Worked example: solution

Apply the law

The totals must match: 15.9 g of reactant becomes 8.3 g of solid plus every gram of escaped gas.

Worked example: solution

Worked example: solution

Worked example 1: solution

Step 1 · Match protons and Z

Z = 14, so the atom holds 14 protons, and 14 protons is what makes it silicon.

Worked example 1: solution

The mass number counts protons and neutrons together: neutrons = 29 − 14 = 15.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Match protons and Z

The tile gives Z = 29: the atom holds 29 protons, and 29 protons is copper.

Worked example 2: solution

Neutrons = 65 − 29 = 36. The tile's 63.55 is an average mass of many atoms; the mass number 65 belongs to this one atom.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: A

C read the mass number as the neutron count: 66. B added instead of subtracting: 66 + 30 = 96. D swapped protons and neutrons: 36 protons is a different element: krypton, not zinc.

Worked example 3: solution

A common first attempt

The mass number counts the nucleus. Electrons never enter it.

Worked example 3: solution

A common first attempt

24 protons means Z = 24: the element is chromium. The neutron and electron counts have no part in the identity.

Worked example 3: solution

A common first attempt

Only the nucleus counts: A = 24 + 28 = 52.

Worked example 3: solution

A common first attempt

Worked example 3: solution

A common first attempt

Worked example 3: the route on the map

Practice 2 · answer: D

A read the mass number as the atomic number: Z = 27 is cobalt, and 27 protons leave no room for 14 neutrons in a nucleus of 27. B read the neutron count as the proton count: Z = 14 is silicon, but neutrons never name the element. C matched the electrons to the neutrons; a neutral atom matches them to the protons, 13.

Worked example 1: solution

Step 1 · Find Z

The lower-left number is the atomic number: Z = 29, so 29 protons. The periodic table agrees: element 29 is copper.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: the first attempt

A common first attempt

Worked example 2: the first attempt

A common first attempt

Worked example 2: the first attempt

A common first attempt

Worked example 2: solution

Step 1 · Find Z

Carbon is element 6 on the periodic table: Z = 6, so 6 protons.

Worked example 2: solution

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A read the mass number as the neutron count; 88 counts protons and neutrons together. C added the two numbers: 88 + 38 = 126, counting the protons twice. D read the atomic number; 38 counts the protons.

Practice 2 · answer: C

A read the mass number as the neutron count; 114 counts protons and neutrons together. B read the atomic number; 48 counts the protons. D added the two numbers: 114 + 48 = 162, counting the protons twice.

Worked example 3: solution

Step 1 · Find Z

The periodic table places gallium at element 31: Z = 31, so 31 protons.

Worked example 3: solution

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 · answer: A

B matched the mass numbers: atom 1 has A = 26 + 32 = 58, like nickel-58, but 26 and 28 protons are two elements. C matched the neutron counts: nickel-58 holds 58 − 28 = 30 neutrons, like atom 3, but neutrons never name the element. D read atom 3's 30 neutrons as its proton count, zinc; subtract instead: 56 − 30 = 26.

Worked example 1: solution

Two products are needed, then one sum.

Step 1 · Percents → fractions

19.9 ÷ 100 = 0.199 and 80.1 ÷ 100 = 0.801. The fractions cover the whole sample: 0.199 + 0.801 = 1.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Dividing by two weights each isotope equally. This sample is not an even split: 75.76% of its atoms carry the lighter mass.

Worked example 2: solution

Two products are needed, then one sum. First the fractions: 75.76 ÷ 100 = 0.7576 and 24.24 ÷ 100 = 0.2424.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A averaged the masses equally: (68.9256 + 70.9247) ÷ 2 = 69.93. C attached the abundances to the wrong isotopes: 68.9256 × 0.39892 + 70.9247 × 0.60108 = 70.13. D used whole percents: 68.9256 × 60.108 + 70.9247 × 39.892 = 6972.

Worked example 3: fractions and contributions

Three products are needed, then one sum.

Step 1 · Percents → fractions

78.99 ÷ 100 = 0.7899, 10.00 ÷ 100 = 0.1000, 11.01 ÷ 100 = 0.1101. Together: 0.7899 + 0.1000 + 0.1101 = 1.

Worked example 3: fractions and contributions

Worked example 3: fractions and contributions

Worked example 3: the sum

Step 3 · Add the contributions

Worked example 3: the sum

Step 3 · Add the contributions

Worked example 3: the route on the map

Practice 2 · answer: D

Practice 2 · answer: D

Worked example 1: solution

Step 1 · Find the period

Chlorine sits in the third row: period 3. That row runs from sodium (Z = 11) to argon (Z = 18): 18 − 11 + 1 = 8 elements.

Worked example 1: solution

Counting columns from the left edge, chlorine lands in column 17: group 17.

Worked example 1: solution

Group 17 is the halogens: fluorine, chlorine, bromine, and iodine.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Thirty-eight protons against twenty is a real difference, and mass roughly doubles. But behavior follows the column: 38 − 20 = 18, exactly one full row, so strontium lands directly under calcium.

Worked example 2: solution

One row below calcium's period 4: period 5. Same column: group 2.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A counted rows inside the halogen column: fluorine's row is period 2, so the fourth halogen down is iodine, period 5. C read group 17 as group 7 and landed in the transition block: manganese sits in period 4, group 7. D overshot by one column: krypton is group 18, a noble gas.

Worked example 3: solution

The ordering rule

By mass, iodine should come first. The table disagrees: position follows atomic number, protons only, and 52 comes before 53. Extra neutrons make tellurium heavier; position ignores them.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 · answer: C

A misreads the geometry: sodium and potassium share a group, a column; period-mates differ. B fails on its own numbers: 39.10 − 22.99 = 16.11, the masses are not close, and mass does not set behavior. D misfiles them: group 1 is a main-group column; the transition block starts at group 3.

Worked example 4: solution

Family → group

The family name fixes the column: alkaline earth metals are group 2.

Worked example 4: solution

The period fixes the row: period 3 runs from sodium (Z = 11) to argon (Z = 18).

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 3 · answer: C

A fits the bench tests, but potassium sits in group 1: one outer electron, so it loses one, never two. B fails at the bench: silicon is a metalloid on the staircase, a semiconductor, not a full conductor. D reverses the electron move: sulfur is a brittle nonmetal, and nonmetals gain electrons.

Practice 4 · answer: A

B ranked by pull instead of size: the atom that pulls hardest is the smallest, so the order came out reversed. C swapped sulfur and oxygen: each row down adds a shell, and sulfur's outer electrons sit one shell beyond oxygen's. D handed the strongest pull to the biggest atom: the biggest atom holds the bonding pair farthest from its nucleus and pulls it least.

Extra practice 1 · answer: D

A carried the class down the column: class shifts down group 15, from nonmetal (N, P) to metalloid (As, Sb) to metal (Bi). B treated row-mates as alike: rubidium and strontium sit far left of the staircase; antimony sits on it. C judged by one bench property: metalloids look metallic, and antimony is only a modest conductor.

Extra practice 2 · answer: A

B read the mass number as the atomic number: element 85 is astatine, period 6, group 17. C found rubidium but swapped the address: period 1 holds only H and He, so no group 5 seat exists there. D took the neutron count as the atomic number: 85 − 37 = 48 is cadmium, and 48 − 37 + 1 = 12 puts it in period 5, group 12.

Extra practice 3 · answer: B

A reversed the electron move: cesium is a metal, and metals lose electrons; gaining one is the halogen move. C carried group 17's fade into group 1: down the alkali column the water reaction escalates. D let the proton gap override the column: 36 = 18 + 18 is exactly two full rows, so cesium stays in group 1 with one outer electron.

Worked example 1: solution

Step 1 · Count the protons

12 protons before, 12 after: the ion is still magnesium. The neutrons also stay at 12.

Worked example 1: solution

Neutral means 12 electrons. Two are lost: 12 − 2 = 10 e⁻.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Each electron carries one negative charge. Adding electrons can only push the total negative.

Worked example 2: solution

16 protons, unchanged: still sulfur. The neutrons stay at 16.

Worked example 2: solution

The nearest noble gas is argon, 18 electrons. Sulfur holds 16 and gains two: 16 + 2 = 18 e⁻.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: A

B: 58 protons is no longer barium; protons never change in chemistry. C: gaining two electrons computes 56 − 58 = 2−, an anion. D: losing two protons changes the element too, and 54 p⁺ against 56 e⁻ computes 54 − 56 = 2−.

Worked example 3: solution

Step 1 · Count the protons

34 protons: selenium. The 36 electrons match krypton's count, but electrons come and go; protons name the element.

Worked example 3: solution

36 electrons are given, two more than the protons: charge 34 − 36 = 2−.

Worked example 3: solution

Protons plus neutrons: 34 + 46 = 80. Electrons never enter the mass number.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 · answer: D

A accepted statement 1: electrons never enter the mass number, so ⁴⁰Ca²⁺ keeps A = 20 + 20 = 40, not 40 − 2 = 38. B read 3− as three electrons lost, 15 − 3 = 12, and rejected statement 2; a negative ion gained them. C found neutrons by subtracting electrons, 85 − 36 = 49, and rejected statement 3; neutrons = A − protons = 48.

Worked example 1: solution

Step 2 · Identify the ions

Potassium has one possible charge, so no Roman numeral will appear.

Worked example 1: solution

The subscript 2 records the balance already:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 2 · Identify the ions

Worked example 2: solution

Neither charge cancels the other one-for-one. The smallest totals that cancel are +6 and −6: three Mg²⁺ with two N³⁻. The criss-cross shortcut writes each ion's charge as the other ion's subscript.

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: C

A read the subscript as a charge: sodium 2+ would give 2(+2) + 1(−2) = +2, not neutral; the 2 counts Na⁺ ions. B counts atoms with Greek prefixes, the naming system for molecular compounds. D writes a numeral for a metal with only one possible charge; numerals mark variable-charge metals only.

Worked example 3: testing the first attempt

A common first attempt

Worked example 3: testing the first attempt

A common first attempt

Worked example 3: solution

Step 2 · Identify the ions Step 3 · Balance the charges to zero

Oxide is fixed at 2−. Copper's charge must come from this formula: one O²⁻ contributes 2−, so the two Cu contribute +2 in total.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 · answer: A

B recycles the 3+ from rust: here 1(+3) + 1(−2) = +1, not neutral; the numeral must balance this formula. C omits the numeral: iron has more than one possible charge, so "iron oxide" cannot separate FeO from Fe₂O₃. D counts atoms with a Greek prefix, the system for molecular compounds.

Worked example 4: solution

Step 2 · Identify the ions

The numeral states tin's charge directly: Sn⁴⁺. Oxide is O²⁻ from periodic position.

Worked example 4: solution

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 3 · answer: D

B balances a different cation: 1(+2) + 2(−1) = 0 holds for Sn²⁺, so SnF₂ is tin(II) fluoride, the stannous fluoride on toothpaste labels. C pairs one of each ion: 1(+4) + 1(−1) = +3, not neutral. A writes the numeral as a count of tin atoms: 4(+4) + 1(−1) = +15.

Practice 4 · answer: A

B is correct: three S²⁻ carry 3(−2) = −6, split over two Fe, 6 ÷ 2 = 3+ each; reading the subscript 2 as the charge gives the wrong numeral. C is correct: aluminum is fixed at 3+, 1(+3) + 1(−3) = 0, so no numeral. D is correct: 1(+2) + 2(−1) = 0, and strontium diiodide would borrow molecular prefixes.

Worked example 1: solution

Step 1 · Identify the ions

The group CO₃ with its charge is on the memorized list: carbonate, CO₃²⁻. The rest is potassium, K⁺.

Worked example 1: solution

Worked example 1: solution

Cation first, then the anion:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: balancing the charges

A common first attempt

Two nitrate ions were intended. Written without parentheses, the subscripts run together and the group disappears.

Worked example 2: balancing the charges

Magnesium forms Mg²⁺. Nitrate is on the memorized list: NO₃⁻.

Worked example 2: balancing the charges

Worked example 2: balancing the charges

Worked example 2: writing the formula

Step 3 · Write the formula or the name

Cation first, and the repeated polyatomic goes in parentheses:

Worked example 2: writing the formula

Worked example 2: the route on the map

Practice 1 · answer: D

Four oxygens is one above the -ate member of the chlorine series: per- + chlorate. C, chlorate, is ClO₃⁻, one oxygen fewer. A, chlorite, is ClO₂⁻, two fewer. B, chloride, is Cl⁻, a monatomic ion with no oxygen at all.

Worked example 3: solution

Step 1 · Identify the ions

Ammonium, the one common polyatomic cation: NH₄⁺. Phosphate: PO₄³⁻.

Worked example 3: solution

Worked example 3: solution

The repeated ion is polyatomic, so it takes the parentheses. Phosphate appears once and needs none.

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 · answer: B

A stops at one of each: 1(+1) + 1(−2) = −1, not zero. C balances its charges, 2(+1) + 1(−2) = 0, but holds sulfite, SO₃²⁻ (the -ite ion, one oxygen fewer). D doubles the wrong ion: 1(+1) + 2(−2) = −3.

Worked example 4: testing the first attempt

A common first attempt

Worked example 4: testing the first attempt

A common first attempt

Worked example 4: solution

Step 1 · Identify the ions

Aluminum forms Al³⁺. Sulfate is on the memorized list: SO₄²⁻.

Worked example 4: solution

Worked example 4: solution

The crossed 3 lands outside the parentheses; the 4 inside never changes.

Worked example 4: solution

Worked example 4: the route on the map

Practice 3 · answer: B

A stops at one anion: 1(+2) + 1(−1) = +1, not neutral. C doubles the cation instead: 2(+2) + 1(−1) = +3. D balances, 1(+2) + 1(−2) = 0, but its anion is carbonate: CaCO₃ is calcium carbonate, a different compound.

Practice 4 · answer: D

A kept the -ate: Fe₂(SO₄)₃ is iron(III) sulfate. B pairs one of each: 1(+3) + 1(−2) = +1; FeSO₃ balances only with Fe²⁺, iron(II) sulfite. C writes each ion's own charge as its own subscript: 3(+3) + 2(−2) = +5. E multiplies the 3 through: Fe₂S₃O₉ has the atoms but no sulfite group left.

Worked example 1: solution

A common first attempt

In an ionic name two element names are enough, because charges fix the ratio. Carbon and oxygen carry no charges to fix it.

Worked example 1: solution

No leading H, no metal: two nonmetals. A molecular compound, so prefixes count the atoms.

Worked example 1: solution

Worked example 1: solution

Monoxide rebuilds one O, dioxide two: each name recovers its own formula.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Classify the compound

Counting prefixes appear only in molecular names: this is a molecular compound of phosphorus and oxygen.

Worked example 2: solution

Each prefix sets its own element's subscript:

Worked example 2: solution

P₄O₁₀ reads back to the same name: the subscripts stay 4 and 10, not a reduced ratio.

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A drops the prefixes, and NO, NO₂, N₂O, and N₂O₃ would all share that name: the formula is lost. C swaps the prefixes: trinitrogen dioxide rebuilds N₃O₂, a different compound. D uses an acid name, but nitrous acid is HNO₂ dissolved in water, and N₂O₃ contains no hydrogen.

Worked example 3: HCl(aq)

Step 1 · Classify the compound

H first and dissolved in water: both are acids. Acid rules, not counting prefixes.

Worked example 3: HCl(aq)

Remove the H from HCl: the anion is chloride, Cl⁻, with no oxygen.

Worked example 3: HCl(aq)

Worked example 3: HNO₃(aq)

A common first attempt

Hydro- means the anion holds no oxygen. Nitrate holds three.

Worked example 3: HNO₃(aq)

Oxygen present, so the anion's suffix maps: -ate → -ic acid.

Worked example 3: HNO₃(aq)

Nitric acid reads back to HNO₃(aq) via nitrate; hydrochloric via chloride.

Worked example 3: HNO₃(aq)

Worked example 3: the route on the map

Practice 2 · answer: C

A names the pure gas, HF(g); the (aq) marks a dissolved acid with its own name. B drops hydro-: without it the name reads as an oxyacid, and fluoride holds no oxygen. D uses counting prefixes, and prefixes never appear in acid names.

Worked example 4: solution

Step 1 · Classify the compound

An acid name, so the formula starts with H and carries (aq).

Worked example 4: solution

Run the suffix map backward: -ous came from -ite. The anion is sulfite, SO₃²⁻.

Worked example 4: solution

Worked example 4: solution

H₂SO₃ minus its hydrogens is sulfite, and -ite returns -ous: sulfurous acid ✓.

Worked example 4: solution

Worked example 4: the route on the map

Practice 3 · answer: D

B writes one H: 1(+1) + 1(−3) = −2, not neutral. C writes two: 2(+1) + 1(−3) = −1. A balances three H on the wrong rung: PO₃³⁻ is phosphite, and its acid is phosphorous acid, one rung down the ladder.

Practice 4 · answer: A

B starts from chlorite, ClO₂⁻, one rung up: chlorous acid is HClO₂(aq). C is the no-oxygen binary acid, HCl(aq); this anion holds one oxygen, so hydro- never appears. D starts from perchlorate, ClO₄⁻, the top rung: perchloric acid is HClO₄(aq).

Worked example 5: solution

Step 1 · Classify the compound

A metal with a polyatomic anion: an ionic salt, carrying attached water. A hydrate.

Worked example 5: solution

Name the salt by the ionic rules, then count the waters with a prefix.

Worked example 5: solution

Worked example 5: solution

Worked example 5: the route on the map

Practice 5 · answer: C

A is correct: two nonmetals take prefixes, di- → 2 N, pent(a)- → 5 O; nitrogen(V) oxide would borrow the ionic numeral. B is correct: iodide holds no oxygen, so hydro- + iod- + -ic. D is correct: three Cl⁻ demand Fe³⁺, 1(+3) + 3(−1) = 0, and hexa- counts the 6 waters.

Worked example 1: solution

Step 1 · Count the carbons Step 2 · Pick the prefix

The count is given: n = 7. Seven carbons take the prefix hept-.

Worked example 1: solution

Every bond is single, so the ending is -ane. hept- + -ane = heptane.

Worked example 1: solution

Worked example 1: solution

Worked example 2: solution

Step 1 · Count the carbons

The subscript on C is the count: n = 4.

Worked example 2: solution

Four carbons take but-; every bond is single, so the ending is -ane. but- + -ane = butane.

Worked example 2: solution

Worked example 2: solution

Guided example: solution

Step 1 · Count the carbons

Every written group holds one carbon. The subscript 4 counts only the CH₂ units inside the parentheses.

Guided example: solution

Eight carbons take oct-. Every bond is single, so the ending is -ane: octane.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: B

A (C₅H₁₀) stops at 2n hydrogens, two H short of the 2n + 2 count. C (C₆H₁₄) is the six-carbon alkane, hexane: one carbon too many. D (C₅H₁₁) has an odd H count no neutral alkane can have. Only B obeys CₙH₂ₙ₊₂.

Practice 1 · answer: B

A (C₅H₁₀) stops at 2n hydrogens, two H short of the 2n + 2 count. C (C₆H₁₄) is the six-carbon alkane, hexane: one carbon too many. D (C₅H₁₁) has an odd H count no neutral alkane can have. Only B obeys CₙH₂ₙ₊₂.

Practice 2 · answer: A

B is the CₙH₂ₙ count, 2×10 = 20, two H short of an alkane. C counted the two ends and the run but missed the two CH₂ written outside the parentheses: 1 + 6 + 1 = 8, octane. D makes both slips: eight carbons and 2×8 = 16 H. E has the right formula but the binary name with no prefixes; hydrocarbons take the alkane ladder.

Practice 2 · answer: A

B is the CₙH₂ₙ count, 2×10 = 20, two H short of an alkane. C counted the two ends and the run but missed the two CH₂ written outside the parentheses: 1 + 6 + 1 = 8, octane. D makes both slips: eight carbons and 2×8 = 16 H. E has the right formula but the binary name with no prefixes; hydrocarbons take the alkane ladder.

Practice 3 · answer: C

A subtracted only one end: 1 + 8 + 1 = 10 C, decane. B put all nine inside the parentheses: 1 + 9 + 1 = 11 C. D ends on CH₂: 3 + 7(2) + 2 = 19 H, an odd count. E dropped both CH₃ caps: 2 + 7(2) + 2 = 18 H, the CₙH₂ₙ count, two short.

Practice 3 · answer: C

A subtracted only one end: 1 + 8 + 1 = 10 C, decane. B put all nine inside the parentheses: 1 + 9 + 1 = 11 C. D ends on CH₂: 3 + 7(2) + 2 = 19 H, an odd count. E dropped both CH₃ caps: 2 + 7(2) + 2 = 18 H, the CₙH₂ₙ count, two short.

Practice 4 · answer: D

A (8) reads (CH₂)₅ as a single CH₂ and drops the subscript: 3 + 2 + 3 = 8. B (14) is the CₙH₂ₙ count, 2×7 = 14, two H short of an alkane. C (15) writes one CH₃ end as CH₂: 3 + 5(2) + 2 = 15, an odd count no neutral alkane can have.

Practice 4 · answer: D

A (8) reads (CH₂)₅ as a single CH₂ and drops the subscript: 3 + 2 + 3 = 8. B (14) is the CₙH₂ₙ count, 2×7 = 14, two H short of an alkane. C (15) writes one CH₃ end as CH₂: 3 + 5(2) + 2 = 15, an odd count no neutral alkane can have.

Practice 5 · answer: C

A (butane) misses both CH₂ outside the parentheses: 1 + 2 + 1 = 4. B (pentane) catches only one of them: 1 + 2 + 1 + 1 = 5. D (hexacarbon tetradecahydride) gets C₆H₁₄ right but uses the binary-molecular prefixes, and E (carbon hydride) uses the no-prefix binary name. Hydrocarbons take their names from the alkane ladder.

Practice 5 · answer: C

A (butane) misses both CH₂ outside the parentheses: 1 + 2 + 1 = 4. B (pentane) catches only one of them: 1 + 2 + 1 + 1 = 5. D (hexacarbon tetradecahydride) gets C₆H₁₄ right but uses the binary-molecular prefixes, and E (carbon hydride) uses the no-prefix binary name. Hydrocarbons take their names from the alkane ladder.

Worked example 1: testing the first attempt

A common first attempt

Worked example 1: testing the first attempt

A common first attempt

Worked example 1: solution

Step 1 · Classify the compound

No leading H. A metal is present: ionic, and copper is a variable-charge metal.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Classify the compound

H first and dissolved in water: an acid. The metal test and the prefix test never run.

Worked example 2: solution

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A borrows the ionic system's numeral, but a numeral reports an ion's charge and ICl₃ holds no ions. C is the fixed-charge ionic style, and without a prefix the name cannot separate ICl₃ from ICl. D swaps the prefix onto iodine: triiodine monochloride rebuilds I₃Cl.

Practice 2 · answer: D

A writes a numeral for a metal with one possible charge; Group 2 is always 2+, so no numeral appears. B counts atoms with molecular prefixes, but the metal makes this ionic. C names the monatomic anion P³⁻; this compound holds the polyatomic group phosphate, PO₄³⁻.

Practice 3 · answer: A

B is the ionic-style read-back, and hydrogen carbonate already names a different species, the ion HCO₃⁻. C adds hydro-, which claims a no-oxygen anion; carbonate holds three. D counts atoms with a prefix, and prefixes never appear in acid names.

Practice 4 · answer: C

A stopped at the raw criss-cross: 2(+4) + 4(−2) = 0 balances, but 2 : 4 reduces to 1 : 2. B read the subscript 2 as lead's charge: 1(+2) + 2(−2) = −2. D borrows molecular prefixes for a compound with a metal. E swaps in sulfite, SO₃²⁻, one oxygen fewer than sulfate.