Chapter 3 · Classification of Elements and Periodicity in Properties

3.2 Electronic Configuration and the Blocks of the Periodic Table

Why the periodic table has the shape it has: periods as shells, groups as identical outer configurations, the s, p, d and f blocks with their general configurations, how to locate any element from its configuration (and back), and where metals, non-metals and metalloids lie.

Mendeleev built his table from chemical behaviour, without knowing why it worked. Section 3.1 replaced atomic mass with atomic number; this section answers the deeper question: why do properties repeat, and why do the periods contain $2$, $8$, $8$, $18$, $18$, $32$ and $32$ elements? The answer is in the electronic configurations of Section 2.4. Elements in the same group have the same arrangement of outer electrons, so they react alike; each period corresponds to filling one new shell; and the table splits naturally into four blocks, named after the subshell — s, p, d or f — that receives the last electron.

Once you see this, the periodic table becomes a map of electronic configurations. You will be able to read off the configuration of an element from its position and, the other way round, place an element from its configuration — a skill tested in almost every Board, JEE and NEET paper. The section ends with the broad division of the elements into metals, non-metals and metalloids, and how metallic character changes across the table.

Periods, Groups and Electronic Configuration

Write the configurations of the first twenty elements side by side and the pattern is clear. Lithium $[\mathrm{He}]\,2s^{1}$, sodium $[\mathrm{Ne}]\,3s^{1}$ and potassium $[\mathrm{Ar}]\,4s^{1}$ all have one electron in an s orbital outside a noble-gas core: they are the alkali metals of group 1. Fluorine $[\mathrm{He}]\,2s^{2}2p^{5}$ and chlorine $[\mathrm{Ne}]\,3s^{2}3p^{5}$ both have seven outer electrons: halogens, group 17. The number of the outermost shell changes from row to row; the pattern of electrons in it repeats.

📐 Rule 3.5 — Periods and groups from configuration
  1. The period number of an element equals the principal quantum number $n$ of its outermost (valence) shell. A new period begins each time an electron enters a new shell, i.e. a new $ns$ orbital.
  2. The number of elements in a period is twice the number of orbitals in the subshells filled during that period. By the Aufbau order (Rule 2.13), period $n$ fills $ns$, then (from period 4) $(n-1)d$, then (from period 6) $(n-2)f$, and finally $np$.
  3. Elements in the same group have the same number of valence electrons in the same type of orbitals — the same outer electronic configuration — and therefore similar chemical properties.
Period$n$Subshells filledOrbitalsElementsFrom – to
11$1s$12H – He
22$2s,\ 2p$48Li – Ne
33$3s,\ 3p$48Na – Ar
44$4s,\ 3d,\ 4p$918K – Kr
55$5s,\ 4d,\ 5p$918Rb – Xe
66$6s,\ 4f,\ 5d,\ 6p$1632Cs – Rn
77$7s,\ 5f,\ 6d,\ 7p$1632Fr – Og
✏️ Example 1 — Why the periods have different lengths
(a) The third shell can hold $2n^{2}=18$ electrons. Why, then, does the third period contain only 8 elements? (b) Show that the fourth period must contain 18 elements. (c) In which period does the $4f$ subshell fill, and how many elements does that add to the period?
(a) Period 3 fills $3s$ (2 electrons) and $3p$ (6): $8$ elements, Na to Ar. The $3d$ subshell has $n+l=5$, higher than $4s$ ($n+l=4$), so after argon the next electron enters $4s$, starting period 4. The $3d$ electrons belong to period 4.
(b) Period 4 fills $4s$ ($1$ orbital), $3d$ ($5$) and $4p$ ($3$): $9$ orbitals × $2$ = $18$ electrons, so $18$ elements, K ($Z=19$) to Kr ($Z=36$). Check: $36-18=18$ ✓.
(c) By the Aufbau order $6s\lt4f\lt5d\lt6p$, the $4f$ subshell fills in period 6, adding $7\times2=14$ elements — the lanthanoids — and raising the period from $18$ to $32$ elements.
✏️ Example 2 — Locating elements from their atomic numbers
Write the configurations of the elements with $Z=17$, $20$, $26$, $35$ and $47$, and give the period and group of each.
Z = 17 $[\mathrm{Ne}]\,3s^{2}3p^{5}$: outermost shell $n=3$ → period 3; last electron in $3p$, $2+5=7$ valence electrons → group 17 (chlorine).
Z = 20 $[\mathrm{Ar}]\,4s^{2}$: period 4, group 2 (calcium).
Z = 26 $[\mathrm{Ar}]\,3d^{6}4s^{2}$: outermost shell $n=4$ → period 4; $6+2=8$ electrons in $3d$ and $4s$ → group 8 (iron).
Z = 35 $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{5}$: period 4, group 17 (bromine). The full $3d^{10}$ does not count toward the group of a p-block element; the group is $10+(2+5)$.
Z = 47 $[\mathrm{Kr}]\,4d^{10}5s^{1}$ (like copper, a filled d subshell): period 5, group $10+1=$ 11 (silver).

The s, p, d and f Blocks

The subshell that receives the last ("differentiating") electron in the Aufbau process divides the table into four blocks. This is the deepest organising idea of the table: each block has its own general configuration and its own characteristic chemistry.

📘 Definition 3.3 — The four blocks
BlockLast electron entersGeneral outer configurationGroupsElements also called
s$ns$$ns^{1-2}$1, 2alkali and alkaline earth metals
p$np$$ns^{2}np^{1-6}$13 – 18with the s-block: representative (main-group) elements
d$(n-1)d$$(n-1)d^{1-10}ns^{0-2}$3 – 12transition elements
f$(n-2)f$$(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}$(below the table)inner transition elements: lanthanoids, actinoids

Helium ($1s^{2}$) is an s-block element by configuration but is placed in group 18, because its filled shell makes it a noble gas. Hydrogen ($1s^{1}$) is placed at the top of group 1 but is unique (Section 3.1). Following the syllabus convention, lanthanum and actinium are counted in the d-block (group 3), and the f-block consists of Ce–Lu ($Z=58$–$71$) and Th–Lr ($Z=90$–$103$).

📐 Rule 3.6 — Group number from the configuration
  • s-block: group = number of $ns$ electrons ($1$ or $2$).
  • p-block: group = $10+$ (number of $ns$ + $np$ electrons). A filled $(n-1)d^{10}$ or $(n-2)f^{14}$ subshell is not counted.
  • d-block: group = number of $(n-1)d$ + $ns$ electrons.
  • f-block: no separate group number; lanthanoids and actinoids belong with group 3 in periods 6 and 7.
In every case the period is the $n$ of the outermost $s$ subshell (Rule 3.5).
Interactive 1 — The Periodic Table as a Map of Configurations

Click any element (or use the slider, or press ▶ Play to walk through $Z=1$ to $118$). Colour the table by block, period, group or metallic character. The panel shows the element's configuration with the valence electrons that fix its block, period and group. Configurations for $Z\gt103$ are predictions.

26
element—
configuration—
block—
period · group—
type—
✏️ Example 3 — Block, period and group from a configuration
Identify the block, period and group of each element, and name it: (a) $[\mathrm{Ne}]\,3s^{2}3p^{4}$; (b) $[\mathrm{Ar}]\,3d^{5}4s^{1}$; (c) $[\mathrm{Kr}]\,5s^{1}$; (d) $[\mathrm{Ar}]\,3d^{10}4s^{2}$; (e) $[\mathrm{Xe}]\,4f^{7}6s^{2}$.
(a) Last electron in $3p$: p-block, period 3, group $10+(2+4)=16$. $Z=10+6=16$: sulphur.
(b) Last electrons in $3d$: d-block, period 4, group $5+1=6$. $Z=18+6=24$: chromium (the half-filled exception of Section 2.4).
(c) s-block, period 5, group 1; $Z=37$: rubidium.
(d) d-block, period 4, group $10+2=12$; $Z=30$: zinc. (Its d subshell is full in the atom and in $\mathrm{Zn^{2+}}$, so zinc lacks typical transition-metal properties such as coloured ions.)
(e) Last electrons in $4f$: f-block (a lanthanoid), period 6; $Z=54+9=63$: europium.
✏️ Example 4 — From position to configuration
Write the outer electronic configuration and atomic number of the element in (a) period 4, group 15; (b) period 5, group 2; (c) period 4, group 8; (d) period 6, group 17; (e) period 5, group 7.
Method Period → $n$; group → block and number of valence electrons; then fill in the inner subshells that the period has already completed.
(a) Group 15: p-block, $15-10=5$ electrons in $4s^{2}4p^{3}$; period 4 has filled $3d$ first: $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{3}$, $Z=18+15=33$ (arsenic).
(b) $[\mathrm{Kr}]\,5s^{2}$, $Z=38$ (strontium).
(c) Group 8 of the d-block: $8$ electrons in $3d+4s$: $[\mathrm{Ar}]\,3d^{6}4s^{2}$, $Z=26$ (iron).
(d) $6s^{2}6p^{5}$; period 6 has already filled $4f^{14}$ and $5d^{10}$: $[\mathrm{Xe}]\,4f^{14}5d^{10}6s^{2}6p^{5}$, $Z=54+14+10+7=85$ (astatine).
(e) $[\mathrm{Kr}]\,4d^{5}5s^{2}$, $Z=36+7=43$ (technetium). (Check with (c): one period lower, $+18$ in $Z$ per period: group 7 of period 4 is Mn, $Z=25$; $25+18=43$ ✓.)
Interactive 2 — Block Builder: Filling the Table Subshell by Subshell

Add the elements in order of atomic number. Each new element's cell is coloured by the subshell that receives its last electron, and the panel keeps a running count of the subshells filled in the current period. Watch the shape of the table emerge: 2, then 8, 8, then 18 when $3d$ appears, and 32 when $4f$ arrives.

21
newest element—
subshell being filled—
this period so far—
period length—

Metals, Non-metals and Metalloids

About four-fifths of the elements are metals. They lie on the left and in the middle of the table — all of the s-block except hydrogen and helium, the whole d- and f-blocks, and the heavier p-block elements such as aluminium, tin, lead and bismuth. Non-metals occupy the top right. Between them runs a zig-zag line, from boron down to astatine, and the elements beside it have intermediate properties.

📘 Definition 3.4 — Metals, non-metals and metalloids
  • Metals are usually solids at room temperature (mercury is a liquid; gallium and caesium melt just above it), lustrous, malleable and ductile, good conductors of heat and electricity; they tend to lose electrons, forming cations and basic oxides.
  • Non-metals are gases or solids (bromine is a liquid) with low melting points (carbon in diamond and graphite is an exception), brittle when solid, and poor conductors (graphite is an exception); they tend to gain or share electrons and form acidic oxides.
  • Metalloids (semi-metals) — boron, silicon, germanium, arsenic, antimony and tellurium — lie along the zig-zag line and show properties of both: silicon and germanium, for example, look metallic but are brittle semiconductors.

Trend: metallic character decreases from left to right across a period and increases down a group; non-metallic character does the opposite.

⚠️ Common mistakes
  • Group ≠ number of valence electrons for the p-block in the 1–18 system. Chlorine has 7 valence electrons but is in group 17 ($=10+7$). In the d-block the group number counts $(n-1)d+ns$ electrons.
  • The period is not the shell of the last electron added. Iron's last electron enters $3d$, but its outermost shell is $n=4$: iron is in period 4. Likewise europium ($4f$) is in period 6.
  • Do not count a full $d^{10}$ in the p-block group. Bromine $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{5}$ is group 17, not 27.
  • Helium is s-block but sits in group 18; do not assign it to group 2.
✏️ Example 5 — Metallic character and block chemistry
(a) Arrange in order of increasing metallic character: (i) Na, Al, Mg, Si; (ii) As, N, Bi, P, Sb. (b) An element X forms a colourless ion $\mathrm{X^{2+}}$, a chloride $\mathrm{XCl_{2}}$ that conducts electricity when molten, and its outer configuration is $ns^{2}$ with no partly filled inner subshell. Which block is it in? (c) An element Y forms $\mathrm{YCl_{2}}$ (pale green as the hydrate) and $\mathrm{YCl_{3}}$ (yellow-brown), and both chlorides are attracted into a magnetic field. Which block is Y most likely in? Explain.
(a)(i) Across period 3 metallic character falls: $\mathrm{Si}\lt\mathrm{Al}\lt\mathrm{Mg}\lt\mathrm{Na}$ (Si a metalloid, Na the most metallic).
(a)(ii) Down group 15 metallic character rises: $\mathrm{N}\lt\mathrm{P}\lt\mathrm{As}\lt\mathrm{Sb}\lt\mathrm{Bi}$ (N, P non-metals; As, Sb metalloids; Bi a metal).
(b) A single, fixed $+2$ state, colourless ion, ionic chloride and an $ns^{2}$ outer configuration: an s-block element of group 2 (e.g. magnesium or calcium).
(c) Two oxidation states, coloured compounds and paramagnetic ions ($\mathrm{Fe^{2+}}$ $3d^{6}$ and $\mathrm{Fe^{3+}}$ $3d^{5}$ have unpaired electrons) are typical of a d-block transition element — this is iron: $\mathrm{FeCl_{2}}$ and $\mathrm{FeCl_{3}}$.
✏️ Example 6 — Superheavy elements and a challenge
(a) Predict the configuration, block, period and group of flerovium ($Z=114$). (b) Element 120 has not yet been made. Predict its block and group, and name one element it should resemble. (c) Challenge (not syllabus). If period 8 filled $8s$, $5g$ ($l=4$), $6f$, $7d$ and $8p$, how many elements would it contain?
(a) Radon is $86$; then $7s^{2}$ ($88$), $5f^{14}$ ($102$), $6d^{10}$ ($112$), $7p^{2}$ ($114$): $[\mathrm{Rn}]\,5f^{14}6d^{10}7s^{2}7p^{2}$ — p-block, period 7, group 14, below lead.
(b) Element 119 begins period 8 ($8s^{1}$); element 120 would be $[\mathrm{Og}]\,8s^{2}$: s-block, group 2, below radium — it should resemble barium and radium.
(c) A g subshell has $2l+1=9$ orbitals. Orbitals: $8s$ $1$ + $5g$ $9$ + $6f$ $7$ + $7d$ $5$ + $8p$ $3$ $=25$; elements $=2\times25=$ 50. (Real superheavy atoms are expected to deviate from this simple picture.)

Exercises

I
From configuration to position. Give the block, period and group, and name the element.
  • 1 $[\mathrm{He}]\,2s^{2}2p^{3}$
  • 2 $[\mathrm{Ar}]\,4s^{1}$
  • 3 $[\mathrm{Ar}]\,3d^{3}4s^{2}$
  • 4 $[\mathrm{Kr}]\,4d^{10}5s^{2}5p^{5}$
  • 5 $[\mathrm{Xe}]\,6s^{2}$
  • 6 $[\mathrm{Ar}]\,3d^{10}4s^{1}$
  • 7 $[\mathrm{Kr}]\,4d^{10}$
  • 8 $[\mathrm{Rn}]\,5f^{3}6d^{1}7s^{2}$
  • 9 $1s^{2}$
1 p-block, period 2, group $10+5=15$; $Z=7$, nitrogen.
2 s-block, period 4, group 1; $Z=19$, potassium.
3 d-block, period 4, group $3+2=5$; $Z=23$, vanadium.
4 p-block, period 5, group $10+7=17$; $Z=53$, iodine.
5 s-block, period 6, group 2; $Z=56$, barium.
6 d-block, period 4, group $10+1=11$; $Z=29$, copper.
7 d-block, period 5 (its outer shell is $n=5$ even though $5s$ is empty), group $10+0=10$; $Z=46$, palladium.
8 f-block (actinoid), period 7; $Z=86+3+1+2=92$, uranium.
9 s-block by configuration, but placed in period 1, group 18; $Z=2$, helium (a noble gas with a filled shell).
II
From position to configuration. Write the noble-gas-core configuration and atomic number of the element in
  • 1 period 3, group 13
  • 2 period 4, group 12
  • 3 period 5, group 16
  • 4 period 6, group 1
  • 5 period 4, group 6 (careful!)
  • 6 period 5, group 11
  • 7 period 6, group 11
  • 8 Elements of the same group in periods 4 and 5 differ in $Z$ by 18; in periods 6 and 7 (s-block) by 32. Find the atomic numbers of the elements directly below selenium ($Z=34$) and directly below barium ($Z=56$).
1 $[\mathrm{Ne}]\,3s^{2}3p^{1}$, $Z=13$ (aluminium).
2 $[\mathrm{Ar}]\,3d^{10}4s^{2}$, $Z=30$ (zinc).
3 $[\mathrm{Kr}]\,4d^{10}5s^{2}5p^{4}$, $Z=36+16=52$ (tellurium).
4 $[\mathrm{Xe}]\,6s^{1}$, $Z=55$ (caesium).
5 Six electrons in $3d+4s$; the half-filled exception gives $[\mathrm{Ar}]\,3d^{5}4s^{1}$, $Z=24$ (chromium), not $3d^{4}4s^{2}$.
6 $[\mathrm{Kr}]\,4d^{10}5s^{1}$, $Z=47$ (silver).
7 $[\mathrm{Xe}]\,4f^{14}5d^{10}6s^{1}$, $Z=54+14+10+1=79$ (gold).
8 Below Se: $34+18=52$ (tellurium). Below Ba: $56+32=88$ (radium).
III
The blocks.
  • 1 Write the general outer electronic configuration of the s-, p-, d- and f-block elements.
  • 2 Why are d-block elements called transition elements? Why are zinc, cadmium and mercury not typical transition elements?
  • 3 Why are the lanthanoids and actinoids placed in separate rows below the table?
  • 4 Of the 118 known elements, how many are in each block?
  • 5 Which block contains (a) the most reactive metals; (b) most elements that form coloured ions; (c) a series whose members are all radioactive; (d) most of the noble gases?
  • 6 Why is helium placed in group 18 although its configuration is $1s^{2}$?
  • 7 Which of Na, Fe, Cl, Ce, Ca and Cu are representative elements?
  • 8 How many unpaired electrons are there in an atom of the element in period 4, group 6, and in the element in period 4, group 15?
1 s: $ns^{1-2}$; p: $ns^{2}np^{1-6}$; d: $(n-1)d^{1-10}ns^{0-2}$; f: $(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}$.
2 They lie between the s- and p-blocks and their properties change gradually ("transition") from the reactive s-block metals to the p-block. Zn, Cd and Hg have $(n-1)d^{10}ns^{2}$: the d subshell is full in the atom and in the common $\mathrm{M^{2+}}$ ion, so they show no coloured ions or variable oxidation states typical of transition metals.
3 They fill the inner $(n-2)f$ subshell and are very similar to each other; placing all 14 of each series inside the main table would make it 32 columns wide. They belong to periods 6 and 7 (with group 3).
4 s: $14$ (groups 1–2, including H and He); p: $36$ (groups 13–18, periods 2–7); d: $40$ (groups 3–12, periods 4–7); f: $28$. Total $118$ ✓.
5 (a) s-block; (b) d-block; (c) f-block (actinoids); (d) p-block (group 18, except He).
6 Its $1s^{2}$ shell is completely filled, like $ns^{2}np^{6}$ of the other noble gases; it is chemically inert, not a reactive metal like the group-2 elements.
7 Na, Cl and Ca (s- and p-block). Fe and Cu are d-block; Ce is f-block.
8 Period 4, group 6: Cr, $[\mathrm{Ar}]\,3d^{5}4s^{1}$, $6$ unpaired. Group 15: As, $4p^{3}$, $3$ unpaired.
IV
Metals, non-metals and metalloids.
  • 1 Name the six metalloids that lie along the zig-zag line.
  • 2 Arrange Mg, Ca, Be, Ba and Sr in order of increasing metallic character.
  • 3 Arrange C, Si, Ge, Sn and Pb in order of increasing metallic character, and classify each.
  • 4 Classify as metal, non-metal or metalloid: Ge, Se, Sb, Sr, Br, Bi.
  • 5 For the element with $Z=34$, give the block, period, group and whether it is a metal or non-metal.
  • 6 A metal in period 3 forms the oxide $\mathrm{X_{2}O_{3}}$. Identify X and write its configuration.
  • 7 Silicon and germanium are the two classic semiconductors. Locate both in the table from their configurations, and explain why they behave alike.
1 Boron, silicon, germanium, arsenic, antimony, tellurium.
2 Metallic character increases down a group: $\mathrm{Be}\lt\mathrm{Mg}\lt\mathrm{Ca}\lt\mathrm{Sr}\lt\mathrm{Ba}$.
3 $\mathrm{C}\lt\mathrm{Si}\lt\mathrm{Ge}\lt\mathrm{Sn}\lt\mathrm{Pb}$: C non-metal; Si, Ge metalloids; Sn, Pb metals.
4 Ge metalloid; Se non-metal; Sb metalloid; Sr metal; Br non-metal; Bi metal.
5 $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{4}$: p-block, period 4, group 16 — selenium, a non-metal.
6 Valence 3 in period 3: group 13, aluminium, $[\mathrm{Ne}]\,3s^{2}3p^{1}$ (oxide $\mathrm{Al_{2}O_{3}}$).
7 Si $[\mathrm{Ne}]\,3s^{2}3p^{2}$: p-block, period 3, group 14. Ge $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{2}$: p-block, period 4, group 14. Same outer configuration $ns^{2}np^{2}$ (same group) and both metalloids on the zig-zag line, so their properties are similar.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET An element has the configuration $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{4}$. It belongs to (A) period 4, group 6, d-block (B) period 4, group 16, p-block (C) period 3, group 16, p-block (D) period 4, group 4, p-block

Read off Outermost shell $n=4$ → period 4; last electron in $4p$ → p-block; group $10+(2+4)=16$. (Selenium.)
Answer (B).

E2. JEE Main Numerical answer. What is the atomic number of the element in period 5, group 14?

Configuration $[\mathrm{Kr}]\,4d^{10}5s^{2}5p^{2}$: $36+10+2+2$.
Answer 50 (tin).

E3. Board · 3 marks On the basis of electronic configuration, how are elements divided into s-, p-, d- and f-blocks? Give the general outer configuration of each block.

1 mark The block is named after the subshell that receives the last (differentiating) electron.
1 mark s-block $ns^{1-2}$ (groups 1, 2); p-block $ns^{2}np^{1-6}$ (groups 13–18).
1 mark d-block $(n-1)d^{1-10}ns^{0-2}$ (groups 3–12); f-block $(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}$ (lanthanoids, actinoids).

E4. JEE Main Numerical answer. How many d-block elements of the fourth period have exactly one electron in their $4s$ orbital in the ground state?

Scan Sc to Zn are $3d^{x}4s^{2}$ except the two exceptions $\mathrm{Cr}$ $3d^{5}4s^{1}$ and $\mathrm{Cu}$ $3d^{10}4s^{1}$.
Answer 2.

E5. JEE Advanced One or more options correct. (A) The element with $Z=56$ is an s-block element of group 2. (B) The element with $Z=33$ is a metalloid. (C) The element with $Z=64$ is a d-block element. (D) The element in period 6, group 14 has the configuration $[\mathrm{Xe}]\,4f^{14}5d^{10}6s^{2}6p^{2}$.

(A) True: $[\mathrm{Xe}]\,6s^{2}$, barium.
(B) True: $Z=33$ is arsenic, $[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{3}$, a metalloid.
(C) False: $Z=64$ is gadolinium, $[\mathrm{Xe}]\,4f^{7}5d^{1}6s^{2}$, a lanthanoid (f-block).
(D) True: $54+14+10+2+2=82$, lead, group $10+4=14$.
Answer (A), (B), (D).

E6. Board · Case study Bromine ($Z=35$) is one of only two elements that are liquid at room temperature. (i) Write its electronic configuration and give its period, group and block. (2 marks) (ii) Is bromine a metal, non-metal or metalloid? (1 mark) (iii) Write the formula of the compound it forms with the element in period 3, group 2. (1 mark)

(i) $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{5}=[\mathrm{Ar}]\,3d^{10}4s^{2}4p^{5}$: period 4, group 17, p-block.
(ii) Non-metal (a halogen).
(iii) Period 3, group 2 is magnesium, valence 2; bromine has valence 1: $\mathrm{MgBr_{2}}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why are the "rare earth" lanthanoids essential for wind turbines and electric cars — and why are they so hard to separate from each other?

Why it matters. The strongest permanent magnets, used in wind-turbine generators, electric-vehicle motors, hard drives and earphones, are made of neodymium, iron and boron, often with a little dysprosium added so they keep working when hot. A few countries supply most of the world's rare earths, and governments in Europe, the USA, Japan and India now treat them as critical materials.

What to do.

  1. Write the configurations of neodymium ($Z=60$) and dysprosium ($Z=66$), and give their block and period.
  2. Count the unpaired electrons in each atom and in the $\mathrm{Nd^{3+}}$ and $\mathrm{Dy^{3+}}$ ions (remove the $6s$ electrons first, then $4f$). Why do many unpaired electrons matter for a magnet?
  3. Compare the outer configurations of neighbouring lanthanoids. Use them to explain why separating Nd from its neighbours Pr and Pm, or Dy from Tb and Ho, takes hundreds of repeated extraction steps.
  4. Find out (from a government critical-minerals report or a recent review) where most rare earths are mined and refined, and one method being developed to recycle them from old magnets.

Tools: Definition 3.3, Interactive 1 (click Nd and Dy) and Section 2.4 for ion configurations.

Hand in: The configurations and unpaired-electron counts, and a half-page summary of why rare earths are "critical".

Going further. Why is the name "rare earths" misleading? Compare the abundance of cerium in the Earth's crust with that of copper.

Instructor note: Nd $[\mathrm{Xe}]\,4f^{4}6s^{2}$ (4 unpaired), Dy $[\mathrm{Xe}]\,4f^{10}6s^{2}$ (4 unpaired); both f-block, period 6. $\mathrm{Nd^{3+}}$ $4f^{3}$ (3 unpaired), $\mathrm{Dy^{3+}}$ $4f^{9}$ (5 unpaired). Neighbouring lanthanoids differ only in the number of inner $4f$ electrons; their outer $6s^{2}$ (and similar sizes, Section 3.3) make their chemistry almost identical, so solvent extraction separates them only slightly at each step. Cerium is about as abundant as copper in the crust.

🧪 Self-check quiz

0/5 correct

Q1. The element $[\mathrm{Ar}]\,3d^{6}4s^{2}$ is in period

Q2. The general outer configuration of the p-block elements is

Q3. The element $[\mathrm{Kr}]\,4d^{10}5s^{2}5p^{3}$ belongs to group

Q4. Which of these is a metalloid?

Q5. Which element belongs to the f-block?

← 3.1 Development of the Periodic Table and the Modern Periodic Law3.3 Periodic Trends in Properties →