Chapter 2 · Structure of Atom

2.4 Electronic Configuration of Atoms

How electrons fill the orbitals of an atom: the Aufbau principle and the (n + l) rule, Pauli's exclusion principle and Hund's rule, configurations of atoms and ions up to krypton, and why chromium and copper break the pattern.

Sodium is a soft metal that reacts violently with water; neon is a gas that reacts with nothing; chlorine is a poisonous gas that grabs electrons from almost anything. They differ by only one or two protons. What makes their chemistry so different is the way their electrons are arranged — in particular, how many electrons sit in the outermost shell and whether a subshell is full, half full or nearly empty.

The distribution of an atom's electrons among its orbitals is called its electronic configuration. Section 2.3 gave us the orbitals and their energies; this section gives the three rules that decide how electrons occupy them in the ground state: build up from the lowest energy (Aufbau), never more than two to an orbital (Pauli), and spread out before pairing (Hund). With them you can write the configuration of any atom up to krypton and beyond, explain the two famous exceptions, chromium and copper, and work out the configurations of ions — the foundation for the periodic table in Chapter 3 and for bonding in Chapter 4.

The Aufbau Principle and the $(n+l)$ Rule

"Aufbau" is German for "building up". Imagine building an atom by adding electrons one at a time to the nucleus; each new electron goes into the lowest-energy orbital still available. Section 2.3 showed that in a many-electron atom the energy of an orbital depends on both $n$ and $l$, so we need a rule for the order.

📐 Rule 2.13 — The Aufbau principle and the $(n+l)$ rule

In the ground state of an atom, electrons occupy orbitals in order of increasing energy: an orbital of higher energy begins to fill only when those of lower energy are full. The order of energies is given by the $(n+l)$ rule:

  1. the subshell with the lower value of $(n+l)$ has the lower energy and fills first;
  2. if two subshells have the same $(n+l)$, the one with the lower $n$ fills first.

This gives the filling order

$$1s\lt2s\lt2p\lt3s\lt3p\lt4s\lt3d\lt4p\lt5s\lt4d\lt5p\lt6s\lt4f\lt5d\lt6p\lt7s\lt5f\lt6d\lt7p$$
✏️ Example 1 — Using the $(n+l)$ rule
(a) Arrange $4d$, $5s$, $3d$, $4p$ and $4s$ in the order in which they are filled. (b) Which fills first, $4f$ or $5d$? $6s$ or $4f$? (c) Why does potassium put its 19th electron into $4s$ rather than $3d$?
(a) n + l $4d$: $4+2=6$; $5s$: $5+0=5$; $3d$: $3+2=5$; $4p$: $4+1=5$; $4s$: $4+0=4$.
(a) order Lowest $(n+l)$ first; ties broken by lower $n$: $4s\,(4)\lt3d\,(5,n=3)\lt4p\,(5,n=4)\lt5s\,(5,n=5)\lt4d\,(6)$.
(b) $4f$ and $5d$ both have $n+l=7$; $4f$ has the lower $n$, so $4f$ first. $6s$ has $n+l=6\lt7$, so $6s$ fills before $4f$.
(c) $4s$ has $n+l=4$, $3d$ has $5$: in potassium the $4s$ orbital is lower in energy, because a $4s$ electron penetrates the inner $[\mathrm{Ar}]$ core much more than a $3d$ electron does (Rule 2.12).

Pauli's Exclusion Principle and Hund's Rule

How many electrons can share one orbital? In 1925 Wolfgang Pauli answered with a principle that holds for every electron in every atom.

📐 Principle 2.14 — Pauli's exclusion principle
No two electrons in an atom can have the same set of all four quantum numbers. Equivalently: an orbital can hold at most two electrons, and they must have opposite spins ($m_{s}=+\tfrac12$ and $-\tfrac12$, drawn $\uparrow\downarrow$). Two electrons in one orbital are said to be paired. It follows that a subshell holds at most $2(2l+1)$ electrons and a shell at most $2n^{2}$ (Rule 2.10).

When a subshell has several degenerate orbitals — three p, five d — a further choice arises. Should the second electron in a p subshell pair up with the first, or go into a different p orbital?

📐 Rule 2.15 — Hund's rule of maximum multiplicity
When electrons occupy degenerate orbitals (orbitals of the same subshell), pairing does not begin until each orbital of that subshell holds one electron. The singly occupied orbitals all have electrons with parallel spins. So the p subshell pairs only on its fourth electron, the d subshell on its sixth, and the f subshell on its eighth.

Hund's rule follows from electron repulsion: two electrons in different orbitals are, on average, farther apart than two in the same orbital, and electrons with parallel spins avoid each other further still. An atom or ion with one or more unpaired electrons is attracted into a magnetic field (it is paramagnetic); one with all electrons paired is weakly repelled (diamagnetic). Measuring this is how chemists count unpaired electrons.

📘 Definition 2.11 — Electronic configuration and orbital diagram

The electronic configuration of an atom lists its occupied subshells with the number of electrons as a superscript, e.g. nitrogen $1s^{2}2s^{2}2p^{3}$. An orbital diagram shows each orbital as a box (or line) and each electron as an arrow, which makes pairing and spin visible:

N:   1s [↑↓]   2s [↑↓]   2p [↑][↑][↑]

A configuration may be shortened by writing the configuration of the preceding noble gas as a noble-gas core in square brackets: sodium $1s^{2}2s^{2}2p^{6}3s^{1}=[\mathrm{Ne}]\,3s^{1}$. Electrons in the outermost shell (and, for the d-block, the $(n-1)d$ electrons) are the valence electrons that take part in bonding; the electrons of the noble-gas core are core electrons.

✏️ Example 2 — Spot the broken rule
Each orbital diagram below is meant to show a ground-state atom. Say which rule (Aufbau, Pauli or Hund) each breaks, if any.
(a) C: 1s [↑↓] 2s [↑↓] 2p [↑↓][ ][ ]   (b) N: 1s [↑↓] 2s [↑↓] 2p [↑][↓][↑]   (c) Be: 1s [↑↓] 2s [↑] 2p [↑][ ][ ]   (d) He: 1s [↑↑]   (e) O: 1s [↑↓] 2s [↑↓] 2p [↑↓][↑][↑]
(a) Hund: the two 2p electrons pair up while empty 2p orbitals are available. Correct: $2p$ [↑][↑][ ].
(b) Hund: each 2p orbital has one electron, but the spins are not all parallel. Correct: [↑][↑][↑].
(c) Aufbau: an electron sits in $2p$ while $2s$ is not full (this is an excited state of Be). Correct: $1s^{2}2s^{2}$.
(d) Pauli: two electrons in one orbital with the same spin would have identical quantum numbers. Correct: [↑↓].
(e) None — the fourth 2p electron must pair, and two orbitals remain singly occupied: oxygen has 2 unpaired electrons.

Electronic Configurations of the Elements

Applying the three rules in turn builds up the configurations of all the elements. The table gives the ground-state configurations up to krypton ($Z=36$); ★ marks the two exceptions explained below.

ZEl.ConfigurationZEl.ConfigurationZEl.Configuration
1H$1s^{1}$13Al$[\mathrm{Ne}]\,3s^{2}3p^{1}$25Mn$[\mathrm{Ar}]\,3d^{5}4s^{2}$
2He$1s^{2}$14Si$[\mathrm{Ne}]\,3s^{2}3p^{2}$26Fe$[\mathrm{Ar}]\,3d^{6}4s^{2}$
3Li$[\mathrm{He}]\,2s^{1}$15P$[\mathrm{Ne}]\,3s^{2}3p^{3}$27Co$[\mathrm{Ar}]\,3d^{7}4s^{2}$
4Be$[\mathrm{He}]\,2s^{2}$16S$[\mathrm{Ne}]\,3s^{2}3p^{4}$28Ni$[\mathrm{Ar}]\,3d^{8}4s^{2}$
5B$[\mathrm{He}]\,2s^{2}2p^{1}$17Cl$[\mathrm{Ne}]\,3s^{2}3p^{5}$29Cu$[\mathrm{Ar}]\,3d^{10}4s^{1}$ ★
6C$[\mathrm{He}]\,2s^{2}2p^{2}$18Ar$[\mathrm{Ne}]\,3s^{2}3p^{6}$30Zn$[\mathrm{Ar}]\,3d^{10}4s^{2}$
7N$[\mathrm{He}]\,2s^{2}2p^{3}$19K$[\mathrm{Ar}]\,4s^{1}$31Ga$[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{1}$
8O$[\mathrm{He}]\,2s^{2}2p^{4}$20Ca$[\mathrm{Ar}]\,4s^{2}$32Ge$[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{2}$
9F$[\mathrm{He}]\,2s^{2}2p^{5}$21Sc$[\mathrm{Ar}]\,3d^{1}4s^{2}$33As$[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{3}$
10Ne$[\mathrm{He}]\,2s^{2}2p^{6}$22Ti$[\mathrm{Ar}]\,3d^{2}4s^{2}$34Se$[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{4}$
11Na$[\mathrm{Ne}]\,3s^{1}$23V$[\mathrm{Ar}]\,3d^{3}4s^{2}$35Br$[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{5}$
12Mg$[\mathrm{Ne}]\,3s^{2}$24Cr$[\mathrm{Ar}]\,3d^{5}4s^{1}$ ★36Kr$[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{6}$

Notice that the $3d$ subshell is written before $4s$ in a configuration, grouping orbitals by shell, even though $4s$ fills first. This is the convention used in the syllabus, and it makes the configurations of ions easier to read (below). Writing $[\mathrm{Ar}]\,4s^{2}3d^{6}$ for iron, in filling order, is also accepted.

Interactive 1 — Aufbau Filler: Building Atoms from H to Kr

Each box is an orbital, arranged by energy from $1s$ at the bottom to $4p$ at the top. Move the slider (or press ▶ Play) to add electrons one at a time: each goes to the lowest free orbital (Aufbau), at most two per box with opposite spins (Pauli), and singly with parallel spins before pairing (Hund). Watch what happens at chromium ($Z=24$) and copper ($Z=29$).

8
element—
configuration—
unpaired electrons—
magnetic behaviour—
last electron enters—
✏️ Example 3 — Writing configurations
Write the full and the noble-gas-core configurations of (a) phosphorus ($Z=15$), (b) calcium ($Z=20$), (c) iron ($Z=26$) and (d) bromine ($Z=35$), and give the number of unpaired electrons in each.
(a) P $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}=[\mathrm{Ne}]\,3s^{2}3p^{3}$. By Hund's rule the three $3p$ electrons occupy separate orbitals: 3 unpaired.
(b) Ca $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}=[\mathrm{Ar}]\,4s^{2}$: 0 unpaired (diamagnetic).
(c) Fe $26$ electrons: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}$ uses $20$, leaving $6$ for $3d$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^{2}=[\mathrm{Ar}]\,3d^{6}4s^{2}$. Five $3d$ electrons go in singly, the sixth pairs: [↑↓][↑][↑][↑][↑] — 4 unpaired.
(d) Br $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{5}$; the $4p$ subshell is [↑↓][↑↓][↑]: 1 unpaired. (Check: $18+10+2+5=35$ ✓.)
✏️ Example 4 — Counting electrons by quantum number
(a) Give a set of four quantum numbers for the last electron added (the "differentiating electron") in sodium and in scandium. (b) How many electrons in an iron atom have $l=0$, $l=1$ and $l=2$? (c) How many electrons in a zinc atom have $m_{l}=0$?
(a) Na: the last electron enters $3s$: $n=3$, $l=0$, $m_{l}=0$, $m_{s}=+\tfrac12$. Sc: it enters $3d$: $n=3$, $l=2$, $m_{l}$ any one of $-2\ldots+2$ (e.g. $-2$), $m_{s}=+\tfrac12$. (By convention the first electron in a subshell is given $m_{s}=+\tfrac12$; the sign itself has no physical significance.)
(b) Fe $=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^{2}$. $l=0$ (s): $2+2+2+2=8$. $l=1$ (p): $6+6=12$. $l=2$ (d): $6$. (Total $26$ ✓.)
(c) Zn $=[\mathrm{Ar}]\,3d^{10}4s^{2}$, every subshell full. $m_{l}=0$ orbitals: $1s,2s,3s,4s$ (one each), one in $2p$, one in $3p$, one in $3d$ — seven orbitals, each with $2$ electrons: 14 electrons.

Stability of Half-filled and Filled Subshells; Configurations of Ions

The Aufbau order predicts $[\mathrm{Ar}]\,3d^{4}4s^{2}$ for chromium and $[\mathrm{Ar}]\,3d^{9}4s^{2}$ for copper. Experiment (spectra and magnetic measurements) shows instead

$$\mathrm{Cr}:\ [\mathrm{Ar}]\,3d^{5}4s^{1},\qquad\mathrm{Cu}:\ [\mathrm{Ar}]\,3d^{10}4s^{1}.$$

In each case one $4s$ electron has moved into $3d$, giving a half-filled ($d^{5}$) or completely filled ($d^{10}$) $3d$ subshell. Because $3d$ and $4s$ are so close in energy, the small extra stability of these arrangements is enough to tip the balance.

📐 Rule 2.16 — Extra stability of half-filled and completely filled subshells

Subshells that are exactly half-filled ($p^{3}$, $d^{5}$, $f^{7}$) or completely filled ($p^{6}$, $d^{10}$, $f^{14}$) are more stable than neighbouring arrangements, for two reasons:

  1. Symmetry. The electron distribution is symmetrical — each orbital of the subshell holds the same number of electrons — which minimises repulsion and lets every electron be held equally by the nucleus.
  2. Exchange energy. Electrons with parallel spins in degenerate orbitals can exchange their positions; every such exchange pair lowers the energy. The number of exchange pairs among $k$ parallel spins is $\dfrac{k(k-1)}{2}$, so it is largest when the subshell holds as many parallel spins as possible — when it is half-filled (or, for $d^{10}$, when each spin set is complete).
✏️ Example 5 — Chromium and copper
(a) Draw the $3d$ and $4s$ orbital diagrams for the predicted and the actual configurations of chromium, and count the exchange pairs in the $3d$ subshell in each. (b) How many unpaired electrons do $\mathrm{Cr}$ and $\mathrm{Cu}$ have? (c) Molybdenum lies directly below chromium. Predict its configuration from that of chromium.
(a) predicted $3d^{4}4s^{2}$: $3d$ [↑][↑][↑][↑][ ] $4s$ [↑↓]. Four parallel $3d$ spins: $\dfrac{4\times3}{2}=6$ exchange pairs.
(a) actual $3d^{5}4s^{1}$: $3d$ [↑][↑][↑][↑][↑] $4s$ [↑]. Five parallel spins: $\dfrac{5\times4}{2}=10$ exchange pairs — four more, and a symmetrical half-filled subshell; moving the electron also removes the pairing repulsion in $4s$.
(b) Cr: $5+1=$ 6 unpaired. Cu ($3d^{10}4s^{1}$): only the $4s$ electron, 1 unpaired.
(c) By analogy $\mathrm{Mo}$: $[\mathrm{Kr}]\,4d^{5}5s^{1}$, which is correct. (Silver, below copper, is likewise $[\mathrm{Kr}]\,4d^{10}5s^{1}$. Further down the table, where $(n-1)d$, $ns$ and $(n-2)f$ lie very close together, there are several more exceptions; you are not expected to memorise them.)

Configurations of ions

To write the configuration of an ion, start from the configuration of the neutral atom. For an anion, add electrons by the Aufbau order. For a cation, remove electrons from the subshell with the highest $n$ first (and, within that shell, the highest $l$) — not simply the last one filled. For the first-row transition elements this means the $4s$ electrons leave before the $3d$ electrons: once $3d$ is occupied, it lies below $4s$ in energy.

⚠️ The 4s-first trap

$\mathrm{Fe}=[\mathrm{Ar}]\,3d^{6}4s^{2}$, so $\mathrm{Fe^{2+}}=[\mathrm{Ar}]\,3d^{6}$ and $\mathrm{Fe^{3+}}=[\mathrm{Ar}]\,3d^{5}$ — not $[\mathrm{Ar}]\,3d^{4}4s^{2}$. "Last in, first out" is wrong for transition-metal ions. Writing the configuration of the atom with $3d$ before $4s$ (by shell) makes the rule easy: remove electrons from the right-hand end.

The half-filled $3d^{5}$ of $\mathrm{Fe^{3+}}$ helps explain why iron(II) compounds are readily oxidised to iron(III) in air, and why $\mathrm{Mn^{2+}}$ ($3d^{5}$) is hard to oxidise.

✏️ Example 6 — Ions
Write the configurations and give the number of unpaired electrons of (a) $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$; (b) $\mathrm{Cu^{+}}$ and $\mathrm{Cu^{2+}}$; (c) $\mathrm{Cl^{-}}$ and $\mathrm{Zn^{2+}}$. Which of these species are diamagnetic? Which noble gas is $\mathrm{Cl^{-}}$ isoelectronic with?
(a) $\mathrm{Fe^{2+}}$: remove both $4s$ electrons: $[\mathrm{Ar}]\,3d^{6}$, 4 unpaired. $\mathrm{Fe^{3+}}$: then one $3d$ electron: $[\mathrm{Ar}]\,3d^{5}$, 5 unpaired (half-filled, the more stable ion).
(b) $\mathrm{Cu}=[\mathrm{Ar}]\,3d^{10}4s^{1}$. $\mathrm{Cu^{+}}=[\mathrm{Ar}]\,3d^{10}$, 0 unpaired. $\mathrm{Cu^{2+}}=[\mathrm{Ar}]\,3d^{9}$, 1 unpaired.
(c) $\mathrm{Cl^{-}}$: $[\mathrm{Ne}]\,3s^{2}3p^{5}$ + one electron $=[\mathrm{Ne}]\,3s^{2}3p^{6}$, the configuration of argon (both have $18$ electrons — they are isoelectronic). $\mathrm{Zn^{2+}}=[\mathrm{Ar}]\,3d^{10}$. Both have 0 unpaired.
Diamagnetic $\mathrm{Cu^{+}}$, $\mathrm{Cl^{-}}$ and $\mathrm{Zn^{2+}}$ (all electrons paired); $\mathrm{Fe^{2+}}$, $\mathrm{Fe^{3+}}$ and $\mathrm{Cu^{2+}}$ are paramagnetic.
Interactive 2 — Ion Builder: Which Electrons Leave First?

Choose an element and a charge. The orbitals are laid out by shell ($3d$ before $4s$). Electrons removed to form a cation are shown as faded dashed arrows; electrons added to form an anion are shown in green. Notice that transition metals lose their $4s$ electrons before any $3d$ electron. Press ▶ Play for a tour of important ions.

species—
configuration—
unpaired electrons—
isoelectronic with—

Exercises

I
The Aufbau principle, Pauli's principle and Hund's rule.
  • 1 Arrange in order of filling: $3d$, $4p$, $5s$, $4s$, $4d$.
  • 2 Find $(n+l)$ for $4f$, $5d$ and $6p$, and hence their order of filling.
  • 3 State Pauli's exclusion principle. What is the maximum number of electrons in one orbital, and why must their spins be opposite?
  • 4 Which rule does the orbital diagram 1s [↑↓] 2s [↑↓] 2p [↑↓][↑][ ] for nitrogen break? Draw the correct diagram.
  • 5 Explain why the $4s$ orbital fills before $3d$.
  • 6 How many unpaired electrons are there in N, O and F atoms?
  • 7 Draw the orbital diagram of a sulphur atom.
  • 8 Which principle is broken by the ground-state configuration $1s^{2}2s^{2}2p^{6}3s^{2}3d^{1}$ proposed for aluminium?
  • 9 Which principle is broken by $1s^{3}$?
1 $(n+l)$: $4s=4$; $3d=4p=5s=5$; $4d=6$. Order: $4s\lt3d\lt4p\lt5s\lt4d$.
2 All three have $n+l=7$; lower $n$ first: $4f\lt5d\lt6p$.
3 No two electrons in an atom can have the same four quantum numbers. Two electrons in one orbital already share $n$, $l$ and $m_{l}$, so they must differ in $m_{s}$ — one $+\tfrac12$, one $-\tfrac12$ — and a third electron is impossible: maximum $2$.
4 Hund's rule (pairing before every 2p orbital has one electron). Correct: 2p [↑][↑][↑].
5 $4s$ has the lower $(n+l)$ value ($4$ against $5$): a $4s$ electron penetrates the inner core much more than a $3d$ electron, so it feels a larger effective nuclear charge and has lower energy in K and Ca.
6 N ($2p^{3}$): $3$; O ($2p^{4}$): $2$; F ($2p^{5}$): $1$.
7 S ($Z=16$): 1s [↑↓] 2s [↑↓] 2p [↑↓][↑↓][↑↓] 3s [↑↓] 3p [↑↓][↑][↑] — two unpaired electrons.
8 The Aufbau principle: $3p$ ($n+l=4$) must be used before $3d$ ($n+l=5$). Correct: $[\mathrm{Ne}]\,3s^{2}3p^{1}$.
9 Pauli's exclusion principle: an orbital holds at most two electrons.
II
Configurations of atoms. Write the full configuration (parts 1–4) or the noble-gas-core configuration (parts 5–7).
  • 1 magnesium ($Z=12$)
  • 2 sulphur ($Z=16$)
  • 3 potassium ($Z=19$)
  • 4 manganese ($Z=25$)
  • 5 cobalt ($Z=27$)
  • 6 gallium ($Z=31$)
  • 7 bromine ($Z=35$)
  • 8 Identify the elements with configurations (a) $[\mathrm{Ar}]\,3d^{10}4s^{1}$ and (b) $[\mathrm{Ne}]\,3s^{2}3p^{3}$.
  • 9 How many unpaired electrons are there in atoms of Mn, Co and Ni?
  • 10 How many electrons with $l=1$ are there in a sulphur atom, and how many with $l=2$ in a zinc atom?
1 $1s^{2}2s^{2}2p^{6}3s^{2}$.
2 $1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$.
3 $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$.
4 $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^{2}$.
5 $[\mathrm{Ar}]\,3d^{7}4s^{2}$.
6 $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{1}$.
7 $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{5}$.
8 (a) $18+10+1=29$: copper. (b) $10+2+3=15$: phosphorus.
9 Mn ($3d^{5}$): $5$; Co ($3d^{7}$: [↑↓][↑↓][↑][↑][↑]): $3$; Ni ($3d^{8}$): $2$.
10 S: $2p^{6}$ and $3p^{4}$: $10$ electrons with $l=1$. Zn: $3d^{10}$: $10$ electrons with $l=2$.
III
Exceptions, stability and ions.
  • 1 Write the configuration of chromium and explain why it differs from the one the Aufbau order predicts.
  • 2 Do the same for copper.
  • 3 Write the configurations of $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$. Which is more stable, and why?
  • 4 Write the configurations of $\mathrm{Cu^{+}}$ and $\mathrm{Cu^{2+}}$. Which is paramagnetic?
  • 5 Write the configuration of $\mathrm{Zn^{2+}}$ and state whether it is para- or diamagnetic.
  • 6 How many unpaired electrons are there in $\mathrm{Mn^{2+}}$? In $\mathrm{Cr^{3+}}$?
  • 7 Which of $\mathrm{K^{+}}$, $\mathrm{Ca^{2+}}$, $\mathrm{Cl^{-}}$, $\mathrm{S^{2-}}$ and $\mathrm{Na^{+}}$ are isoelectronic with argon?
  • 8 How many exchange pairs are there among the parallel spins of a $d^{5}$ subshell? Of a $d^{4}$ subshell?
  • 9 Predict the configurations of silver ($Z=47$) and molybdenum ($Z=42$) from those of copper and chromium.
1 $[\mathrm{Ar}]\,3d^{5}4s^{1}$ rather than $3d^{4}4s^{2}$. $3d$ and $4s$ are close in energy, and a half-filled $3d^{5}$ subshell is extra stable (symmetrical distribution, maximum exchange energy: $10$ exchange pairs instead of $6$).
2 $[\mathrm{Ar}]\,3d^{10}4s^{1}$ rather than $3d^{9}4s^{2}$: a completely filled $3d^{10}$ subshell is extra stable.
3 $\mathrm{Fe^{2+}}=[\mathrm{Ar}]\,3d^{6}$; $\mathrm{Fe^{3+}}=[\mathrm{Ar}]\,3d^{5}$. $\mathrm{Fe^{3+}}$ is more stable because $3d^{5}$ is half-filled.
4 $\mathrm{Cu^{+}}=[\mathrm{Ar}]\,3d^{10}$ (diamagnetic); $\mathrm{Cu^{2+}}=[\mathrm{Ar}]\,3d^{9}$ — one unpaired electron, so $\mathrm{Cu^{2+}}$ is paramagnetic.
5 $[\mathrm{Ar}]\,3d^{10}$; all electrons paired: diamagnetic.
6 $\mathrm{Mn^{2+}}=[\mathrm{Ar}]\,3d^{5}$: $5$. $\mathrm{Cr^{3+}}$: from $[\mathrm{Ar}]\,3d^{5}4s^{1}$ remove $4s^{1}$ and two $3d$ electrons: $[\mathrm{Ar}]\,3d^{3}$, $3$ unpaired.
7 $\mathrm{K^{+}}$, $\mathrm{Ca^{2+}}$, $\mathrm{Cl^{-}}$ and $\mathrm{S^{2-}}$ all have $18$ electrons; $\mathrm{Na^{+}}$ has $10$ (isoelectronic with neon).
8 $d^{5}$: $\dfrac{5\times4}{2}=10$; $d^{4}$: $\dfrac{4\times3}{2}=6$.
9 Ag: $[\mathrm{Kr}]\,4d^{10}5s^{1}$; Mo: $[\mathrm{Kr}]\,4d^{5}5s^{1}$ (both correct — the same stabilisation operates one period lower).
IV
Harder problems.
  • 1 Give a possible set of four quantum numbers for the 19th electron of potassium and for the 21st electron of scandium.
  • 2 How many electrons in a zinc atom have $m_{l}=0$? How many have $m_{s}=-\tfrac12$?
  • 3 An element of the fourth period has three unpaired electrons, all in its $4p$ subshell. Identify it and write its configuration.
  • 4 Predict the configurations of rubidium ($Z=37$) and strontium ($Z=38$). Which subshell begins to fill at yttrium ($Z=39$)?
1 K, 19th electron in $4s$: $(4,0,0,+\tfrac12)$. Sc, 21st electron in $3d$: $(3,2,-2,+\tfrac12)$ — any $m_{l}$ from $-2$ to $+2$ is acceptable.
2 $m_{l}=0$: $14$ (Example 4). $m_{s}=-\tfrac12$: every subshell of Zn is full, so exactly half the $30$ electrons: $15$.
3 $4p^{3}$: arsenic, $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{3}$ ($Z=33$).
4 Rb: $[\mathrm{Kr}]\,5s^{1}$; Sr: $[\mathrm{Kr}]\,5s^{2}$. Next by the $(n+l)$ rule comes $4d$ ($n+l=6$, lower $n$ than $5p$): yttrium is $[\mathrm{Kr}]\,4d^{1}5s^{2}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET Which ion has the largest number of unpaired electrons? (A) $\mathrm{Fe^{3+}}$ (B) $\mathrm{Co^{2+}}$ (C) $\mathrm{Ni^{2+}}$ (D) $\mathrm{Cu^{2+}}$

Configurations Remove $4s$ first: $\mathrm{Fe^{3+}}$ $3d^{5}$ (5 unpaired), $\mathrm{Co^{2+}}$ $3d^{7}$ (3), $\mathrm{Ni^{2+}}$ $3d^{8}$ (2), $\mathrm{Cu^{2+}}$ $3d^{9}$ (1).
Answer (A).

E2. JEE Main Numerical answer. What is the total number of unpaired electrons in one $\mathrm{Cr}$ atom and one $\mathrm{Cu^{2+}}$ ion taken together?

Cr $[\mathrm{Ar}]\,3d^{5}4s^{1}$: $6$ unpaired (not $4$, which the Aufbau prediction $3d^{4}4s^{2}$ would give).
Cu²⁺ $[\mathrm{Ar}]\,3d^{9}$: $1$ unpaired.
Answer $6+1=$ 7.

E3. Board · 3 marks Write the electronic configurations of chromium ($Z=24$) and copper ($Z=29$). Explain why they do not follow the Aufbau order.

1 mark $\mathrm{Cr}$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^{1}$; $\mathrm{Cu}$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{1}$.
1 mark The $3d$ and $4s$ orbitals have very similar energies, so a small extra stabilisation can move one electron from $4s$ to $3d$.
1 mark Half-filled ($d^{5}$) and completely filled ($d^{10}$) subshells are extra stable because of their symmetrical electron distribution and larger exchange energy (more pairs of parallel-spin electrons that can exchange).

E4. JEE Main NEET Assertion–reason. Assertion (A): In potassium the 19th electron enters the $4s$ orbital and not $3d$. Reason (R): The $(n+l)$ value of $4s$ is smaller than that of $3d$. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Check A is true: $\mathrm{K}=[\mathrm{Ar}]\,4s^{1}$. R is true: $4+0=4\lt3+2=5$, and by the $(n+l)$ rule the lower value means lower energy, which is exactly why the electron goes to $4s$.
Answer (A).

E5. JEE Advanced One or more options correct. (A) $\mathrm{Fe^{2+}}$ has four unpaired electrons. (B) $\mathrm{Sc^{3+}}$ is diamagnetic. (C) When titanium forms $\mathrm{Ti^{2+}}$, the electrons are removed from $3d$ before $4s$. (D) $\mathrm{Cu^{+}}$ and $\mathrm{Zn^{2+}}$ have the same electronic configuration.

(A) True: $[\mathrm{Ar}]\,3d^{6}$ = [↑↓][↑][↑][↑][↑].
(B) True: $\mathrm{Sc^{3+}}=[\mathrm{Ar}]$, all paired.
(C) False: $\mathrm{Ti}=[\mathrm{Ar}]\,3d^{2}4s^{2}$ loses its $4s$ electrons first: $\mathrm{Ti^{2+}}=[\mathrm{Ar}]\,3d^{2}$.
(D) True: both are $[\mathrm{Ar}]\,3d^{10}$ ($28$ electrons; they are isoelectronic).
Answer (A), (B), (D).

E6. Board · Case study Liquid oxygen poured between the poles of a strong magnet sticks there, while liquid nitrogen runs straight through. (The magnetism of the $\mathrm{O_{2}}$ molecule is explained in Chapter 4; here consider the atoms.) (i) Draw the orbital diagrams of N ($Z=7$) and O ($Z=8$) atoms and give the number of unpaired electrons in each. (2 marks) (ii) Which rule decides how the $2p$ electrons are arranged? (1 mark) (iii) Is an atom with unpaired electrons paramagnetic or diamagnetic? (1 mark)

(i) N: 1s [↑↓] 2s [↑↓] 2p [↑][↑][↑] — $3$ unpaired. O: 1s [↑↓] 2s [↑↓] 2p [↑↓][↑][↑] — $2$ unpaired.
(ii) Hund's rule of maximum multiplicity: degenerate orbitals are singly occupied, with parallel spins, before pairing begins.
(iii) Paramagnetic — it is attracted into a magnetic field.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why do hospitals inject compounds of gadolinium before some MRI scans — and why are chemists now testing manganese and iron compounds to replace them?

Why it matters. Magnetic resonance imaging (MRI) is used tens of millions of times a year worldwide. "Contrast agents" brighten blood vessels and tumours in the images; they work because their metal ions have many unpaired electrons, which affect the nearby water molecules that MRI detects. Concern that tiny amounts of gadolinium stay in the body has driven research, in Europe, the USA and Asia, into alternatives.

What to do.

  1. Gadolinium ($Z=64$) has the configuration $[\mathrm{Xe}]\,4f^{7}5d^{1}6s^{2}$, where $[\mathrm{Xe}]$ is the configuration of xenon ($Z=54$). Check the electron count. Which subshell is exactly half-filled, and which rule from this section does the configuration illustrate?
  2. Write the configuration of $\mathrm{Gd^{3+}}$ (remove the outermost electrons first: $6s$, then $5d$) and count its unpaired electrons.
  3. Write the configurations of $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ and count their unpaired electrons. Rank the three ions by number of unpaired electrons.
  4. Look up (for example on the US Food and Drug Administration or European Medicines Agency websites, or in a recent review) why gadolinium agents are wrapped in large organic molecules ("chelates"), and one manganese- or iron-based agent now being tested.

Tools: The configuration table and Interactive 2 (for $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$); a reliable medical or regulatory source.

Hand in: Your three configurations with orbital diagrams for the partly filled subshells, and a half-page summary of the benefits and risks of gadolinium agents.

Going further. Free $\mathrm{Gd^{3+}}$ ions are toxic partly because they are similar in size to $\mathrm{Ca^{2+}}$. Find out how a "chelate" stops the ion from being released.

Instructor note: $54+7+1+2=64$ ✓; $4f^{7}$ is half-filled (the analogue of Cr's $3d^{5}$: the "expected" $4f^{8}6s^{2}$ is not adopted). $\mathrm{Gd^{3+}}=[\mathrm{Xe}]\,4f^{7}$, $7$ unpaired; $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ are both $[\mathrm{Ar}]\,3d^{5}$, $5$ unpaired. Ranking: $\mathrm{Gd^{3+}}\gt\mathrm{Mn^{2+}}=\mathrm{Fe^{3+}}$. Students may find the manganese agent studied under the name Mn-PyC3A, and iron-oxide nanoparticle agents.

🧪 Self-check quiz

0/5 correct

Q1. After the $4s$ subshell is full, the next electron enters

Q2. Which principle rules out the configuration $1s^{3}$?

Q3. The ground-state configuration of copper ($Z=29$) is

Q4. How many unpaired electrons does $\mathrm{Fe^{2+}}$ have?

Q5. When an iron atom forms $\mathrm{Fe^{2+}}$, the two electrons are removed from

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