirror-image arrow shows the cone's other side. Five burnt-orange dashed horizontal lines mark the allowed heights $L_{z} = m_{l} ℏ$ for $m_{l} = - 2, - 1, 0, + 1, + 2$ , each labelled at the right. The horizontal axis is the sideways component of $L$ (in units of $ℏ$ ); the vertical axis is $L_{z}$ (in units of $ℏ$ ). The key visual point: no arrow reaches the top of the $z$ -axis — the longest allowed height ( $2ℏ$ ) is shorter than the arrow ( $6 ℏ \approx 2.449ℏ$ ), so every arrow tilts.

Step 1 — Get $l$ . "d" means $l = 2$ . Why this step? The subshell letter is only a nickname; every rule below is written in terms of the number $l$ , so we must translate the letter back using the code defined at the top (d ↔ $l = 2$ ) before any formula will accept it.

Step 2 — Magnitude of the angular-momentum arrow. $∣ L ∣ = l (l + 1) ℏ = 2 \cdot 3 ℏ = 6 ℏ \approx 2.449 ℏ.$ Why this step? The total length is set by $l$ alone, in units of $ℏ$ (defined above). Look at the figure: every arrow has the same fixed length $6 ℏ$ .

Step 3 — Allowed projections. $L_{z} = m_{l} ℏ$ with $m_{l} = - 2, - 1, 0, + 1, + 2$ : $L_{z} \in {- 2, - 1, 0, + 1, + 2} ℏ.$ Why this step? $m_{l}$ is the height of the arrow's tip on the $z$ -axis (the dashed levels in the figure). Five allowed heights = five orientations.

Step 4 — Answer the forecast. The biggest $L_{z}$ is $2ℏ$ , but the arrow's length is $6 ℏ \approx 2.449ℏ$ . Since $2ℏ < 6 ℏ$ , the arrow can never lie flat along $z$ — it always tilts. That is why the tips trace cones, not a single line. Why this matters: the projection can never equal the full length; there is always "sideways" angular momentum we cannot pin down. This is the quantum-mechanical refusal to let $L$ point exactly along an axis.

Verify: Largest tilt gap: $6 - 2 \approx 0.449 > 0$ , confirming no full alignment ✓. Number of orientations $= 2 l + 1 = 5$ ✓. ✓

Example 7 — Case H: how the ranges grow for large $n$

Worked example For the

$n = 5$ shell, find the highest subshell letter, the largest possible $m_{l}$ , and the electron capacity. Forecast: Each range stretches as $n$ grows. Predict the capacity from $2 n^{2}$ first.

Step 1 — Highest $l$ . $l_{m a x} = n - 1 = 4$ . From the letter code, the letter after f ( $l = 3$ ) is g ( $l = 4$ ). Why this step? Range ceilings scale with $n$ ; at $n = 5$ we reach the g-subshell (real, though not occupied in ground-state atoms of the periodic table).

Step 2 — Largest $m_{l}$ . In the g-subshell, $m_{l}$ runs $- 4 \to + 4$ , so the largest is $+ 4$ , giving $2 l + 1 = 9$ orbitals. Why this step? The most tilted orbital orientations grow with $l_{m a x}$ .

Step 3 — Capacity. $2 n^{2} = 2 (5)^{2} = 50$ electrons.

Verify: Long way: $2 + 6 + 10 + 14 + 18 = 50$ ✓ (s,p,d,f,g widths). This shows the limiting trend: as $n \to$ large, both the number of subshells ( $= n$ ) and the widest $m_{l}$ range ( $= 2 n - 1$ ) keep growing without bound. ✓

Example 8 — Case I: a real-world Zeeman word problem

Worked example A hydrogen atom sits in a magnetic field. One of its electrons is in a

2p subshell. Into how many distinct energy levels does that subshell split, and why weren't they split before the field was on? Forecast: Guess the number of split levels from the count of orbitals in 2p.

Step 1 — Identify the subshell. 2p means $n = 2$ , $l = 1$ .

Step 2 — Count the orbitals. $2 l + 1 = 2 (1) + 1 = 3$ , with $m_{l} = - 1, 0, + 1$ . Why this step? Each distinct $m_{l}$ is a distinct orientation of orbital angular momentum in space.

Step 3 — Apply the field. A magnetic field defines a $z$ -axis. An orbital's energy shift depends on its projection $L_{z} = m_{l} ℏ$ , so each $m_{l}$ now sits at a different energy. Three $m_{l}$ values → 3 energy levels (see Zeeman effect). Why this step? This is literally why $m_{l}$ is named the magnetic quantum number.

Step 4 — Why not before? With no field there is no special direction, so all three orientations have identical energy — they are degenerate. The field breaks that tie.

Verify: Splitting count $= 2 l + 1 = 3$ ✓. Sanity check with the parent's mistake box: " $m_{l}$ affects energy only in a magnetic field" — consistent ✓. ✓

Example 9 — Case J: the reverse (exam) twist

Worked example A certain

subshell contains exactly 10 electrons when full. (a) Identify $l$ and its letter. (b) What is the smallest shell $n$ this subshell can appear in? Forecast: Work the rules backwards: from a count to the quantum numbers. Guess the letter before reading.

Step 1 — Turn the count into orbitals. Each orbital holds 2 electrons, so orbitals $= 10/2 = 5$ . Why this step? Capacity of a subshell is $2 (2 l + 1)$ ; stripping the factor of 2 (the two spins) recovers the raw orbital count $2 l + 1$ , which is what actually pins down $l$ .

Step 2 — Solve for $l$ (answer to part a). Set $2 l + 1 = 5$ . Then $2 l = 4$ , so $l = 2$ . From the letter code, $l = 2$ is the d subshell. Why this step? $l$ is uniquely fixed once we know the orbital count — there is exactly one $l$ giving 5 orbitals.

Step 3 — Find the smallest $n$ (answer to part b). The nesting rule is $l \leq n - 1$ , i.e. $2 \leq n - 1$ , which rearranges to $n \geq 3$ . The smallest whole number satisfying this is $n = 3$ , giving the 3d subshell. Why this step? The $l$ -box must fit inside the $n$ -box; $n = 3$ is the first shell tall enough to hold $l = 2$ . Any smaller shell ( $n = 1$ or $2$ ) has an $l$ -ceiling of $0$ or $1$ , too low for $l = 2$ .

Answers: (a) $l = 2$ , the d subshell. (b) smallest shell is $n = 3$ → the 3d subshell.

Verify: 3d full capacity $= 2 (2 \cdot 2 + 1) = 10$ ✓ — matches the given count. Nesting check: $l = 2 \leq n - 1 = 2$ ✓ (the boundary case fits exactly). And $n = 2$ would fail since its $l$ -ceiling is $1 < 2$ ✓. ✓

Recall Quick self-test across the matrix

Which case class does each belong to, and is it valid? $(3, 3, 0, + \frac{1}{2})$ ::: Case C — invalid, $l = 3 > n - 1 = 2$ $(1, 0, 0, - \frac{1}{2})$ ::: Case B — valid, the second helium electron $(2, 1, + 2, + \frac{1}{2})$ ::: Case D — invalid, $m_{l} = + 2 > l = 1$ $(0, 0, 0, + \frac{1}{2})$ ::: Case E — invalid, $n$ cannot be $0$ How many electrons in the $n = 4$ shell? ::: Case F — $2 n^{2} = 32$ $∣ L ∣$ for a 3d electron? ::: Case G — $6 ℏ$ A subshell holding 14 electrons has which $l$ ? ::: Case J — $2 (2 l + 1) = 14 \Rightarrow l = 3$ (f)

Connections

Parent topic — the four numbers and their rules
Pauli exclusion principle — why each valid four-number address holds at most one electron
Aufbau principle and electron configuration — using these counts to fill atoms
Zeeman effect — Example 8's field-induced splitting of $m_{l}$ levels
Shapes of atomic orbitals (s, p, d) — the geometry behind Example 6 and the letter code
Hund's rule — how spins ( $m_{s}$ ) distribute across the orbitals we counted