Hund's rule of maximum multiplicity: Exercises — Chemistry deep dive

Before starting, keep three tools in your pocket (all built in the parent Hund's rule of maximum multiplicity note):

Intuition Where does

$μ = n (n + 2) μ_{B}$ come from? A spinning electron is a tiny magnet. Quantum mechanics says the magnet strength attached to a total spin $S$ is $μ = 2 S (S + 1) μ_{B},$ where the leading $2$ is the electron's " $g$ -factor" (spin makes twice the magnetism you'd naively expect) and $S (S + 1)$ is the length of the spin vector — quantum spins never point perfectly along one axis, so their length is $S (S + 1)$ , not just $S$ . WHY switch to $n$ ? Every unpaired electron adds $+ \frac{1}{2}$ to $S$ , so $S = \frac{n}{2}$ . Substitute:

= 2\sqrt{\tfrac{n(n+2)}{4}}\;\mu_B = \sqrt{n(n+2)}\;\mu_B.$$ The two "$2$"s cancel through the $\tfrac14$, leaving the clean **spin-only** formula. It is called *spin-only* because we ignored any magnetism from orbital motion — an excellent approximation for most first-row transition-metal ions. See [[Magnetic Moment of Atoms]].

Whenever you draw a subshell below, I mean a row of boxes (one box per orbital), with arrows ↑ (spin up) or ↓ (spin down). A box with ↑↓ is a paired orbital.

Figure 1 below shows exactly how to read these box diagrams — glance at it once and the rest of the page is just applying it.

Figure — Hund's rule of maximum multiplicity

Figure 1 — Reading orbital-box diagrams. Top (blue): a $p$ subshell filled Hund-style, three boxes each with a single red up-arrow (parallel). Bottom (yellow): after all three are singly filled, the fourth electron pairs into the left box as a green down-arrow (↑↓). The legend on the right fixes the convention: red arrow = up-spin ( $+ \frac{1}{2}$ ), green arrow = down-spin ( $- \frac{1}{2}$ ). Everywhere below, a plain ↑ means "red up-arrow" and a plain ↓ means "green down-arrow".

Level 1 — Recognition

(Can you read the rule off a picture?)

L1.1

State how many unpaired electrons are in the ground-state $2 p^{2}$ configuration, and write its multiplicity.

Recall Solution

WHAT: Two electrons, three degenerate $p$ boxes ( $p_{x}, p_{y}, p_{z}$ ). WHY singly: Hund says fill empty boxes one-at-a-time, parallel, before doubling up. So: ↑ ↑ _ (exactly the top row of Figure 1, minus one arrow). No box has a pair yet.

Unpaired $n = 2$ .
Multiplicity $= n + 1 = 2 + 1 = 3$ .

L1.2

Which of these obeys Hund's rule for $2 p^{3}$ ? (a) ↑↓ ↑ _ (b) ↑ ↑ ↑ (c) ↑↓ ↓ _

Recall Solution

(b) is correct: three boxes, three electrons, each alone and parallel (all up) — this is the top row of Figure 1. (a) pairs one box while a box is still empty — forbidden by Hund. (c) has anti-parallel spins, so it does not maximize total spin.

For (b): $n = 3$ , multiplicity $= 4$ . Answer: (b).

Level 2 — Application

(Plug real atoms into the formulas.)

L2.1

For the oxygen atom ( $Z = 8$ , $1 s^{2} 2 s^{2} 2 p^{4}$ ): find $n$ , $S$ , multiplicity, and $μ$ .

Recall Solution

WHAT: four $p$ electrons in three boxes. Fill singly first (↑ ↑ ↑), then the 4th must pair (no empty degenerate box left): ↑↓ ↑ ↑ — precisely the bottom row of Figure 1.

Unpaired $n = 2$ .
$S = n /2 = 1$ .
Multiplicity $= n + 1 = 3$ (a "triplet").
$μ = n (n + 2) = 2 \cdot 4 = 8 \approx 2.83 μ_{B}$ .

L2.2

Manganese is $[Ar] 3 d^{5} 4 s^{2}$ . For the $3 d^{5}$ set, find the number of unpaired electrons and the number of exchange pairs.

Recall Solution

WHAT: five $d$ boxes, five electrons → each box gets exactly one, all parallel: ↑ ↑ ↑ ↑ ↑ (a half-filled shell).

Unpaired $n = 5$ .
Exchange pairs $= (2 5) = \frac{5 \cdot 4}{2} = 10$ (all counted within the one $3 d$ subshell).
Multiplicity $= 6$ (a "sextet").

L2.3

Find $μ$ for $F e^{2 +}$ ( $[Ar] 3 d^{6}$ ).

Recall Solution

Five boxes: fill singly (↑ ↑ ↑ ↑ ↑), the 6th pairs → ↑↓ ↑ ↑ ↑ ↑.

Unpaired $n = 4$ .
$μ = 4 \cdot 6 = 24 \approx 4.90 μ_{B}$ .

Level 3 — Analysis

(Compare configurations; explain the "why".)

L3.1

Compute the change in exchange pairs when going $d^{4} \to d^{5}$ while keeping all spins parallel. Interpret the sign.

Recall Solution

$d^{4}$ : $n = 4$ , pairs $= (2 4) = \frac{4 \cdot 3}{2} = 6$ .
$d^{5}$ : $n = 5$ , pairs $= (2 5) = 10$ .
Change $= 10 - 6 = + 4$ pairs. Interpret: each pair lowers energy by the exchange integral $K$ (defined at the top of this page), so going $6 \to 10$ pairs gains an extra $- 4 K$ of stabilization. A big discount for just one more parallel electron — this is why the half-filled shell is a stability landmark. See Exchange Energy and Half-filled Stability.

L3.2

Chromium could be $[Ar] 3 d^{4} 4 s^{2}$ (naïve Aufbau) or the observed $[Ar] 3 d^{5} 4 s^{1}$ . Using unpaired-electron count and exchange pairs, argue which is more stable.

Recall Solution

Read the answer straight off Figure 2, which draws both configurations as box rows. Promoting one electron from $4 s$ into the last empty $3 d$ box takes $3 d^{4}$ (top) to a fully half-filled $3 d^{5}$ (bottom):

$3 d^{4}$ : 4 unpaired → $(2 4) = 6$ exchange pairs.
$3 d^{5}$ : 5 unpaired → $(2 5) = 10$ exchange pairs.

Why only the $3 d$ electrons count: exchange pairs are counted within one degenerate subshell. The lone $4 s$ electron in $3 d^{5} 4 s^{1}$ is parallel to the $3 d$ electrons, but $4 s$ and $3 d$ are different energies, so it forms no exchange pair with the $3 d$ set.

Moving one electron from $4 s$ to $3 d$ costs a little promotion energy but buys 4 extra exchange pairs ( $6 \to 10$ , worth $- 4 K$ ) plus a fully half-filled, spread-out $d$ shell (less repulsion). The exchange gain wins, so nature picks $[Ar] 3 d^{5} 4 s^{1}$ . (This is the Aufbau Principle "exception" — really Hund + exchange overriding it.)

Figure 2 — Why Cr is $[Ar] 3 d^{5} 4 s^{1}$ . Top (blue): $3 d^{4}$ with one empty box → 4 unpaired, 6 exchange pairs. Bottom (yellow): promote one electron so all five $d$ boxes are singly filled → 5 unpaired, 10 exchange pairs. The green arrow marks the trade: promoting one electron gains +4 exchange pairs ( $- 4 K$ ), which outweighs the promotion cost.

Level 4 — Synthesis

(Combine Aufbau + Pauli + Hund + magnetism in one problem.)

L4.1

Build the ground-state configuration of sulphur ( $Z = 16$ ) from scratch, then report unpaired count, multiplicity, $μ$ , and whether it is paramagnetic.

Recall Solution

Step 1 — Aufbau (which subshells, in order): $1 s 2 s 2 p 3 s 3 p$ → fill 16 electrons: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4} .$ Step 2 — Pauli: each orbital holds ≤2, opposite spins. All inner shells are closed. Step 3 — Hund on the open $3 p^{4}$ : three boxes, 4 electrons → ↑↓ ↑ ↑.

Unpaired $n = 2$ .
Multiplicity $= 3$ .
$μ = 2 \cdot 4 = 8 \approx 2.83 μ_{B}$ .
Unpaired electrons present ⇒ paramagnetic (Paramagnetism and Diamagnetism). (Note $3 p^{4}$ mirrors oxygen's $2 p^{4}$ — same box picture, same answers.)

L4.2

A transition-metal ion $M^{3 +}$ has $μ = 3.87 μ_{B}$ . Find the number of unpaired electrons and identify a $d^{n}$ config consistent with it.

Recall Solution

WHY $μ$ first: the measured magnetic moment counts unpaired spins, so invert the formula. $μ = n (n + 2) \Rightarrow n (n + 2) = μ^{2} = 3.8 7^{2} \approx 15.0.$ Try $n = 3$ : $3 \cdot 5 = 15$ . ✔ So $n = 3$ unpaired electrons. A $d^{3}$ configuration (↑ ↑ ↑ _ _) fits, e.g. $C r^{3 +} [Ar] 3 d^{3}$ — and with only three $d$ electrons in five boxes, Hund gives all three singly and parallel no matter what, so the unpaired count is $3$ regardless of surroundings.

Level 5 — Mastery

(Edge cases, zero inputs, and full reasoning chains.)

L5.1

Give the multiplicity and $μ$ of a fully-filled subshell such as $2 p^{6}$ , and of the noble gas neon overall. Explain why the answer is a special ( $n = 0$ ) case.

Recall Solution

$2 p^{6}$ : all boxes ↑↓ ↑↓ ↑↓ — every electron paired.

Unpaired $n = 0$ .
$S = 0$ , multiplicity $= n + 1 = 1$ (a "singlet").
$μ = 0 \cdot 2 = 0 μ_{B}$ → diamagnetic. Neon $1 s^{2} 2 s^{2} 2 p^{6}$ has all shells closed, so $n = 0$ throughout: multiplicity $1$ , $μ = 0$ . This is the zero-input edge case of Hund's rule: the three $2 p$ orbitals were still degenerate (same energy) while filling, but once every one is doubly occupied there is nothing left to spread out. So "degenerate orbitals" (the same-energy $p$ set) and " $n = 0$ " (no unpaired electrons) are two different facts — Hund still applies, it simply produces the trivial answer because no orbital is partially filled.

L5.2

For each $d^{n}$ ( $n = 1 \dots 10$ , high-spin), tabulate unpaired electrons $u$ and exchange pairs $P = (2 u)$ . Where does $P$ peak, and why does that predict half-filled stability?

Recall Solution

Fill 5 boxes singly (up to $d^{5}$ ), then start pairing (down-spins) for $d^{6}$ – $d^{10}$ . The box column below spells out all five orbitals explicitly (↑↓ = paired, ↑ = single, _ = empty):

$d^{n}$	boxes (5 orbitals)	unpaired $u$	pairs $P = (2 u)$
$d^{1}$	↑ _ _ _ _	1	0
$d^{2}$	↑ ↑ _ _ _	2	1
$d^{3}$	↑ ↑ ↑ _ _	3	3
$d^{4}$	↑ ↑ ↑ ↑ _	4	6
$d^{5}$	↑ ↑ ↑ ↑ ↑	5	10
$d^{6}$	↑↓ ↑ ↑ ↑ ↑	4	6
$d^{7}$	↑↓ ↑↓ ↑ ↑ ↑	3	3
$d^{8}$	↑↓ ↑↓ ↑↓ ↑ ↑	2	1
$d^{9}$	↑↓ ↑↓ ↑↓ ↑↓ ↑	1	0
$d^{10}$	↑↓ ↑↓ ↑↓ ↑↓ ↑↓	0	0

Peak: $P = 10$ at $d^{5}$ (half-filled). Beyond it, every added electron must pair, reducing $u$ and therefore $P$ . The symmetric rise-then-fall around $d^{5}$ is exactly why the half-filled shell is a stability maximum — and why $d^{10}$ (all paired, $P = 0$ but a closed, symmetric shell) is the other special landmark. Figure 3 plots this $P$ -versus- $d^{n}$ curve so you can see the single sharp peak at $d^{5}$ . See Magnetic Moment of Atoms.

Figure 3 — Exchange pairs peak at the half-filled shell. Horizontal axis: number of $d$ electrons ( $d^{n}$ , $n = 1$ – $10$ ). Vertical axis: exchange pairs $P = (2 u)$ (a pure count, no units). The yellow curve rises to a single maximum of 10 at $d^{5}$ (green dashed line), then falls symmetrically back to $0$ at $d^{10}$ — the visual signature of half-filled stability.

L5.3

$C u^{+}$ vs $C u^{2 +}$ : give $d^{n}$ , unpaired count and $μ$ for each. Which is diamagnetic?

Recall Solution

Copper atom is $[Ar] 3 d^{10} 4 s^{1}$ .

$C u^{+}$ : lose the $4 s$ electron → $[Ar] 3 d^{10}$ . Unpaired $n = 0$ , $μ = 0$ → diamagnetic.
$C u^{2 +}$ : $[Ar] 3 d^{9}$ → ↑↓ ↑↓ ↑↓ ↑↓ ↑. Unpaired $n = 1$ , $μ = 1 \cdot 3 = 3 \approx 1.73 μ_{B}$ → paramagnetic.

Active Recall

Recall Rapid fire (forecast, then reveal)

Multiplicity of $2 p^{2}$ ? ::: 3 (n=2) Unpaired electrons in $F e^{2 +}$ ( $3 d^{6}$ )? ::: 4 Exchange pairs in $d^{5}$ ? ::: 10 $μ$ for $n = 3$ unpaired? ::: $15 \approx 3.87 μ_{B}$ Where does $n (n + 2)$ come from? ::: $μ = 2 S (S + 1)$ with $S = n /2$ Multiplicity and $μ$ of $2 p^{6}$ ? ::: 1 and 0 Is $C u^{+}$ ( $d^{10}$ ) para- or diamagnetic? ::: Diamagnetic Extra exchange pairs gained going $d^{4} \to d^{5}$ ? ::: 4 (from 6 to 10)

Connections

Hund's rule of maximum multiplicity — the parent rule this set drills.
Aufbau Principle — Step 1 of every L4/L5 build (which subshell).
Pauli Exclusion Principle — caps boxes at 2, opposite spins.
Electron Spin Quantum Number — where ↑/↓ and $S = n /2$ come from.
Exchange Energy and Half-filled Stability — the $(2 n)$ argument for Cr, Mn.
Paramagnetism and Diamagnetism — unpaired ⇒ paramagnetic (L4.1, L5.3).
Magnetic Moment of Atoms — the $μ = 2 S (S + 1) = n (n + 2)$ tool.