Chemistry 2e · Representative Metals, Metalloids, and Nonmetals
Occurrence and Preparation of the Representative Metals
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The representative (main-group) metals are the metals of the s-block and p-block: the alkali metals (group 1), the alkaline earth metals (group 2), and the p-block metals aluminum, gallium, indium, thallium, tin, lead, and bismuth — all ready electron losers and cation formers.
The key fact: most are too reactive to exist uncombined. Only the least reactive metals — chiefly silver, gold, and platinum — occur native. Everything else is locked up in ores: minerals rich enough in the metal to make extraction worthwhile. Aluminum, the most abundant crustal metal (about 8% by mass), is found in bauxite (mostly Al₂O₃·xH₂O); iron (about 5%) is mined as hematite (Fe₂O₃) and magnetite (Fe₃O₄); active metals like sodium and calcium occur as halides, carbonates, and sulfates.
Metallurgy — extracting and refining a metal from its Ore A mineral rich enough in a metal to make extraction economical Full entry → — is a chemistry of moving electrons: the ore holds the metal oxidized, and the process must supply electrons to reduce it. The method chosen (carbon Smelting Reducing an oxide ore with carbon (coke) at high temperature Full entry →, reaction with a more active metal, or Electrolysis Using electric current to force a nonspontaneous reduction Full entry →) follows the metal's position in the activity series.
Why this matters
- Everything built around you is metallurgy: steel, aluminum frames, copper wiring, and battery lithium all begin as minerals that had to be reduced — which is why iron is cheap and aluminum was once more valuable than gold.
- Energy and climate: aluminum electrolysis is among the most energy-intensive industrial processes; recycling saves much of that energy.
- Predicting reactivity: the activity series lets you predict displacement reactions, whether a metal reacts with acid, and whether it can be won by carbon or needs electrolysis.
- Exams: expect questions walking through "ore → concentrate → roast → reduce → refine" for a given metal, or asking why sodium cannot be made by heating its oxide with carbon.
The college version
Core Concepts
Which elements are the representative metals
Groups 1 and 2 make up the s-block metals; the p-block metals sit in the lower-right corner of the metal/nonmetal "staircase": Al, Ga, In, Tl (group 13), Sn, Pb (group 14), and Bi (group 15). These elements lose electrons to form cations with charges equal to their group number (s-block: Na⁺, Ca²⁺) or group number minus 10 (p-block: Al³⁺, Sn²⁺/Sn⁴⁺, Pb²⁺/Pb⁴⁺, Bi³⁺). Sn and Pb's two common states (2+ and 4+) reflect the Inert pair effect Heavy p-block metals using only their p electrons in bonding Full entry →: the s² pair of valence electrons bonds less readily in heavier elements.
Native metals versus combined metals
A metal found uncombined occurs native. Only metals near the bottom of the activity series — silver, gold, platinum — are reliably found native, because their electrode potentials are so unfavorable for oxidation that air and water cannot corrode them. Every other representative metal occurs combined: as oxides (bauxite Al₂O₃·xH₂O, hematite Fe₂O₃), sulfides (galena PbS, sphalerite ZnS), carbonates (limestone CaCO₃), and halides (halite NaCl). General rule: the more active the metal, the more tightly it is bound, and the more energy is needed to free it.
The metallurgical sequence
Most extraction schemes follow four stages:
- Concentration (beneficiation): crushing, flotation, and magnetic separation raise the metal content and remove worthless rock (Gangue The worthless rock mixed with the ore Full entry →).
- Conversion: sulfide ores are roasted in air to the oxide, releasing SO₂: 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g). Oxides are easier to reduce than sulfides; carbonates are converted by heating (Calcination Heating a carbonate ore to drive off CO₂ Full entry →).
- Reduction to the free metal — the central step, discussed next.
- Refining: the crude metal is purified by electrolysis, distillation, or other means.
Choosing the reducing agent: the activity series
The reducing agent must be stronger than the metal being extracted:
- Carbon (smelting): for moderately active metals, coke reduces the oxide at high temperature — for zinc, ZnO(s) + C(s) → Zn(g) + CO(g). Works for Fe, Zn, Sn, and Pb; the CO escapes as gas, driving the equilibrium forward.
- A more active metal: for metals carbon cannot reduce (e.g., titanium), magnesium or sodium is used. The thermite reaction — Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s) — welds rails.
- Electrolysis: for the most active metals (Na, K, Ca, Mg, Al), no chemical reducing agent is strong enough, so electrical energy supplies the electrons: molten NaCl in the Downs cell gives Na and Cl₂; molten Al₂O₃ in cryolite (Na₃AlF₆) in the Hall–Héroult cell gives Al and O₂.
Why electrolysis of melts, not solutions
Water is easier to reduce than Na⁺, K⁺, Ca²⁺, Mg²⁺, and Al³⁺, so electrolyzing an aqueous solution of these ions produces H₂ at the cathode, not the metal. That is why industry melts the compound (often with a flux like cryolite to lower the melting point) and runs electrolysis with no water present. Active metals require molten (fused) electrolysis; only metals below hydrogen can be obtained from aqueous solutions.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Roasting | Smelting | Roasting heats a sulfide in air to make the oxide; smelting reduces the oxide with carbon. |
| Native metal | Any metal in an ore | Native means uncombined (Au, Ag, Pt); almost all other metals are found as compounds. |
| Electrolysis of melts | Electrolysis of aqueous solutions | Melts are required for Na, K, Ca, Mg, Al because water is reduced before those ions. |
| An ore | A mineral | A mineral is any naturally occurring solid with defined composition; an ore is one worth mining. |
| Carbon reduction for all metals | Carbon reduction for active metals | Carbon can reduce Zn, Fe, Sn, Pb but not Na, K, Ca, Mg, Al. |
| Al's scarcity | Al's abundance | Al is the most abundant crustal metal; its former high price came from extraction energy, not scarcity. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Metals like gold and silver are so calm that they sit around as shiny nuggets, but most metals are "hyper" — they grab onto oxygen so tightly that they hide inside rocks. To get the metal out, miners crush the rock and chemists use tricks to free it: cook it with carbon, react it with a hungrier metal, or use electricity like a magnet to drag the metal out. The more "hyper" the metal, the more electricity it takes to free it.
Worked example
Example 1: Smelting zinc — how much metal from how much ore?
Sphalerite is roasted and the zinc oxide is reduced with carbon:
ZnO(s) + C(s) → Zn(g) + CO(g)
Problem: How many grams of zinc can be produced from 250.0 g of ZnO with excess carbon? Molar masses: M(Zn) = 65.38 g mol-1, M(O) = 16.00 g mol-1, so M(ZnO) = 65.38 + 16.00 = 81.38 g mol-1.
Plan: n = mM, apply the 1:1 mole ratio, then m = n × M.
Substitution (dimensional analysis):
n(ZnO) = 250.0 g × 1 mol81.38 g = 3.072 mol
n(Zn) = 3.072 mol ZnO × 1 mol Zn1 mol ZnO = 3.072 mol Zn
m(Zn) = 3.072 mol × 65.38 g1 mol = 200.9 g
Answer: about 201 g of zinc.
Example 2: Electrolysis — how much aluminum does a Hall–Héroult cell produce?
At the cathode of a Hall–Héroult cell, molten Al³⁺ gains electrons:
Al3+ + 3e- → Al
Problem: A cell operates at 1.50 × 104 A for 2.00 h. What mass of aluminum is deposited? (Faraday constant F = 96,485 C mol-1; M(Al) = 26.98 g mol-1.)
Plan: Q = I × t; n(e-) = QF; n(Al) = n(e-)3; m = n × M.
Substitution:
Q = (1.50 × 104 A)(2.00 h) × 3600 s1 h = 1.08 × 108 C
n(e-) = 1.08 × 108 C96,485 C mol-1 = 1.12 × 103 mol
n(Al) = 1.12 × 103 mol e- × 1 mol Al3 mol e- = 3.73 × 102 mol
m(Al) = 3.73 × 102 mol × 26.98 g mol-1 = 1.01 × 104 g ≈ 10.1 kg
Answer: about 10.1 kg of aluminum per cell over two hours.
Key takeaways
- Representative metals = s-block (groups 1–2) plus p-block metals (Al–Tl, Sn–Pb, Bi).
- Most metals occur combined; only Au, Ag, Pt (occasionally Cu, Hg) occur native.
- Extraction follows the activity series: carbon smelting (Fe, Zn, Sn, Pb) → active-metal reduction (Ti) → electrolysis of melts (Na, K, Ca, Mg, Al).
- Sulfide ores are roasted to oxides before reduction; the SO₂ produced is a pollution concern.
- Active metals cannot be won from aqueous solution — water is reduced first; melts are required (Downs cell for Na; Hall–Héroult cell for Al with cryolite).
- Al is the most abundant crustal metal (~8% by mass) yet was once costly — extraction needs huge amounts of energy.
- Inert pair effect: heavier p-block metals show stable low states (Sn²⁺, Pb²⁺, Tl⁺) alongside their group states.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Why do most representative metals occur as compounds rather than free elements?
Show answer
Most representative metals are strong reducing agents — they lose electrons readily — so in nature they are oxidized and locked into compounds. Only metals too unreactive to be oxidized (Au, Ag, Pt) remain native.
Arrange Na, Zn, Au, and Al by extraction method (carbon, electrolysis, or none — native).
Show answer
Au: none (native). Zn: carbon smelting. Al: electrolysis of the molten oxide. Na: electrolysis of molten NaCl.
Why must aluminum be made by electrolyzing molten Al₂O₃ rather than an aqueous aluminum salt solution?
Show answer
In aqueous solution, water is reduced to H₂ more easily than Al³⁺, so no metal forms. Melting the compound (cryolite lowers the melting point) removes water and lets Al³⁺ be reduced directly.
A blast furnace runs Fe2O3 + 3CO → 2Fe + 3CO2. How many grams of iron form from 400.0 g of Fe₂O₃?
Show answer
n(Fe2O3) = 400.0 / 159.69 = 2.505 mol; the ratio gives 5.010 mol Fe; m = 5.010 × 55.85 = 279.8 g ≈ 280 g.
What is the inert pair effect, and which lead oxidation states does it explain?
Show answer
In heavy p-block metals the s² electrons bond poorly, so a stable state two less than the group number appears — Pb²⁺ (beside Pb⁴⁺), Sn²⁺, Tl⁺.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Ore
- A mineral rich enough in a metal to make extraction economical
- Gangue
- The worthless rock mixed with the ore
- Roasting
- Heating a sulfide ore in air to convert it to the oxide
- Smelting
- Reducing an oxide ore with carbon (coke) at high temperature
- Calcination
- Heating a carbonate ore to drive off CO₂
- Electrolysis
- Using electric current to force a nonspontaneous reduction
- Native metal
- A metal found uncombined in nature
- Inert pair effect
- Heavy p-block metals using only their p electrons in bonding
Sources & references
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