PART I · SHORT RESPONSE · SOLUTIONS第一部分 · 短答题 · 详解AP MCQ + ON/BC/AB short answer · 23 marksAP 选择题 + 安/卑/阿省考短答 · 共 23 分
Section A · Worked SolutionsA 部分 · 详细解答
Q1EASY易🇺🇸 US美AP-style MCQAP 风格选择题§1 Ions and Charges离子与电荷[3 marks][3 分]
Sulfur (S) is in Group 16 of the periodic table. Which ion does sulfur typically form?硫（S）位于元素周期表第 16 族。硫通常形成哪种离子？
Answer:答案：(C) $\text{S}^{2-}$
Determine the ion from group number由族序数确定离子M1·A1·A1
Group 16 elements have 6 valence electrons. To reach a full octet they gain 2 electrons, forming a 2- ion: $\text{S} + 2e^- \rightarrow \text{S}^{2-}$. This is the sulfide ion. The charge equals $16 - 18 = -2$ (group number minus 18 for nonmetals).第 16 族元素有 6 个价电子。为达到满八隅体，需得到 2 个电子，形成 2- 离子：$\text{S} + 2e^- \rightarrow \text{S}^{2-}$。这就是硫离子。电荷 $= 16 - 18 = -2$（非金属族序数减 18）。
Why the distractors fail.干扰项分析。
(A) $\text{S}^{2+}$:sulfur is a nonmetal; it gains electrons rather than losing them to form anions, not cations.硫是非金属，得电子而非失电子，形成阴离子而非阳离子。
(B) $\text{S}^{6+}$:sulfur can exhibit +6 in polyatomic ions such as sulfate, but as a simple monatomic ion it does not form $\text{S}^{6+}$.硫在多原子离子（如硫酸根）中可显 +6，但单原子离子不形成 $\text{S}^{6+}$。
(D) $\text{S}^{1-}$:would require gaining only 1 electron, leaving sulfur one short of an octet. Nonmetals in Group 16 always gain 2 electrons.仅得 1 个电子，距八隅体仍差 1 个。第 16 族非金属总是得 2 个电子。
Group number predicts ion charge for main-group elements.主族元素的族序数可预测离子电荷。For nonmetals (Groups 15-17), the ion charge is (group number - 18): Group 15 gives 3-, Group 16 gives 2-, Group 17 gives 1-. For metals in Groups 1-2, the charge equals the group number: Group 1 gives 1+, Group 2 gives 2+. Transition metals are variable and must be determined from context. Knowing these patterns makes ion identification fast and reliable on any exam.对于非金属（第 15-17 族），离子电荷为（族序数 - 18）：第 15 族为 3-，第 16 族为 2-，第 17 族为 1-。对于第 1-2 族金属，电荷等于族序数：第 1 族为 1+，第 2 族为 2+。过渡金属电荷可变，须从化合物语境推导。掌握这些规律，可在任何考试中快速可靠地判断离子电荷。
Na is sodium (Group 1, fixed 1+ charge). Cl forms Cl$^-$ (chloride). Rule: cation name + anion root + "-ide." Result: sodium chloride. No Roman numeral needed because Na has only one possible charge.Na 为钠（第 1 族，固定 1+ 电荷）。Cl 形成 Cl$^-$（氯离子）。规则：阳离子名称 + 阴离子词根 + "-ide"。结果：氯化钠。因 Na 只有一种电荷，无需罗马数字。
(b) MgO A1
Mg is magnesium (Group 2, fixed 2+ charge). O forms O$^{2-}$ (oxide). Result: magnesium oxide.Mg 为镁（第 2 族，固定 2+ 电荷）。O 形成 O$^{2-}$（氧离子）。结果：氧化镁。
(c) Al$_2$O$_3$ A1
Al is aluminum (fixed 3+ charge). O$^{2-}$ is oxide. Subscripts confirm charge balance: $2(3+) + 3(2-) = 0$. Result: aluminum oxide. No prefix used for ionic compounds.Al 为铝（固定 3+ 电荷）。O$^{2-}$ 为氧离子。下标验证电荷平衡：$2(3+) + 3(2-) = 0$。结果：氧化铝。离子化合物命名不使用数量前缀。
(d) CaF$_2$ A1
Ca is calcium (fixed 2+ charge). F forms F$^-$ (fluoride). Result: calcium fluoride. The subscript 2 reflects charge balance, not a naming prefix.Ca 为钙（固定 2+ 电荷）。F 形成 F$^-$（氟离子）。结果：氟化钙。下标 2 体现电荷平衡，不是命名前缀。
Ionic naming rule: cation name + anion stem + "-ide." Subscripts are never spoken.离子化合物命名规则：阳离子名 + 阴离子词根 + "-ide"。下标不读出来。All four compounds here contain main-group metals with fixed charges, so no Roman numeral is required. The anion stems are: chlor- (Cl), ox- (O), fluor- (F). A common mistake is to say "dialuminum trioxide" or "calcium difluoride" as if naming a molecular compound. Subscripts in ionic formulas arise from charge balancing; they carry no naming information. Save Greek prefixes for molecular compounds only.以上四种化合物的阳离子均为电荷固定的主族金属，无需罗马数字。阴离子词根分别为：chlor-（Cl）、ox-（O）、fluor-（F）。常见错误是套用共价化合物命名，说成"二氧化三铝"或"二氟化钙"。离子化合物中下标来自电荷平衡，不体现在名称里。希腊数字前缀仅用于共价化合物。
Which formula correctly represents the sulfate ion?哪个化学式正确表示了硫酸根离子？
Answer:答案：(B) $\text{SO}_4^{2-}$
Recall the sulfate ion formula记忆硫酸根离子的化学式M1·A1·A1
The sulfate ion is $\text{SO}_4^{2-}$: one sulfur atom bonded to four oxygen atoms, overall charge 2-. Sulfur is in the +6 oxidation state within this ion: $+6 + 4(-2) = -2$.硫酸根离子为 $\text{SO}_4^{2-}$：一个硫原子与四个氧原子键合，总电荷为 2-。该离子中硫的氧化态为 +6：$+6 + 4(-2) = -2$。
Why the distractors fail.干扰项分析。
(A) $\text{SO}_3^{2-}$:this is the sulfite ion (three oxygens), not sulfate.这是亚硫酸根离子（三个氧），不是硫酸根。
(C) $\text{SO}_4^{-}$:correct atom count but wrong charge; sulfate carries a 2- charge, not 1-.原子数正确但电荷错误；硫酸根电荷为 2-，不是 1-。
(D) $\text{S}_2\text{O}_4^{2-}$:this is the dithionite ion, an uncommon reducing agent; not sulfate.这是连二亚硫酸根离子，是一种不常见的还原剂，不是硫酸根。
Sulfate vs. sulfite: the "-ate" ion has one more oxygen than the "-ite" ion.硫酸根 vs. 亚硫酸根："-ate" 离子比 "-ite" 离子多一个氧。For the sulfur oxyanion pair: sulfate $= \text{SO}_4^{2-}$ (4 O, "-ate") and sulfite $= \text{SO}_3^{2-}$ (3 O, "-ite"). Both carry a 2- charge. The same "-ate vs. -ite" pattern holds for nitrogen: nitrate $\text{NO}_3^-$ vs. nitrite $\text{NO}_2^-$, and for chlorine: perchlorate $\text{ClO}_4^-$ / chlorate $\text{ClO}_3^-$ / chlorite $\text{ClO}_2^-$ / hypochlorite $\text{ClO}^-$. Memorising the most-oxygen form ("-ate") and counting down is the most reliable strategy.硫的含氧酸根：硫酸根 $= \text{SO}_4^{2-}$（4 个 O，"-ate"）和亚硫酸根 $= \text{SO}_3^{2-}$（3 个 O，"-ite"），均带 2- 电荷。同样的 "-ate vs. -ite" 规律适用于氮（硝酸根 $\text{NO}_3^-$ vs. 亚硝酸根 $\text{NO}_2^-$）和氯（高氯酸根 $\text{ClO}_4^-$ / 氯酸根 $\text{ClO}_3^-$ / 亚氯酸根 $\text{ClO}_2^-$ / 次氯酸根 $\text{ClO}^-$）。记住含氧最多的形式（"-ate"），再往下数，是最可靠的策略。
MCQ: Apply Greek prefixes to $\text{P}_4\text{O}_{10}$选择题：对 $\text{P}_4\text{O}_{10}$ 套用希腊数字前缀M1·A1·A1
Molecular compounds use Greek prefixes: mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-, nona-, deca-. $\text{P}_4\text{O}_{10}$: 4 phosphorus atoms (tetra-) + 10 oxygen atoms (deca-). The name is tetraphosphorus decoxide. The prefix "deca-" contracts to "dec-" before "oxide" because "oxide" begins with a vowel (vowel elision rule for the "a" in deca-): deca + oxide = decoxide.共价化合物使用希腊数字前缀：mono-、di-、tri-、tetra-、penta-、hexa-、hepta-、octa-、nona-、deca-。$\text{P}_4\text{O}_{10}$：4 个磷原子（tetra-）+ 10 个氧原子（deca-）。名称为 tetraphosphorus decoxide。前缀 "deca-" 在 "oxide" 前缩写为 "dec-"，因为 "oxide" 以元音开头（元音省略规则：deca + oxide = decoxide）。
(e) Write the formula for "dinitrogen trioxide"写出"三氧化二氮"的化学式M1·A1
Di- = 2 nitrogen, tri- = 3 oxygen. Formula: $\text{N}_2\text{O}_3$. First element named first in the formula; Greek prefixes map directly to subscripts.di- = 2 个氮，tri- = 3 个氧。化学式：$\text{N}_2\text{O}_3$。名称中第一个元素在化学式中写在前面；希腊数字前缀直接对应下标。
Why the other MCQ options fail.其他选项错误原因。
(A) phosphorus(IV) oxide:Roman numerals are used for transition metal ionic compounds, not for molecular compounds.罗马数字用于过渡金属离子化合物，不用于共价化合物。
(C) tetraphosphorus decaoxide:keeps the full "deca-" before "oxide" without applying vowel elision; IUPAC drops the terminal "a" of deca- before "oxide."在 "oxide" 前保留完整的 "deca-"，未套用元音省略规则；IUPAC 规定在 "oxide" 前省去 deca- 末尾的 "a"。
(D) phosphorus tetraoxide:omits the tetra- prefix on phosphorus and misapplies it to oxygen; both elements need prefixes.漏写磷的 tetra- 前缀，并错误地将其用于氧；两种元素都需要前缀。
Vowel elision: drop the terminal vowel of a Greek prefix before "oxide" or "iodide."元音省略：在 "oxide" 或 "iodide" 前，省去希腊数字前缀末尾的元音。Only "mono-" and the prefixes ending in a vowel (mono, tetra, hepta, octa, nona, deca) apply elision when the element name starts with a vowel. So "mon-oxide" (monoxide), "dec-oxide" (decoxide), but "trioxide" and "dioxide" retain the prefix unchanged because di- and tri- do not end in a vowel that conflicts. In practice, test items most often target "monoxide" and "decoxide."只有 "mono-" 和以元音结尾的前缀（mono、tetra、hepta、octa、nona、deca）在元素名以元音开头时才省略末尾元音。因此 "mon-oxide"（monoxide）、"dec-oxide"（decoxide），而 "trioxide" 和 "dioxide" 保持不变，因为 di- 和 tri- 末尾不是会产生冲突的元音。实际考试中最常考 "monoxide" 和 "decoxide"。
Deduce metal charge and write the full IUPAC name for (a) CuBr$_2$, (b) Fe$_2$O$_3$, (c) MnO$_2$.推导金属离子电荷，写出 (a) CuBr$_2$、(b) Fe$_2$O$_3$、(c) MnO$_2$ 的完整 IUPAC 名称。
Each Br$^-$ carries a 1- charge; two bromides give a total anion charge of 2-. For the formula to be neutral, Cu must be 2+. Roman numeral required (Cu has multiple possible charges: +1 and +2). Name: copper(II) bromide.每个 Br$^-$ 电荷为 1-；两个溴离子合计阴离子电荷为 2-。化合物呈中性，故 Cu 必须为 2+。需要罗马数字（Cu 有 +1 和 +2 两种可能电荷）。名称：溴化铜(II)。
(b) Fe$_2$O$_3$ M1·A1·A1
Each O$^{2-}$ carries 2-; three oxide ions give a total anion charge of 6-. Two Fe atoms must together supply 6+, so each Fe $= 6/2 = 3+$. Roman numeral required. Name: iron(III) oxide.每个 O$^{2-}$ 电荷为 2-；三个氧离子合计阴离子电荷为 6-。两个 Fe 原子共需提供 6+，故每个 Fe $= 6/2 = 3+$。需要罗马数字。名称：氧化铁(III)。
(c) MnO$_2$ M1·A1
Two O$^{2-}$ give 4- total. One Mn must be 4+. Roman numeral required (Mn commonly forms +2, +4, +7). Name: manganese(IV) oxide.两个 O$^{2-}$ 合计 4-。一个 Mn 必须为 4+。需要罗马数字（Mn 常见 +2、+4、+7）。名称：氧化锰(IV)。
Charge deduction for transition metals: let anion total charge = metal total charge in magnitude.过渡金属电荷推导：令阴离子总电荷大小等于金属总电荷大小。Step 1: count the anion(s) and multiply by their charge to get the total negative charge. Step 2: divide by the number of metal atoms to find the metal ion charge. Step 3: add a Roman numeral in parentheses after the metal name whenever the metal can have more than one ionic charge. Main-group metals (Na, Mg, Al, Ca) have fixed charges and never need Roman numerals; transition metals almost always do. A student who writes "copper bromide" without the Roman numeral loses the mark because it is ambiguous between CuBr (copper(I)) and CuBr$_2$ (copper(II)).第一步：数出阴离子个数，乘以其电荷，得到总负电荷。第二步：除以金属原子数，得到金属离子电荷。第三步：只要金属可能有多种离子电荷，就在金属名称后用括号标注罗马数字。主族金属（Na、Mg、Al、Ca）电荷固定，不需要罗马数字；过渡金属几乎总是需要。若学生写"copper bromide"而没有罗马数字，则因含义不明确（可能是 CuBr 即铜(I)，也可能是 CuBr$_2$ 即铜(II)）而丢分。
Q7MEDIUM中🇺🇸 US美AP-feeder FRQAP 衔接简答题§3 Polyatomic ions in compounds含多原子离子的化合物[8 marks][8 分]
Name each compound; identify the polyatomic ion and its charge.命名以下各化合物；指出多原子离子及其电荷。
Polyatomic ion: sulfate $\text{SO}_4^{2-}$ (charge 2-). Cation: Na$^+$. Two Na$^+$ balance one SO$_4^{2-}$. Name: sodium sulfate. Na is a main-group metal; no Roman numeral needed.多原子离子：硫酸根 $\text{SO}_4^{2-}$（电荷 2-）。阳离子：Na$^+$。两个 Na$^+$ 平衡一个 SO$_4^{2-}$。名称：硫酸钠。Na 为主族金属，不需罗马数字。
(b) $\text{Fe(NO}_3)_3$ M1·A1
Polyatomic ion: nitrate $\text{NO}_3^-$ (charge 1-). Three nitrate ions give 3- total; Fe must be 3+. Roman numeral required. Name: iron(III) nitrate.多原子离子：硝酸根 $\text{NO}_3^-$（电荷 1-）。三个硝酸根合计 3-；Fe 必须为 3+。需要罗马数字。名称：硝酸铁(III)。
(c) $\text{NH}_4\text{Cl}$ M1·A1
Polyatomic ion: ammonium $\text{NH}_4^+$ (charge 1+), which is the cation here. Anion: Cl$^-$ (chloride). Name: ammonium chloride. Ammonium is the only common polyatomic cation.多原子离子：铵根 $\text{NH}_4^+$（电荷 1+），此处为阳离子。阴离子：Cl$^-$（氯离子）。名称：氯化铵。铵根是唯一常见的多原子阳离子。
Polyatomic ions are named as a unit; the "-ate"/"-ite" suffix travels with the ion into the compound name unchanged.多原子离子作为整体命名；"-ate"/"-ite" 后缀随离子直接带入化合物名称，保持不变。Unlike simple anions (Cl$^-$ becomes chloride), polyatomic oxyanions keep their full name: sulfate, nitrate, carbonate, phosphate. A key exam trap is part (b): students may forget that Fe(NO$_3$)$_3$ has three nitrate ions, deduce Fe = 3+ correctly, but then write "iron nitrate" without the Roman numeral. Since iron can be +2 or +3, the Roman numeral is mandatory. Part (c) tests whether students recognise $\text{NH}_4^+$ as a polyatomic cation rather than treating N and H separately.与简单阴离子不同（Cl$^-$ 变为 chloride），多原子含氧酸根保留完整名称：硫酸根、硝酸根、碳酸根、磷酸根。(b) 是常见考试陷阱：学生可能正确推导出 Fe = 3+，却漏写罗马数字，写成"iron nitrate"。由于铁可为 +2 或 +3，罗马数字是必须的。(c) 考查学生是否将 $\text{NH}_4^+$ 识别为多原子阳离子，而不是将 N 和 H 分开处理。
Ions: Al$^{3+}$ and SO$_4^{2-}$. Criss-cross: the magnitude of Al's charge (3) becomes the subscript of SO$_4$; the magnitude of SO$_4$'s charge (2) becomes the subscript of Al. Raw result: Al$_2$(SO$_4$)$_3$. Check charge balance: $2(3+) + 3(2-) = +6 - 6 = 0$. The polyatomic group must be enclosed in parentheses when its subscript is greater than 1. Formula: $\text{Al}_2(\text{SO}_4)_3$.离子：Al$^{3+}$ 和 SO$_4^{2-}$。交叉法：Al 的电荷大小（3）作为 SO$_4$ 的下标；SO$_4$ 的电荷大小（2）作为 Al 的下标。初步结果：Al$_2$(SO$_4$)$_3$。电荷平衡验证：$2(3+) + 3(2-) = +6 - 6 = 0$。多原子基团的下标大于 1 时必须加括号。化学式：$\text{Al}_2(\text{SO}_4)_3$。
(b) iron(III) phosphate磷酸铁(III)M1·A1·A1
Ions: Fe$^{3+}$ and PO$_4^{3-}$. Criss-cross: 3 crosses to PO$_4$ subscript, 3 crosses to Fe subscript, giving Fe$_3$(PO$_4$)$_3$. The ratio 3:3 reduces to 1:1. Simplified formula: $\text{FePO}_4$. Check: $1(3+) + 1(3-) = 0$. Always reduce to the simplest whole-number ratio for ionic formulas.离子：Fe$^{3+}$ 和 PO$_4^{3-}$。交叉法：3 交叉给 PO$_4$ 下标，3 交叉给 Fe 下标，得 Fe$_3$(PO$_4$)$_3$。比例 3:3 约为 1:1。化简后化学式：$\text{FePO}_4$。验证：$1(3+) + 1(3-) = 0$。离子化合物化学式须约简为最简整数比。
(c) calcium nitrate硝酸钙M1·A1
Ions: Ca$^{2+}$ and NO$_3^-$. Criss-cross: 2 goes to NO$_3$ subscript, 1 goes to Ca subscript. Raw: Ca$_1$(NO$_3$)$_2$. Since Ca subscript is 1, omit it. Parentheses required around NO$_3$ because the subscript 2 applies to the whole group. Formula: $\text{Ca(NO}_3)_2$. Check: $1(2+) + 2(1-) = 0$.离子：Ca$^{2+}$ 和 NO$_3^-$。交叉法：2 给 NO$_3$ 下标，1 给 Ca 下标。初步：Ca$_1$(NO$_3$)$_2$。Ca 下标为 1 时省略不写。NO$_3$ 下标为 2，必须加括号。化学式：$\text{Ca(NO}_3)_2$。验证：$1(2+) + 2(1-) = 0$。
Criss-cross three-step rule: (1) write ions with charges, (2) swap the charge magnitudes as subscripts, (3) reduce and add parentheses as needed.交叉法三步规则：(1) 写出离子及其电荷，(2) 将电荷大小交叉作为下标，(3) 约简并按需加括号。The most common errors are: forgetting to reduce (leaving Fe$_3$(PO$_4$)$_3$ instead of FePO$_4$); omitting parentheses around polyatomic ions when subscript is greater than 1 (writing CaNO$_3$2 instead of Ca(NO$_3$)$_2$); and applying criss-cross to a pair that are already balanced (e.g., MgO: Mg$^{2+}$ and O$^{2-}$ already 1:1, no cross needed). Always verify with a charge-balance check at the end.最常见的错误有：忘记约简（写成 Fe$_3$(PO$_4$)$_3$ 而不是 FePO$_4$）；多原子离子下标大于 1 时忘记加括号（写成 CaNO$_3$2 而不是 Ca(NO$_3$)$_2$）；对已经平衡的离子对错用交叉法（如 MgO：Mg$^{2+}$ 和 O$^{2-}$ 已是 1:1，不需交叉）。最后务必用电荷平衡验证。
Molecular compounds and acids: justify your reasoning in each case.共价化合物与酸：每小问均需写出判断依据。
Answer:答案：(a)dinitrogen trioxide; no Roman numeral because N$_2$O$_3$ is a molecular compound三氧化二氮；不用罗马数字，因为 N$_2$O$_3$ 是共价化合物
· (b)H$_2$SO$_3$; anion is sulfite SO$_3^{2-}$, charge 2-H$_2$SO$_3$；阴离子为亚硫酸根 SO$_3^{2-}$，电荷 2-
· (c)vowel-elision rule: "mono-" drops its "o" before "oxide"元音省略规则："mono-" 在 "oxide" 前省去末尾 "o"
· (d)ClO$_4^-$ is perchlorate, not chlorate; correct name is perchloric acidClO$_4^-$ 是高氯酸根，不是氯酸根；正确名称为高氯酸
(a) Name N$_2$O$_3$; explain no Roman numeral命名 N$_2$O$_3$；解释不用罗马数字M1·A1
N$_2$O$_3$ is a molecular (covalent) compound formed between two nonmetals. Molecular compounds use Greek prefixes: 2 nitrogen (di-) + 3 oxygen (tri-) = dinitrogen trioxide. Roman numerals indicate the charge on a metal ion in ionic compounds; nitrogen is a nonmetal and has no ionic charge, so Roman numerals are never used for molecular compound names.N$_2$O$_3$ 是由两种非金属形成的共价化合物。共价化合物使用希腊数字前缀：2 个氮（di-）+ 3 个氧（tri-）= 三氧化二氮。罗马数字表示离子化合物中金属离子的电荷；氮是非金属，没有离子电荷，因此共价化合物命名中绝不使用罗马数字。
(b) Sulfurous acid H$_2$SO$_3$: identify the anion亚硫酸 H$_2$SO$_3$：确定阴离子M1·A1
The formula is given as H$_2$SO$_3$. Sulfurous acid contains the sulfite ion $\text{SO}_3^{2-}$ (charge 2-). Two H$^+$ ions (each 1+) exactly balance the 2- charge of one sulfite ion, confirming the formula. The "-ous acid" suffix confirms the anion is "-ite" (sulfite, $\text{SO}_3^{2-}$), not "-ate" (sulfate, $\text{SO}_4^{2-}$).化学式已给出为 H$_2$SO$_3$。亚硫酸含有亚硫酸根 $\text{SO}_3^{2-}$（电荷 2-）。两个 H$^+$ 离子（各 1+）恰好平衡一个亚硫酸根的 2- 电荷，验证了化学式正确。"-ous acid" 后缀确认阴离子为 "-ite"（亚硫酸根 $\text{SO}_3^{2-}$），而非 "-ate"（硫酸根 $\text{SO}_4^{2-}$）。
(c) Why CO is "carbon monoxide," not "carbon monooxide"为何 CO 叫 "carbon monoxide" 而非 "carbon monooxide"M1·A1
The vowel-elision rule: when a Greek prefix ends in a vowel (here, "mono-" ends in "o") and the element name begins with a vowel (here, "oxide" begins with "o"), the terminal vowel of the prefix is dropped to avoid an awkward double-vowel. "Mono-" + "oxide" becomes "monoxide." This is a phonetic convention in IUPAC nomenclature; the same applies to "decoxide" (not "decaoxide").元音省略规则：当希腊数字前缀以元音结尾（此处 "mono-" 以 "o" 结尾），而元素名称以元音开头（此处 "oxide" 以 "o" 开头），则省去前缀末尾的元音，以避免双元音的拼写。"mono-" + "oxide" 变为 "monoxide"。这是 IUPAC 命名法中的语音惯例；同样适用于 "decoxide"（而非 "decaoxide"）。
(d) Correct the student's error for HClO$_4$(aq)纠正学生关于 HClO$_4$(aq) 的错误M1·A1
The student confused $\text{ClO}_4^-$ (perchlorate, 4 oxygens) with $\text{ClO}_3^-$ (chlorate, 3 oxygens). HClO$_4$(aq) contains perchlorate, so the correct name is perchloric acid (not "hydrogen chlorate"). The prefix "per-" signals one more oxygen than the "-ate" ion (chlorate $\text{ClO}_3^-$ vs. perchlorate $\text{ClO}_4^-$).该学生将 $\text{ClO}_4^-$（高氯酸根，4 个氧）与 $\text{ClO}_3^-$（氯酸根，3 个氧）混淆。HClO$_4$(aq) 含高氯酸根，正确名称为高氯酸（不是 "hydrogen chlorate"）。前缀 "per-" 表示比 "-ate" 离子多一个氧（氯酸根 $\text{ClO}_3^-$ vs. 高氯酸根 $\text{ClO}_4^-$）。
The chlorine oxyanion series is the most-tested four-ion family; the "per-...-ate" and "hypo-...-ite" extensions follow a strict pattern.氯的含氧酸根系列是最常考的四离子系列；"per-...-ate" 和 "hypo-...-ite" 扩展遵循严格规律。Counting down from the most-oxygen form: perchlorate ClO$_4^-$ (per- + ate) $\rightarrow$ chlorate ClO$_3^-$ (ate) $\rightarrow$ chlorite ClO$_2^-$ (ite) $\rightarrow$ hypochlorite ClO$^-$ (hypo- + ite). Corresponding acids: perchloric, chloric, chlorous, hypochlorous. Each step down removes one oxygen and alternates between the "-ic"/"-ous" naming pattern. Memorise the middle pair (chlorate/chlorite) and extend outward using "per-" and "hypo-" prefixes.从含氧最多的形式往下数：高氯酸根 ClO$_4^-$（per- + ate）$\rightarrow$ 氯酸根 ClO$_3^-$（ate）$\rightarrow$ 亚氯酸根 ClO$_2^-$（ite）$\rightarrow$ 次氯酸根 ClO$^-$（hypo- + ite）。对应酸：高氯酸、氯酸、亚氯酸、次氯酸。每步减少一个氧，在 "-ic"/"-ous" 命名模式间交替。记住中间两个（氯酸根/亚氯酸根），再用 "per-" 和 "hypo-" 向外延伸。
Ions: Cu$^{2+}$ and CO$_3^{2-}$ (carbonate, charge 2-). Criss-cross: 2 crosses to CO$_3$ subscript, 2 crosses to Cu subscript, giving Cu$_2$(CO$_3$)$_2$. The ratio 2:2 reduces to 1:1. Simplified formula: $\text{CuCO}_3$. No parentheses needed when the subscript for the polyatomic group is 1. Check: $1(2+) + 1(2-) = 0$.离子：Cu$^{2+}$ 和 CO$_3^{2-}$（碳酸根，电荷 2-）。交叉法：2 给 CO$_3$ 下标，2 给 Cu 下标，得 Cu$_2$(CO$_3$)$_2$。比例 2:2 约为 1:1。化简后化学式：$\text{CuCO}_3$。多原子基团下标为 1 时不需括号。验证：$1(2+) + 1(2-) = 0$。
(b) iron(III) sulfate硫酸铁(III)M1·A1·A1
Ions: Fe$^{3+}$ and SO$_4^{2-}$ (sulfate, charge 2-). Criss-cross: magnitude of Fe charge (3) becomes SO$_4$ subscript; magnitude of SO$_4$ charge (2) becomes Fe subscript. Result: Fe$_2$(SO$_4$)$_3$. Check: $2(3+) + 3(2-) = +6 - 6 = 0$. No reduction possible (2 and 3 share no common factor). Parentheses required because SO$_4$ subscript is 3. Formula: $\text{Fe}_2(\text{SO}_4)_3$.离子：Fe$^{3+}$ 和 SO$_4^{2-}$（硫酸根，电荷 2-）。交叉法：Fe 电荷大小（3）作为 SO$_4$ 下标；SO$_4$ 电荷大小（2）作为 Fe 下标。结果：Fe$_2$(SO$_4$)$_3$。验证：$2(3+) + 3(2-) = +6 - 6 = 0$。2 和 3 无公因数，无法约简。SO$_4$ 下标为 3，必须加括号。化学式：$\text{Fe}_2(\text{SO}_4)_3$。
Always reduce criss-cross results to lowest terms before checking parentheses; never skip the charge-balance verification.交叉法所得结果必须先约为最简整数比，再检查括号；不可跳过电荷平衡验证。Part (a) and part (b) test the same skill but produce different outcomes: Cu$^{2+}$/CO$_3^{2-}$ reduces to 1:1 (no parentheses needed), while Fe$^{3+}$/SO$_4^{2-}$ yields 2:3 (parentheses needed). Part (c) introduces a scenario where both ions are polyatomic. A systematic approach: (1) write ions with charges, (2) criss-cross magnitudes, (3) divide subscripts by their GCD, (4) add parentheses around any polyatomic group with subscript greater than 1, (5) verify. This five-step protocol eliminates all common errors.(a) 和 (b) 考查相同技能，但结果不同：Cu$^{2+}$/CO$_3^{2-}$ 约为 1:1（不需括号），而 Fe$^{3+}$/SO$_4^{2-}$ 得 2:3（需要括号）。(c) 是两种多原子离子均出现的情景。系统方法：(1) 写出离子及电荷，(2) 交叉电荷大小，(3) 下标除以最大公因数，(4) 对下标大于 1 的多原子基团加括号，(5) 验证。这五步流程可消除所有常见错误。
Answer:答案：(a)copper(II) sulfate pentahydrate; 5 water molecules per formula unit五水合硫酸铜(II)；每个化学式单元含 5 个水分子
· (b)sodium carbonate decahydrate; washing soda十水合碳酸钠；洗涤碱
· (c)(i) NaCl, sodium chloride; (ii) NaHCO$_3$, sodium hydrogen carbonate; (iii) CaO, calcium oxide(i) NaCl，氯化钠；(ii) NaHCO$_3$，碳酸氢钠；(iii) CaO，氧化钙
(a) CuSO$_4 \cdot 5$H$_2$O M1·A1·A1
The anhydrous salt is CuSO$_4$: copper(II) sulfate. The $\cdot 5$H$_2$O notation indicates that 5 water molecules are incorporated into each formula unit of the crystal lattice through coordinate bonds or hydrogen bonding. The Greek prefix for 5 is "penta-." Full name: copper(II) sulfate pentahydrate. The water molecules are lost on heating (anhydrous CuSO$_4$ is white; hydrated form is the familiar blue crystals).无水盐为 CuSO$_4$：硫酸铜(II)。$\cdot 5$H$_2$O 表示每个化学式单元的晶格中通过配位键或氢键结合了 5 个水分子。5 的希腊数字前缀为 "penta-"。完整名称：五水合硫酸铜(II)。加热时水分子脱离（无水 CuSO$_4$ 为白色；水合形式为熟悉的蓝色晶体）。
(b) Na$_2$CO$_3 \cdot 10$H$_2$O M1·A1·A1
The anhydrous salt is Na$_2$CO$_3$: sodium carbonate. "Deca-" = 10. Full IUPAC name: sodium carbonate decahydrate. Common name: washing soda (used in laundry and as a water softener, to be distinguished from baking soda, NaHCO$_3$).无水盐为 Na$_2$CO$_3$：碳酸钠。"deca-" = 10。完整 IUPAC 名称：十水合碳酸钠。俗名：洗涤碱（用于洗涤和软化水，注意与小苏打 NaHCO$_3$ 区分）。
(c) Trivial name matching俗名对应M1·A1·A1
(i) Table salt: NaCl, IUPAC name sodium chloride. The most abundant ionic compound in everyday use. (ii) Baking soda: NaHCO$_3$, IUPAC name sodium hydrogen carbonate (also called sodium bicarbonate). It decomposes on heating to release CO$_2$, causing baked goods to rise. (iii) Quicklime: CaO, IUPAC name calcium oxide. Produced by heating limestone (CaCO$_3$); reacts vigorously with water to give slaked lime Ca(OH)$_2$.(i) 食盐：NaCl，IUPAC 名称氯化钠。日常使用最广泛的离子化合物。(ii) 小苏打：NaHCO$_3$，IUPAC 名称碳酸氢钠（又称重碳酸钠）。加热时分解释放 CO$_2$，使烘焙食品膨胀。(iii) 生石灰：CaO，IUPAC 名称氧化钙。由石灰石（CaCO$_3$）加热制得；与水剧烈反应生成熟石灰 Ca(OH)$_2$。
Hydrate names use Greek prefixes for the water count; common names are anchored to function or historical usage.水合物名称使用希腊数字前缀表示水分子数；俗名源于功能或历史用法。Hydrate prefixes follow the same series as molecular compounds: mono- (1), di- (2), tri- (3), tetra- (4), penta- (5), hexa- (6), hepta- (7), octa- (8), nona- (9), deca- (10). In exam contexts, hydrate nomenclature bridges ionic naming (base salt) with molecular prefix rules (water count). Trivial names that appear on AP and provincial exams: table salt (NaCl), baking soda (NaHCO$_3$), washing soda (Na$_2$CO$_3 \cdot 10$H$_2$O), quicklime (CaO), slaked lime (Ca(OH)$_2$), limestone (CaCO$_3$), gypsum (CaSO$_4 \cdot 2$H$_2$O). Memorise the formula alongside each trivial name.水合物前缀与共价化合物相同：mono-（1）、di-（2）、tri-（3）、tetra-（4）、penta-（5）、hexa-（6）、hepta-（7）、octa-（8）、nona-（9）、deca-（10）。考试中，水合物命名将离子命名（基础盐）与分子前缀规则（水分子数）结合。AP 和省考中出现的俗名：食盐（NaCl）、小苏打（NaHCO$_3$）、洗涤碱（Na$_2$CO$_3 \cdot 10$H$_2$O）、生石灰（CaO）、熟石灰（Ca(OH)$_2$）、石灰石（CaCO$_3$）、石膏（CaSO$_4 \cdot 2$H$_2$O）。每个俗名都要连同化学式一起记忆。
Q12HARD难🇺🇸 US美AP-feeder FRQAP 衔接简答题§1-§7 Comprehensive - naming and formula writing综合 - 命名与化学式书写[10 marks][10 分]
Comprehensive question spanning all seven sections. Show reasoning at every step.综合题，涵盖全部七节内容。每步均需写出推理过程。
(a) Classify and name SO$_2$, MgCl$_2$, HNO$_3$(aq)分类并命名 SO$_2$、MgCl$_2$、HNO$_3$(aq)M1·A1·A1
SO$_2$: two nonmetals bonded covalently. Classification: molecular compound. Name: sulfur dioxide (1 S = "sulfur" with no prefix since it is the first element; 2 O = "di-" + "ox-" + "-ide" = "dioxide"). MgCl$_2$: metal (Mg, Group 2, fixed 2+) bonded to a nonmetal (Cl, 1-). Classification: ionic compound. Name: magnesium chloride (no Roman numeral; Mg is always 2+). HNO$_3$(aq): hydrogen combined with the nitrate oxyanion, dissolved in water. Classification: acid. The anion is nitrate NO$_3^-$ ("-ate"); replace "-ate" with "-ic acid" to give nitric acid.SO$_2$：两种非金属以共价键结合。分类：共价化合物。名称：二氧化硫（1 个 S = "sulfur"，作为第一个元素不加前缀；2 个 O = "di-" + "ox-" + "-ide" = "dioxide"）。MgCl$_2$：金属（Mg，第 2 族，固定 2+）与非金属（Cl，1-）结合。分类：离子化合物。名称：氯化镁（不需罗马数字；Mg 始终为 2+）。HNO$_3$(aq)：氢与硝酸根含氧酸根结合，溶于水。分类：酸。阴离子为硝酸根 NO$_3^-$（"-ate"）；将 "-ate" 替换为 "-ic acid" 得硝酸。
(b) Name Mn$_2$O$_3$ with charge calculation命名 Mn$_2$O$_3$，写出电荷计算M1·A1·A1
Total anion charge from 3 oxide ions: $3 \times (2-) = 6-$. Two Mn atoms must supply $6+$ total: $\text{Mn charge} = 6/2 = 3+$. Mn is a transition metal with variable charge, so Roman numeral is required. Name: manganese(III) oxide.3 个氧离子的总阴离子电荷：$3 \times (2-) = 6-$。两个 Mn 原子共需提供 $6+$：$\text{Mn 电荷} = 6/2 = 3+$。Mn 为可变电荷过渡金属，必须标注罗马数字。名称：氧化锰(III)。
(c) Write the formula for perchloric acid写出高氯酸的化学式M1·A1
Perchloric acid contains the perchlorate ion ClO$_4^-$ (charge 1-). To balance one perchlorate ion (1-), exactly one H$^+$ ion (1+) is needed. Formula: $\text{HClO}_4$. In aqueous solution: $\text{HClO}_4\text{(aq)}$.高氯酸含高氯酸根离子 ClO$_4^-$（电荷 1-）。平衡一个高氯酸根（1-），恰好需要一个 H$^+$（1+）。化学式：$\text{HClO}_4$。水溶液写作 $\text{HClO}_4\text{(aq)}$。
(d) Identify and correct the error in "iron(II) phosphate"找出并纠正"磷酸铁(II)"的错误M1·A1
Iron(II) means Fe$^{2+}$. The student wrote FePO$_4$, which implies: $1(+?) + 1(3-) = 0$, so the metal must be 3+, not 2+. FePO$_4$ is actually iron(III) phosphate. For iron(II) phosphate: Fe$^{2+}$ and PO$_4^{3-}$. Criss-cross: 2 goes to PO$_4$ subscript, 3 goes to Fe subscript. Raw: Fe$_3$(PO$_4$)$_2$. GCD of 3 and 2 is 1; no reduction. Check: $3(2+) + 2(3-) = 6 - 6 = 0$. Correct formula: $\text{Fe}_3(\text{PO}_4)_2$.磷酸铁(II) 意味着 Fe$^{2+}$。学生写的 FePO$_4$ 意味着：$1(+?) + 1(3-) = 0$，故金属必须为 3+ 而非 2+。FePO$_4$ 实际上是磷酸铁(III)。正确的磷酸铁(II)：Fe$^{2+}$ 和 PO$_4^{3-}$。交叉法：2 给 PO$_4$ 下标，3 给 Fe 下标。初步：Fe$_3$(PO$_4$)$_2$。3 和 2 的最大公因数为 1，无法约简。验证：$3(2+) + 2(3-) = 6 - 6 = 0$。正确化学式：$\text{Fe}_3(\text{PO}_4)_2$。
Classifying a compound before naming it prevents the single biggest category of nomenclature errors.命名前先分类化合物，可防止命名错误中最大的一类。The three compound types have mutually exclusive naming rules: (1) ionic compounds: cation name + anion name, Roman numeral if the metal has variable charge, no prefixes; (2) molecular compounds: Greek prefix + first element, Greek prefix + second element + "-ide", vowel elision applies; (3) acids: "hydro-...-ic" for binary acids; "-ate" anion gives "-ic acid", "-ite" anion gives "-ous acid" for oxoacids. Part (d) illustrates a high-value trap: the student's formula FePO$_4$ is not wrong as a formula in isolation, but it represents the wrong compound (iron(III), not iron(II) phosphate). Reading the Roman numeral carefully before writing the formula is the fix.三类化合物的命名规则互斥：(1) 离子化合物：阳离子名 + 阴离子名，电荷可变的金属需加罗马数字，不用前缀；(2) 共价化合物：希腊前缀 + 第一个元素，希腊前缀 + 第二个元素 + "-ide"，适用元音省略；(3) 酸：二元酸用 "hydro-...-ic"；含氧酸中 "-ate" 阴离子对应 "-ic acid"，"-ite" 阴离子对应 "-ous acid"。(d) 展示了一个高价值陷阱：学生写的 FePO$_4$ 单独看并不是错误的化学式，但它代表的是错误的化合物（磷酸铁(III) 而非磷酸铁(II)）。在书写化学式前仔细阅读罗马数字是避免这类错误的关键。