Chemistry · Structure of Atom

Aufbau, Pauli, Hund's Rules

The ground-state arrangement of every electron in every atom is fixed by just three rules working together: the Aufbau principle, the Pauli exclusion principle and Hund's rule of maximum multiplicity. NCERT Class 11 Chemistry (Unit 2, Section 2.6.4) sets these out as the foundation for writing electronic configurations, and the half-filled and fully-filled stability that follows from them produces the chromium and copper anomalies that NEET examiners return to year after year.

The three rules at a glance

An atom in its ground state is the lowest-energy arrangement of its electrons. Reaching that arrangement is not guesswork — it follows a fixed procedure. The filling of electrons into the orbitals of different atoms takes place according to the Aufbau principle, which is itself built on the Pauli exclusion principle, Hund's rule of maximum multiplicity, and the relative energies of the orbitals. Each rule answers a different question, and only when all three act together does a unique configuration emerge.

It helps to keep the division of labour straight before the detail arrives. Aufbau fixes the order in which orbitals are filled; Pauli fixes the capacity of each orbital and subshell; Hund fixes how electrons distribute within a set of equal-energy orbitals.

RuleGovernsOne-line statement
Aufbau principleOrder of fillingOrbitals fill in order of increasing energy; lowest available orbital first.
Pauli exclusionCapacity per orbitalNo two electrons share all four quantum numbers; max two per orbital, opposite spins.
Hund's ruleDistribution within a subshellEach degenerate orbital gets one electron before any pairing begins; singles stay parallel.

The Aufbau principle and the (n+l) rule

The word aufbau is German for "building up." The building up of orbitals means the filling up of orbitals with electrons. The principle states that in the ground state of the atoms, the orbitals are filled in order of their increasing energies — electrons first occupy the lowest energy orbital available to them and enter higher energy orbitals only after the lower ones are filled.

That immediately raises a question: how do we rank orbital energies? In a multi-electron atom the energy of an orbital depends on both the principal quantum number $n$ and the azimuthal quantum number $l$, because electrons in different orbitals experience different effective nuclear charge through shielding. The exact dependence is complicated, but one simple rule captures it for examination purposes — the (n+l) rule:

The lower the value of $(n+l)$ for an orbital, the lower is its energy. If two orbitals have the same value of $(n+l)$, the orbital with the lower value of $n$ has the lower energy.

Worked across the early orbitals, the $(n+l)$ values explain the famous "out of sequence" filling. The case that confuses most students is 4s versus 3d: even though 3d carries the smaller principal quantum number, 4s is filled first.

Orbitalnl(n + l)Filled before
1s1012s
2s2022p
2p2133s (lower n wins the tie)
3s3033p
3p3144s (lower n wins the tie)
4s4043d
3d3254p (lower n wins the tie)
4p4155s
5s5054d
4d4265p

Reading the table, $4s$ has $(n+l)=4+0=4$ while $3d$ has $(n+l)=3+2=5$, so $4s$ is the lower-energy orbital and fills first. This is exactly why the valence electron of potassium, given a choice between 3d and 4s, is found in 4s. The tie cases — 2p versus 3s, 3p versus 4s, 3d versus 4p — are all settled by the second clause of the rule: when $(n+l)$ matches, lower $n$ wins.

Order of filling and the diagonal diagram

Applying the $(n+l)$ rule to every orbital produces a single useful sequence that NCERT recommends committing to memory:

$1s,\ 2s,\ 2p,\ 3s,\ 3p,\ 4s,\ 3d,\ 4p,\ 5s,\ 4d,\ 5p,\ 6s,\ 4f,\ 5d,\ 6p,\ 7s\ \dots$

Rather than re-derive every $(n+l)$ value in the exam hall, the order is read off a diagonal diagram. The subshells are written in rows by shell, and parallel arrows are drawn through them from top-right to bottom-left; following the arrows in turn gives the filling sequence.

1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p Follow each arrow from top-right to bottom-left; finish one arrow before starting the next. 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s …
Figure 1. The Aufbau diagonal diagram. Subshells are arranged in rows by principal quantum number; the parallel teal arrows, traced from top-right to bottom-left, reproduce the (n+l) filling order. With respect to the placement of outermost valence electrons it is remarkably accurate for all atoms, though it should be treated as a useful guide rather than an exact law.

NCERT is careful to caution that there is no single ordering of orbital energies that is universally correct for every atom. The sequence above is an extremely useful guide, but in many cases orbitals are close in energy and small changes in atomic structure can alter the order of filling. Exceptions, as the chromium and copper cases below show, do occur.

n Build the foundation

The (n+l) rule rests entirely on what n and l mean. If the four quantum numbers feel shaky, revisit Quantum Numbers before going further.

The Pauli exclusion principle

Aufbau tells us which orbital to fill next, but not how many electrons that orbital can hold. That limit comes from the exclusion principle, given by the Austrian scientist Wolfgang Pauli in 1926. According to this principle, no two electrons in an atom can have the same set of four quantum numbers. An equivalent and more usable statement is: only two electrons may exist in the same orbital, and these electrons must have opposite spin.

The reasoning is direct. An orbital is fixed by a single set of $n$, $l$ and $m_l$. Two electrons sharing that orbital therefore already agree on three of the four quantum numbers, so to remain distinct they must differ in the fourth — the spin quantum number $m_s$, taking $+\tfrac{1}{2}$ and $-\tfrac{1}{2}$. A third electron would be forced to repeat a complete set, which the principle forbids.

This restriction is what hands every subshell its capacity. Since the 1s subshell comprises one orbital, it holds at most two electrons; the p and d subshells, with three and five orbitals, hold a maximum of 6 and 10 respectively. In general the maximum number of electrons in the shell with principal quantum number $n$ is $2n^2$.

SubshelllOrbitalsMax electrons
s012
p136
d2510
f3714
one orbital ALLOWED ms = +½ and −½ one orbital FORBIDDEN repeats a full set of n, l, ml, ms Two electrons may share an orbital only with opposite spins. Parallel spins, or a third electron, would duplicate all four quantum numbers.
Figure 2. The Pauli exclusion principle in one orbital. An orbital is fixed by a single set of n, l and ml, so the two electrons it can hold must differ in the fourth quantum number — taking opposite spins ms = +½ and −½ (left, allowed). Two parallel-spin electrons, or a third electron, would force a complete repeat of all four quantum numbers and is forbidden (right).
NEET Trap

"Same orbital" still means a difference

A 2016 NEET item asked how two electrons occupying the same orbital are distinguished. The answer is the spin quantum number — not the magnetic, azimuthal or principal quantum number, all of which the two electrons share. Two electrons in one orbital are identical in $n$, $l$ and $m_l$; only $m_s$ tells them apart.

Same orbital ⇒ same n, l, ml; the distinguishing label is ms (spin).

Hund's rule of maximum multiplicity

Pauli caps an orbital at two electrons, but when a subshell has several orbitals of equal energy — degenerate orbitals such as the three p, five d or seven f orbitals — a further question arises: do incoming electrons pair up in one orbital, or spread out singly? Hund's rule of maximum multiplicity settles this. It states that pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital belonging to that subshell has got one electron each, i.e. is singly occupied. The single electrons keep parallel spins as far as possible.

A useful consequence follows directly. Since there are three p, five d and seven f orbitals, pairing begins with the entry of the 4th, 6th and 8th electron into the p, d and f subshells respectively. Before those electrons arrive, every orbital of the subshell holds a single, parallel-spin electron.

NEET Trap

Nitrogen's configuration is not 2px12py12pz1 with mixed spins

A 2018 NEET item flagged "the electronic configuration of N atom is $1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^1$" as the wrong statement — not because the orbital occupancy is wrong, but because in degenerate orbitals all unpaired electrons must show the same (parallel) spin. The occupancy is right; the implied spin pairing is the error Hund's rule rules out.

In a degenerate set, single electrons occupy separate orbitals with parallel spins — never paired prematurely.

Orbital diagrams in practice

Electronic configuration can be written two ways: the $s^a p^b d^c$ notation, or the orbital (box-and-arrow) diagram in which each orbital is a box and each electron an arrow — up arrow for one spin, down arrow for the opposite. The orbital diagram is more informative because the arrows display all four quantum numbers, spin included. Nitrogen and oxygen show Hund's rule cleanly: nitrogen ($1s^2\,2s^2\,2p^3$) keeps its three 2p electrons in separate boxes with parallel spin, while oxygen's fourth 2p electron is forced to pair.

N (Z = 7)
1s² 2s² 2p³
1s
2s
2p
O (Z = 8)
1s² 2s² 2p⁴
1s
2s
2p
Figure 3. Orbital diagrams for nitrogen and oxygen. Nitrogen's three 2p electrons stay singly occupied with parallel (up) spins, giving three unpaired electrons. Oxygen's fourth 2p electron is the first to pair — Hund's rule means pairing in the p subshell begins only with the 4th electron — leaving two unpaired electrons.
Worked example

How many unpaired electrons are present in ground-state oxygen, and which rule fixes the answer?

Oxygen ($Z=8$) has the configuration $1s^2\,2s^2\,2p^4$. By Hund's rule, the first three 2p electrons singly occupy the three 2p orbitals; the fourth must pair with one of them. That leaves two singly occupied 2p orbitals, so oxygen has two unpaired electrons. Hund's rule fixes the distribution; Pauli ensures the paired electrons in the doubly-filled 2p orbital carry opposite spin.

Half-filled and fully-filled stability

Hund's rule already hints at something deeper: a set of degenerate orbitals that is exactly half-filled or completely filled carries extra stability. The ground-state configuration of an atom always corresponds to the state of lowest total electronic energy, and NCERT identifies two physical reasons why $p^3$, $p^6$, $d^5$, $d^{10}$, $f^7$ and $f^{14}$ sets sit at unusually low energy.

CauseWhy it lowers the energy
Symmetrical distributionHalf-filled and completely filled subshells have a symmetrical distribution of electrons. Electrons in the same subshell have equal energy but different spatial distribution, so they shield one another relatively little and are held more strongly by the nucleus.
Exchange energyWhenever two or more electrons with the same spin occupy degenerate orbitals, they can exchange positions, releasing exchange energy. The number of possible exchanges is maximum when the subshell is exactly half-filled or completely filled — so the exchange energy, and the stability, is greatest there.

The exchange energy is, in fact, the underlying basis of Hund's rule itself — electrons that enter orbitals of equal energy adopt parallel spins as far as possible precisely because doing so maximises the number of exchanges. The extra stability of half-filled and completely filled subshells is therefore due to relatively small shielding, smaller coulombic repulsion energy, and larger exchange energy.

The chromium and copper anomaly

The clearest payoff of half-filled and fully-filled stability appears in the first transition series. Working through scandium to zinc, the five 3d orbitals fill progressively — but chromium and copper break the simple pattern. Their position would suggest $3d^4\,4s^2$ and $3d^9\,4s^2$, yet they are actually $3d^5\,4s^1$ and $3d^{10}\,4s^1$.

The explanation is exactly the stability argument above. In these atoms the 4s and 3d subshells differ only slightly in energy, so an electron shifts from the lower-energy 4s to the higher-energy 3d when that shift leaves the 3d subshell either exactly half-filled or completely filled. Chromium gains a half-filled $3d^5$ alongside a half-filled $4s^1$; copper gains a completely filled $3d^{10}$ alongside a half-filled $4s^1$.

Cr (Z = 24)
[Ar] 3d⁵ 4s¹
3d
4s
Cu (Z = 29)
[Ar] 3d¹⁰ 4s¹
3d
4s
Figure 4. Anomalous configurations of chromium and copper. Chromium adopts a half-filled 3d⁵ with a half-filled 4s¹ (six unpaired electrons); copper adopts a completely filled 3d¹⁰ with a half-filled 4s¹ (one unpaired electron). One electron has shifted from 4s into 3d in each case to reach the extra-stable arrangement.
NEET Trap

Memorise Cr and Cu, then count unpaired electrons carefully

The two exceptions every NEET candidate must know cold: chromium is $\ce{[Ar]}\,3d^5\,4s^1$ (not $3d^4\,4s^2$) and copper is $\ce{[Ar]}\,3d^{10}\,4s^1$ (not $3d^9\,4s^2$). Errors usually appear when counting unpaired electrons: Cr has 6 unpaired electrons (five in 3d plus one in 4s), whereas Cu has only 1 (the lone 4s electron, since 3d is fully paired).

Cr: 3d⁵4s¹ → 6 unpaired. Cu: 3d¹⁰4s¹ → 1 unpaired. NCERT itself adds the caution that further exceptions do exist.

One important caveat keeps these anomalies in proportion. NCERT writes its own warning beside the chromium and copper cases — "caution: exceptions do exist" — because the half-filled and fully-filled stability argument is a guideline, not an inviolable law. It accounts neatly for Cr and Cu in the 3d series, and similar effects appear among the heavier transition and inner-transition metals, but the diagonal Aufbau order remains the default starting point for every atom.

Quick Recap

Aufbau, Pauli, Hund's Rules

  • Aufbau principle: in the ground state, orbitals fill in order of increasing energy — lowest available orbital first.
  • (n+l) rule: lower $(n+l)$ means lower energy; for equal $(n+l)$, lower $n$ fills first. This puts 4s ($n+l=4$) before 3d ($n+l=5$).
  • Filling order: $1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,7s$, read off the diagonal diagram.
  • Pauli exclusion: no two electrons share all four quantum numbers; max two per orbital with opposite spins; shell capacity $=2n^2$.
  • Hund's rule: each degenerate orbital takes one parallel-spin electron before any pairing; p, d and f pairing starts at the 4th, 6th and 8th electron.
  • Stability: half-filled and fully-filled sets ($p^3,p^6,d^5,d^{10},f^7,f^{14}$) are extra-stable from symmetry and exchange energy — giving Cr $= 3d^5 4s^1$ and Cu $= 3d^{10}4s^1$.

NEET PYQ Snapshot — Aufbau, Pauli, Hund's Rules

Real NEET questions on spin distinction, orbital occupancy and the filling rules, drawn from the official papers.

NEET 2016

Two electrons occupying the same orbital are distinguished by:

  1. Magnetic quantum number
  2. Azimuthal quantum number
  3. Spin quantum number
  4. Principal quantum number
Answer: (3) Spin quantum number

Two electrons in the same orbital already share $n$, $l$ and $m_l$. By the Pauli exclusion principle they cannot share all four quantum numbers, so they must differ in the spin quantum number $m_s$ ($+\tfrac{1}{2}$ and $-\tfrac{1}{2}$).

NEET 2018

Which one is a wrong statement?

  1. Total orbital angular momentum of electron in 's' orbital is equal to zero
  2. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
  3. The electronic configuration of N atom is $1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^1$
  4. The value of $m$ for $d_{z^2}$ is zero
Answer: (3)

Statement (3) is the wrong one: in degenerate orbitals all unpaired electrons must have the same (parallel) spin per Hund's rule. The occupancy shown is correct, but the configuration is conventionally faulted because it does not show the parallel spins the rule demands.

Concept · Aufbau / (n+l)

In potassium ($Z=19$), the last electron enters which orbital, and why?

  1. 3d, because it has the lower principal quantum number
  2. 4s, because $(n+l)=4$ is lower than $(n+l)=5$ for 3d
  3. 4p, because it follows 3p directly
  4. 3p, because 3p is not yet full
Answer: (2) 4s

For 4s, $(n+l)=4+0=4$; for 3d, $(n+l)=3+2=5$. The lower $(n+l)$ orbital is lower in energy, so by the Aufbau principle the 19th electron occupies 4s, giving K $= \ce{[Ar]}\,4s^1$.

Concept · Cr/Cu anomaly

The number of unpaired electrons in ground-state chromium ($Z=24$) is:

  1. 4
  2. 5
  3. 6
  4. 2
Answer: (3) 6

Chromium adopts the extra-stable $\ce{[Ar]}\,3d^5\,4s^1$ rather than $3d^4\,4s^2$. The five 3d orbitals are singly occupied (5 unpaired) and the 4s holds one electron (1 unpaired), giving 6 unpaired electrons in total.

FAQs — Aufbau, Pauli, Hund's Rules

The questions students most often get wrong on the three filling rules and the Cr/Cu trap.

What is the difference between the Aufbau principle and the (n+l) rule?

The Aufbau principle is the general statement that, in the ground state, orbitals are filled in order of increasing energy — electrons occupy the lowest available orbital first. The (n+l) rule is the practical tool used to rank those energies: the lower the value of (n+l) for an orbital, the lower its energy, and if two orbitals have the same (n+l) value, the one with the lower n is filled first. So the (n+l) rule is the operational recipe that lets you apply the Aufbau principle. Together they give the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s.

Why is 4s filled before 3d even though 3d belongs to a lower shell?

By the (n+l) rule, the 4s orbital has (n+l) = 4+0 = 4, while the 3d orbital has (n+l) = 3+2 = 5. The orbital with the lower (n+l) value is lower in energy, so 4s (4) fills before 3d (5). This is why the valence electron of potassium goes into 4s and not 3d, even though 3d carries the smaller principal quantum number. The same logic explains why 5s precedes 4d and 6s precedes 4f.

What does the Pauli exclusion principle actually forbid?

The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers. Equivalently, only two electrons may occupy the same orbital, and those two must have opposite spin. The two electrons in a filled orbital share the same n, l and m-l values but differ in the spin quantum number. This restriction is what fixes the capacity of every subshell: 2 electrons in s, 6 in p, 10 in d, and a maximum of 2n-squared electrons in the shell of principal quantum number n.

Why do nitrogen's 2p electrons stay unpaired with parallel spins?

Hund's rule of maximum multiplicity states that pairing of electrons in the orbitals of a subshell does not begin until each orbital of that subshell holds one electron. Nitrogen has the configuration 1s2 2s2 2p3, so its three 2p electrons each occupy a separate 2p orbital singly, all with parallel spin, giving three unpaired electrons. Pairing in the p subshell starts only with the fourth electron, which is why oxygen (2p4) has its first paired 2p orbital and two unpaired electrons.

Why is chromium 3d5 4s1 and not 3d4 4s2?

Because exactly half-filled and completely filled subshells carry extra stability (lower energy). In chromium the 4s and 3d subshells differ only slightly in energy, so one electron shifts from 4s to 3d, giving 3d5 4s1 — a half-filled 3d and a half-filled 4s — instead of 3d4 4s2. The same effect makes copper 3d10 4s1 (a completely filled 3d) rather than 3d9 4s2. The stability arises from the symmetrical distribution of electrons and from the larger exchange energy of half-filled and fully-filled sets.

How do I write an orbital diagram correctly?

In the orbital (box-and-arrow) diagram each orbital of a subshell is drawn as a box and each electron as an arrow — an up arrow for one spin, a down arrow for the opposite spin. Fill boxes in order of increasing energy (Aufbau), place no more than two arrows per box and make them opposite (Pauli), and within a degenerate subshell add one arrow to each box before any pairing, keeping the single arrows parallel (Hund). The advantage of this notation over the s-a p-b d-c form is that the arrows display all four quantum numbers, including spin.