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

Every reveal line has the shape Prompt ::: Answer with a reason. The reason is the point.

True or false — justify

Hund's rule decides which subshell (like

3 d

4 s

) fills first.

False. That ordering is set by the Aufbau Principle (

n + l

rule); Hund only arranges electrons within one already-chosen set of degenerate orbitals.

A subshell with more electrons always has higher spin multiplicity.

False. Multiplicity depends only on unpaired electrons

n

(

= 2 S + 1 = n + 1

). Filled

2 p^{6}

has

6

electrons but

n = 0

, so multiplicity

= 1

— lower than

2 p^{3}

's multiplicity of

4

Two electrons placed in different degenerate orbitals repel less than two crammed into the same orbital.

True. Different orbitals occupy different regions of space, so the average electron–electron distance is larger and Coulomb repulsion is smaller — one of the two reasons Hund's rule holds.

"Parallel spins" in Hund's rule means the electrons literally spin the same way physically.

Loosely true but subtle. It means they share the same spin quantum number

m_{s}

(both

+ \frac{1}{2}

), which lets them "exchange" and gain exchange stabilization; the classical picture of spinning balls is only a crutch.

Exchange energy is a form of electrostatic potential energy between charges.

False. Exchange energy is a purely quantum effect arising from the indistinguishability of same-spin electrons — it has no classical charge-repulsion analogue. See Exchange Energy and Half-filled Stability.

Once one electron pairs up in an orbital, Hund's rule has been violated.

False. Pairing is required once every degenerate orbital already holds one electron (e.g. oxygen

2 p^{4}

). Hund's rule is only violated if you pair before all orbitals are singly filled.

Increasing the number of parallel-spin electrons always increases the count of exchange pairs.

True. For

n

parallel electrons the pairs are

(2 n) = n (n - 1) /2

, which strictly increases with

n

— so more parallel spins means more stabilization.

A ground-state configuration must have the maximum possible total spin

S

True (this is Hund's first rule). Among allowed configurations, nature picks the one with the largest

S = n /2

, i.e. the most unpaired parallel electrons.

Spot the error

Claim: "Carbon

2 p^{2}

is written ↑↓ in

p_{x}

, leaving

p_{y}, p_{z}

empty." What's wrong?

Wrong — the two

2 p

electrons must go singly into different orbitals (↑ in

p_{x}

, ↑ in

p_{y}

), giving

n = 2

unpaired, not

0

. Pairing in

p_{x}

violates maximum multiplicity.

Claim: "Nitrogen

2 p^{3}

has multiplicity

2 S + 1 = 2 (3) + 1 = 7

." Find the slip.

The slip is

S = n

; actually

S = n /2

. With

n = 3

S = \frac{3}{2}

, so multiplicity

= 2 (\frac{3}{2}) + 1 = 4

, a quartet — not

7

Claim: "Oxygen

2 p^{4}

has

n = 4

unpaired electrons." Correct it.

No — after filling three orbitals singly (↑↑↑), the 4th electron pairs, leaving ↑↓ ↑ ↑, so only

n = 2

are unpaired. Multiplicity

= 3

, a triplet.

Claim: "Chromium is

[Ar] 3 d^{4} 4 s^{2}

because

4 s

fills before

3 d

." Why is the real answer different?

The real ground state is

[Ar] 3 d^{5} 4 s^{1}

: promoting one

4 s

electron gives a half-filled

3 d^{5}

with

10

exchange pairs instead of

6

, and that exchange stabilization outweighs the promotion cost.

Claim: "

F e^{2 +}

(

3 d^{6}

) has multiplicity

6

because it has

6

d

-electrons." Fix it.

Multiplicity counts unpaired, not total, electrons.

3 d^{6}

fills all five orbitals singly then pairs one: ↑↓ ↑ ↑ ↑ ↑, so

n = 4

, multiplicity

= 2 S + 1 = 5

Claim: "Since Hund maximises spin, an electron will jump to a higher-energy orbital to stay unpaired." Where does this reasoning break?

Hund's rule only distributes electrons among orbitals of the same energy (degenerate). It never moves an electron up to a genuinely higher subshell — the Aufbau Principle governs that.

Why questions

Why do electrons enter degenerate orbitals singly before pairing?

Two reasons: single occupancy keeps electrons in separate regions (less Coulomb repulsion), and it maximises the number of parallel-spin pairs, gaining exchange energy

- K

per pair.

Why is the multiplicity formula

2 S + 1

rather than just

S

A total spin

S

can orient in

2 S + 1

distinct ways (

m_{S} = - S, \dots, + S

in integer steps). Multiplicity literally counts those orientations, which is why maximizing it = maximizing

S

= maximizing unpaired electrons.

Why does

O_{2}

being paramagnetic trace back to Hund's rule?

Hund's rule forces oxygen's

2 p^{4}

(and

O_{2}

's antibonding

π^{*}

orbitals) to keep two unpaired parallel electrons; unpaired electrons make the molecule paramagnetic.

Why does going from

d^{4}

d^{5}

(keeping spins parallel) give a large stability jump?

Exchange pairs rise from

(2 4) = 6

(2 5) = 10

— four extra

- K

discounts — which is why half-filled

d^{5}

shells are unusually stable.

Why does maximum multiplicity relate to less electron–electron repulsion at all?

Maximum multiplicity forces single occupancy of separate orbitals first; electrons in different spatial orbitals stay farther apart on average, so Coulomb repulsion drops as a side benefit.

Why can't two same-spin electrons ever occupy the same orbital, forcing Hund's single-filling?

The Pauli Exclusion Principle forbids two electrons with identical quantum numbers; same orbital + same spin would be identical, so same-spin electrons must go into separate orbitals.

Edge cases

What is the multiplicity of a completely filled subshell such as

2 p^{6}

3 d^{10}

1

(a singlet). All electrons are paired, so

n = 0

S = 0

, and

2 S + 1 = 1

— no net spin, hence diamagnetic.

What is the multiplicity of a single lone electron, e.g. hydrogen

1 s^{1}

2

(a doublet). One unpaired electron gives

n = 1

S = \frac{1}{2}

, so

2 S + 1 = 2

How many exchange pairs exist when there is only

n = 1

parallel-spin electron?

Zero.

(2 1) = 0

— you need at least two same-spin electrons to form an exchange pair, so a single electron gets no exchange stabilization.

Does Hund's rule apply when there is exactly one electron to place in a degenerate set (e.g.

2 p^{1}

Trivially yes but vacuously — it simply occupies one orbital (

n = 1

), and there's no pairing decision to make. The rule only "bites" from the second electron onward.

For

3 d^{5}

3 d^{6}

, which has higher multiplicity, and why does adding an electron lower it?

3 d^{5}

has

n = 5

(multiplicity

6

);

3 d^{6}

must pair one electron, dropping to

n = 4

(multiplicity

5

). Adding the 6th electron destroys one unpaired spin, so multiplicity falls.

At half-filling (

p^{3}

d^{5}

f^{7}

), what is special about the unpaired count?

It is maximal for that subshell — every orbital holds exactly one parallel-spin electron, giving the largest

n

(and largest exchange stabilization) achievable before pairing must begin.

Connections

Aufbau Principle — sets subshell order; Hund arranges within a subshell.
Pauli Exclusion Principle — why same-spin electrons cannot share an orbital.
Electron Spin Quantum Number — origin of $m_{s}$ and total spin $S$ .
Exchange Energy and Half-filled Stability — the deep "why" of $d^{5}$ , $d^{10}$ .
Paramagnetism and Diamagnetism — unpaired ⇒ paramagnetic; paired ⇒ diamagnetic.
Magnetic Moment of Atoms — $μ = n (n + 2) μ_{B}$ .