Unit 4 — End of Topic Tests (reviewer copy: all keys and model answers visible)

[
 {
  "value": "SO₃²⁻",
  "misconception": "you gave sulfate only three oxygen atoms — sulfate is SO₄²⁻, with four",
  "feedback": "This answer is wrong — you gave sulfate only three oxygen atoms. The stem counts four oxygen atoms around the sulfur, and the group shares a 2− charge: SO₄²⁻."
 },
 {
  "value": "SO₄³⁻",
  "misconception": "you gave sulfate phosphate's 3− charge — sulfate carries 2−",
  "feedback": "This answer is wrong — you gave sulfate phosphate's 3− charge. The ion described is the 2− ion built on sulfur, so its formula is SO₄²⁻."
 }
]

{
 "keyExplanation": "The sulfur-based 2− polyatomic ion is sulfate: one sulfur atom and four oxygen atoms sharing a 2− charge, SO₄²⁻.",
 "anticipatedWrong": [
  {
   "value": "SO₃²⁻",
   "misconception": "you gave sulfate only three oxygen atoms — sulfate is SO₄²⁻, with four",
   "feedback": "This answer is wrong — you gave sulfate only three oxygen atoms. The stem counts four oxygen atoms around the sulfur, and the group shares a 2− charge: SO₄²⁻."
  },
  {
   "value": "SO₄³⁻",
   "misconception": "you gave sulfate phosphate's 3− charge — sulfate carries 2−",
   "feedback": "This answer is wrong — you gave sulfate phosphate's 3− charge. The ion described is the 2− ion built on sulfur, so its formula is SO₄²⁻."
  }
 ]
}

[
 {
  "value": "carbonate",
  "misconception": "you named the carbon-and-oxygen lookalike — carbonate is CO₃²⁻ and carries no hydrogen",
  "feedback": "This answer is wrong — carbonate is the carbon-and-oxygen lookalike, CO₃²⁻, and it carries no hydrogen. C₂H₃O₂⁻ — two carbon, three hydrogen, and two oxygen atoms with a 1− charge — is the acetate ion."
 },
 {
  "value": "hydrogen carbonate",
  "misconception": "you named the one-extra-hydrogen ion — hydrogen carbonate is HCO₃⁻",
  "feedback": "This answer is wrong — hydrogen carbonate is HCO₃⁻, a carbonate ion carrying one extra hydrogen atom. C₂H₃O₂⁻ — two carbon, three hydrogen, and two oxygen atoms with a 1− charge — is the acetate ion."
 }
]

{
 "keyExplanation": "C₂H₃O₂⁻ — two carbon, three hydrogen, and two oxygen atoms with a 1− charge — is the acetate ion.",
 "anticipatedWrong": [
  {
   "value": "carbonate",
   "misconception": "you named the carbon-and-oxygen lookalike — carbonate is CO₃²⁻ and carries no hydrogen",
   "feedback": "This answer is wrong — carbonate is the carbon-and-oxygen lookalike, CO₃²⁻, and it carries no hydrogen. C₂H₃O₂⁻ — two carbon, three hydrogen, and two oxygen atoms with a 1− charge — is the acetate ion."
  },
  {
   "value": "hydrogen carbonate",
   "misconception": "you named the one-extra-hydrogen ion — hydrogen carbonate is HCO₃⁻",
   "feedback": "This answer is wrong — hydrogen carbonate is HCO₃⁻, a carbonate ion carrying one extra hydrogen atom. C₂H₃O₂⁻ — two carbon, three hydrogen, and two oxygen atoms with a 1− charge — is the acetate ion."
  }
 ]
}

[
 {
  "value": "Sr₂S₂",
  "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
  "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Sr₂S₂ divide by 2, and a formula gives the simplest whole-number ratio: SrS. Check: (2+) + (2−) = 0."
 },
 {
  "value": "SrS₂",
  "misconception": "you kept sulfur's crossed subscript but reduced strontium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept sulfur's crossed subscript but reduced strontium's. Reduction divides BOTH subscripts together: Sr₂S₂ becomes SrS, and (2+) + (2−) = 0."
 },
 {
  "value": "Sr₂S",
  "misconception": "you kept strontium's crossed subscript but reduced sulfur's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept strontium's crossed subscript but reduced sulfur's. Reduction divides BOTH subscripts together: Sr₂S₂ becomes SrS, and (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Criss-crossing the two 2s gives Sr₂S₂. Both subscripts divide by 2, and a formula gives the simplest whole-number ratio, so the final formula is SrS: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Sr₂S₂",
   "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
   "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Sr₂S₂ divide by 2, and a formula gives the simplest whole-number ratio: SrS. Check: (2+) + (2−) = 0."
  },
  {
   "value": "SrS₂",
   "misconception": "you kept sulfur's crossed subscript but reduced strontium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept sulfur's crossed subscript but reduced strontium's. Reduction divides BOTH subscripts together: Sr₂S₂ becomes SrS, and (2+) + (2−) = 0."
  },
  {
   "value": "Sr₂S",
   "misconception": "you kept strontium's crossed subscript but reduced sulfur's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept strontium's crossed subscript but reduced sulfur's. Reduction divides BOTH subscripts together: Sr₂S₂ becomes SrS, and (2+) + (2−) = 0."
  }
 ]
}

[
 "Each hydrogen carbonate block HCO₃ holds 1 carbon atom.",
 "The 2 outside the parentheses doubles the block: 2 × 1 = 2.",
 "Sr(HCO₃)₂ contains 2 carbon atoms."
]

[
 {
  "value": 1,
  "misconception": "you forgot the outer subscript — the 2 outside the parentheses doubles the one carbon in each block",
  "feedback": "This answer is wrong — you forgot the outer subscript. Each HCO₃ block holds one carbon atom, and the 2 outside the parentheses doubles the whole block: 2 × 1 = 2 carbon atoms."
 },
 {
  "value": 6,
  "misconception": "you multiplied by a subscript that belongs to oxygen — the 3 in O₃ counts oxygen atoms only",
  "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the 3 in O₃ counts oxygen atoms only. Each HCO₃ block holds one carbon atom, and the outer 2 doubles it: 2 × 1 = 2 carbon atoms."
 }
]

{
 "keyExplanation": "Each HCO₃ block holds 1 carbon atom, and the 2 outside the parentheses doubles the block: 2 × 1 = 2 carbon atoms. The Sr outside the parentheses is not multiplied and carries no carbon.",
 "anticipatedWrong": [
  {
   "value": 1,
   "misconception": "you forgot the outer subscript — the 2 outside the parentheses doubles the one carbon in each block",
   "feedback": "This answer is wrong — you forgot the outer subscript. Each HCO₃ block holds one carbon atom, and the 2 outside the parentheses doubles the whole block: 2 × 1 = 2 carbon atoms."
  },
  {
   "value": 6,
   "misconception": "you multiplied by a subscript that belongs to oxygen — the 3 in O₃ counts oxygen atoms only",
   "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the 3 in O₃ counts oxygen atoms only. Each HCO₃ block holds one carbon atom, and the outer 2 doubles it: 2 × 1 = 2 carbon atoms."
  }
 ]
}

[
 {
  "value": "CaClO₃",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Ca²⁺ takes two ClO₃⁻ ions, and the repeated chlorate goes in parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "CaClO₃₂",
  "misconception": "you left out the parentheses — the subscript must repeat the whole chlorate block",
  "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole chlorate block, so it goes outside parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Ca²⁺ needs two ClO₃⁻ ions: (2+) + 2 × (1−) = 0. Chlorate repeats, so parentheses: Ca(ClO₃)₂.",
 "anticipatedWrong": [
  {
   "value": "CaClO₃",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Ca²⁺ takes two ClO₃⁻ ions, and the repeated chlorate goes in parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "CaClO₃₂",
   "misconception": "you left out the parentheses — the subscript must repeat the whole chlorate block",
   "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole chlorate block, so it goes outside parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "K₂(CO₃)",
  "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and carbonate appears once here: K₂CO₃. Check: 2 × (1+) + (2−) = 0."
 },
 {
  "value": "KCO₃",
  "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, whose charges give (1+) + (2−) = 1−, not zero. One 2− ion needs two 1+ ions: K₂CO₃. Check: 2 × (1+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Two K⁺ ions cancel one CO₃²⁻ ion. Carbonate appears only once, so no parentheses: K₂CO₃. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "K₂(CO₃)",
   "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and carbonate appears once here: K₂CO₃. Check: 2 × (1+) + (2−) = 0."
  },
  {
   "value": "KCO₃",
   "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, whose charges give (1+) + (2−) = 1−, not zero. One 2− ion needs two 1+ ions: K₂CO₃. Check: 2 × (1+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "Mg₂N₃",
  "misconception": "you swapped the counts — magnesium takes the 3 and nitrogen takes the 2",
  "feedback": "This answer is wrong — you swapped the counts: magnesium takes the 3 and nitrogen takes the 2. Write the ions: Mg²⁺ and N³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Mg₃N₂. Check: 3 × (2+) + 2 × (3−) = 0."
 },
 {
  "value": "Mg₃(N)₂",
  "misconception": "you added parentheses, but N³⁻ is a single-atom ion — parentheses are only for repeated polyatomic ions",
  "feedback": "This answer is wrong — you added parentheses, but N³⁻ is a single-atom ion; parentheses are only for repeated polyatomic ions. Write the ions: Mg²⁺ and N³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Mg₃N₂. Check: 3 × (2+) + 2 × (3−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Mg²⁺ and N³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Mg₃N₂, with no parentheses because nitride is a single atom. Check: 3 × (2+) + 2 × (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Mg₂N₃",
   "misconception": "you swapped the counts — magnesium takes the 3 and nitrogen takes the 2",
   "feedback": "This answer is wrong — you swapped the counts: magnesium takes the 3 and nitrogen takes the 2. Write the ions: Mg²⁺ and N³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Mg₃N₂. Check: 3 × (2+) + 2 × (3−) = 0."
  },
  {
   "value": "Mg₃(N)₂",
   "misconception": "you added parentheses, but N³⁻ is a single-atom ion — parentheses are only for repeated polyatomic ions",
   "feedback": "This answer is wrong — you added parentheses, but N³⁻ is a single-atom ion; parentheses are only for repeated polyatomic ions. Write the ions: Mg²⁺ and N³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Mg₃N₂. Check: 3 × (2+) + 2 × (3−) = 0."
  }
 ]
}

[
 "Inside the parentheses: each NH₄ block holds 1 nitrogen atom and 4 hydrogen atoms — 5 atoms.",
 "The 2 outside doubles the block: 2 × 5 = 10 atoms.",
 "The S outside the parentheses counts once: 10 + 1 = 11.",
 "(NH₄)₂S contains 11 atoms in total."
]

[
 {
  "value": 6,
  "misconception": "you forgot to double the ammonium block — the 2 outside multiplies the N and all four H atoms",
  "feedback": "This answer is wrong — you forgot to double the ammonium block; the 2 outside multiplies the N and all four H atoms. Each NH₄ block holds 5 atoms, the 2 outside doubles that to 10, and the single sulfur makes 11 atoms."
 },
 {
  "value": 10,
  "misconception": "you left out the sulfur — atoms outside the parentheses still count",
  "feedback": "This answer is wrong — you left out the sulfur; atoms outside the parentheses still count. The doubled NH₄ blocks give 10 atoms, and the single S brings the total to 11 atoms."
 }
]

{
 "keyExplanation": "Each NH₄ block holds 1 nitrogen and 4 hydrogen atoms; the 2 outside doubles the block to 2 N and 8 H. Adding the single sulfur: 2 + 8 + 1 = 11 atoms.",
 "anticipatedWrong": [
  {
   "value": 6,
   "misconception": "you forgot to double the ammonium block — the 2 outside multiplies the N and all four H atoms",
   "feedback": "This answer is wrong — you forgot to double the ammonium block; the 2 outside multiplies the N and all four H atoms. Each NH₄ block holds 5 atoms, the 2 outside doubles that to 10, and the single sulfur makes 11 atoms."
  },
  {
   "value": 10,
   "misconception": "you left out the sulfur — atoms outside the parentheses still count",
   "feedback": "This answer is wrong — you left out the sulfur; atoms outside the parentheses still count. The doubled NH₄ blocks give 10 atoms, and the single S brings the total to 11 atoms."
  }
 ]
}

[
 {
  "value": "Ba₂(PO₄)₃",
  "misconception": "you swapped the counts — the 3− ion's charge number becomes barium's subscript, and the 2+ ion's becomes phosphate's",
  "feedback": "This answer is wrong — you swapped the counts. The 3− ion's charge number becomes barium's subscript, and the 2+ ion's becomes phosphate's: three 2+ ions give 6+ and two 3− ions give 6−, so Ba₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
 },
 {
  "value": "BaPO₄",
  "misconception": "you wrote a 1:1 ratio — 2+ and 3− do not cancel one for one",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, and 2+ and 3− do not cancel one for one. Three 2+ ions give 6+ and two 3− ions give 6−, so Ba₃(PO₄)₂ — parentheses because phosphate repeats. Check: 3 × (2+) + 2 × (3−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Ba²⁺ and PO₄³⁻. Three 2+ ions give 6+ and two 3− ions give 6−, so Ba₃(PO₄)₂ — parentheses because phosphate repeats. Check: 3 × (2+) + 2 × (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ba₂(PO₄)₃",
   "misconception": "you swapped the counts — the 3− ion's charge number becomes barium's subscript, and the 2+ ion's becomes phosphate's",
   "feedback": "This answer is wrong — you swapped the counts. The 3− ion's charge number becomes barium's subscript, and the 2+ ion's becomes phosphate's: three 2+ ions give 6+ and two 3− ions give 6−, so Ba₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
  },
  {
   "value": "BaPO₄",
   "misconception": "you wrote a 1:1 ratio — 2+ and 3− do not cancel one for one",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, and 2+ and 3− do not cancel one for one. Three 2+ ions give 6+ and two 3− ions give 6−, so Ba₃(PO₄)₂ — parentheses because phosphate repeats. Check: 3 × (2+) + 2 × (3−) = 0."
  }
 ]
}

[
 {
  "value": "NH₄PO₄",
  "misconception": "you wrote a 1:1 ratio — one 3− ion needs three 1+ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 3− ion needs three 1+ ions, and the repeated ion is polyatomic, so the parentheses wrap NH₄: (NH₄)₃PO₄. Check: 3 × (1+) + (3−) = 0."
 },
 {
  "value": "NH₄(PO₄)₃",
  "misconception": "you put the parentheses around the wrong ion — phosphate appears once; the ammonium is what repeats",
  "feedback": "This answer is wrong — you put the parentheses around the wrong ion. Phosphate appears once; the ammonium is what repeats: three 1+ ions cancel one 3− ion, so (NH₄)₃PO₄. Check: 3 × (1+) + (3−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: NH₄⁺ and PO₄³⁻. Three 1+ ions cancel one 3− ion, and the repeated ion is polyatomic, so the parentheses wrap NH₄: (NH₄)₃PO₄. Check: 3 × (1+) + (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "NH₄PO₄",
   "misconception": "you wrote a 1:1 ratio — one 3− ion needs three 1+ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 3− ion needs three 1+ ions, and the repeated ion is polyatomic, so the parentheses wrap NH₄: (NH₄)₃PO₄. Check: 3 × (1+) + (3−) = 0."
  },
  {
   "value": "NH₄(PO₄)₃",
   "misconception": "you put the parentheses around the wrong ion — phosphate appears once; the ammonium is what repeats",
   "feedback": "This answer is wrong — you put the parentheses around the wrong ion. Phosphate appears once; the ammonium is what repeats: three 1+ ions cancel one 3− ion, so (NH₄)₃PO₄. Check: 3 × (1+) + (3−) = 0."
  }
 ]
}

[
 {
  "value": "PO₄²⁻",
  "misconception": "you gave phosphate sulfate's 2− charge — phosphate carries 3−",
  "feedback": "This answer is wrong — you gave phosphate sulfate's 2− charge. Phosphate is the 3− member of the trio: one phosphorus atom and four oxygen atoms sharing a 3− charge, PO₄³⁻."
 },
 {
  "value": "PO₃³⁻",
  "misconception": "you gave phosphate only three oxygen atoms — phosphate is PO₄³⁻, with four",
  "feedback": "This answer is wrong — you gave phosphate only three oxygen atoms. Phosphate carries four oxygen atoms around its phosphorus: PO₄³⁻."
 }
]

{
 "keyExplanation": "The phosphate ion is one phosphorus atom and four oxygen atoms sharing a 3− charge: PO₄³⁻.",
 "anticipatedWrong": [
  {
   "value": "PO₄²⁻",
   "misconception": "you gave phosphate sulfate's 2− charge — phosphate carries 3−",
   "feedback": "This answer is wrong — you gave phosphate sulfate's 2− charge. Phosphate is the 3− member of the trio: one phosphorus atom and four oxygen atoms sharing a 3− charge, PO₄³⁻."
  },
  {
   "value": "PO₃³⁻",
   "misconception": "you gave phosphate only three oxygen atoms — phosphate is PO₄³⁻, with four",
   "feedback": "This answer is wrong — you gave phosphate only three oxygen atoms. Phosphate carries four oxygen atoms around its phosphorus: PO₄³⁻."
  }
 ]
}

[
 {
  "value": "ClO₄⁻",
  "misconception": "you gave chlorate four oxygen atoms — chlorate has three, like nitrate",
  "feedback": "This answer is wrong — ClO₄⁻ gives chlorate four oxygen atoms. Chlorate has three, like nitrate: one chlorine atom and three oxygen atoms sharing a 1− charge, ClO₃⁻."
 },
 {
  "value": "Cl⁻",
  "misconception": "you wrote the chloride ion, Cl⁻ — chlorate carries three oxygen atoms",
  "feedback": "This answer is wrong — Cl⁻ is the chloride ion, Cl⁻. The chlorate ion is one chlorine atom and three oxygen atoms sharing a 1− charge: ClO₃⁻."
 }
]

{
 "keyExplanation": "The chlorate ion is one chlorine atom and three oxygen atoms sharing a 1− charge: ClO₃⁻.",
 "anticipatedWrong": [
  {
   "value": "ClO₄⁻",
   "misconception": "you gave chlorate four oxygen atoms — chlorate has three, like nitrate",
   "feedback": "This answer is wrong — ClO₄⁻ gives chlorate four oxygen atoms. Chlorate has three, like nitrate: one chlorine atom and three oxygen atoms sharing a 1− charge, ClO₃⁻."
  },
  {
   "value": "Cl⁻",
   "misconception": "you wrote the chloride ion, Cl⁻ — chlorate carries three oxygen atoms",
   "feedback": "This answer is wrong — Cl⁻ is the chloride ion, Cl⁻. The chlorate ion is one chlorine atom and three oxygen atoms sharing a 1− charge: ClO₃⁻."
  }
 ]
}

[
 {
  "value": "Mg₂Se₂",
  "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
  "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Mg₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: MgSe. Check: (2+) + (2−) = 0."
 },
 {
  "value": "MgSe₂",
  "misconception": "you kept selenium's crossed subscript but reduced magnesium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced magnesium's. Reduction divides BOTH subscripts together: Mg₂Se₂ becomes MgSe, and (2+) + (2−) = 0."
 },
 {
  "value": "Mg₂Se",
  "misconception": "you kept magnesium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept magnesium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Mg₂Se₂ becomes MgSe, and (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Criss-crossing the two 2s gives Mg₂Se₂. Both subscripts divide by 2, and a formula gives the simplest whole-number ratio, so the final formula is MgSe: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Mg₂Se₂",
   "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
   "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Mg₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: MgSe. Check: (2+) + (2−) = 0."
  },
  {
   "value": "MgSe₂",
   "misconception": "you kept selenium's crossed subscript but reduced magnesium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced magnesium's. Reduction divides BOTH subscripts together: Mg₂Se₂ becomes MgSe, and (2+) + (2−) = 0."
  },
  {
   "value": "Mg₂Se",
   "misconception": "you kept magnesium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept magnesium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Mg₂Se₂ becomes MgSe, and (2+) + (2−) = 0."
  }
 ]
}

[
 "KC₂H₃O₂ has no parentheses, so every subscript counts exactly as written.",
 "The C₂ means 2 carbon atoms.",
 "KC₂H₃O₂ contains 2 carbon atoms."
]

[
 {
  "value": 4,
  "misconception": "you multiplied by a subscript that belongs to oxygen — the trailing 2 in O₂ counts oxygen atoms only",
  "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the trailing 2 in O₂ counts oxygen atoms only. KC₂H₃O₂ has no parentheses, so the C₂ counts exactly as written: 2 carbon atoms."
 },
 {
  "value": 6,
  "misconception": "you applied the 3 that belongs to hydrogen to the carbons — with no parentheses, each subscript counts only the symbol to its left",
  "feedback": "This answer is wrong — you applied the subscript 3 that belongs to hydrogen to the carbons. KC₂H₃O₂ has no parentheses, so each subscript counts only the symbol to its left: the C₂ means 2 carbon atoms."
 }
]

{
 "keyExplanation": "KC₂H₃O₂ has no parentheses, so every subscript counts exactly as written: the C₂ means 2 carbon atoms.",
 "anticipatedWrong": [
  {
   "value": 4,
   "misconception": "you multiplied by a subscript that belongs to oxygen — the trailing 2 in O₂ counts oxygen atoms only",
   "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the trailing 2 in O₂ counts oxygen atoms only. KC₂H₃O₂ has no parentheses, so the C₂ counts exactly as written: 2 carbon atoms."
  },
  {
   "value": 6,
   "misconception": "you applied the 3 that belongs to hydrogen to the carbons — with no parentheses, each subscript counts only the symbol to its left",
   "feedback": "This answer is wrong — you applied the subscript 3 that belongs to hydrogen to the carbons. KC₂H₃O₂ has no parentheses, so each subscript counts only the symbol to its left: the C₂ means 2 carbon atoms."
  }
 ]
}

[
 {
  "value": "BaOH",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Ba²⁺ takes two OH⁻ ions, and the repeated hydroxide goes in parentheses: Ba(OH)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "BaOH₂",
  "misconception": "you left out the parentheses — the subscript must repeat the whole hydroxide block",
  "feedback": "This answer is wrong — you left out the parentheses. Written that way, the 2 hangs on the last atom alone instead of repeating the whole hydroxide block; wrap the block and put the 2 outside: Ba(OH)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Ba²⁺ needs two OH⁻ ions: (2+) + 2 × (1−) = 0. Hydroxide repeats, so parentheses: Ba(OH)₂.",
 "anticipatedWrong": [
  {
   "value": "BaOH",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Ba²⁺ takes two OH⁻ ions, and the repeated hydroxide goes in parentheses: Ba(OH)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "BaOH₂",
   "misconception": "you left out the parentheses — the subscript must repeat the whole hydroxide block",
   "feedback": "This answer is wrong — you left out the parentheses. Written that way, the 2 hangs on the last atom alone instead of repeating the whole hydroxide block; wrap the block and put the 2 outside: Ba(OH)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "Na₂(SO₄)",
  "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and sulfate appears once here: Na₂SO₄. Check: 2 × (1+) + (2−) = 0."
 },
 {
  "value": "NaSO₄",
  "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, whose charges give (1+) + (2−) = 1−, not zero. One 2− ion needs two 1+ ions: Na₂SO₄. Check: 2 × (1+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Two Na⁺ ions cancel one SO₄²⁻ ion. Sulfate appears only once, so no parentheses: Na₂SO₄. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Na₂(SO₄)",
   "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and sulfate appears once here: Na₂SO₄. Check: 2 × (1+) + (2−) = 0."
  },
  {
   "value": "NaSO₄",
   "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, whose charges give (1+) + (2−) = 1−, not zero. One 2− ion needs two 1+ ions: Na₂SO₄. Check: 2 × (1+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "CaBr",
  "misconception": "you wrote a 1:1 ratio — one 1− ion cannot cancel a 2+ ion",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 1− ion cannot cancel a 2+ ion. Write the ions: Ca²⁺ and Br⁻. One 2+ ion needs two 1− ions: CaBr₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "Ca₂Br",
  "misconception": "you swapped the counts — calcium's 2 becomes bromide's subscript, not its own",
  "feedback": "This answer is wrong — you swapped the counts: in the criss-cross, calcium's charge number 2 becomes bromide's subscript, not calcium's. Write the ions: Ca²⁺ and Br⁻. One 2+ ion needs two 1− ions: CaBr₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Ca²⁺ and Br⁻. One 2+ ion needs two 1− ions to cancel: CaBr₂, with no parentheses because bromide is a single atom, and 1 and 2 share no common factor, so nothing reduces. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "CaBr",
   "misconception": "you wrote a 1:1 ratio — one 1− ion cannot cancel a 2+ ion",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 1− ion cannot cancel a 2+ ion. Write the ions: Ca²⁺ and Br⁻. One 2+ ion needs two 1− ions: CaBr₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "Ca₂Br",
   "misconception": "you swapped the counts — calcium's 2 becomes bromide's subscript, not its own",
   "feedback": "This answer is wrong — you swapped the counts: in the criss-cross, calcium's charge number 2 becomes bromide's subscript, not calcium's. Write the ions: Ca²⁺ and Br⁻. One 2+ ion needs two 1− ions: CaBr₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 "Inside the parentheses: each HCO₃ block holds 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms — 5 atoms.",
 "The 2 outside doubles the block: 2 × 5 = 10 atoms.",
 "The Mg outside the parentheses counts once: 10 + 1 = 11.",
 "Mg(HCO₃)₂ contains 11 atoms in total."
]

[
 {
  "value": 6,
  "misconception": "you forgot to double the HCO₃ block — the 2 outside multiplies the H, the C, and all three O atoms",
  "feedback": "This answer is wrong — you forgot to double the HCO₃ block; the 2 outside multiplies the H, the C, and all three O atoms. Each block holds 5 atoms, the 2 outside doubles that to 10, and the single magnesium makes 11 atoms."
 },
 {
  "value": 10,
  "misconception": "you left out the magnesium — atoms outside the parentheses still count",
  "feedback": "This answer is wrong — you left out the magnesium; atoms outside the parentheses still count. The doubled HCO₃ blocks give 10 atoms, and the single Mg brings the total to 11 atoms."
 }
]

{
 "keyExplanation": "Each HCO₃ block holds 1 hydrogen, 1 carbon, and 3 oxygen atoms; the 2 outside doubles the block to 2 H, 2 C, and 6 O. Adding the single magnesium: 2 + 2 + 6 + 1 = 11 atoms.",
 "anticipatedWrong": [
  {
   "value": 6,
   "misconception": "you forgot to double the HCO₃ block — the 2 outside multiplies the H, the C, and all three O atoms",
   "feedback": "This answer is wrong — you forgot to double the HCO₃ block; the 2 outside multiplies the H, the C, and all three O atoms. Each block holds 5 atoms, the 2 outside doubles that to 10, and the single magnesium makes 11 atoms."
  },
  {
   "value": 10,
   "misconception": "you left out the magnesium — atoms outside the parentheses still count",
   "feedback": "This answer is wrong — you left out the magnesium; atoms outside the parentheses still count. The doubled HCO₃ blocks give 10 atoms, and the single Mg brings the total to 11 atoms."
  }
 ]
}

[
 {
  "value": "Sr₂(CO₃)₂",
  "misconception": "you skipped the reduce step — criss-cross gives Sr₂(CO₃)₂, and 2:2 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Sr₂(CO₃)₂, but the subscripts 2 and 2 share a common factor and reduce to 1 and 1. Carbonate then appears once, so no parentheses: SrCO₃. Check: (2+) + (2−) = 0."
 },
 {
  "value": "Sr(CO₃)",
  "misconception": "you kept the parentheses — after reducing, carbonate appears only once, and parentheses are only for a repeated polyatomic ion",
  "feedback": "This answer is wrong — you kept the parentheses. After 2:2 reduces to 1:1, carbonate appears only once, and parentheses are only for a repeated polyatomic ion: SrCO₃. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Sr²⁺ and CO₃²⁻. Criss-cross gives Sr₂(CO₃)₂ — reduce 2:2 to 1:1. Carbonate then appears once, so no parentheses: SrCO₃. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Sr₂(CO₃)₂",
   "misconception": "you skipped the reduce step — criss-cross gives Sr₂(CO₃)₂, and 2:2 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Sr₂(CO₃)₂, but the subscripts 2 and 2 share a common factor and reduce to 1 and 1. Carbonate then appears once, so no parentheses: SrCO₃. Check: (2+) + (2−) = 0."
  },
  {
   "value": "Sr(CO₃)",
   "misconception": "you kept the parentheses — after reducing, carbonate appears only once, and parentheses are only for a repeated polyatomic ion",
   "feedback": "This answer is wrong — you kept the parentheses. After 2:2 reduces to 1:1, carbonate appears only once, and parentheses are only for a repeated polyatomic ion: SrCO₃. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "(NH₄)₂NO₃",
  "misconception": "you doubled the ammonium — the charges are 1+ and 1−, so one of each ion balances",
  "feedback": "This answer is wrong — you doubled the ammonium. NH₄⁺ is 1+ and NO₃⁻ is 1−, so one of each balances: NH₄NO₃, no parentheses needed."
 },
 {
  "value": "NH₄(NO₃)₂",
  "misconception": "you doubled the nitrate — parentheses with a subscript only appear when more than one polyatomic ion is needed",
  "feedback": "This answer is wrong — you doubled the nitrate. The charges 1+ and 1− already cancel one-to-one, so the formula is NH₄NO₃ with no parentheses."
 },
 {
  "value": "(NH₄)NO₃",
  "misconception": "you wrapped a count-1 polyatomic ion in parentheses — parentheses appear only when the subscript outside them is 2 or more",
  "feedback": "This answer is wrong — parentheses are used only when more than one of a polyatomic ion is needed. Both counts here are 1, so the formula is written NH₄NO₃ with no parentheses."
 }
]

{
 "keyExplanation": "Both counts are 1 — one NH₄⁺ balances one NO₃⁻ — so no parentheses are needed anywhere: NH₄NO₃.",
 "anticipatedWrong": [
  {
   "value": "(NH₄)₂NO₃",
   "misconception": "you doubled the ammonium — the charges are 1+ and 1−, so one of each ion balances",
   "feedback": "This answer is wrong — you doubled the ammonium. NH₄⁺ is 1+ and NO₃⁻ is 1−, so one of each balances: NH₄NO₃, no parentheses needed."
  },
  {
   "value": "NH₄(NO₃)₂",
   "misconception": "you doubled the nitrate — parentheses with a subscript only appear when more than one polyatomic ion is needed",
   "feedback": "This answer is wrong — you doubled the nitrate. The charges 1+ and 1− already cancel one-to-one, so the formula is NH₄NO₃ with no parentheses."
  },
  {
   "value": "(NH₄)NO₃",
   "misconception": "you wrapped a count-1 polyatomic ion in parentheses — parentheses appear only when the subscript outside them is 2 or more",
   "feedback": "This answer is wrong — parentheses are used only when more than one of a polyatomic ion is needed. Both counts here are 1, so the formula is written NH₄NO₃ with no parentheses."
  }
 ]
}

[
 {
  "value": "carbonate",
  "misconception": "you matched the 2− charge — the ion shown has a sulfur center; carbonate is CO₃²⁻, built on carbon",
  "feedback": "This answer is wrong — you matched the 2− charge. Carbonate is CO₃²⁻, built on carbon with three oxygens. The ion shown, SO₄²⁻, has a sulfur center: it is sulfate."
 },
 {
  "value": "phosphate",
  "misconception": "you matched the oxygen count — phosphate is PO₄³⁻, a 3− ion; the 2− ion shown has a sulfur center",
  "feedback": "This answer is wrong — you matched the oxygen count. Phosphate is PO₄³⁻ and carries 3−. The ion shown, SO₄²⁻, has a sulfur center and carries 2−: it is sulfate."
 }
]

{
 "keyExplanation": "SO₄²⁻ — one sulfur atom and four oxygen atoms sharing a 2− charge — is the sulfate ion.",
 "anticipatedWrong": [
  {
   "value": "carbonate",
   "misconception": "you matched the 2− charge — the ion shown has a sulfur center; carbonate is CO₃²⁻, built on carbon",
   "feedback": "This answer is wrong — you matched the 2− charge. Carbonate is CO₃²⁻, built on carbon with three oxygens. The ion shown, SO₄²⁻, has a sulfur center: it is sulfate."
  },
  {
   "value": "phosphate",
   "misconception": "you matched the oxygen count — phosphate is PO₄³⁻, a 3− ion; the 2− ion shown has a sulfur center",
   "feedback": "This answer is wrong — you matched the oxygen count. Phosphate is PO₄³⁻ and carries 3−. The ion shown, SO₄²⁻, has a sulfur center and carries 2−: it is sulfate."
  }
 ]
}

[
 {
  "value": "carbonate",
  "misconception": "you ignored the extra H at the front of the formula — HCO₃⁻ is carbonate plus one hydrogen",
  "feedback": "This answer is wrong — you ignored the extra H at the front of the formula. CO₃²⁻ alone is carbonate; the shown ion HCO₃⁻ keeps one H attached and carries 1−: it is the hydrogen carbonate ion."
 },
 {
  "value": "hydroxide",
  "misconception": "you matched the H alone and ignored the C and the three O atoms — hydroxide is OH⁻, just two atoms",
  "feedback": "This answer is wrong — you matched the H alone and ignored the C and the three O atoms. Hydroxide is OH⁻, a two-atom ion; the shown five-atom ion HCO₃⁻ is hydrogen carbonate."
 }
]

{
 "keyExplanation": "HCO₃⁻ — a carbonate ion carrying one extra hydrogen atom, with a 1− charge — is the hydrogen carbonate ion.",
 "anticipatedWrong": [
  {
   "value": "carbonate",
   "misconception": "you ignored the extra H at the front of the formula — HCO₃⁻ is carbonate plus one hydrogen",
   "feedback": "This answer is wrong — you ignored the extra H at the front of the formula. CO₃²⁻ alone is carbonate; the shown ion HCO₃⁻ keeps one H attached and carries 1−: it is the hydrogen carbonate ion."
  },
  {
   "value": "hydroxide",
   "misconception": "you matched the H alone and ignored the C and the three O atoms — hydroxide is OH⁻, just two atoms",
   "feedback": "This answer is wrong — you matched the H alone and ignored the C and the three O atoms. Hydroxide is OH⁻, a two-atom ion; the shown five-atom ion HCO₃⁻ is hydrogen carbonate."
  }
 ]
}

[
 {
  "value": "Ca₂Se₂",
  "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
  "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Ca₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: CaSe. Check: (2+) + (2−) = 0."
 },
 {
  "value": "CaSe₂",
  "misconception": "you kept selenium's crossed subscript but reduced calcium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced calcium's. Reduction divides BOTH subscripts together: Ca₂Se₂ becomes CaSe, and (2+) + (2−) = 0."
 },
 {
  "value": "Ca₂Se",
  "misconception": "you kept calcium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept calcium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Ca₂Se₂ becomes CaSe, and (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Criss-crossing the two 2s gives Ca₂Se₂. Both subscripts divide by 2, and a formula gives the simplest whole-number ratio, so the final formula is CaSe: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ca₂Se₂",
   "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
   "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Ca₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: CaSe. Check: (2+) + (2−) = 0."
  },
  {
   "value": "CaSe₂",
   "misconception": "you kept selenium's crossed subscript but reduced calcium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced calcium's. Reduction divides BOTH subscripts together: Ca₂Se₂ becomes CaSe, and (2+) + (2−) = 0."
  },
  {
   "value": "Ca₂Se",
   "misconception": "you kept calcium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept calcium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Ca₂Se₂ becomes CaSe, and (2+) + (2−) = 0."
  }
 ]
}

[
 "Each chlorate block ClO₃ holds 1 chlorine atom.",
 "The 2 outside the parentheses doubles the block: 2 × 1 = 2.",
 "Ba(ClO₃)₂ contains 2 chlorine atoms."
]

[
 {
  "value": 1,
  "misconception": "you forgot the outer subscript — the 2 outside the parentheses doubles the one chlorine in each block",
  "feedback": "This answer is wrong — you forgot the outer subscript. Each ClO₃ block holds one chlorine atom, and the 2 outside the parentheses doubles the whole block: 2 × 1 = 2 chlorine atoms."
 },
 {
  "value": 6,
  "misconception": "you multiplied by a subscript that belongs to oxygen — the 3 in O₃ counts oxygen atoms only",
  "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the 3 in O₃ counts oxygen atoms only. Each ClO₃ block holds one chlorine atom, and the outer 2 doubles it: 2 × 1 = 2 chlorine atoms."
 }
]

{
 "keyExplanation": "Each ClO₃ block holds 1 chlorine atom, and the 2 outside the parentheses doubles the block: 2 × 1 = 2 chlorine atoms. The Ba outside the parentheses is not multiplied and carries no chlorine.",
 "anticipatedWrong": [
  {
   "value": 1,
   "misconception": "you forgot the outer subscript — the 2 outside the parentheses doubles the one chlorine in each block",
   "feedback": "This answer is wrong — you forgot the outer subscript. Each ClO₃ block holds one chlorine atom, and the 2 outside the parentheses doubles the whole block: 2 × 1 = 2 chlorine atoms."
  },
  {
   "value": 6,
   "misconception": "you multiplied by a subscript that belongs to oxygen — the 3 in O₃ counts oxygen atoms only",
   "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the 3 in O₃ counts oxygen atoms only. Each ClO₃ block holds one chlorine atom, and the outer 2 doubles it: 2 × 1 = 2 chlorine atoms."
  }
 ]
}

[
 {
  "value": "MgC₂H₃O₂",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Mg²⁺ takes two C₂H₃O₂⁻ ions, and the repeated acetate goes in parentheses: Mg(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "MgC₂H₃O₂₂",
  "misconception": "you left out the parentheses — the subscript must repeat the whole acetate block",
  "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole acetate block, so it goes outside parentheses: Mg(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Mg²⁺ needs two C₂H₃O₂⁻ ions: (2+) + 2 × (1−) = 0. Acetate repeats, so parentheses: Mg(C₂H₃O₂)₂.",
 "anticipatedWrong": [
  {
   "value": "MgC₂H₃O₂",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Mg²⁺ takes two C₂H₃O₂⁻ ions, and the repeated acetate goes in parentheses: Mg(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "MgC₂H₃O₂₂",
   "misconception": "you left out the parentheses — the subscript must repeat the whole acetate block",
   "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole acetate block, so it goes outside parentheses: Mg(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "Li₃(PO₄)",
  "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and phosphate appears once here: Li₃PO₄. Check: 3 × (1+) + (3−) = 0."
 },
 {
  "value": "LiPO₄",
  "misconception": "you wrote a 1:1 ratio — one 3− ion needs three 1+ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, whose charges give (1+) + (3−) = 2−, not zero. One 3− ion needs three 1+ ions: Li₃PO₄. Check: 3 × (1+) + (3−) = 0."
 }
]

{
 "keyExplanation": "Three Li⁺ ions cancel one PO₄³⁻ ion. Phosphate appears only once, so no parentheses: Li₃PO₄. Check: 3 × (1+) + (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Li₃(PO₄)",
   "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and phosphate appears once here: Li₃PO₄. Check: 3 × (1+) + (3−) = 0."
  },
  {
   "value": "LiPO₄",
   "misconception": "you wrote a 1:1 ratio — one 3− ion needs three 1+ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, whose charges give (1+) + (3−) = 2−, not zero. One 3− ion needs three 1+ ions: Li₃PO₄. Check: 3 × (1+) + (3−) = 0."
  }
 ]
}

[
 {
  "value": "Mg₂S₂",
  "misconception": "you skipped the reduce step — criss-cross gives Mg₂S₂, and 2:2 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Mg₂S₂, but the subscripts 2 and 2 share a common factor and reduce to 1 and 1: MgS. Check: (2+) + (2−) = 0."
 },
 {
  "value": "Mg₂S",
  "misconception": "you doubled only one side — one 2+ ion already cancels one 2− ion",
  "feedback": "This answer is wrong — you doubled only one side, but one 2+ ion already cancels one 2− ion, so the simplest ratio is 1:1: MgS. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Mg²⁺ and S²⁻. Criss-cross gives Mg₂S₂ — but 2 and 2 reduce to 1 and 1: one 2+ ion already cancels one 2− ion, so the formula is MgS. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Mg₂S₂",
   "misconception": "you skipped the reduce step — criss-cross gives Mg₂S₂, and 2:2 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Mg₂S₂, but the subscripts 2 and 2 share a common factor and reduce to 1 and 1: MgS. Check: (2+) + (2−) = 0."
  },
  {
   "value": "Mg₂S",
   "misconception": "you doubled only one side — one 2+ ion already cancels one 2− ion",
   "feedback": "This answer is wrong — you doubled only one side, but one 2+ ion already cancels one 2− ion, so the simplest ratio is 1:1: MgS. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 "Inside the parentheses: each ClO₃ block holds 1 chlorine atom and 3 oxygen atoms — 4 atoms.",
 "The 2 outside doubles the block: 2 × 4 = 8 atoms.",
 "The Sr outside the parentheses counts once: 8 + 1 = 9.",
 "Sr(ClO₃)₂ contains 9 atoms in total."
]

[
 {
  "value": 5,
  "misconception": "you forgot to double the ClO₃ block — the 2 outside multiplies the Cl and all three O atoms",
  "feedback": "This answer is wrong — you forgot to double the ClO₃ block; the 2 outside multiplies the Cl and all three O atoms. Each block holds 4 atoms, the 2 outside doubles that to 8, and the single strontium makes 9 atoms."
 },
 {
  "value": 8,
  "misconception": "you left out the strontium — atoms outside the parentheses still count",
  "feedback": "This answer is wrong — you left out the strontium; atoms outside the parentheses still count. The doubled ClO₃ blocks give 8 atoms, and the single Sr brings the total to 9 atoms."
 }
]

{
 "keyExplanation": "Each ClO₃ block holds 1 chlorine and 3 oxygen atoms; the 2 outside doubles the block to 2 Cl and 6 O. Adding the single strontium: 2 + 6 + 1 = 9 atoms.",
 "anticipatedWrong": [
  {
   "value": 5,
   "misconception": "you forgot to double the ClO₃ block — the 2 outside multiplies the Cl and all three O atoms",
   "feedback": "This answer is wrong — you forgot to double the ClO₃ block; the 2 outside multiplies the Cl and all three O atoms. Each block holds 4 atoms, the 2 outside doubles that to 8, and the single strontium makes 9 atoms."
  },
  {
   "value": 8,
   "misconception": "you left out the strontium — atoms outside the parentheses still count",
   "feedback": "This answer is wrong — you left out the strontium; atoms outside the parentheses still count. The doubled ClO₃ blocks give 8 atoms, and the single Sr brings the total to 9 atoms."
  }
 ]
}

[
 {
  "value": "Sr₂(PO₄)₃",
  "misconception": "you swapped the counts — the 3− ion's charge number becomes Sr's subscript, and the 2+ ion's becomes phosphate's",
  "feedback": "This answer is wrong — you swapped the counts. The 3− ion's charge number becomes Sr's subscript, and the 2+ ion's becomes phosphate's: three 2+ ions give 6+ and two 3− ions give 6−, so Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
 },
 {
  "value": "Sr₃PO₄₂",
  "misconception": "you left out the parentheses — the subscript must repeat the whole phosphate block",
  "feedback": "This answer is wrong — you left out the parentheses. Phosphate repeats, and its count of 2 must repeat the whole PO₄ block, so the block is wrapped in parentheses: Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Sr²⁺ and PO₄³⁻. Three 2+ ions give 6+ and two 3− ions give 6−, so Sr₃(PO₄)₂ — parentheses because phosphate repeats. Check: 3 × (2+) + 2 × (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Sr₂(PO₄)₃",
   "misconception": "you swapped the counts — the 3− ion's charge number becomes Sr's subscript, and the 2+ ion's becomes phosphate's",
   "feedback": "This answer is wrong — you swapped the counts. The 3− ion's charge number becomes Sr's subscript, and the 2+ ion's becomes phosphate's: three 2+ ions give 6+ and two 3− ions give 6−, so Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
  },
  {
   "value": "Sr₃PO₄₂",
   "misconception": "you left out the parentheses — the subscript must repeat the whole phosphate block",
   "feedback": "This answer is wrong — you left out the parentheses. Phosphate repeats, and its count of 2 must repeat the whole PO₄ block, so the block is wrapped in parentheses: Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
  }
 ]
}

[
 {
  "value": "NH₄(C₂H₃O₂)",
  "misconception": "you added parentheses around a block that appears only once",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Use parentheses only when the formula contains more than one of that polyatomic ion; here one 1+ ion cancels one 1− ion, so no parentheses: NH₄C₂H₃O₂. Check: (1+) + (1−) = 0."
 },
 {
  "value": "(NH₄)₂C₂H₃O₂",
  "misconception": "you doubled only one side — one 1+ ion already cancels one 1− ion",
  "feedback": "This answer is wrong — you doubled only one side. One 1+ ion already cancels one 1− ion, and neither polyatomic ion repeats, so no parentheses: NH₄C₂H₃O₂. Check: (1+) + (1−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: NH₄⁺ and C₂H₃O₂⁻. One 1+ ion cancels one 1− ion, and neither polyatomic ion appears more than once, so no parentheses: NH₄C₂H₃O₂. Check: (1+) + (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "NH₄(C₂H₃O₂)",
   "misconception": "you added parentheses around a block that appears only once",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Use parentheses only when the formula contains more than one of that polyatomic ion; here one 1+ ion cancels one 1− ion, so no parentheses: NH₄C₂H₃O₂. Check: (1+) + (1−) = 0."
  },
  {
   "value": "(NH₄)₂C₂H₃O₂",
   "misconception": "you doubled only one side — one 1+ ion already cancels one 1− ion",
   "feedback": "This answer is wrong — you doubled only one side. One 1+ ion already cancels one 1− ion, and neither polyatomic ion repeats, so no parentheses: NH₄C₂H₃O₂. Check: (1+) + (1−) = 0."
  }
 ]
}

[
 {
  "value": "CO₄²⁻",
  "misconception": "you gave carbonate sulfate's four oxygen atoms — carbonate is CO₃²⁻, with three",
  "feedback": "This answer is wrong — you gave carbonate sulfate's four oxygen atoms. The described ion has three oxygens around its carbon, sharing 2−: CO₃²⁻, the carbonate ion."
 },
 {
  "value": "CO₃³⁻",
  "misconception": "you gave carbonate phosphate's 3− charge — carbonate carries 2−",
  "feedback": "This answer is wrong — you gave carbonate phosphate's 3− charge. The described ion shares a 2− charge, so the formula is CO₃²⁻ — the carbonate ion."
 }
]

{
 "keyExplanation": "The carbon-based 2− polyatomic ion is carbonate: one carbon atom and three oxygen atoms sharing a 2− charge, CO₃²⁻.",
 "anticipatedWrong": [
  {
   "value": "CO₄²⁻",
   "misconception": "you gave carbonate sulfate's four oxygen atoms — carbonate is CO₃²⁻, with three",
   "feedback": "This answer is wrong — you gave carbonate sulfate's four oxygen atoms. The described ion has three oxygens around its carbon, sharing 2−: CO₃²⁻, the carbonate ion."
  },
  {
   "value": "CO₃³⁻",
   "misconception": "you gave carbonate phosphate's 3− charge — carbonate carries 2−",
   "feedback": "This answer is wrong — you gave carbonate phosphate's 3− charge. The described ion shares a 2− charge, so the formula is CO₃²⁻ — the carbonate ion."
  }
 ]
}

[
 {
  "value": "2−",
  "misconception": "you carried carbonate's 2− across — chlorate is a 1− ion, built like nitrate with chlorine in nitrogen's place",
  "feedback": "This answer is wrong — you carried carbonate's 2− across. Chlorate is ClO₃⁻: its four atoms share a single 1− charge, so the answer is 1−."
 },
 {
  "value": "3−",
  "misconception": "you read the subscript 3 as the charge — the 3 counts oxygen atoms; the ion's charge is 1−",
  "feedback": "This answer is wrong — you read the subscript 3 as the charge. The 3 in ClO₃⁻ counts the oxygen atoms; the whole group shares a 1− charge."
 },
 {
  "value": "1+",
  "misconception": "you flipped the sign — chlorate is a negative ion",
  "feedback": "This answer is wrong — you flipped the sign. Chlorate is a negative polyatomic ion: ClO₃⁻ carries 1−, not 1+."
 }
]

{
 "keyExplanation": "The chlorate ion is ClO₃⁻ — one chlorine atom and three oxygen atoms sharing a 1− charge. All three ions of this lesson — hydrogen carbonate, acetate, and chlorate — carry 1−, like nitrate and hydroxide.",
 "anticipatedWrong": [
  {
   "value": "2−",
   "misconception": "you carried carbonate's 2− across — chlorate is a 1− ion, built like nitrate with chlorine in nitrogen's place",
   "feedback": "This answer is wrong — you carried carbonate's 2− across. Chlorate is ClO₃⁻: its four atoms share a single 1− charge, so the answer is 1−."
  },
  {
   "value": "3−",
   "misconception": "you read the subscript 3 as the charge — the 3 counts oxygen atoms; the ion's charge is 1−",
   "feedback": "This answer is wrong — you read the subscript 3 as the charge. The 3 in ClO₃⁻ counts the oxygen atoms; the whole group shares a 1− charge."
  },
  {
   "value": "1+",
   "misconception": "you flipped the sign — chlorate is a negative ion",
   "feedback": "This answer is wrong — you flipped the sign. Chlorate is a negative polyatomic ion: ClO₃⁻ carries 1−, not 1+."
  }
 ]
}

[
 {
  "value": "Sr₂Se₂",
  "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
  "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Sr₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: SrSe. Check: (2+) + (2−) = 0."
 },
 {
  "value": "SrSe₂",
  "misconception": "you kept selenium's crossed subscript but reduced strontium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced strontium's. Reduction divides BOTH subscripts together: Sr₂Se₂ becomes SrSe, and (2+) + (2−) = 0."
 },
 {
  "value": "Sr₂Se",
  "misconception": "you kept strontium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept strontium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Sr₂Se₂ becomes SrSe, and (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Criss-crossing the two 2s gives Sr₂Se₂. Both subscripts divide by 2, and a formula gives the simplest whole-number ratio, so the final formula is SrSe: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Sr₂Se₂",
   "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
   "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Sr₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: SrSe. Check: (2+) + (2−) = 0."
  },
  {
   "value": "SrSe₂",
   "misconception": "you kept selenium's crossed subscript but reduced strontium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced strontium's. Reduction divides BOTH subscripts together: Sr₂Se₂ becomes SrSe, and (2+) + (2−) = 0."
  },
  {
   "value": "Sr₂Se",
   "misconception": "you kept strontium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept strontium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Sr₂Se₂ becomes SrSe, and (2+) + (2−) = 0."
  }
 ]
}

[
 "Each acetate block C₂H₃O₂ holds 2 carbon atoms.",
 "The 2 outside the parentheses doubles the block: 2 × 2 = 4.",
 "Sr(C₂H₃O₂)₂ contains 4 carbon atoms."
]

[
 {
  "value": 2,
  "misconception": "you forgot the outer subscript — each block holds two carbons and the outer 2 doubles them",
  "feedback": "This answer is wrong — you forgot the outer subscript. Each C₂H₃O₂ block holds two carbon atoms, and the 2 outside the parentheses doubles the whole block: 2 × 2 = 4 carbon atoms."
 },
 {
  "value": 6,
  "misconception": "you multiplied by a subscript that belongs to hydrogen — the 3 in H₃ counts hydrogen atoms only",
  "feedback": "This answer is wrong — you multiplied by a subscript that belongs to hydrogen; the 3 in H₃ counts hydrogen atoms only. Each block holds two carbon atoms, and the outer 2 doubles them: 2 × 2 = 4 carbon atoms."
 }
]

{
 "keyExplanation": "Each C₂H₃O₂ block holds 2 carbon atoms, and the 2 outside the parentheses doubles the block: 2 × 2 = 4 carbon atoms. The Sr outside the parentheses is not multiplied and carries no carbon.",
 "anticipatedWrong": [
  {
   "value": 2,
   "misconception": "you forgot the outer subscript — each block holds two carbons and the outer 2 doubles them",
   "feedback": "This answer is wrong — you forgot the outer subscript. Each C₂H₃O₂ block holds two carbon atoms, and the 2 outside the parentheses doubles the whole block: 2 × 2 = 4 carbon atoms."
  },
  {
   "value": 6,
   "misconception": "you multiplied by a subscript that belongs to hydrogen — the 3 in H₃ counts hydrogen atoms only",
   "feedback": "This answer is wrong — you multiplied by a subscript that belongs to hydrogen; the 3 in H₃ counts hydrogen atoms only. Each block holds two carbon atoms, and the outer 2 doubles them: 2 × 2 = 4 carbon atoms."
  }
 ]
}

[
 {
  "value": "MgClO₃",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Mg²⁺ takes two ClO₃⁻ ions, and the repeated chlorate goes in parentheses: Mg(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "MgClO₃₂",
  "misconception": "you left out the parentheses — the subscript must repeat the whole chlorate block",
  "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole chlorate block, so it goes outside parentheses: Mg(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Mg²⁺ needs two ClO₃⁻ ions: (2+) + 2 × (1−) = 0. Chlorate repeats, so parentheses: Mg(ClO₃)₂.",
 "anticipatedWrong": [
  {
   "value": "MgClO₃",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Mg²⁺ takes two ClO₃⁻ ions, and the repeated chlorate goes in parentheses: Mg(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "MgClO₃₂",
   "misconception": "you left out the parentheses — the subscript must repeat the whole chlorate block",
   "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole chlorate block, so it goes outside parentheses: Mg(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "K(HCO₃)",
  "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and hydrogen carbonate appears once here: KHCO₃. Check: (1+) + (1−) = 0."
 },
 {
  "value": "K(HCO₃)₂",
  "misconception": "you doubled the polyatomic ion — one 1+ ion needs only one 1− ion",
  "feedback": "This answer is wrong — you doubled the hydrogen carbonate ion, and the charges no longer cancel: (1+) + 2 × (1−) = 1−, not zero. One 1+ ion needs only one 1− ion: KHCO₃. Check: (1+) + (1−) = 0."
 }
]

{
 "keyExplanation": "One K⁺ ion cancels one HCO₃⁻ ion. Hydrogen carbonate appears only once, so no parentheses — and no additional ion-count subscripts or parentheses are needed: KHCO₃. Check: (1+) + (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "K(HCO₃)",
   "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and hydrogen carbonate appears once here: KHCO₃. Check: (1+) + (1−) = 0."
  },
  {
   "value": "K(HCO₃)₂",
   "misconception": "you doubled the polyatomic ion — one 1+ ion needs only one 1− ion",
   "feedback": "This answer is wrong — you doubled the hydrogen carbonate ion, and the charges no longer cancel: (1+) + 2 × (1−) = 1−, not zero. One 1+ ion needs only one 1− ion: KHCO₃. Check: (1+) + (1−) = 0."
  }
 ]
}

[
 {
  "value": "NaS",
  "misconception": "you wrote a 1:1 ratio — one 1+ ion cannot cancel a 2− ion",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 1+ ion cannot cancel a 2− ion. Write the ions: Na⁺ and S²⁻. Two 1+ ions cancel one 2− ion: Na₂S. Check: 2 × (1+) + (2−) = 0."
 },
 {
  "value": "NaS₂",
  "misconception": "you swapped the counts — the 2 counts the 1+ ions, because sulfide's charge number crosses onto sodium",
  "feedback": "This answer is wrong — you swapped the counts: in the criss-cross, sulfide's charge number 2 becomes sodium's subscript, not sulfide's. Write the ions: Na⁺ and S²⁻. Two 1+ ions cancel one 2− ion: Na₂S. Check: 2 × (1+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Na⁺ and S²⁻. Two 1+ ions cancel one 2− ion: Na₂S, with no parentheses because sulfide is a single atom, and 2 and 1 share no common factor, so nothing reduces. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "NaS",
   "misconception": "you wrote a 1:1 ratio — one 1+ ion cannot cancel a 2− ion",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 1+ ion cannot cancel a 2− ion. Write the ions: Na⁺ and S²⁻. Two 1+ ions cancel one 2− ion: Na₂S. Check: 2 × (1+) + (2−) = 0."
  },
  {
   "value": "NaS₂",
   "misconception": "you swapped the counts — the 2 counts the 1+ ions, because sulfide's charge number crosses onto sodium",
   "feedback": "This answer is wrong — you swapped the counts: in the criss-cross, sulfide's charge number 2 becomes sodium's subscript, not sulfide's. Write the ions: Na⁺ and S²⁻. Two 1+ ions cancel one 2− ion: Na₂S. Check: 2 × (1+) + (2−) = 0."
  }
 ]
}

[
 "Inside the parentheses: each HCO₃ block holds 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms — 5 atoms.",
 "The 2 outside doubles the block: 2 × 5 = 10 atoms.",
 "The Ca outside the parentheses counts once: 10 + 1 = 11.",
 "Ca(HCO₃)₂ contains 11 atoms in total."
]

[
 {
  "value": 6,
  "misconception": "you forgot to double the HCO₃ block — the 2 outside multiplies the H, the C, and all three O atoms",
  "feedback": "This answer is wrong — you forgot to double the HCO₃ block; the 2 outside multiplies the H, the C, and all three O atoms. Each block holds 5 atoms, the 2 outside doubles that to 10, and the single calcium makes 11 atoms."
 },
 {
  "value": 10,
  "misconception": "you left out the calcium — atoms outside the parentheses still count",
  "feedback": "This answer is wrong — you left out the calcium; atoms outside the parentheses still count. The doubled HCO₃ blocks give 10 atoms, and the single Ca brings the total to 11 atoms."
 }
]

{
 "keyExplanation": "Each HCO₃ block holds 1 hydrogen, 1 carbon, and 3 oxygen atoms; the 2 outside doubles the block to 2 H, 2 C, and 6 O. Adding the single calcium: 2 + 2 + 6 + 1 = 11 atoms.",
 "anticipatedWrong": [
  {
   "value": 6,
   "misconception": "you forgot to double the HCO₃ block — the 2 outside multiplies the H, the C, and all three O atoms",
   "feedback": "This answer is wrong — you forgot to double the HCO₃ block; the 2 outside multiplies the H, the C, and all three O atoms. Each block holds 5 atoms, the 2 outside doubles that to 10, and the single calcium makes 11 atoms."
  },
  {
   "value": 10,
   "misconception": "you left out the calcium — atoms outside the parentheses still count",
   "feedback": "This answer is wrong — you left out the calcium; atoms outside the parentheses still count. The doubled HCO₃ blocks give 10 atoms, and the single Ca brings the total to 11 atoms."
  }
 ]
}

[
 {
  "value": "Ca₂(SO₄)₂",
  "misconception": "you skipped the reduce step — criss-cross gives Ca₂(SO₄)₂, and 2:2 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ca₂(SO₄)₂, but the subscripts 2 and 2 share a common factor and reduce to 1 and 1. Sulfate then appears once, so no parentheses: CaSO₄. Check: (2+) + (2−) = 0."
 },
 {
  "value": "Ca(SO₄)",
  "misconception": "you kept the parentheses — after reducing, sulfate appears only once, and parentheses are only for a repeated polyatomic ion",
  "feedback": "This answer is wrong — you kept the parentheses. After 2:2 reduces to 1:1, sulfate appears only once, and parentheses are only for a repeated polyatomic ion: CaSO₄. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Ca²⁺ and SO₄²⁻. Criss-cross gives Ca₂(SO₄)₂ — reduce 2:2 to 1:1. Sulfate then appears once, so no parentheses: CaSO₄. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ca₂(SO₄)₂",
   "misconception": "you skipped the reduce step — criss-cross gives Ca₂(SO₄)₂, and 2:2 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ca₂(SO₄)₂, but the subscripts 2 and 2 share a common factor and reduce to 1 and 1. Sulfate then appears once, so no parentheses: CaSO₄. Check: (2+) + (2−) = 0."
  },
  {
   "value": "Ca(SO₄)",
   "misconception": "you kept the parentheses — after reducing, sulfate appears only once, and parentheses are only for a repeated polyatomic ion",
   "feedback": "This answer is wrong — you kept the parentheses. After 2:2 reduces to 1:1, sulfate appears only once, and parentheses are only for a repeated polyatomic ion: CaSO₄. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "NH₄(HCO₃)",
  "misconception": "you added parentheses around a block that appears only once",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Use parentheses only when the formula contains more than one of that polyatomic ion; here one 1+ ion cancels one 1− ion, so no parentheses: NH₄HCO₃. Check: (1+) + (1−) = 0."
 },
 {
  "value": "(NH₄)₂HCO₃",
  "misconception": "you balanced against a 2− charge — the supplied ion HCO₃⁻ carries 1−",
  "feedback": "This answer is wrong — you balanced against a 2− charge. The supplied ion HCO₃⁻ carries 1−, so one 1+ ion already cancels one 1− ion, and no parentheses are needed: NH₄HCO₃. Check: (1+) + (1−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: NH₄⁺ and HCO₃⁻. One 1+ ion cancels one 1− ion, and neither polyatomic ion appears more than once, so no parentheses: NH₄HCO₃. Check: (1+) + (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "NH₄(HCO₃)",
   "misconception": "you added parentheses around a block that appears only once",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Use parentheses only when the formula contains more than one of that polyatomic ion; here one 1+ ion cancels one 1− ion, so no parentheses: NH₄HCO₃. Check: (1+) + (1−) = 0."
  },
  {
   "value": "(NH₄)₂HCO₃",
   "misconception": "you balanced against a 2− charge — the supplied ion HCO₃⁻ carries 1−",
   "feedback": "This answer is wrong — you balanced against a 2− charge. The supplied ion HCO₃⁻ carries 1−, so one 1+ ion already cancels one 1− ion, and no parentheses are needed: NH₄HCO₃. Check: (1+) + (1−) = 0."
  }
 ]
}

[
 {
  "value": "2−",
  "misconception": "you gave phosphate the 2− that carbonate and sulfate carry — phosphate is the trio's 3− member",
  "feedback": "This answer is wrong — you gave phosphate the 2− that carbonate and sulfate carry. Phosphate is PO₄³⁻: its group shares a 3− charge, so the answer is 3−."
 },
 {
  "value": "3+",
  "misconception": "you flipped the sign — phosphate is a negative ion",
  "feedback": "This answer is wrong — you flipped the sign. Phosphate is a negative polyatomic ion: PO₄³⁻ carries 3−, not 3+."
 }
]

{
 "keyExplanation": "The phosphate ion is PO₄³⁻ — one phosphorus atom and four oxygen atoms sharing a 3− charge. Its charge is 3−: in the taught trio, carbonate and sulfate carry 2−, and phosphate is the 3− one.",
 "anticipatedWrong": [
  {
   "value": "2−",
   "misconception": "you gave phosphate the 2− that carbonate and sulfate carry — phosphate is the trio's 3− member",
   "feedback": "This answer is wrong — you gave phosphate the 2− that carbonate and sulfate carry. Phosphate is PO₄³⁻: its group shares a 3− charge, so the answer is 3−."
  },
  {
   "value": "3+",
   "misconception": "you flipped the sign — phosphate is a negative ion",
   "feedback": "This answer is wrong — you flipped the sign. Phosphate is a negative polyatomic ion: PO₄³⁻ carries 3−, not 3+."
  }
 ]
}

[
 {
  "value": "carbonate",
  "misconception": "you dropped the hydrogen — carbonate is CO₃²⁻, with a 2− charge and no hydrogen",
  "feedback": "This answer is wrong — carbonate is CO₃²⁻. HCO₃⁻ is carbonate plus one hydrogen, and its charge drops from carbonate's 2− to 1−: this ion is hydrogen carbonate."
 },
 {
  "value": "acetate",
  "misconception": "you named the carbon-and-hydrogen lookalike — acetate is C₂H₃O₂⁻, with two carbons and three hydrogens",
  "feedback": "This answer is wrong — acetate is C₂H₃O₂⁻: two carbon, three hydrogen, and two oxygen atoms. HCO₃⁻ — one H on a carbonate ion with a 1− charge — is the hydrogen carbonate ion."
 }
]

{
 "keyExplanation": "HCO₃⁻ — a carbonate ion carrying one extra hydrogen atom, with a 1− charge — is the hydrogen carbonate ion.",
 "anticipatedWrong": [
  {
   "value": "carbonate",
   "misconception": "you dropped the hydrogen — carbonate is CO₃²⁻, with a 2− charge and no hydrogen",
   "feedback": "This answer is wrong — carbonate is CO₃²⁻. HCO₃⁻ is carbonate plus one hydrogen, and its charge drops from carbonate's 2− to 1−: this ion is hydrogen carbonate."
  },
  {
   "value": "acetate",
   "misconception": "you named the carbon-and-hydrogen lookalike — acetate is C₂H₃O₂⁻, with two carbons and three hydrogens",
   "feedback": "This answer is wrong — acetate is C₂H₃O₂⁻: two carbon, three hydrogen, and two oxygen atoms. HCO₃⁻ — one H on a carbonate ion with a 1− charge — is the hydrogen carbonate ion."
  }
 ]
}

[
 {
  "value": "Ba₂Se₂",
  "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
  "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Ba₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: BaSe. Check: (2+) + (2−) = 0."
 },
 {
  "value": "BaSe₂",
  "misconception": "you kept selenium's crossed subscript but reduced barium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced barium's. Reduction divides BOTH subscripts together: Ba₂Se₂ becomes BaSe, and (2+) + (2−) = 0."
 },
 {
  "value": "Ba₂Se",
  "misconception": "you kept barium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
  "feedback": "This answer is wrong — you kept barium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Ba₂Se₂ becomes BaSe, and (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Criss-crossing the two 2s gives Ba₂Se₂. Both subscripts divide by 2, and a formula gives the simplest whole-number ratio, so the final formula is BaSe: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ba₂Se₂",
   "misconception": "you stopped after criss-crossing — both subscripts divide by 2, and the formula must show the simplest whole-number ratio",
   "feedback": "This answer is wrong — you stopped after criss-crossing. Both subscripts of Ba₂Se₂ divide by 2, and a formula gives the simplest whole-number ratio: BaSe. Check: (2+) + (2−) = 0."
  },
  {
   "value": "BaSe₂",
   "misconception": "you kept selenium's crossed subscript but reduced barium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept selenium's crossed subscript but reduced barium's. Reduction divides BOTH subscripts together: Ba₂Se₂ becomes BaSe, and (2+) + (2−) = 0."
  },
  {
   "value": "Ba₂Se",
   "misconception": "you kept barium's crossed subscript but reduced selenium's — reduction divides BOTH subscripts together",
   "feedback": "This answer is wrong — you kept barium's crossed subscript but reduced selenium's. Reduction divides BOTH subscripts together: Ba₂Se₂ becomes BaSe, and (2+) + (2−) = 0."
  }
 ]
}

[
 "Each hydrogen carbonate block HCO₃ holds 1 carbon atom.",
 "The 2 outside the parentheses doubles the block: 2 × 1 = 2.",
 "Ba(HCO₃)₂ contains 2 carbon atoms."
]

[
 {
  "value": 1,
  "misconception": "you forgot the outer subscript — the 2 outside the parentheses doubles the one carbon in each block",
  "feedback": "This answer is wrong — you forgot the outer subscript. Each HCO₃ block holds one carbon atom, and the 2 outside the parentheses doubles the whole block: 2 × 1 = 2 carbon atoms."
 },
 {
  "value": 6,
  "misconception": "you multiplied by a subscript that belongs to oxygen — the 3 in O₃ counts oxygen atoms only",
  "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the 3 in O₃ counts oxygen atoms only. Each HCO₃ block holds one carbon atom, and the outer 2 doubles it: 2 × 1 = 2 carbon atoms."
 }
]

{
 "keyExplanation": "Each HCO₃ block holds 1 carbon atom, and the 2 outside the parentheses doubles the block: 2 × 1 = 2 carbon atoms. The Ba outside the parentheses is not multiplied and carries no carbon.",
 "anticipatedWrong": [
  {
   "value": 1,
   "misconception": "you forgot the outer subscript — the 2 outside the parentheses doubles the one carbon in each block",
   "feedback": "This answer is wrong — you forgot the outer subscript. Each HCO₃ block holds one carbon atom, and the 2 outside the parentheses doubles the whole block: 2 × 1 = 2 carbon atoms."
  },
  {
   "value": 6,
   "misconception": "you multiplied by a subscript that belongs to oxygen — the 3 in O₃ counts oxygen atoms only",
   "feedback": "This answer is wrong — you multiplied by a subscript that belongs to oxygen; the 3 in O₃ counts oxygen atoms only. Each HCO₃ block holds one carbon atom, and the outer 2 doubles it: 2 × 1 = 2 carbon atoms."
  }
 ]
}

[
 {
  "value": "CaC₂H₃O₂",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Ca²⁺ takes two C₂H₃O₂⁻ ions, and the repeated acetate goes in parentheses: Ca(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "CaC₂H₃O₂₂",
  "misconception": "you left out the parentheses — the subscript must repeat the whole acetate block",
  "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole acetate block, so it goes outside parentheses: Ca(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Ca²⁺ needs two C₂H₃O₂⁻ ions: (2+) + 2 × (1−) = 0. Acetate repeats, so parentheses: Ca(C₂H₃O₂)₂.",
 "anticipatedWrong": [
  {
   "value": "CaC₂H₃O₂",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions before the charges cancel. Ca²⁺ takes two C₂H₃O₂⁻ ions, and the repeated acetate goes in parentheses: Ca(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "CaC₂H₃O₂₂",
   "misconception": "you left out the parentheses — the subscript must repeat the whole acetate block",
   "feedback": "This answer is wrong — you left out the parentheses. A bare subscript tacked on the end changes only the last atom's count; the 2 must repeat the whole acetate block, so it goes outside parentheses: Ca(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "Rb(OH)",
  "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and hydroxide appears once here: RbOH. Check: (1+) + (1−) = 0."
 },
 {
  "value": "Rb(OH)₂",
  "misconception": "you doubled the polyatomic ion — one 1+ ion needs only one 1− ion",
  "feedback": "This answer is wrong — you doubled the hydroxide ion, and the charges no longer cancel: (1+) + 2 × (1−) = 1−, not zero. One 1+ ion needs only one 1− ion: RbOH. Check: (1+) + (1−) = 0."
 }
]

{
 "keyExplanation": "One Rb⁺ ion cancels one OH⁻ ion. Hydroxide appears only once, so no parentheses — and no subscripts are needed at all: RbOH. Check: (1+) + (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Rb(OH)",
   "misconception": "you added parentheses around a block that appears only once — parentheses are needed only when the polyatomic ion repeats",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are needed only when the polyatomic ion repeats, and hydroxide appears once here: RbOH. Check: (1+) + (1−) = 0."
  },
  {
   "value": "Rb(OH)₂",
   "misconception": "you doubled the polyatomic ion — one 1+ ion needs only one 1− ion",
   "feedback": "This answer is wrong — you doubled the hydroxide ion, and the charges no longer cancel: (1+) + 2 × (1−) = 1−, not zero. One 1+ ion needs only one 1− ion: RbOH. Check: (1+) + (1−) = 0."
  }
 ]
}

[
 {
  "value": "Ca₂P₃",
  "misconception": "you swapped the counts — calcium takes the 3 and phosphide takes the 2",
  "feedback": "This answer is wrong — you swapped the counts: calcium takes the 3 and phosphide takes the 2. Write the ions: Ca²⁺ and P³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Ca₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
 },
 {
  "value": "Ca₃(P)₂",
  "misconception": "you added parentheses, but P³⁻ is a single-atom ion — parentheses are only for repeated polyatomic ions",
  "feedback": "This answer is wrong — you added parentheses, but P³⁻ is a single-atom ion; parentheses are only for repeated polyatomic ions. Write the ions: Ca²⁺ and P³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Ca₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Ca²⁺ and P³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Ca₃P₂, with no parentheses because phosphide is a single atom. Check: 3 × (2+) + 2 × (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ca₂P₃",
   "misconception": "you swapped the counts — calcium takes the 3 and phosphide takes the 2",
   "feedback": "This answer is wrong — you swapped the counts: calcium takes the 3 and phosphide takes the 2. Write the ions: Ca²⁺ and P³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Ca₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
  },
  {
   "value": "Ca₃(P)₂",
   "misconception": "you added parentheses, but P³⁻ is a single-atom ion — parentheses are only for repeated polyatomic ions",
   "feedback": "This answer is wrong — you added parentheses, but P³⁻ is a single-atom ion; parentheses are only for repeated polyatomic ions. Write the ions: Ca²⁺ and P³⁻. Three 2+ ions give 6+ and two 3− ions give 6−: Ca₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
  }
 ]
}

[
 "Inside the parentheses: each ClO₃ block holds 1 chlorine atom and 3 oxygen atoms — 4 atoms.",
 "The 2 outside doubles the block: 2 × 4 = 8 atoms.",
 "The Ba outside the parentheses counts once: 8 + 1 = 9.",
 "Ba(ClO₃)₂ contains 9 atoms in total."
]

[
 {
  "value": 5,
  "misconception": "you forgot to double the ClO₃ block — the 2 outside multiplies the Cl and all three O atoms",
  "feedback": "This answer is wrong — you forgot to double the ClO₃ block; the 2 outside multiplies the Cl and all three O atoms. Each block holds 4 atoms, the 2 outside doubles that to 8, and the single barium makes 9 atoms."
 },
 {
  "value": 8,
  "misconception": "you left out the barium — atoms outside the parentheses still count",
  "feedback": "This answer is wrong — you left out the barium; atoms outside the parentheses still count. The doubled ClO₃ blocks give 8 atoms, and the single Ba brings the total to 9 atoms."
 }
]

{
 "keyExplanation": "Each ClO₃ block holds 1 chlorine and 3 oxygen atoms; the 2 outside doubles the block to 2 Cl and 6 O. Adding the single barium: 2 + 6 + 1 = 9 atoms.",
 "anticipatedWrong": [
  {
   "value": 5,
   "misconception": "you forgot to double the ClO₃ block — the 2 outside multiplies the Cl and all three O atoms",
   "feedback": "This answer is wrong — you forgot to double the ClO₃ block; the 2 outside multiplies the Cl and all three O atoms. Each block holds 4 atoms, the 2 outside doubles that to 8, and the single barium makes 9 atoms."
  },
  {
   "value": 8,
   "misconception": "you left out the barium — atoms outside the parentheses still count",
   "feedback": "This answer is wrong — you left out the barium; atoms outside the parentheses still count. The doubled ClO₃ blocks give 8 atoms, and the single Ba brings the total to 9 atoms."
  }
 ]
}

[
 {
  "value": "Al₃(CO₃)₂",
  "misconception": "you swapped the counts — the 2− ion's charge number becomes Al's subscript, and the 3+ ion's becomes carbonate's",
  "feedback": "This answer is wrong — you swapped the counts. The 2− ion's charge number becomes Al's subscript, and the 3+ ion's becomes carbonate's: two 3+ ions give 6+ and three 2− ions give 6−, so Al₂(CO₃)₃. Check: 2 × (3+) + 3 × (2−) = 0."
 },
 {
  "value": "Al₂CO₃₃",
  "misconception": "you left out the parentheses — the subscript must repeat the whole carbonate block",
  "feedback": "This answer is wrong — you left out the parentheses. Carbonate repeats, and its count of 3 must repeat the whole CO₃ block, so the block is wrapped in parentheses: Al₂(CO₃)₃. Check: 2 × (3+) + 3 × (2−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: Al³⁺ and CO₃²⁻. Two 3+ ions give 6+ and three 2− ions give 6−, and carbonate repeats, so parentheses: Al₂(CO₃)₃. Check: 2 × (3+) + 3 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Al₃(CO₃)₂",
   "misconception": "you swapped the counts — the 2− ion's charge number becomes Al's subscript, and the 3+ ion's becomes carbonate's",
   "feedback": "This answer is wrong — you swapped the counts. The 2− ion's charge number becomes Al's subscript, and the 3+ ion's becomes carbonate's: two 3+ ions give 6+ and three 2− ions give 6−, so Al₂(CO₃)₃. Check: 2 × (3+) + 3 × (2−) = 0."
  },
  {
   "value": "Al₂CO₃₃",
   "misconception": "you left out the parentheses — the subscript must repeat the whole carbonate block",
   "feedback": "This answer is wrong — you left out the parentheses. Carbonate repeats, and its count of 3 must repeat the whole CO₃ block, so the block is wrapped in parentheses: Al₂(CO₃)₃. Check: 2 × (3+) + 3 × (2−) = 0."
  }
 ]
}

[
 {
  "value": "NH₄CO₃",
  "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2− ion needs two 1+ ions, and the repeated ion is polyatomic, so parentheses wrap NH₄: (NH₄)₂CO₃. Check: 2 × (1+) + (2−) = 0."
 },
 {
  "value": "NH₄(CO₃)₂",
  "misconception": "you put the parentheses around the wrong ion — CO₃²⁻ appears once; it is the NH₄ that repeats",
  "feedback": "This answer is wrong — you put the parentheses around the wrong ion. CO₃²⁻ appears once; the NH₄ is what repeats: two 1+ ions cancel one 2− ion, so (NH₄)₂CO₃. Check: 2 × (1+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Write the ions: NH₄⁺ and CO₃²⁻. Two 1+ ions cancel one 2− ion, and the repeated ion is polyatomic, so parentheses wrap NH₄: (NH₄)₂CO₃. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "NH₄CO₃",
   "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2− ion needs two 1+ ions, and the repeated ion is polyatomic, so parentheses wrap NH₄: (NH₄)₂CO₃. Check: 2 × (1+) + (2−) = 0."
  },
  {
   "value": "NH₄(CO₃)₂",
   "misconception": "you put the parentheses around the wrong ion — CO₃²⁻ appears once; it is the NH₄ that repeats",
   "feedback": "This answer is wrong — you put the parentheses around the wrong ion. CO₃²⁻ appears once; the NH₄ is what repeats: two 1+ ions cancel one 2− ion, so (NH₄)₂CO₃. Check: 2 × (1+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "sodium phosphate",
  "misconception": "you used the polyatomic ion's name — phosphate is PO₄³⁻; a bare P ion is phosphide",
  "feedback": "This answer is wrong — you used the polyatomic ion's name. Phosphate is PO₄³⁻; the bare P ion in Na₃P is phosphide, so the name is sodium phosphide."
 },
 {
  "value": "sodium phosphorus",
  "misconception": "you kept the element's name — phosphorus's ending changes to -ide",
  "feedback": "This answer is wrong — you kept the element's name. In the compound's name phosphorus's ending changes to -ide: Na₃P is sodium phosphide."
 }
]

{
 "keyExplanation": "Metal first, unchanged: sodium. The single-atom nonmetal ion takes -ide: phosphorus becomes phosphide. Na₃P is sodium phosphide.",
 "anticipatedWrong": [
  {
   "value": "sodium phosphate",
   "misconception": "you used the polyatomic ion's name — phosphate is PO₄³⁻; a bare P ion is phosphide",
   "feedback": "This answer is wrong — you used the polyatomic ion's name. Phosphate is PO₄³⁻; the bare P ion in Na₃P is phosphide, so the name is sodium phosphide."
  },
  {
   "value": "sodium phosphorus",
   "misconception": "you kept the element's name — phosphorus's ending changes to -ide",
   "feedback": "This answer is wrong — you kept the element's name. In the compound's name phosphorus's ending changes to -ide: Na₃P is sodium phosphide."
  }
 ]
}

[
 "The 4 after the raised dot counts whole water molecules.",
 "Each water molecule H₂O carries 2 hydrogen atoms: 4 × 2 = 8.",
 "Sr(NO₃)₂·4H₂O contains 8 hydrogen atoms."
]

[
 {
  "value": 4,
  "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 4 waters carries 2 hydrogens",
  "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 4 waters carries 2 hydrogens. The 4 after the dot counts whole water molecules, and each water molecule has 2 hydrogen atoms: 4 × 2 = 8 hydrogen atoms."
 },
 {
  "value": 2,
  "misconception": "you counted one water molecule's hydrogens — the 4 in front multiplies the whole H₂O",
  "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 4 in front multiplies the whole H₂O. Four water molecules each carry 2 hydrogen atoms: 4 × 2 = 8 hydrogen atoms."
 }
]

{
 "keyExplanation": "The 4 after the dot counts whole water molecules, and each water molecule has 2 hydrogen atoms: 4 × 2 = 8 hydrogen atoms.",
 "anticipatedWrong": [
  {
   "value": 4,
   "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 4 waters carries 2 hydrogens",
   "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 4 waters carries 2 hydrogens. The 4 after the dot counts whole water molecules, and each water molecule has 2 hydrogen atoms: 4 × 2 = 8 hydrogen atoms."
  },
  {
   "value": 2,
   "misconception": "you counted one water molecule's hydrogens — the 4 in front multiplies the whole H₂O",
   "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 4 in front multiplies the whole H₂O. Four water molecules each carry 2 hydrogen atoms: 4 × 2 = 8 hydrogen atoms."
  }
 ]
}

[
 {
  "value": "carbonate calcium",
  "misconception": "you swapped the order — the positive ion's name always comes first",
  "feedback": "This answer is wrong — you swapped the order. The positive ion's name always comes first: CaCO₃ is calcium carbonate."
 },
 {
  "value": "calcium carbon trioxide",
  "misconception": "you pulled the subscript into the name and rebuilt the block as element names — a polyatomic ion's name is never rebuilt",
  "feedback": "This answer is wrong — you pulled the subscript into the name and broke the block into element names. CO₃ is one polyatomic ion, the carbonate ion, and its name never changes: CaCO₃ is calcium carbonate."
 }
]

{
 "keyExplanation": "The block CO₃ is the carbonate ion, and it keeps its own name, unchanged: CaCO₃ is calcium carbonate — the subscript stays out of the name.",
 "anticipatedWrong": [
  {
   "value": "carbonate calcium",
   "misconception": "you swapped the order — the positive ion's name always comes first",
   "feedback": "This answer is wrong — you swapped the order. The positive ion's name always comes first: CaCO₃ is calcium carbonate."
  },
  {
   "value": "calcium carbon trioxide",
   "misconception": "you pulled the subscript into the name and rebuilt the block as element names — a polyatomic ion's name is never rebuilt",
   "feedback": "This answer is wrong — you pulled the subscript into the name and broke the block into element names. CO₃ is one polyatomic ion, the carbonate ion, and its name never changes: CaCO₃ is calcium carbonate."
  }
 ]
}

[
 {
  "value": "Ba₂P₃",
  "misconception": "you swapped the counts — barium takes the 3 and phosphorus takes the 2",
  "feedback": "This answer is wrong — you swapped the counts. Barium is Ba²⁺ and phosphide is P³⁻, so it takes three 2+ ions to give 6+ and two 3− ions to give 6−: barium takes the 3 and phosphorus takes the 2, Ba₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
 },
 {
  "value": "Ba₃(PO₄)₂",
  "misconception": "you wrote the polyatomic phosphate — phosphide is the bare P³⁻ ion",
  "feedback": "This answer is wrong — you wrote the polyatomic phosphate. The -ide ending names the bare ion from the element: phosphide is P³⁻, from phosphorus in Group 15. Three Ba²⁺ ions balance two P³⁻ ions: Ba₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
 }
]

{
 "keyExplanation": "Barium is Ba²⁺ (Group 2) and phosphide is P³⁻ (Group 15). Three 2+ ions give 6+ and two 3− ions give 6−: Ba₃P₂. Check: 3 × (2+) + 2 × (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ba₂P₃",
   "misconception": "you swapped the counts — barium takes the 3 and phosphorus takes the 2",
   "feedback": "This answer is wrong — you swapped the counts. Barium is Ba²⁺ and phosphide is P³⁻, so it takes three 2+ ions to give 6+ and two 3− ions to give 6−: barium takes the 3 and phosphorus takes the 2, Ba₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
  },
  {
   "value": "Ba₃(PO₄)₂",
   "misconception": "you wrote the polyatomic phosphate — phosphide is the bare P³⁻ ion",
   "feedback": "This answer is wrong — you wrote the polyatomic phosphate. The -ide ending names the bare ion from the element: phosphide is P³⁻, from phosphorus in Group 15. Three Ba²⁺ ions balance two P³⁻ ions: Ba₃P₂. Check: 3 × (2+) + 2 × (3−) = 0."
  }
 ]
}

[
 {
  "value": "3+",
  "misconception": "you assumed iron is always 3+ - here two I⁻ ions total 2−, so the iron is 2+",
  "feedback": "This answer is wrong - you assumed iron is always 3+. Iron forms more than one ion, so the formula decides: the two I⁻ ions total 2−, and the single iron ion must contribute 2+, written 2+."
 },
 {
  "value": "1+",
  "misconception": "you gave one iodide's charge - the two iodide ions together set the total",
  "feedback": "This answer is wrong - you gave one iodide's charge. The two I⁻ ions together total 2−, so the iron ion must contribute 2+, written 2+."
 }
]

{
 "keyExplanation": "Iodide, from Group 17, is I⁻; two of them total 2−. One iron ion must contribute 2+. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "3+",
   "misconception": "you assumed iron is always 3+ - here two I⁻ ions total 2−, so the iron is 2+",
   "feedback": "This answer is wrong - you assumed iron is always 3+. Iron forms more than one ion, so the formula decides: the two I⁻ ions total 2−, and the single iron ion must contribute 2+, written 2+.",
   "check": {
    "expr": "3",
    "expected": 3
   }
  },
  {
   "value": "1+",
   "misconception": "you gave one iodide's charge - the two iodide ions together set the total",
   "feedback": "This answer is wrong - you gave one iodide's charge. The two I⁻ ions together total 2−, so the iron ion must contribute 2+, written 2+.",
   "check": {
    "expr": "1*1",
    "expected": 1
   }
  }
 ]
}

[
 {
  "value": "2+",
  "misconception": "you used cobalt's subscript as its charge — the 2 counts atoms; the charge comes from the balance",
  "feedback": "This answer is wrong — you used cobalt's subscript as its charge. The 2 counts atoms; the charge comes from the balance: three O²⁻ total 6−, shared by two cobalt ions, so each cobalt ion is 3+."
 },
 {
  "value": "6+",
  "misconception": "you forgot to share the total — 6+ is split between two cobalt ions",
  "feedback": "This answer is wrong — you forgot to share the total. The 6+ that balances three O²⁻ is split between two cobalt ions: 6+ ÷ 2 = 3+ each."
 }
]

{
 "keyExplanation": "Oxide is O²⁻, so three oxide ions total 6−. The total positive charge must be 6+, shared by two cobalt ions: 6+ ÷ 2 = 3+ each. Check: 2 × (3+) + 3 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "2+",
   "misconception": "you used cobalt's subscript as its charge — the 2 counts atoms; the charge comes from the balance",
   "feedback": "This answer is wrong — you used cobalt's subscript as its charge. The 2 counts atoms; the charge comes from the balance: three O²⁻ total 6−, shared by two cobalt ions, so each cobalt ion is 3+."
  },
  {
   "value": "6+",
   "misconception": "you forgot to share the total — 6+ is split between two cobalt ions",
   "feedback": "This answer is wrong — you forgot to share the total. The 6+ that balances three O²⁻ is split between two cobalt ions: 6+ ÷ 2 = 3+ each."
  }
 ]
}

[
 {
  "value": "manganese chloride(II)",
  "misconception": "you used the numeral to count the chlorides — the numeral is the manganese ion's 2+ charge, found from the balance",
  "feedback": "This answer is wrong — you used the numeral to count the chlorides. The numeral is the manganese ion's 2+ charge, found from the balance, and it sits right after the metal's name: MnCl₂ is manganese(II) chloride."
 },
 {
  "value": "manganese chloride",
  "misconception": "you left out the numeral — the name must show the metal ion's charge",
  "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two Cl⁻ ions total 2−, so the manganese ion is 2+ and MnCl₂ is manganese(II) chloride."
 }
]

{
 "keyExplanation": "Two Cl⁻ ions total 2−, so the manganese ion is 2+ — manganese(II). MnCl₂ is manganese(II) chloride.",
 "anticipatedWrong": [
  {
   "value": "manganese chloride(II)",
   "misconception": "you used the numeral to count the chlorides — the numeral is the manganese ion's 2+ charge, found from the balance",
   "feedback": "This answer is wrong — you used the numeral to count the chlorides. The numeral is the manganese ion's 2+ charge, found from the balance, and it sits right after the metal's name: MnCl₂ is manganese(II) chloride."
  },
  {
   "value": "manganese chloride",
   "misconception": "you left out the numeral — the name must show the metal ion's charge",
   "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two Cl⁻ ions total 2−, so the manganese ion is 2+ and MnCl₂ is manganese(II) chloride."
  }
 ]
}

[
 {
  "value": "iron(II) oxide",
  "misconception": "you used iron's subscript as the numeral — the 2 counts atoms; three O²⁻ give 6−, shared by two irons as 3+ each",
  "feedback": "This answer is wrong — you used iron's subscript as the numeral. The 2 counts atoms; three O²⁻ give 6−, shared by two irons as 3+ each, so Fe₂O₃ is iron(III) oxide."
 },
 {
  "value": "iron(VI) oxide",
  "misconception": "you gave the total positive charge — the numeral is the charge on one iron ion, 6+ ÷ 2 = 3+",
  "feedback": "This answer is wrong — you gave the total positive charge. The numeral is the charge on one iron ion, 6+ ÷ 2 = 3+, so Fe₂O₃ is iron(III) oxide."
 }
]

{
 "keyExplanation": "Three O²⁻ ions total 6−; two iron ions share the balancing 6+, so each is 3+ — iron(III). Fe₂O₃ is iron(III) oxide.",
 "anticipatedWrong": [
  {
   "value": "iron(II) oxide",
   "misconception": "you used iron's subscript as the numeral — the 2 counts atoms; three O²⁻ give 6−, shared by two irons as 3+ each",
   "feedback": "This answer is wrong — you used iron's subscript as the numeral. The 2 counts atoms; three O²⁻ give 6−, shared by two irons as 3+ each, so Fe₂O₃ is iron(III) oxide."
  },
  {
   "value": "iron(VI) oxide",
   "misconception": "you gave the total positive charge — the numeral is the charge on one iron ion, 6+ ÷ 2 = 3+",
   "feedback": "This answer is wrong — you gave the total positive charge. The numeral is the charge on one iron ion, 6+ ÷ 2 = 3+, so Fe₂O₃ is iron(III) oxide."
  }
 ]
}

[
 {
  "value": "Cr₃S₂",
  "misconception": "you swapped the counts — chromium takes the 2 and sulfur takes the 3",
  "feedback": "This answer is wrong — you swapped the counts. Chromium takes the 2 and sulfur takes the 3: 2 × (3+) cancels 3 × (2−), so chromium(III) sulfide is Cr₂S₃."
 },
 {
  "value": "CrS₃",
  "misconception": "you turned the numeral into sulfur's count directly — balance the charges: 2 × (3+) cancels 3 × (2−)",
  "feedback": "This answer is wrong — you turned the numeral into sulfur's count directly. Balance the charges: 2 × (3+) cancels 3 × (2−), so chromium(III) sulfide is Cr₂S₃."
 }
]

{
 "keyExplanation": "Chromium(III) is Cr³⁺ and sulfide is S²⁻. Two 3+ ions give 6+ and three 2− ions give 6−: Cr₂S₃. Check: 2 × (3+) + 3 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Cr₃S₂",
   "misconception": "you swapped the counts — chromium takes the 2 and sulfur takes the 3",
   "feedback": "This answer is wrong — you swapped the counts. Chromium takes the 2 and sulfur takes the 3: 2 × (3+) cancels 3 × (2−), so chromium(III) sulfide is Cr₂S₃."
  },
  {
   "value": "CrS₃",
   "misconception": "you turned the numeral into sulfur's count directly — balance the charges: 2 × (3+) cancels 3 × (2−)",
   "feedback": "This answer is wrong — you turned the numeral into sulfur's count directly. Balance the charges: 2 × (3+) cancels 3 × (2−), so chromium(III) sulfide is Cr₂S₃."
  }
 ]
}

[
 {
  "value": "Cu₂I₂",
  "misconception": "you doubled a pair that already cancels — one 1+ ion cancels one 1− ion",
  "feedback": "This answer is wrong — you doubled a pair that already cancels. Copper(I) is Cu⁺ and iodide is I⁻, so one 1+ ion cancels one 1− ion and the ratio is already 1:1: CuI. Check: (1+) + (1−) = 0."
 },
 {
  "value": "CuI₂",
  "misconception": "you gave copper a 2+ charge — the numeral (I) says this copper ion is 1+",
  "feedback": "This answer is wrong — you gave copper a 2+ charge. The numeral (I) says this copper ion is 1+, and one 1+ ion cancels one 1− iodide exactly: CuI. Check: (1+) + (1−) = 0."
 }
]

{
 "keyExplanation": "Copper(I) is Cu⁺ and iodide is I⁻. One 1+ cancels one 1−: CuI. Check: (1+) + (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Cu₂I₂",
   "misconception": "you doubled a pair that already cancels — one 1+ ion cancels one 1− ion",
   "feedback": "This answer is wrong — you doubled a pair that already cancels. Copper(I) is Cu⁺ and iodide is I⁻, so one 1+ ion cancels one 1− ion and the ratio is already 1:1: CuI. Check: (1+) + (1−) = 0."
  },
  {
   "value": "CuI₂",
   "misconception": "you gave copper a 2+ charge — the numeral (I) says this copper ion is 1+",
   "feedback": "This answer is wrong — you gave copper a 2+ charge. The numeral (I) says this copper ion is 1+, and one 1+ ion cancels one 1− iodide exactly: CuI. Check: (1+) + (1−) = 0."
  }
 ]
}

[
 {
  "value": "KCO₃",
  "misconception": "you dropped the hydrogen — hydrogen carbonate is the whole block HCO₃⁻",
  "feedback": "This answer is wrong — you dropped the hydrogen. Hydrogen carbonate is the whole block HCO₃⁻, and one K⁺ balances its 1− charge: KHCO₃. Check: (1+) + (1−) = 0."
 },
 {
  "value": "K₂HCO₃",
  "misconception": "you used carbonate's 2− charge — hydrogen carbonate carries 1−, so one K⁺ balances it",
  "feedback": "This answer is wrong — you used carbonate's 2− charge. Hydrogen carbonate carries 1−, so one K⁺ balances it: KHCO₃. Check: (1+) + (1−) = 0."
 }
]

{
 "keyExplanation": "Potassium is K⁺ and hydrogen carbonate is HCO₃⁻. One 1+ cancels one 1−: KHCO₃. Check: (1+) + (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "KCO₃",
   "misconception": "you dropped the hydrogen — hydrogen carbonate is the whole block HCO₃⁻",
   "feedback": "This answer is wrong — you dropped the hydrogen. Hydrogen carbonate is the whole block HCO₃⁻, and one K⁺ balances its 1− charge: KHCO₃. Check: (1+) + (1−) = 0."
  },
  {
   "value": "K₂HCO₃",
   "misconception": "you used carbonate's 2− charge — hydrogen carbonate carries 1−, so one K⁺ balances it",
   "feedback": "This answer is wrong — you used carbonate's 2− charge. Hydrogen carbonate carries 1−, so one K⁺ balances it: KHCO₃. Check: (1+) + (1−) = 0."
  }
 ]
}

[
 {
  "value": "AlOH",
  "misconception": "you wrote a 1:1 ratio — one 3+ ion needs three 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 3+ ion needs three 1− ions. Aluminum is Al³⁺ and hydroxide is OH⁻, and the repeated hydroxide goes in parentheses: Al(OH)₃. Check: (3+) + 3 × (1−) = 0."
 },
 {
  "value": "AlOH₃",
  "misconception": "you left out the parentheses — AlOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions",
  "feedback": "This answer is wrong — you left out the parentheses. AlOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions. One Al³⁺ needs three OH⁻ ions, each kept whole: Al(OH)₃. Check: (3+) + 3 × (1−) = 0."
 }
]

{
 "keyExplanation": "Aluminum is Al³⁺ and hydroxide is OH⁻. One 3+ ion needs three 1− ions, and hydroxide repeats, so parentheses: Al(OH)₃. Check: (3+) + 3 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "AlOH",
   "misconception": "you wrote a 1:1 ratio — one 3+ ion needs three 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 3+ ion needs three 1− ions. Aluminum is Al³⁺ and hydroxide is OH⁻, and the repeated hydroxide goes in parentheses: Al(OH)₃. Check: (3+) + 3 × (1−) = 0."
  },
  {
   "value": "AlOH₃",
   "misconception": "you left out the parentheses — AlOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions",
   "feedback": "This answer is wrong — you left out the parentheses. AlOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions. One Al³⁺ needs three OH⁻ ions, each kept whole: Al(OH)₃. Check: (3+) + 3 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "copper nitrate trihydrate",
  "misconception": "you left out the Roman numeral — two NO₃⁻ ions total 2−, so the copper is 2+",
  "feedback": "This answer is wrong — you left out the Roman numeral. Two NO₃⁻ ions total 2−, so the copper ion is 2+ and the name must show it: copper(II) nitrate. Three waters: trihydrate. Cu(NO₃)₂·3H₂O is copper(II) nitrate trihydrate."
 },
 {
  "value": "copper(II) nitrate dihydrate",
  "misconception": "you counted the nitrates for the prefix — the prefix counts the three waters",
  "feedback": "This answer is wrong — you counted the nitrates for the prefix. The prefix counts water molecules only: three waters means trihydrate. Cu(NO₃)₂·3H₂O is copper(II) nitrate trihydrate."
 }
]

{
 "keyExplanation": "Two nitrate ions total 2−, so the copper ion is 2+ — copper(II) nitrate. Three waters: trihydrate. Cu(NO₃)₂·3H₂O is copper(II) nitrate trihydrate.",
 "anticipatedWrong": [
  {
   "value": "copper nitrate trihydrate",
   "misconception": "you left out the Roman numeral — two NO₃⁻ ions total 2−, so the copper is 2+",
   "feedback": "This answer is wrong — you left out the Roman numeral. Two NO₃⁻ ions total 2−, so the copper ion is 2+ and the name must show it: copper(II) nitrate. Three waters: trihydrate. Cu(NO₃)₂·3H₂O is copper(II) nitrate trihydrate."
  },
  {
   "value": "copper(II) nitrate dihydrate",
   "misconception": "you counted the nitrates for the prefix — the prefix counts the three waters",
   "feedback": "This answer is wrong — you counted the nitrates for the prefix. The prefix counts water molecules only: three waters means trihydrate. Cu(NO₃)₂·3H₂O is copper(II) nitrate trihydrate."
  }
 ]
}

[
 {
  "value": "Ni₂SO₄·6H₂O",
  "misconception": "you turned the numeral into a subscript — the (II) is the nickel ion's 2+ charge, which one SO₄²⁻ cancels",
  "feedback": "This answer is wrong — you turned the numeral into a subscript. The (II) is the nickel ion's 2+ charge, which one SO₄²⁻ cancels: the ionic part is NiSO₄. Hexa- means 6 waters: NiSO₄·6H₂O."
 },
 {
  "value": "NiSO₄",
  "misconception": "you dropped the water — hexahydrate adds ·6H₂O after the compound",
  "feedback": "This answer is wrong — you dropped the water. Hexahydrate adds ·6H₂O after the compound: NiSO₄·6H₂O."
 }
]

{
 "keyExplanation": "Nickel(II) is Ni²⁺ and sulfate is SO₄²⁻ — one 2+ cancels one 2−, so NiSO₄. Hexa- means 6 waters: NiSO₄·6H₂O.",
 "anticipatedWrong": [
  {
   "value": "Ni₂SO₄·6H₂O",
   "misconception": "you turned the numeral into a subscript — the (II) is the nickel ion's 2+ charge, which one SO₄²⁻ cancels",
   "feedback": "This answer is wrong — you turned the numeral into a subscript. The (II) is the nickel ion's 2+ charge, which one SO₄²⁻ cancels: the ionic part is NiSO₄. Hexa- means 6 waters: NiSO₄·6H₂O."
  },
  {
   "value": "NiSO₄",
   "misconception": "you dropped the water — hexahydrate adds ·6H₂O after the compound",
   "feedback": "This answer is wrong — you dropped the water. Hexahydrate adds ·6H₂O after the compound: NiSO₄·6H₂O."
  }
 ]
}

[
 {
  "value": "nitrogen hydride nitrate",
  "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
  "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so NH₄NO₃ is ammonium nitrate."
 },
 {
  "value": "ammonium nitride",
  "misconception": "you used -ide for a polyatomic ion — nitride is the bare N ion; NO₃⁻ is the polyatomic nitrate, which keeps its roster name",
  "feedback": "This answer is wrong — you used -ide for a polyatomic ion. Nitride is the bare N ion; the NO₃⁻ block here is the polyatomic nitrate, which keeps its roster name, so NH₄NO₃ is ammonium nitrate."
 }
]

{
 "keyExplanation": "NH₄ is the ammonium ion and NO₃ is the nitrate ion — both polyatomic, both keeping their own names: ammonium nitrate.",
 "anticipatedWrong": [
  {
   "value": "nitrogen hydride nitrate",
   "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
   "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so NH₄NO₃ is ammonium nitrate."
  },
  {
   "value": "ammonium nitride",
   "misconception": "you used -ide for a polyatomic ion — nitride is the bare N ion; NO₃⁻ is the polyatomic nitrate, which keeps its roster name",
   "feedback": "This answer is wrong — you used -ide for a polyatomic ion. Nitride is the bare N ion; the NO₃⁻ block here is the polyatomic nitrate, which keeps its roster name, so NH₄NO₃ is ammonium nitrate."
  }
 ]
}

[
 {
  "value": "lithium phosphide",
  "misconception": "you used the single-atom ion's name — phosphide is bare P³⁻; the block PO₄ is phosphate",
  "feedback": "This answer is wrong — you used the single-atom ion's name. Phosphide is the bare P³⁻ ion; the block PO₄ in Li₃PO₄ is the polyatomic phosphate ion, and it keeps its own name: lithium phosphate."
 },
 {
  "value": "lithium(I) phosphate",
  "misconception": "you added a numeral — lithium, in Group 1, is fixed-charge",
  "feedback": "This answer is wrong — you added a numeral. Lithium, in Group 1, is a fixed-charge metal, so its name never carries a Roman numeral: Li₃PO₄ is lithium phosphate."
 }
]

{
 "keyExplanation": "No dot; lithium is a fixed-charge Group 1 metal, and the block PO₄ is the polyatomic phosphate ion: lithium phosphate.",
 "anticipatedWrong": [
  {
   "value": "lithium phosphide",
   "misconception": "you used the single-atom ion's name — phosphide is bare P³⁻; the block PO₄ is phosphate",
   "feedback": "This answer is wrong — you used the single-atom ion's name. Phosphide is the bare P³⁻ ion; the block PO₄ in Li₃PO₄ is the polyatomic phosphate ion, and it keeps its own name: lithium phosphate."
  },
  {
   "value": "lithium(I) phosphate",
   "misconception": "you added a numeral — lithium, in Group 1, is fixed-charge",
   "feedback": "This answer is wrong — you added a numeral. Lithium, in Group 1, is a fixed-charge metal, so its name never carries a Roman numeral: Li₃PO₄ is lithium phosphate."
  }
 ]
}

[
 {
  "value": "FeOH₃",
  "misconception": "you left out the parentheses — FeOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions",
  "feedback": "This answer is wrong — you left out the parentheses. FeOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions: iron(III) hydroxide is Fe(OH)₃. Check: (3+) + 3 × (1−) = 0."
 },
 {
  "value": "FeOH",
  "misconception": "you wrote a 1:1 ratio — the 3+ iron ion needs three 1− hydroxide ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. The 3+ iron ion needs three 1− hydroxide ions, and the repeating hydroxide is wrapped in parentheses: Fe(OH)₃. Check: (3+) + 3 × (1−) = 0."
 }
]

{
 "keyExplanation": "Iron(III) is Fe³⁺ and hydroxide is OH⁻. One 3+ ion needs three 1− ions, and hydroxide repeats, so parentheses: Fe(OH)₃. Check: (3+) + 3 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "FeOH₃",
   "misconception": "you left out the parentheses — FeOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions",
   "feedback": "This answer is wrong — you left out the parentheses. FeOH₃ would read as one oxygen and three hydrogens, not three hydroxide ions: iron(III) hydroxide is Fe(OH)₃. Check: (3+) + 3 × (1−) = 0."
  },
  {
   "value": "FeOH",
   "misconception": "you wrote a 1:1 ratio — the 3+ iron ion needs three 1− hydroxide ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. The 3+ iron ion needs three 1− hydroxide ions, and the repeating hydroxide is wrapped in parentheses: Fe(OH)₃. Check: (3+) + 3 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "cobalt(II) sulfate",
  "misconception": "you dropped the water word — the raised dot must show up as a prefix plus 'hydrate'",
  "feedback": "This answer is wrong — you dropped the water word. The raised dot must show up as a prefix plus 'hydrate': seven waters make it heptahydrate, so CoSO₄·7H₂O is cobalt(II) sulfate heptahydrate."
 },
 {
  "value": "cobalt sulfate heptahydrate",
  "misconception": "you skipped the numeral — one SO₄²⁻ makes the cobalt 2+",
  "feedback": "This answer is wrong — you skipped the numeral. Cobalt is a transition metal, and one SO₄²⁻ makes the cobalt 2+, so the name must show it: CoSO₄·7H₂O is cobalt(II) sulfate heptahydrate."
 }
]

{
 "keyExplanation": "A raised dot: name the part before it first. One sulfate ion is 2−, so the cobalt is 2+ — cobalt(II) sulfate. Seven waters: heptahydrate. CoSO₄·7H₂O is cobalt(II) sulfate heptahydrate.",
 "anticipatedWrong": [
  {
   "value": "cobalt(II) sulfate",
   "misconception": "you dropped the water word — the raised dot must show up as a prefix plus 'hydrate'",
   "feedback": "This answer is wrong — you dropped the water word. The raised dot must show up as a prefix plus 'hydrate': seven waters make it heptahydrate, so CoSO₄·7H₂O is cobalt(II) sulfate heptahydrate."
  },
  {
   "value": "cobalt sulfate heptahydrate",
   "misconception": "you skipped the numeral — one SO₄²⁻ makes the cobalt 2+",
   "feedback": "This answer is wrong — you skipped the numeral. Cobalt is a transition metal, and one SO₄²⁻ makes the cobalt 2+, so the name must show it: CoSO₄·7H₂O is cobalt(II) sulfate heptahydrate."
  }
 ]
}

[
 {
  "value": "aluminum sulfate",
  "misconception": "you used the polyatomic ion's name — sulfate is SO₄²⁻; a bare S ion is sulfide",
  "feedback": "This answer is wrong — you used the polyatomic ion's name. Sulfate is SO₄²⁻; the bare S ion in Al₂S₃ is sulfide, so the name is aluminum sulfide."
 },
 {
  "value": "dialuminum trisulfide",
  "misconception": "you added counting words — the 2 and 3 never enter the name",
  "feedback": "This answer is wrong — you added counting words. Ionic names never count atoms — the 2 and 3 are recovered from the charges whenever the formula is needed — so Al₂S₃ is just aluminum sulfide."
 }
]

{
 "keyExplanation": "Metal first, unchanged: aluminum. Nonmetal with -ide: sulfur becomes sulfide. Al₂S₃ is aluminum sulfide — the subscripts stay out of the name.",
 "anticipatedWrong": [
  {
   "value": "aluminum sulfate",
   "misconception": "you used the polyatomic ion's name — sulfate is SO₄²⁻; a bare S ion is sulfide",
   "feedback": "This answer is wrong — you used the polyatomic ion's name. Sulfate is SO₄²⁻; the bare S ion in Al₂S₃ is sulfide, so the name is aluminum sulfide."
  },
  {
   "value": "dialuminum trisulfide",
   "misconception": "you added counting words — the 2 and 3 never enter the name",
   "feedback": "This answer is wrong — you added counting words. Ionic names never count atoms — the 2 and 3 are recovered from the charges whenever the formula is needed — so Al₂S₃ is just aluminum sulfide."
  }
 ]
}

[
 "The 3 after the raised dot counts whole water molecules.",
 "Each water molecule H₂O carries 2 hydrogen atoms: 3 × 2 = 6.",
 "LiI·3H₂O contains 6 hydrogen atoms."
]

[
 {
  "value": 3,
  "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 3 waters carries 2 hydrogens",
  "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 3 waters carries 2 hydrogens. The 3 after the dot counts whole water molecules: 3 × 2 = 6 hydrogen atoms."
 },
 {
  "value": 2,
  "misconception": "you counted one water molecule's hydrogens — the 3 in front multiplies the whole H₂O",
  "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 3 in front multiplies the whole H₂O. Three water molecules each carry 2 hydrogen atoms: 3 × 2 = 6 hydrogen atoms."
 },
 {
  "value": 9,
  "misconception": "you counted every atom in the water molecules — each water holds 2 hydrogens and 1 oxygen, but the question asks for hydrogen only",
  "feedback": "This answer is wrong — you counted every atom in the water molecules, but the question asks for hydrogen only. Each of the 3 waters carries 2 hydrogen atoms: 3 × 2 = 6 hydrogen atoms."
 }
]

{
 "keyExplanation": "The 3 after the dot counts whole water molecules, and each water molecule has 2 hydrogen atoms: 3 × 2 = 6 hydrogen atoms.",
 "anticipatedWrong": [
  {
   "value": 3,
   "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 3 waters carries 2 hydrogens",
   "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 3 waters carries 2 hydrogens. The 3 after the dot counts whole water molecules: 3 × 2 = 6 hydrogen atoms."
  },
  {
   "value": 2,
   "misconception": "you counted one water molecule's hydrogens — the 3 in front multiplies the whole H₂O",
   "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 3 in front multiplies the whole H₂O. Three water molecules each carry 2 hydrogen atoms: 3 × 2 = 6 hydrogen atoms."
  },
  {
   "value": 9,
   "misconception": "you counted every atom in the water molecules — each water holds 2 hydrogens and 1 oxygen, but the question asks for hydrogen only",
   "feedback": "This answer is wrong — you counted every atom in the water molecules, but the question asks for hydrogen only. Each of the 3 waters carries 2 hydrogen atoms: 3 × 2 = 6 hydrogen atoms."
  }
 ]
}

[
 {
  "value": "sodium carbonate",
  "misconception": "you dropped the hydrogen from the block — HCO₃ is hydrogen carbonate, not carbonate",
  "feedback": "This answer is wrong — you dropped the hydrogen from the block. HCO₃ is hydrogen carbonate, not carbonate, so NaHCO₃ is sodium hydrogen carbonate."
 },
 {
  "value": "sodium hydrocarbonate",
  "misconception": "you merged the ion's name — it is 'hydrogen carbonate', two separate words",
  "feedback": "This answer is wrong — you merged the ion's name. It is 'hydrogen carbonate', two separate words: NaHCO₃ is sodium hydrogen carbonate."
 }
]

{
 "keyExplanation": "The block HCO₃ is the hydrogen carbonate ion, and it keeps its full name: NaHCO₃ is sodium hydrogen carbonate.",
 "anticipatedWrong": [
  {
   "value": "sodium carbonate",
   "misconception": "you dropped the hydrogen from the block — HCO₃ is hydrogen carbonate, not carbonate",
   "feedback": "This answer is wrong — you dropped the hydrogen from the block. HCO₃ is hydrogen carbonate, not carbonate, so NaHCO₃ is sodium hydrogen carbonate."
  },
  {
   "value": "sodium hydrocarbonate",
   "misconception": "you merged the ion's name — it is 'hydrogen carbonate', two separate words",
   "feedback": "This answer is wrong — you merged the ion's name. It is 'hydrogen carbonate', two separate words: NaHCO₃ is sodium hydrogen carbonate."
  }
 ]
}

[
 {
  "value": "Ba₂O₂",
  "misconception": "you skipped the reduce step — criss-cross gives Ba₂O₂, and 2:2 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ba₂O₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: BaO. Check: (2+) + (2−) = 0."
 },
 {
  "value": "BaO₂",
  "misconception": "you gave oxygen a 1− charge — oxide, from Group 16, is O²⁻, so one 2+ cancels one 2−",
  "feedback": "This answer is wrong — you gave oxygen a 1− charge. Oxide comes from Group 16, so it is O²⁻, and one Ba²⁺ cancels one O²⁻: BaO. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Barium is Ba²⁺ (Group 2) and oxide is O²⁻ (Group 16). One 2+ cancels one 2−: BaO. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ba₂O₂",
   "misconception": "you skipped the reduce step — criss-cross gives Ba₂O₂, and 2:2 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ba₂O₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: BaO. Check: (2+) + (2−) = 0."
  },
  {
   "value": "BaO₂",
   "misconception": "you gave oxygen a 1− charge — oxide, from Group 16, is O²⁻, so one 2+ cancels one 2−",
   "feedback": "This answer is wrong — you gave oxygen a 1− charge. Oxide comes from Group 16, so it is O²⁻, and one Ba²⁺ cancels one O²⁻: BaO. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "1+",
  "misconception": "you stopped at one fluoride's charge - three F⁻ ions total 3−",
  "feedback": "This answer is wrong - you stopped at one fluoride's charge. The three F⁻ ions together total 3−, so the cobalt ion must contribute 3+, written 3+."
 },
 {
  "value": "3−",
  "misconception": "you flipped the sign - the metal ion is the positive one",
  "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the three F⁻ ions total 3−, and the cobalt ion balances them with 3+, written 3+."
 }
]

{
 "keyExplanation": "Fluoride, from Group 17, is F⁻; three of them total 3−. One cobalt ion must contribute 3+. Check: (3+) + 3 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "1+",
   "misconception": "you stopped at one fluoride's charge - three F⁻ ions total 3−",
   "feedback": "This answer is wrong - you stopped at one fluoride's charge. The three F⁻ ions together total 3−, so the cobalt ion must contribute 3+, written 3+.",
   "check": {
    "expr": "1*1",
    "expected": 1
   }
  },
  {
   "value": "3−",
   "misconception": "you flipped the sign - the metal ion is the positive one",
   "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the three F⁻ ions total 3−, and the cobalt ion balances them with 3+, written 3+.",
   "check": {
    "expr": "-3",
    "expected": -3
   }
  }
 ]
}

[
 {
  "value": "2+",
  "misconception": "you forgot to share the total — 2+ is split between two silver ions",
  "feedback": "This answer is wrong — you forgot to share the total. One O²⁻ makes the total positive charge 2+, and that 2+ is split between two silver ions, so each silver ion is 1+."
 },
 {
  "value": "1-",
  "misconception": "you flipped the sign — the metal ion balances the negative oxide, so its charge is positive",
  "feedback": "This answer is wrong — you flipped the sign. The oxide ion carries the negative charge (2−); the two silver ions supply the balancing POSITIVE charge, 1+ each. Check: 2 × (1+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Oxide, from Group 16, is O²⁻; one of them totals 2−. The total positive charge must be 2+, shared by two silver ions: 2+ ÷ 2 = 1+ each. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "2+",
   "misconception": "you forgot to share the total — 2+ is split between two silver ions",
   "feedback": "This answer is wrong — you forgot to share the total. One O²⁻ makes the total positive charge 2+, and that 2+ is split between two silver ions, so each silver ion is 1+."
  },
  {
   "value": "1-",
   "misconception": "you flipped the sign — the metal ion balances the negative oxide, so its charge is positive",
   "feedback": "This answer is wrong — you flipped the sign. The oxide ion carries the negative charge (2−); the two silver ions supply the balancing POSITIVE charge, 1+ each. Check: 2 × (1+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "nickel(I) sulfide",
  "misconception": "you gave the wrong charge — one S²⁻ needs one 2+ nickel ion",
  "feedback": "This answer is wrong — you gave the wrong charge. Sulfide is S²⁻, so the single nickel ion must be 2+, not 1+: NiS is nickel(II) sulfide."
 },
 {
  "value": "nickel sulfide",
  "misconception": "you left out the numeral — the name must show the metal ion's charge",
  "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: one S²⁻ needs one 2+ nickel ion, so NiS is nickel(II) sulfide."
 }
]

{
 "keyExplanation": "Sulfide is S²⁻, so the single nickel ion must be 2+ — nickel(II). NiS is nickel(II) sulfide.",
 "anticipatedWrong": [
  {
   "value": "nickel(I) sulfide",
   "misconception": "you gave the wrong charge — one S²⁻ needs one 2+ nickel ion",
   "feedback": "This answer is wrong — you gave the wrong charge. Sulfide is S²⁻, so the single nickel ion must be 2+, not 1+: NiS is nickel(II) sulfide."
  },
  {
   "value": "nickel sulfide",
   "misconception": "you left out the numeral — the name must show the metal ion's charge",
   "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: one S²⁻ needs one 2+ nickel ion, so NiS is nickel(II) sulfide."
  }
 ]
}

[
 {
  "value": "manganese(II) oxide",
  "misconception": "you used the manganese subscript as the numeral — the 2 counts atoms; the numeral is the charge on one metal ion",
  "feedback": "This answer is wrong — you used the manganese subscript as the numeral. The 2 counts manganese atoms; three O²⁻ add up to 6−, and the two manganese ions split that 6+ between them as 3+ each, so Mn₂O₃ is manganese(III) oxide."
 },
 {
  "value": "manganese oxide",
  "misconception": "you left out the numeral — manganese forms more than one ion, so the name must show which one",
  "feedback": "This answer is wrong — you left out the numeral. Manganese forms more than one ion, so the name must show which one: here the balance makes each manganese ion 3+, and Mn₂O₃ is manganese(III) oxide."
 }
]

{
 "keyExplanation": "Three O²⁻ ions add up to 6−. Two manganese ions split that 6+ balance between them, so each one is 3+ — manganese(III). Mn₂O₃ is manganese(III) oxide.",
 "anticipatedWrong": [
  {
   "value": "manganese(II) oxide",
   "misconception": "you used the manganese subscript as the numeral — the 2 counts atoms; the numeral is the charge on one metal ion",
   "feedback": "This answer is wrong — you used the manganese subscript as the numeral. The 2 counts manganese atoms; three O²⁻ add up to 6−, and the two manganese ions split that 6+ between them as 3+ each, so Mn₂O₃ is manganese(III) oxide."
  },
  {
   "value": "manganese oxide",
   "misconception": "you left out the numeral — manganese forms more than one ion, so the name must show which one",
   "feedback": "This answer is wrong — you left out the numeral. Manganese forms more than one ion, so the name must show which one: here the balance makes each manganese ion 3+, and Mn₂O₃ is manganese(III) oxide."
  }
 ]
}

[
 {
  "value": "Ni₂F",
  "misconception": "you turned the numeral into nickel's subscript — the II is the charge on one nickel ion",
  "feedback": "This answer is wrong — you turned the numeral into nickel's subscript. The II is the charge on one nickel ion: one 2+ ion needs two 1− ions, so nickel(II) fluoride is NiF₂."
 },
 {
  "value": "NiF",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2+ ion needs two 1− ions, so nickel(II) fluoride is NiF₂."
 }
]

{
 "keyExplanation": "Nickel(II) is Ni²⁺ and fluoride is F⁻. One 2+ ion needs two 1− ions: NiF₂. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ni₂F",
   "misconception": "you turned the numeral into nickel's subscript — the II is the charge on one nickel ion",
   "feedback": "This answer is wrong — you turned the numeral into nickel's subscript. The II is the charge on one nickel ion: one 2+ ion needs two 1− ions, so nickel(II) fluoride is NiF₂."
  },
  {
   "value": "NiF",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2+ ion needs two 1− ions, so nickel(II) fluoride is NiF₂."
  }
 ]
}

[
 {
  "value": "Ti₂O₄",
  "misconception": "you skipped the reduce step — criss-cross gives Ti₂O₄, and 2:4 reduces to 1:2",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ti₂O₄, but a formula shows the simplest whole-number ratio of ions, and 2:4 reduces to 1:2: TiO₂. Check: (4+) + 2 × (2−) = 0."
 },
 {
  "value": "TiO₄",
  "misconception": "you turned the numeral into oxygen's count — the IV is the titanium ion's 4+ charge, and 4+ needs two 2− ions",
  "feedback": "This answer is wrong — you turned the numeral into oxygen's count. The IV is the titanium ion's 4+ charge, not a subscript: oxide is O²⁻, and one 4+ ion needs two 2− ions: TiO₂. Check: (4+) + 2 × (2−) = 0."
 }
]

{
 "keyExplanation": "Titanium(IV) is Ti⁴⁺ and oxide is O²⁻. One 4+ ion needs two 2− ions (criss-cross gives Ti₂O₄, reduced to TiO₂). Check: (4+) + 2 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ti₂O₄",
   "misconception": "you skipped the reduce step — criss-cross gives Ti₂O₄, and 2:4 reduces to 1:2",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ti₂O₄, but a formula shows the simplest whole-number ratio of ions, and 2:4 reduces to 1:2: TiO₂. Check: (4+) + 2 × (2−) = 0."
  },
  {
   "value": "TiO₄",
   "misconception": "you turned the numeral into oxygen's count — the IV is the titanium ion's 4+ charge, and 4+ needs two 2− ions",
   "feedback": "This answer is wrong — you turned the numeral into oxygen's count. The IV is the titanium ion's 4+ charge, not a subscript: oxide is O²⁻, and one 4+ ion needs two 2− ions: TiO₂. Check: (4+) + 2 × (2−) = 0."
  }
 ]
}

[
 {
  "value": "Sr₂(CO₃)₂",
  "misconception": "you skipped the reduce step — 2+ and 2− cancel one for one, so no subscripts are needed",
  "feedback": "This answer is wrong — you skipped the reduce step. A formula shows the simplest whole-number ratio of ions, and 2+ and 2− cancel one for one, so no subscripts are needed: SrCO₃. Check: (2+) + (2−) = 0."
 },
 {
  "value": "Sr(CO₃)₂",
  "misconception": "you doubled only the carbonate — one 2+ ion already cancels one 2− ion",
  "feedback": "This answer is wrong — you doubled only the carbonate. One 2+ ion already cancels one 2− ion, so the ratio is 1:1 and the single carbonate needs no parentheses: SrCO₃. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Strontium is Sr²⁺ and carbonate is CO₃²⁻. One 2+ cancels one 2−, and the single carbonate block needs no parentheses: SrCO₃. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Sr₂(CO₃)₂",
   "misconception": "you skipped the reduce step — 2+ and 2− cancel one for one, so no subscripts are needed",
   "feedback": "This answer is wrong — you skipped the reduce step. A formula shows the simplest whole-number ratio of ions, and 2+ and 2− cancel one for one, so no subscripts are needed: SrCO₃. Check: (2+) + (2−) = 0."
  },
  {
   "value": "Sr(CO₃)₂",
   "misconception": "you doubled only the carbonate — one 2+ ion already cancels one 2− ion",
   "feedback": "This answer is wrong — you doubled only the carbonate. One 2+ ion already cancels one 2− ion, so the ratio is 1:1 and the single carbonate needs no parentheses: SrCO₃. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "Sr₂(PO₄)₃",
  "misconception": "you swapped the counts — strontium takes the 3 and phosphate takes the 2",
  "feedback": "This answer is wrong — you swapped the counts. Three 2+ ions give 6+ and two 3− ions give 6−, so strontium takes the 3 and phosphate takes the 2: Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
 },
 {
  "value": "SrPO₄",
  "misconception": "you wrote a 1:1 ratio — 2+ and 3− do not cancel one for one",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but 2+ and 3− do not cancel one for one. Three 2+ ions give 6+ and two 3− ions give 6−, and phosphate repeats, so parentheses: Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
 }
]

{
 "keyExplanation": "The strontium ion is Sr²⁺, and phosphate is PO₄³⁻. Three 2+ ions give 6+ and two 3− ions give 6−, and phosphate repeats, so parentheses: Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Sr₂(PO₄)₃",
   "misconception": "you swapped the counts — strontium takes the 3 and phosphate takes the 2",
   "feedback": "This answer is wrong — you swapped the counts. Three 2+ ions give 6+ and two 3− ions give 6−, so strontium takes the 3 and phosphate takes the 2: Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
  },
  {
   "value": "SrPO₄",
   "misconception": "you wrote a 1:1 ratio — 2+ and 3− do not cancel one for one",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but 2+ and 3− do not cancel one for one. Three 2+ ions give 6+ and two 3− ions give 6−, and phosphate repeats, so parentheses: Sr₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
  }
 ]
}

[
 {
  "value": "iron(III) sulfate heptahydrate",
  "misconception": "you gave iron a 3+ numeral — one SO₄²⁻ needs one 2+ iron ion",
  "feedback": "This answer is wrong — you gave iron a 3+ numeral. One SO₄²⁻ ion is 2−, so it needs one 2+ iron ion to balance: iron(II) sulfate. Seven waters: heptahydrate. FeSO₄·7H₂O is iron(II) sulfate heptahydrate."
 },
 {
  "value": "iron sulfate heptahydrate",
  "misconception": "you left out the Roman numeral — iron forms more than one ion",
  "feedback": "This answer is wrong — you left out the Roman numeral. Iron forms more than one ion, so the name must show the charge: one SO₄²⁻ is 2−, so the iron ion is 2+ — iron(II) sulfate. Seven waters: heptahydrate. FeSO₄·7H₂O is iron(II) sulfate heptahydrate."
 }
]

{
 "keyExplanation": "One sulfate ion is 2−, so the iron ion is 2+ — iron(II) sulfate. Seven waters: heptahydrate. FeSO₄·7H₂O is iron(II) sulfate heptahydrate.",
 "anticipatedWrong": [
  {
   "value": "iron(III) sulfate heptahydrate",
   "misconception": "you gave iron a 3+ numeral — one SO₄²⁻ needs one 2+ iron ion",
   "feedback": "This answer is wrong — you gave iron a 3+ numeral. One SO₄²⁻ ion is 2−, so it needs one 2+ iron ion to balance: iron(II) sulfate. Seven waters: heptahydrate. FeSO₄·7H₂O is iron(II) sulfate heptahydrate."
  },
  {
   "value": "iron sulfate heptahydrate",
   "misconception": "you left out the Roman numeral — iron forms more than one ion",
   "feedback": "This answer is wrong — you left out the Roman numeral. Iron forms more than one ion, so the name must show the charge: one SO₄²⁻ is 2−, so the iron ion is 2+ — iron(II) sulfate. Seven waters: heptahydrate. FeSO₄·7H₂O is iron(II) sulfate heptahydrate."
  }
 ]
}

[
 {
  "value": "AlNO₃·9H₂O",
  "misconception": "you wrote a 1:1 ratio in the ionic part — Al³⁺ needs three NO₃⁻ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio in the ionic part. Al³⁺ needs three NO₃⁻ ions before the charges cancel, wrapped in parentheses: Al(NO₃)₃. Nona- means 9 waters: Al(NO₃)₃·9H₂O."
 },
 {
  "value": "Al(NO₃)₃·4H₂O",
  "misconception": "you read nona- as 4 — nona- means 9; tetra- is 4",
  "feedback": "This answer is wrong — you read nona- as 4. Nona- means 9 waters (tetra- is 4): Al(NO₃)₃·9H₂O."
 }
]

{
 "keyExplanation": "Aluminum is Al³⁺ and nitrate is NO₃⁻, so the ionic part is Al(NO₃)₃ — check: (3+) + 3 × (1−) = 0. Nona- means 9 waters: Al(NO₃)₃·9H₂O.",
 "anticipatedWrong": [
  {
   "value": "AlNO₃·9H₂O",
   "misconception": "you wrote a 1:1 ratio in the ionic part — Al³⁺ needs three NO₃⁻ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio in the ionic part. Al³⁺ needs three NO₃⁻ ions before the charges cancel, wrapped in parentheses: Al(NO₃)₃. Nona- means 9 waters: Al(NO₃)₃·9H₂O."
  },
  {
   "value": "Al(NO₃)₃·4H₂O",
   "misconception": "you read nona- as 4 — nona- means 9; tetra- is 4",
   "feedback": "This answer is wrong — you read nona- as 4. Nona- means 9 waters (tetra- is 4): Al(NO₃)₃·9H₂O."
  }
 ]
}

[
 {
  "value": "nitrogen hydride iodide",
  "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
  "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so NH₄I is ammonium iodide."
 },
 {
  "value": "ammonium iodate",
  "misconception": "you gave the bare ion an -ate name — there is no oxygen block here; a lone iodine is the single-atom I⁻ ion, iodide",
  "feedback": "This answer is wrong — you gave the bare ion an -ate name. There is no oxygen block here: a lone iodine is the single-atom I⁻ ion, and the -ide ending is only for single-atom negative ions, so NH₄I is ammonium iodide."
 }
]

{
 "keyExplanation": "NH₄ is the ammonium ion, and it leads the name. Iodine here is a single-atom negative ion, so it takes the -ide ending: NH₄I is ammonium iodide.",
 "anticipatedWrong": [
  {
   "value": "nitrogen hydride iodide",
   "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
   "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so NH₄I is ammonium iodide."
  },
  {
   "value": "ammonium iodate",
   "misconception": "you gave the bare ion an -ate name — there is no oxygen block here; a lone iodine is the single-atom I⁻ ion, iodide",
   "feedback": "This answer is wrong — you gave the bare ion an -ate name. There is no oxygen block here: a lone iodine is the single-atom I⁻ ion, and the -ide ending is only for single-atom negative ions, so NH₄I is ammonium iodide."
  }
 ]
}

[
 {
  "value": "chromium hydroxide",
  "misconception": "you skipped the numeral — three OH⁻ ions total 3−, so the chromium is 3+",
  "feedback": "This answer is wrong — you skipped the numeral. Chromium is a transition metal, so its name must show the charge: three OH⁻ ions total 3−, making the chromium ion 3+ — chromium(III) hydroxide."
 },
 {
  "value": "chromium hydroxide(III)",
  "misconception": "you placed the numeral after the wrong word — it sits right after the metal's name",
  "feedback": "This answer is wrong — you placed the numeral after the wrong word. The Roman numeral sits right after the metal's name, because it is the metal's charge: chromium(III) hydroxide."
 }
]

{
 "keyExplanation": "No dot; chromium is a transition metal, and three OH⁻ ions total 3−, so the chromium ion is 3+ — chromium(III). The polyatomic hydroxide keeps its name: chromium(III) hydroxide.",
 "anticipatedWrong": [
  {
   "value": "chromium hydroxide",
   "misconception": "you skipped the numeral — three OH⁻ ions total 3−, so the chromium is 3+",
   "feedback": "This answer is wrong — you skipped the numeral. Chromium is a transition metal, so its name must show the charge: three OH⁻ ions total 3−, making the chromium ion 3+ — chromium(III) hydroxide."
  },
  {
   "value": "chromium hydroxide(III)",
   "misconception": "you placed the numeral after the wrong word — it sits right after the metal's name",
   "feedback": "This answer is wrong — you placed the numeral after the wrong word. The Roman numeral sits right after the metal's name, because it is the metal's charge: chromium(III) hydroxide."
  }
 ]
}

[
 {
  "value": "MgC₂H₃O₂·4H₂O",
  "misconception": "you wrote a 1:1 ratio in the ionic part — Mg²⁺ needs two acetate ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio in the ionic part. Mg²⁺ needs two acetate ions, wrapped in parentheses: Mg(C₂H₃O₂)₂·4H₂O. Check on the ionic part: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "Mg(C₂H₃O₂)₄·2H₂O",
  "misconception": "you attached the water count to the acetate — tetra- counts water molecules only",
  "feedback": "This answer is wrong — you attached the water count to the acetate. Tetra- counts water molecules only: the ionic part balances as Mg(C₂H₃O₂)₂, and the 4 waters follow the raised dot: Mg(C₂H₃O₂)₂·4H₂O."
 }
]

{
 "keyExplanation": "Magnesium is Mg²⁺ and acetate is C₂H₃O₂⁻, so the ionic part is Mg(C₂H₃O₂)₂ — check: (2+) + 2 × (1−) = 0. Tetra- means 4 waters: Mg(C₂H₃O₂)₂·4H₂O.",
 "anticipatedWrong": [
  {
   "value": "MgC₂H₃O₂·4H₂O",
   "misconception": "you wrote a 1:1 ratio in the ionic part — Mg²⁺ needs two acetate ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio in the ionic part. Mg²⁺ needs two acetate ions, wrapped in parentheses: Mg(C₂H₃O₂)₂·4H₂O. Check on the ionic part: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "Mg(C₂H₃O₂)₄·2H₂O",
   "misconception": "you attached the water count to the acetate — tetra- counts water molecules only",
   "feedback": "This answer is wrong — you attached the water count to the acetate. Tetra- counts water molecules only: the ionic part balances as Mg(C₂H₃O₂)₂, and the 4 waters follow the raised dot: Mg(C₂H₃O₂)₂·4H₂O."
  }
 ]
}

[
 {
  "value": "titanium chloride hexahydrate",
  "misconception": "you skipped the numeral — titanium is a transition metal, and three Cl⁻ ions make it 3+",
  "feedback": "This answer is wrong — you skipped the numeral. Titanium is a transition metal, and three Cl⁻ ions total 3−, so the titanium is 3+ and the name must show it: TiCl₃·6H₂O is titanium(III) chloride hexahydrate."
 },
 {
  "value": "titanium(IV) chloride hexahydrate",
  "misconception": "you gave the wrong charge — three Cl⁻ ions total 3−, so this titanium ion is 3+",
  "feedback": "This answer is wrong — you gave the wrong charge. The numeral must be computed from THIS formula: three Cl⁻ ions total 3−, so the titanium here is 3+, not 4+: TiCl₃·6H₂O is titanium(III) chloride hexahydrate."
 }
]

{
 "keyExplanation": "A raised dot: name the part before it first. Three Cl⁻ ions total 3−, so the titanium is 3+ — titanium(III) chloride. Six waters: hexahydrate. TiCl₃·6H₂O is titanium(III) chloride hexahydrate.",
 "anticipatedWrong": [
  {
   "value": "titanium chloride hexahydrate",
   "misconception": "you skipped the numeral — titanium is a transition metal, and three Cl⁻ ions make it 3+",
   "feedback": "This answer is wrong — you skipped the numeral. Titanium is a transition metal, and three Cl⁻ ions total 3−, so the titanium is 3+ and the name must show it: TiCl₃·6H₂O is titanium(III) chloride hexahydrate."
  },
  {
   "value": "titanium(IV) chloride hexahydrate",
   "misconception": "you gave the wrong charge — three Cl⁻ ions total 3−, so this titanium ion is 3+",
   "feedback": "This answer is wrong — you gave the wrong charge. The numeral must be computed from THIS formula: three Cl⁻ ions total 3−, so the titanium here is 3+, not 4+: TiCl₃·6H₂O is titanium(III) chloride hexahydrate."
  }
 ]
}

[
 {
  "value": "magnesium nitrate",
  "misconception": "you used the polyatomic ion's name — nitrate is NO₃⁻; a bare N ion is nitride",
  "feedback": "This answer is wrong — you used the polyatomic ion's name. Nitrate is NO₃⁻; the bare N ion in Mg₃N₂ is nitride, so the name is magnesium nitride."
 },
 {
  "value": "magnesium nitrogen",
  "misconception": "you kept the element's name — nitrogen's ending changes to -ide",
  "feedback": "This answer is wrong — you kept the element's name. In the compound's name nitrogen's ending changes to -ide: Mg₃N₂ is magnesium nitride."
 }
]

{
 "keyExplanation": "Metal first, unchanged: magnesium. Nonmetal with -ide: nitrogen becomes nitride. Mg₃N₂ is magnesium nitride.",
 "anticipatedWrong": [
  {
   "value": "magnesium nitrate",
   "misconception": "you used the polyatomic ion's name — nitrate is NO₃⁻; a bare N ion is nitride",
   "feedback": "This answer is wrong — you used the polyatomic ion's name. Nitrate is NO₃⁻; the bare N ion in Mg₃N₂ is nitride, so the name is magnesium nitride."
  },
  {
   "value": "magnesium nitrogen",
   "misconception": "you kept the element's name — nitrogen's ending changes to -ide",
   "feedback": "This answer is wrong — you kept the element's name. In the compound's name nitrogen's ending changes to -ide: Mg₃N₂ is magnesium nitride."
  }
 ]
}

[
 "The 4 after the raised dot counts whole water molecules.",
 "Each water molecule H₂O holds 3 atoms — 2 hydrogens and 1 oxygen: 4 × 3 = 12.",
 "The built-in water in MnCl₂·4H₂O contains 12 atoms."
]

[
 {
  "value": 4,
  "misconception": "you counted the water molecules themselves, not their atoms — each of the 4 waters holds 3 atoms",
  "feedback": "This answer is wrong — you counted the water molecules themselves, not their atoms. Each of the 4 waters holds 3 atoms (two hydrogens and one oxygen): 4 × 3 = 12 atoms."
 },
 {
  "value": 3,
  "misconception": "you counted the atoms of one water molecule — the 4 in front multiplies the whole H₂O",
  "feedback": "This answer is wrong — you counted the atoms of one water molecule; the 4 in front multiplies the whole H₂O. Four water molecules each hold 3 atoms: 4 × 3 = 12 atoms."
 },
 {
  "value": 8,
  "misconception": "you counted only the hydrogen atoms — each water molecule also carries one oxygen atom",
  "feedback": "This answer is wrong — you counted only the hydrogen atoms. Each water molecule holds 3 atoms — two hydrogens AND one oxygen — so the 4 waters contribute 4 × 3 = 12 atoms."
 }
]

{
 "keyExplanation": "The 4 after the dot counts whole water molecules, and each water molecule holds 3 atoms — two hydrogens and one oxygen: 4 × 3 = 12 atoms.",
 "anticipatedWrong": [
  {
   "value": 4,
   "misconception": "you counted the water molecules themselves, not their atoms — each of the 4 waters holds 3 atoms",
   "feedback": "This answer is wrong — you counted the water molecules themselves, not their atoms. Each of the 4 waters holds 3 atoms (two hydrogens and one oxygen): 4 × 3 = 12 atoms."
  },
  {
   "value": 3,
   "misconception": "you counted the atoms of one water molecule — the 4 in front multiplies the whole H₂O",
   "feedback": "This answer is wrong — you counted the atoms of one water molecule; the 4 in front multiplies the whole H₂O. Four water molecules each hold 3 atoms: 4 × 3 = 12 atoms."
  },
  {
   "value": 8,
   "misconception": "you counted only the hydrogen atoms — each water molecule also carries one oxygen atom",
   "feedback": "This answer is wrong — you counted only the hydrogen atoms. Each water molecule holds 3 atoms — two hydrogens AND one oxygen — so the 4 waters contribute 4 × 3 = 12 atoms."
  }
 ]
}

[
 {
  "value": "lithium oxide",
  "misconception": "you dropped the hydrogen from the block — OH is the hydroxide ion, not the bare oxide ion",
  "feedback": "This answer is wrong — you dropped the hydrogen from the block. OH is the hydroxide ion, not the bare oxide ion, so LiOH is lithium hydroxide."
 },
 {
  "value": "lithium hydrogen oxide",
  "misconception": "you broke the block into separate atom names — a polyatomic ion's name is never rebuilt",
  "feedback": "This answer is wrong — you broke the block into separate atom names. OH is one polyatomic ion, hydroxide — its name happens to end in -ide already and is used as it is: LiOH is lithium hydroxide."
 }
]

{
 "keyExplanation": "The block OH is the hydroxide ion. Hydroxide's name happens to end in -ide already; use it as it is: LiOH is lithium hydroxide.",
 "anticipatedWrong": [
  {
   "value": "lithium oxide",
   "misconception": "you dropped the hydrogen from the block — OH is the hydroxide ion, not the bare oxide ion",
   "feedback": "This answer is wrong — you dropped the hydrogen from the block. OH is the hydroxide ion, not the bare oxide ion, so LiOH is lithium hydroxide."
  },
  {
   "value": "lithium hydrogen oxide",
   "misconception": "you broke the block into separate atom names — a polyatomic ion's name is never rebuilt",
   "feedback": "This answer is wrong — you broke the block into separate atom names. OH is one polyatomic ion, hydroxide — its name happens to end in -ide already and is used as it is: LiOH is lithium hydroxide."
  }
 ]
}

[
 {
  "value": "Al₃N₃",
  "misconception": "you skipped the reduce step — criss-cross gives Al₃N₃, and 3:3 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Al₃N₃, but a formula shows the simplest whole-number ratio of ions, and 3:3 reduces to 1:1: AlN. Check: (3+) + (3−) = 0."
 },
 {
  "value": "Al₃N₂",
  "misconception": "you borrowed nitride's usual partner count — here one 3+ ion cancels one 3− ion exactly",
  "feedback": "This answer is wrong — you borrowed nitride's usual partner count. Here the charges already match: aluminum is Al³⁺ and nitride is N³⁻, so one 3+ ion cancels one 3− ion exactly and the ratio is 1:1: AlN. Check: (3+) + (3−) = 0."
 }
]

{
 "keyExplanation": "Aluminum is Al³⁺ (Group 13) and nitride is N³⁻ (Group 15). One 3+ cancels one 3−, so the simplest ratio is 1:1: AlN. Check: (3+) + (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Al₃N₃",
   "misconception": "you skipped the reduce step — criss-cross gives Al₃N₃, and 3:3 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Al₃N₃, but a formula shows the simplest whole-number ratio of ions, and 3:3 reduces to 1:1: AlN. Check: (3+) + (3−) = 0."
  },
  {
   "value": "Al₃N₂",
   "misconception": "you borrowed nitride's usual partner count — here one 3+ ion cancels one 3− ion exactly",
   "feedback": "This answer is wrong — you borrowed nitride's usual partner count. Here the charges already match: aluminum is Al³⁺ and nitride is N³⁻, so one 3+ ion cancels one 3− ion exactly and the ratio is 1:1: AlN. Check: (3+) + (3−) = 0."
  }
 ]
}

[
 {
  "value": "1+",
  "misconception": "you gave one chloride's charge - the two chloride ions together set the total",
  "feedback": "This answer is wrong - you gave one chloride's charge. The two Cl⁻ ions together total 2−, so the zinc ion must contribute 2+, written 2+."
 },
 {
  "value": "2−",
  "misconception": "you flipped the sign - the metal ion is the positive one",
  "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the two Cl⁻ ions total 2−, and the zinc ion balances them with 2+, written 2+."
 }
]

{
 "keyExplanation": "Chloride, from Group 17, is Cl⁻; two of them total 2−. One zinc ion must contribute 2+. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "1+",
   "misconception": "you gave one chloride's charge - the two chloride ions together set the total",
   "feedback": "This answer is wrong - you gave one chloride's charge. The two Cl⁻ ions together total 2−, so the zinc ion must contribute 2+, written 2+.",
   "check": {
    "expr": "1*1",
    "expected": 1
   }
  },
  {
   "value": "2−",
   "misconception": "you flipped the sign - the metal ion is the positive one",
   "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the two Cl⁻ ions total 2−, and the zinc ion balances them with 2+, written 2+.",
   "check": {
    "expr": "-2",
    "expected": -2
   }
  }
 ]
}

[
 {
  "value": "2+",
  "misconception": "you assumed nickel is always 2+ — the charge comes from the balance",
  "feedback": "This answer is wrong — you assumed nickel is always 2+. The charge comes from the balance: three O²⁻ total 6−, and that 6+ is shared by two nickel ions, so each nickel ion is 3+."
 },
 {
  "value": "6+",
  "misconception": "you forgot to share the total — 6+ is split between two nickel ions",
  "feedback": "This answer is wrong — you forgot to share the total. The 6+ that balances three O²⁻ is split between two nickel ions: 6+ ÷ 2 = 3+ each."
 }
]

{
 "keyExplanation": "Oxide, from Group 16, is O²⁻; three of them total 6−. The total positive charge must be 6+, shared by two nickel ions: 6+ ÷ 2 = 3+ each. Check: 2 × (3+) + 3 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "2+",
   "misconception": "you assumed nickel is always 2+ — the charge comes from the balance",
   "feedback": "This answer is wrong — you assumed nickel is always 2+. The charge comes from the balance: three O²⁻ total 6−, and that 6+ is shared by two nickel ions, so each nickel ion is 3+."
  },
  {
   "value": "6+",
   "misconception": "you forgot to share the total — 6+ is split between two nickel ions",
   "feedback": "This answer is wrong — you forgot to share the total. The 6+ that balances three O²⁻ is split between two nickel ions: 6+ ÷ 2 = 3+ each."
  }
 ]
}

[
 {
  "value": "manganese(I) bromide",
  "misconception": "you counted the single manganese atom for the numeral — the numeral is the ion's 2+ charge, found from the two Br⁻",
  "feedback": "This answer is wrong — you counted atoms for the numeral. The numeral is the charge on one manganese ion: two Br⁻ ions total 2−, so it is 2+ and MnBr₂ is manganese(II) bromide."
 },
 {
  "value": "manganese bromide",
  "misconception": "you left out the numeral — the name must show the metal ion's charge",
  "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two Br⁻ ions total 2−, so the manganese ion is 2+ and MnBr₂ is manganese(II) bromide."
 }
]

{
 "keyExplanation": "Two Br⁻ ions total 2−, so the manganese ion is 2+ — manganese(II). MnBr₂ is manganese(II) bromide.",
 "anticipatedWrong": [
  {
   "value": "manganese(I) bromide",
   "misconception": "you counted the single manganese atom for the numeral — the numeral is the ion's 2+ charge, found from the two Br⁻",
   "feedback": "This answer is wrong — you counted atoms for the numeral. The numeral is the charge on one manganese ion: two Br⁻ ions total 2−, so it is 2+ and MnBr₂ is manganese(II) bromide."
  },
  {
   "value": "manganese bromide",
   "misconception": "you left out the numeral — the name must show the metal ion's charge",
   "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two Br⁻ ions total 2−, so the manganese ion is 2+ and MnBr₂ is manganese(II) bromide."
  }
 ]
}

[
 {
  "value": "cobalt(II) oxide",
  "misconception": "you used cobalt's subscript as the numeral — the 2 counts atoms; the numeral is the charge on one metal ion",
  "feedback": "This answer is wrong — you used cobalt's subscript as the numeral. The 2 counts cobalt atoms; the three oxide ions make 6−, so each of the two cobalt ions must be 3+ — Co₂O₃ is cobalt(III) oxide."
 },
 {
  "value": "cobalt(VI) oxide",
  "misconception": "you gave the total positive charge — the numeral is the charge on one cobalt ion, not the whole 6+",
  "feedback": "This answer is wrong — you gave the total positive charge. The numeral is the charge on one cobalt ion — the 6+ balance splits across two ions, 3+ apiece — so Co₂O₃ is cobalt(III) oxide."
 }
]

{
 "keyExplanation": "Each oxide ion is 2−, and three of them make 6− altogether. Two cobalt ions provide the matching 6+, so each cobalt is 3+ — cobalt(III). Co₂O₃ is cobalt(III) oxide.",
 "anticipatedWrong": [
  {
   "value": "cobalt(II) oxide",
   "misconception": "you used cobalt's subscript as the numeral — the 2 counts atoms; the numeral is the charge on one metal ion",
   "feedback": "This answer is wrong — you used cobalt's subscript as the numeral. The 2 counts cobalt atoms; the three oxide ions make 6−, so each of the two cobalt ions must be 3+ — Co₂O₃ is cobalt(III) oxide."
  },
  {
   "value": "cobalt(VI) oxide",
   "misconception": "you gave the total positive charge — the numeral is the charge on one cobalt ion, not the whole 6+",
   "feedback": "This answer is wrong — you gave the total positive charge. The numeral is the charge on one cobalt ion — the 6+ balance splits across two ions, 3+ apiece — so Co₂O₃ is cobalt(III) oxide."
  }
 ]
}

[
 {
  "value": "Fe₃F",
  "misconception": "you turned the numeral into iron's subscript — the III is the charge on one iron ion",
  "feedback": "This answer is wrong — you turned the numeral into iron's subscript. The III is the charge on one iron ion, and three 1− fluoride ions balance it: iron(III) fluoride is FeF₃."
 },
 {
  "value": "FeF₂",
  "misconception": "you gave iron a 2+ charge — the numeral (III) says this iron ion is 3+",
  "feedback": "This answer is wrong — you gave iron a 2+ charge. The numeral (III) says this iron ion is 3+, and three 1− ions balance it: iron(III) fluoride is FeF₃."
 }
]

{
 "keyExplanation": "Iron(III) is Fe³⁺ and fluoride is F⁻. Three 1− ions balance one 3+ ion: FeF₃. Check: (3+) + 3 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Fe₃F",
   "misconception": "you turned the numeral into iron's subscript — the III is the charge on one iron ion",
   "feedback": "This answer is wrong — you turned the numeral into iron's subscript. The III is the charge on one iron ion, and three 1− fluoride ions balance it: iron(III) fluoride is FeF₃."
  },
  {
   "value": "FeF₂",
   "misconception": "you gave iron a 2+ charge — the numeral (III) says this iron ion is 3+",
   "feedback": "This answer is wrong — you gave iron a 2+ charge. The numeral (III) says this iron ion is 3+, and three 1− ions balance it: iron(III) fluoride is FeF₃."
  }
 ]
}

[
 {
  "value": "Mn₂O₂",
  "misconception": "you skipped the reduce step — criss-cross gives Mn₂O₂, and 2:2 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Mn₂O₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: MnO. Check: (2+) + (2−) = 0."
 },
 {
  "value": "Mn₂O",
  "misconception": "you turned the numeral into a subscript — the II is the manganese ion's 2+ charge",
  "feedback": "This answer is wrong — you turned the numeral into a subscript. The II is the manganese ion's 2+ charge, and the subscripts come from the balancing: one 2+ ion cancels one 2− oxide exactly, a 1:1 ratio: MnO. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Manganese(II) is Mn²⁺ and oxide is O²⁻. One 2+ cancels one 2−: MnO. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Mn₂O₂",
   "misconception": "you skipped the reduce step — criss-cross gives Mn₂O₂, and 2:2 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Mn₂O₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: MnO. Check: (2+) + (2−) = 0."
  },
  {
   "value": "Mn₂O",
   "misconception": "you turned the numeral into a subscript — the II is the manganese ion's 2+ charge",
   "feedback": "This answer is wrong — you turned the numeral into a subscript. The II is the manganese ion's 2+ charge, and the subscripts come from the balancing: one 2+ ion cancels one 2− oxide exactly, a 1:1 ratio: MnO. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "Ba(CO₃)",
  "misconception": "you added parentheses around a block that appears only once",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are only needed when the polyatomic ion repeats; a single carbonate is written bare: BaCO₃. Check: (2+) + (2−) = 0."
 },
 {
  "value": "Ba(CO₃)₂",
  "misconception": "you doubled only the carbonate — one 2+ ion already cancels one 2− ion",
  "feedback": "This answer is wrong — you doubled only the carbonate. One 2+ ion already cancels one 2− ion, so the ratio is 1:1 and the single carbonate needs no parentheses: BaCO₃. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Barium is Ba²⁺ and carbonate is CO₃²⁻. One 2+ cancels one 2−, and the single carbonate block needs no parentheses: BaCO₃. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ba(CO₃)",
   "misconception": "you added parentheses around a block that appears only once",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are only needed when the polyatomic ion repeats; a single carbonate is written bare: BaCO₃. Check: (2+) + (2−) = 0."
  },
  {
   "value": "Ba(CO₃)₂",
   "misconception": "you doubled only the carbonate — one 2+ ion already cancels one 2− ion",
   "feedback": "This answer is wrong — you doubled only the carbonate. One 2+ ion already cancels one 2− ion, so the ratio is 1:1 and the single carbonate needs no parentheses: BaCO₃. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "(NH₄)₂SO₄",
  "misconception": "you wrote the polyatomic sulfate — sulfide is the bare S²⁻ ion",
  "feedback": "This answer is wrong — you wrote the polyatomic sulfate. The -ide ending names the bare ion: sulfide is S²⁻, from sulfur. Two 1+ ions cancel one 2−, and the repeated ion is polyatomic, so parentheses: (NH₄)₂S. Check: 2 × (1+) + (2−) = 0."
 },
 {
  "value": "NH₄S",
  "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2− ion needs two 1+ ions. Ammonium is NH₄⁺ and sulfide is S²⁻, so two ammonium ions balance each sulfide ion, kept whole in parentheses: (NH₄)₂S. Check: 2 × (1+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Ammonium is NH₄⁺ and sulfide is the bare S²⁻ ion. Two 1+ ions cancel one 2−, and the repeated ion is polyatomic, so parentheses: (NH₄)₂S. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "(NH₄)₂SO₄",
   "misconception": "you wrote the polyatomic sulfate — sulfide is the bare S²⁻ ion",
   "feedback": "This answer is wrong — you wrote the polyatomic sulfate. The -ide ending names the bare ion: sulfide is S²⁻, from sulfur. Two 1+ ions cancel one 2−, and the repeated ion is polyatomic, so parentheses: (NH₄)₂S. Check: 2 × (1+) + (2−) = 0."
  },
  {
   "value": "NH₄S",
   "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2− ion needs two 1+ ions. Ammonium is NH₄⁺ and sulfide is S²⁻, so two ammonium ions balance each sulfide ion, kept whole in parentheses: (NH₄)₂S. Check: 2 × (1+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "magnesium(II) chloride hexahydrate",
  "misconception": "you added a Roman numeral to a fixed-charge metal — magnesium is always 2+, so no numeral",
  "feedback": "This answer is wrong — you added a Roman numeral to a fixed-charge metal. Magnesium is a Group 2 metal and is always 2+, so its name never carries a numeral: magnesium chloride. Six waters: hexahydrate. MgCl₂·6H₂O is magnesium chloride hexahydrate."
 },
 {
  "value": "magnesium chloride dihydrate",
  "misconception": "you counted the chlorides for the prefix — the prefix counts the six waters",
  "feedback": "This answer is wrong — you counted the chlorides for the prefix. The prefix counts water molecules only: six waters means hexahydrate. MgCl₂·6H₂O is magnesium chloride hexahydrate."
 }
]

{
 "keyExplanation": "Magnesium is a fixed-charge Group 2 metal, so no numeral: magnesium chloride. Six waters: hexahydrate. MgCl₂·6H₂O is magnesium chloride hexahydrate.",
 "anticipatedWrong": [
  {
   "value": "magnesium(II) chloride hexahydrate",
   "misconception": "you added a Roman numeral to a fixed-charge metal — magnesium is always 2+, so no numeral",
   "feedback": "This answer is wrong — you added a Roman numeral to a fixed-charge metal. Magnesium is a Group 2 metal and is always 2+, so its name never carries a numeral: magnesium chloride. Six waters: hexahydrate. MgCl₂·6H₂O is magnesium chloride hexahydrate."
  },
  {
   "value": "magnesium chloride dihydrate",
   "misconception": "you counted the chlorides for the prefix — the prefix counts the six waters",
   "feedback": "This answer is wrong — you counted the chlorides for the prefix. The prefix counts water molecules only: six waters means hexahydrate. MgCl₂·6H₂O is magnesium chloride hexahydrate."
  }
 ]
}

[
 {
  "value": "SrNO₃₂·4H₂O",
  "misconception": "you left out the parentheses — two whole nitrate ions need the block wrapped",
  "feedback": "This answer is wrong — you left out the parentheses. Two whole nitrate ions need the block wrapped: the ionic part is Sr(NO₃)₂. Tetra- means 4 waters: Sr(NO₃)₂·4H₂O."
 },
 {
  "value": "Sr(NO₃)₄·2H₂O",
  "misconception": "you attached the water count to the nitrate — tetra- counts water molecules only",
  "feedback": "This answer is wrong — you attached the water count to the nitrate. Tetra- counts water molecules only: the ionic part stays Sr(NO₃)₂, and tetra- adds 4 waters: Sr(NO₃)₂·4H₂O."
 }
]

{
 "keyExplanation": "Strontium is Sr²⁺ and nitrate is NO₃⁻, so the ionic part is Sr(NO₃)₂ — check: (2+) + 2 × (1−) = 0. Tetra- means 4 waters: Sr(NO₃)₂·4H₂O.",
 "anticipatedWrong": [
  {
   "value": "SrNO₃₂·4H₂O",
   "misconception": "you left out the parentheses — two whole nitrate ions need the block wrapped",
   "feedback": "This answer is wrong — you left out the parentheses. Two whole nitrate ions need the block wrapped: the ionic part is Sr(NO₃)₂. Tetra- means 4 waters: Sr(NO₃)₂·4H₂O."
  },
  {
   "value": "Sr(NO₃)₄·2H₂O",
   "misconception": "you attached the water count to the nitrate — tetra- counts water molecules only",
   "feedback": "This answer is wrong — you attached the water count to the nitrate. Tetra- counts water molecules only: the ionic part stays Sr(NO₃)₂, and tetra- adds 4 waters: Sr(NO₃)₂·4H₂O."
  }
 ]
}

[
 {
  "value": "ammonium carbonate",
  "misconception": "you dropped the hydrogen from the block — HCO₃ is hydrogen carbonate, not carbonate",
  "feedback": "This answer is wrong — you dropped the hydrogen from the block. HCO₃ is hydrogen carbonate, not carbonate, so NH₄HCO₃ is ammonium hydrogen carbonate."
 },
 {
  "value": "ammonium hydrocarbonate",
  "misconception": "you merged the ion's name — it is 'hydrogen carbonate', two separate words",
  "feedback": "This answer is wrong — you merged the ion's name. It is 'hydrogen carbonate', two separate words: NH₄HCO₃ is ammonium hydrogen carbonate."
 }
]

{
 "keyExplanation": "NH₄ is the ammonium ion and HCO₃ is the hydrogen carbonate ion — both polyatomic, both keeping their own names: NH₄HCO₃ is ammonium hydrogen carbonate.",
 "anticipatedWrong": [
  {
   "value": "ammonium carbonate",
   "misconception": "you dropped the hydrogen from the block — HCO₃ is hydrogen carbonate, not carbonate",
   "feedback": "This answer is wrong — you dropped the hydrogen from the block. HCO₃ is hydrogen carbonate, not carbonate, so NH₄HCO₃ is ammonium hydrogen carbonate."
  },
  {
   "value": "ammonium hydrocarbonate",
   "misconception": "you merged the ion's name — it is 'hydrogen carbonate', two separate words",
   "feedback": "This answer is wrong — you merged the ion's name. It is 'hydrogen carbonate', two separate words: NH₄HCO₃ is ammonium hydrogen carbonate."
  }
 ]
}

[
 {
  "value": "cobalt acetate",
  "misconception": "you skipped the numeral — two acetate ions total 2−, so the cobalt is 2+",
  "feedback": "This answer is wrong — you skipped the numeral. Cobalt is a transition metal, so its name must show the charge: the two acetate ions total 2−, making the cobalt ion 2+ — cobalt(II) acetate."
 },
 {
  "value": "cobalt(I) acetate",
  "misconception": "you gave the wrong charge — two C₂H₃O₂⁻ ions total 2−, so the cobalt is 2+",
  "feedback": "This answer is wrong — you gave the wrong charge. Each acetate ion is C₂H₃O₂⁻, so the two of them total 2−, and one cobalt ion must balance all of it: the cobalt is 2+ — cobalt(II) acetate."
 }
]

{
 "keyExplanation": "No dot; cobalt is a transition metal, and two acetate ions total 2−, so the cobalt ion is 2+ — cobalt(II). The polyatomic acetate keeps its name: cobalt(II) acetate.",
 "anticipatedWrong": [
  {
   "value": "cobalt acetate",
   "misconception": "you skipped the numeral — two acetate ions total 2−, so the cobalt is 2+",
   "feedback": "This answer is wrong — you skipped the numeral. Cobalt is a transition metal, so its name must show the charge: the two acetate ions total 2−, making the cobalt ion 2+ — cobalt(II) acetate."
  },
  {
   "value": "cobalt(I) acetate",
   "misconception": "you gave the wrong charge — two C₂H₃O₂⁻ ions total 2−, so the cobalt is 2+",
   "feedback": "This answer is wrong — you gave the wrong charge. Each acetate ion is C₂H₃O₂⁻, so the two of them total 2−, and one cobalt ion must balance all of it: the cobalt is 2+ — cobalt(II) acetate."
  }
 ]
}

[
 {
  "value": "Cr₃(SO₄)₂",
  "misconception": "you swapped the counts — chromium takes the 2 and sulfate takes the 3",
  "feedback": "This answer is wrong — you swapped the counts. Chromium takes the 2 and sulfate takes the 3: 2 × (3+) cancels 3 × (2−), so chromium(III) sulfate is Cr₂(SO₄)₃."
 },
 {
  "value": "CrSO₄",
  "misconception": "you wrote a 1:1 ratio — 3+ and 2− do not cancel one for one",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. 3+ and 2− do not cancel one for one: two 3+ ions give 6+ and three 2− ions give 6−, so chromium(III) sulfate is Cr₂(SO₄)₃. Check: 2 × (3+) + 3 × (2−) = 0."
 }
]

{
 "keyExplanation": "Chromium(III) is Cr³⁺ and sulfate is SO₄²⁻. Two 3+ ions give 6+ and three 2− ions give 6−, with sulfate wrapped in parentheses: Cr₂(SO₄)₃. Check: 2 × (3+) + 3 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Cr₃(SO₄)₂",
   "misconception": "you swapped the counts — chromium takes the 2 and sulfate takes the 3",
   "feedback": "This answer is wrong — you swapped the counts. Chromium takes the 2 and sulfate takes the 3: 2 × (3+) cancels 3 × (2−), so chromium(III) sulfate is Cr₂(SO₄)₃."
  },
  {
   "value": "CrSO₄",
   "misconception": "you wrote a 1:1 ratio — 3+ and 2− do not cancel one for one",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. 3+ and 2− do not cancel one for one: two 3+ ions give 6+ and three 2− ions give 6−, so chromium(III) sulfate is Cr₂(SO₄)₃. Check: 2 × (3+) + 3 × (2−) = 0."
  }
 ]
}

[
 {
  "value": "iron(III) acetate tetrahydrate",
  "misconception": "you gave the wrong charge — two 1− acetate ions total 2−, so this iron ion is 2+",
  "feedback": "This answer is wrong — you gave the wrong charge. Iron can be 2+ or 3+, so the numeral must be computed from THIS formula: two acetate ions total 2−, so the iron here is 2+: Fe(C₂H₃O₂)₂·4H₂O is iron(II) acetate tetrahydrate."
 },
 {
  "value": "iron(II) acetate",
  "misconception": "you dropped the water word — the raised dot must show up as a prefix plus 'hydrate'",
  "feedback": "This answer is wrong — you dropped the water word. The raised dot must show up as a prefix plus 'hydrate': four waters make it tetrahydrate, so Fe(C₂H₃O₂)₂·4H₂O is iron(II) acetate tetrahydrate."
 }
]

{
 "keyExplanation": "A raised dot: name the part before it first. Two acetate ions total 2−, so the iron is 2+ — iron(II) acetate; the polyatomic acetate keeps its own name. Four waters: tetrahydrate. Fe(C₂H₃O₂)₂·4H₂O is iron(II) acetate tetrahydrate.",
 "anticipatedWrong": [
  {
   "value": "iron(III) acetate tetrahydrate",
   "misconception": "you gave the wrong charge — two 1− acetate ions total 2−, so this iron ion is 2+",
   "feedback": "This answer is wrong — you gave the wrong charge. Iron can be 2+ or 3+, so the numeral must be computed from THIS formula: two acetate ions total 2−, so the iron here is 2+: Fe(C₂H₃O₂)₂·4H₂O is iron(II) acetate tetrahydrate."
  },
  {
   "value": "iron(II) acetate",
   "misconception": "you dropped the water word — the raised dot must show up as a prefix plus 'hydrate'",
   "feedback": "This answer is wrong — you dropped the water word. The raised dot must show up as a prefix plus 'hydrate': four waters make it tetrahydrate, so Fe(C₂H₃O₂)₂·4H₂O is iron(II) acetate tetrahydrate."
  }
 ]
}

[
 {
  "value": "bromide potassium",
  "misconception": "you swapped the order — the metal's name always comes first",
  "feedback": "This answer is wrong — you swapped the order. The metal's name always comes first: KBr is potassium bromide."
 },
 {
  "value": "potassium bromine",
  "misconception": "you kept the element's name — in a compound the nonmetal takes -ide: bromide, not bromine",
  "feedback": "This answer is wrong — you kept the element's name. In the compound's name the nonmetal takes the -ide ending — bromine becomes bromide — so KBr is potassium bromide."
 }
]

{
 "keyExplanation": "Metal first, unchanged: potassium. Nonmetal with -ide: bromine becomes bromide. KBr is potassium bromide.",
 "anticipatedWrong": [
  {
   "value": "bromide potassium",
   "misconception": "you swapped the order — the metal's name always comes first",
   "feedback": "This answer is wrong — you swapped the order. The metal's name always comes first: KBr is potassium bromide."
  },
  {
   "value": "potassium bromine",
   "misconception": "you kept the element's name — in a compound the nonmetal takes -ide: bromide, not bromine",
   "feedback": "This answer is wrong — you kept the element's name. In the compound's name the nonmetal takes the -ide ending — bromine becomes bromide — so KBr is potassium bromide."
  }
 ]
}

[
 "The 9 after the raised dot counts whole water molecules.",
 "Each water molecule H₂O carries 2 hydrogen atoms: 9 × 2 = 18.",
 "Fe(NO₃)₃·9H₂O contains 18 hydrogen atoms."
]

[
 {
  "value": 9,
  "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 9 waters carries 2 hydrogens",
  "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 9 waters carries 2 hydrogens. The 9 after the dot counts whole water molecules: 9 × 2 = 18 hydrogen atoms."
 },
 {
  "value": 2,
  "misconception": "you counted one water molecule's hydrogens — the 9 in front multiplies the whole H₂O",
  "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 9 in front multiplies the whole H₂O. Nine water molecules each carry 2 hydrogen atoms: 9 × 2 = 18 hydrogen atoms."
 },
 {
  "value": 27,
  "misconception": "you counted every atom in the water molecules — each water holds 2 hydrogens and 1 oxygen, but the question asks for hydrogen only",
  "feedback": "This answer is wrong — you counted every atom in the water molecules, but the question asks for hydrogen only. Each of the 9 waters carries 2 hydrogen atoms: 9 × 2 = 18 hydrogen atoms."
 }
]

{
 "keyExplanation": "The 9 after the dot counts whole water molecules, and each water molecule has 2 hydrogen atoms: 9 × 2 = 18 hydrogen atoms.",
 "anticipatedWrong": [
  {
   "value": 9,
   "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 9 waters carries 2 hydrogens",
   "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 9 waters carries 2 hydrogens. The 9 after the dot counts whole water molecules: 9 × 2 = 18 hydrogen atoms."
  },
  {
   "value": 2,
   "misconception": "you counted one water molecule's hydrogens — the 9 in front multiplies the whole H₂O",
   "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 9 in front multiplies the whole H₂O. Nine water molecules each carry 2 hydrogen atoms: 9 × 2 = 18 hydrogen atoms."
  },
  {
   "value": 27,
   "misconception": "you counted every atom in the water molecules — each water holds 2 hydrogens and 1 oxygen, but the question asks for hydrogen only",
   "feedback": "This answer is wrong — you counted every atom in the water molecules, but the question asks for hydrogen only. Each of the 9 waters carries 2 hydrogen atoms: 9 × 2 = 18 hydrogen atoms."
  }
 ]
}

[
 {
  "value": "magnesium nitride",
  "misconception": "you used the single-atom ion's name — nitride is bare N³⁻; the block NO₃ is nitrate",
  "feedback": "This answer is wrong — you used the single-atom ion's name. Nitride is the bare N³⁻ ion; the block NO₃ is the nitrate ion, so Mg(NO₃)₂ is magnesium nitrate."
 },
 {
  "value": "magnesium dinitrate",
  "misconception": "you pulled the subscript into the name — parentheses and subscripts never enter the name",
  "feedback": "This answer is wrong — you pulled the subscript into the name. Parentheses and subscripts never enter the name: Mg(NO₃)₂ is magnesium nitrate."
 }
]

{
 "keyExplanation": "The block in the parentheses, NO₃, is the nitrate ion. Metal first, then the ion's own name: Mg(NO₃)₂ is magnesium nitrate — the subscript 2 stays out of the name.",
 "anticipatedWrong": [
  {
   "value": "magnesium nitride",
   "misconception": "you used the single-atom ion's name — nitride is bare N³⁻; the block NO₃ is nitrate",
   "feedback": "This answer is wrong — you used the single-atom ion's name. Nitride is the bare N³⁻ ion; the block NO₃ is the nitrate ion, so Mg(NO₃)₂ is magnesium nitrate."
  },
  {
   "value": "magnesium dinitrate",
   "misconception": "you pulled the subscript into the name — parentheses and subscripts never enter the name",
   "feedback": "This answer is wrong — you pulled the subscript into the name. Parentheses and subscripts never enter the name: Mg(NO₃)₂ is magnesium nitrate."
  }
 ]
}

[
 {
  "value": "LiS",
  "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2− ion needs two 1+ ions before the charges cancel. Lithium is Li⁺ and sulfide is S²⁻, so two lithium ions balance each sulfide ion: Li₂S. Check: 2 × (1+) + (2−) = 0."
 },
 {
  "value": "Li₂SO₄",
  "misconception": "you wrote the polyatomic sulfate — sulfide is the bare S²⁻ ion",
  "feedback": "This answer is wrong — you wrote the polyatomic sulfate. The -ide ending names the bare ion from the element: sulfide is S²⁻, from sulfur in Group 16. Two Li⁺ ions balance one S²⁻ ion: Li₂S. Check: 2 × (1+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Lithium is Li⁺ (Group 1) and sulfide is S²⁻ (Group 16). Two 1+ ions cancel one 2−: Li₂S. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "LiS",
   "misconception": "you wrote a 1:1 ratio — one 2− ion needs two 1+ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2− ion needs two 1+ ions before the charges cancel. Lithium is Li⁺ and sulfide is S²⁻, so two lithium ions balance each sulfide ion: Li₂S. Check: 2 × (1+) + (2−) = 0."
  },
  {
   "value": "Li₂SO₄",
   "misconception": "you wrote the polyatomic sulfate — sulfide is the bare S²⁻ ion",
   "feedback": "This answer is wrong — you wrote the polyatomic sulfate. The -ide ending names the bare ion from the element: sulfide is S²⁻, from sulfur in Group 16. Two Li⁺ ions balance one S²⁻ ion: Li₂S. Check: 2 × (1+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "2+",
  "misconception": "you gave one oxide's charge - the two oxide ions together set the total",
  "feedback": "This answer is wrong - you gave one oxide's charge. Each O²⁻ carries 2−, and the two of them total 4−, so the tin ion must contribute 4+, written 4+."
 },
 {
  "value": "4−",
  "misconception": "you flipped the sign - the metal ion is the positive one",
  "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the two O²⁻ ions total 4−, and the tin ion balances them with 4+, written 4+."
 }
]

{
 "keyExplanation": "Oxide, from Group 16, is O²⁻; two of them total 4−. One tin ion must contribute 4+. Check: (4+) + 2 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "2+",
   "misconception": "you gave one oxide's charge - the two oxide ions together set the total",
   "feedback": "This answer is wrong - you gave one oxide's charge. Each O²⁻ carries 2−, and the two of them total 4−, so the tin ion must contribute 4+, written 4+.",
   "check": {
    "expr": "1*2",
    "expected": 2
   }
  },
  {
   "value": "4−",
   "misconception": "you flipped the sign - the metal ion is the positive one",
   "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the two O²⁻ ions total 4−, and the tin ion balances them with 4+, written 4+.",
   "check": {
    "expr": "-4",
    "expected": -4
   }
  }
 ]
}

[
 {
  "value": "2+",
  "misconception": "you forgot to share the charge — the 2+ that balances the sulfide is split between two silver ions",
  "feedback": "This answer is wrong — you forgot to share the charge. One S²⁻ needs 2+ in total, and that 2+ is split between two silver ions: 2+ ÷ 2 = 1+ each."
 },
 {
  "value": "1−",
  "misconception": "you flipped the sign — the metal ion is the positive one",
  "feedback": "This answer is wrong — you flipped the sign. The metal ion is the positive one: the sulfide ion is 2−, and each silver ion is 1+."
 }
]

{
 "keyExplanation": "Sulfide, from Group 16, is S²⁻; one of them makes the total negative charge 2−. The total positive charge must be 2+, shared by two silver ions: 2+ ÷ 2 = 1+ each. Check: 2 × (1+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "2+",
   "misconception": "you forgot to share the charge — the 2+ that balances the sulfide is split between two silver ions",
   "feedback": "This answer is wrong — you forgot to share the charge. One S²⁻ needs 2+ in total, and that 2+ is split between two silver ions: 2+ ÷ 2 = 1+ each."
  },
  {
   "value": "1−",
   "misconception": "you flipped the sign — the metal ion is the positive one",
   "feedback": "This answer is wrong — you flipped the sign. The metal ion is the positive one: the sulfide ion is 2−, and each silver ion is 1+."
  }
 ]
}

[
 {
  "value": "manganese(I) fluoride",
  "misconception": "you counted the single manganese atom for the numeral — the numeral is the ion's 2+ charge, found from the two F⁻",
  "feedback": "This answer is wrong — you counted atoms for the numeral. The numeral is the charge on one manganese ion: two F⁻ ions total 2−, so it is 2+ and MnF₂ is manganese(II) fluoride."
 },
 {
  "value": "manganese fluoride",
  "misconception": "you left out the numeral — the name must show the metal ion's charge",
  "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two F⁻ ions total 2−, so the manganese ion is 2+ and MnF₂ is manganese(II) fluoride."
 }
]

{
 "keyExplanation": "Two F⁻ ions total 2−, so the manganese ion is 2+ — manganese(II). MnF₂ is manganese(II) fluoride.",
 "anticipatedWrong": [
  {
   "value": "manganese(I) fluoride",
   "misconception": "you counted the single manganese atom for the numeral — the numeral is the ion's 2+ charge, found from the two F⁻",
   "feedback": "This answer is wrong — you counted atoms for the numeral. The numeral is the charge on one manganese ion: two F⁻ ions total 2−, so it is 2+ and MnF₂ is manganese(II) fluoride."
  },
  {
   "value": "manganese fluoride",
   "misconception": "you left out the numeral — the name must show the metal ion's charge",
   "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two F⁻ ions total 2−, so the manganese ion is 2+ and MnF₂ is manganese(II) fluoride."
  }
 ]
}

[
 {
  "value": "nickel(II) oxide",
  "misconception": "you used nickel's subscript as the numeral — the 2 counts atoms; the numeral is the charge on one metal ion",
  "feedback": "This answer is wrong — you used nickel's subscript as the numeral. The 2 counts nickel atoms; balancing three O²⁻ (6− in all) takes 3+ from each of the two nickel ions, so Ni₂O₃ is nickel(III) oxide."
 },
 {
  "value": "nickel(VI) oxide",
  "misconception": "you gave the total positive charge — the numeral is the charge on one nickel ion, not the whole 6+",
  "feedback": "This answer is wrong — you gave the total positive charge. The numeral is the charge on one nickel ion — the 6+ total splits across two ions, 3+ apiece — so Ni₂O₃ is nickel(III) oxide."
 }
]

{
 "keyExplanation": "Three oxide ions carry 6− altogether, and the two nickel ions share the matching positive charge equally: 3+ each — nickel(III). Ni₂O₃ is nickel(III) oxide.",
 "anticipatedWrong": [
  {
   "value": "nickel(II) oxide",
   "misconception": "you used nickel's subscript as the numeral — the 2 counts atoms; the numeral is the charge on one metal ion",
   "feedback": "This answer is wrong — you used nickel's subscript as the numeral. The 2 counts nickel atoms; balancing three O²⁻ (6− in all) takes 3+ from each of the two nickel ions, so Ni₂O₃ is nickel(III) oxide."
  },
  {
   "value": "nickel(VI) oxide",
   "misconception": "you gave the total positive charge — the numeral is the charge on one nickel ion, not the whole 6+",
   "feedback": "This answer is wrong — you gave the total positive charge. The numeral is the charge on one nickel ion — the 6+ total splits across two ions, 3+ apiece — so Ni₂O₃ is nickel(III) oxide."
  }
 ]
}

[
 {
  "value": "Co₂I",
  "misconception": "you turned the numeral into cobalt's subscript — the II is the charge on one cobalt ion",
  "feedback": "This answer is wrong — you turned the numeral into cobalt's subscript. The II is the charge on one cobalt ion: one 2+ ion needs two 1− ions, so cobalt(II) iodide is CoI₂."
 },
 {
  "value": "CoI",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2+ ion needs two 1− ions, so cobalt(II) iodide is CoI₂."
 }
]

{
 "keyExplanation": "Cobalt(II) is Co²⁺ and iodide is I⁻. One 2+ ion needs two 1− ions: CoI₂. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Co₂I",
   "misconception": "you turned the numeral into cobalt's subscript — the II is the charge on one cobalt ion",
   "feedback": "This answer is wrong — you turned the numeral into cobalt's subscript. The II is the charge on one cobalt ion: one 2+ ion needs two 1− ions, so cobalt(II) iodide is CoI₂."
  },
  {
   "value": "CoI",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2+ ion needs two 1− ions, so cobalt(II) iodide is CoI₂."
  }
 ]
}

[
 {
  "value": "Cr₃N₃",
  "misconception": "you skipped the reduce step — criss-cross gives Cr₃N₃, and 3:3 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Cr₃N₃, but a formula shows the simplest whole-number ratio of ions, and 3:3 reduces to 1:1: CrN. Check: (3+) + (3−) = 0."
 },
 {
  "value": "Cr₃N",
  "misconception": "you turned the numeral into a subscript — the III is the chromium ion's 3+ charge",
  "feedback": "This answer is wrong — you turned the numeral into a subscript. The III is the chromium ion's 3+ charge, and the subscripts come from the balancing: one 3+ ion cancels one 3− nitride exactly, a 1:1 ratio: CrN. Check: (3+) + (3−) = 0."
 }
]

{
 "keyExplanation": "Chromium(III) is Cr³⁺ and nitride is N³⁻. One 3+ cancels one 3−, so the simplest ratio is 1:1: CrN. Check: (3+) + (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Cr₃N₃",
   "misconception": "you skipped the reduce step — criss-cross gives Cr₃N₃, and 3:3 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Cr₃N₃, but a formula shows the simplest whole-number ratio of ions, and 3:3 reduces to 1:1: CrN. Check: (3+) + (3−) = 0."
  },
  {
   "value": "Cr₃N",
   "misconception": "you turned the numeral into a subscript — the III is the chromium ion's 3+ charge",
   "feedback": "This answer is wrong — you turned the numeral into a subscript. The III is the chromium ion's 3+ charge, and the subscripts come from the balancing: one 3+ ion cancels one 3− nitride exactly, a 1:1 ratio: CrN. Check: (3+) + (3−) = 0."
  }
 ]
}

[
 {
  "value": "Na(OH)",
  "misconception": "you added parentheses around a block that appears only once",
  "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are only needed when the polyatomic ion repeats; a single hydroxide is written bare: NaOH. Check: (1+) + (1−) = 0."
 },
 {
  "value": "Na(OH)₂",
  "misconception": "you doubled a pair that already cancels — one 1+ ion cancels one 1− ion",
  "feedback": "This answer is wrong — you doubled a pair that already cancels. One 1+ ion cancels one 1− ion exactly, so nothing repeats and no parentheses are needed: NaOH. Check: (1+) + (1−) = 0."
 }
]

{
 "keyExplanation": "Sodium is Na⁺ and hydroxide is OH⁻. One 1+ cancels one 1−, and the single hydroxide block needs no parentheses: NaOH. Check: (1+) + (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Na(OH)",
   "misconception": "you added parentheses around a block that appears only once",
   "feedback": "This answer is wrong — you added parentheses around a block that appears only once. Parentheses are only needed when the polyatomic ion repeats; a single hydroxide is written bare: NaOH. Check: (1+) + (1−) = 0."
  },
  {
   "value": "Na(OH)₂",
   "misconception": "you doubled a pair that already cancels — one 1+ ion cancels one 1− ion",
   "feedback": "This answer is wrong — you doubled a pair that already cancels. One 1+ ion cancels one 1− ion exactly, so nothing repeats and no parentheses are needed: NaOH. Check: (1+) + (1−) = 0."
  }
 ]
}

[
 {
  "value": "CaClO₃",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions. One Ca²⁺ needs two ClO₃⁻ ions, and the repeated chlorate goes in parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "CaCl₂",
  "misconception": "you dropped the oxygens — chlorate is the whole ClO₃⁻ block, not the bare chloride ion",
  "feedback": "This answer is wrong — you dropped the oxygens. Chlorate is the whole ClO₃⁻ block; the bare Cl⁻ ion would be chloride, a different compound entirely. One Ca²⁺ needs two ClO₃⁻ ions, kept whole in parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Calcium forms Ca²⁺, and the chlorate ion is ClO₃⁻. One 2+ ion needs two 1− ions, and the chlorate ion repeats, so parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "CaClO₃",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions. One Ca²⁺ needs two ClO₃⁻ ions, and the repeated chlorate goes in parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "CaCl₂",
   "misconception": "you dropped the oxygens — chlorate is the whole ClO₃⁻ block, not the bare chloride ion",
   "feedback": "This answer is wrong — you dropped the oxygens. Chlorate is the whole ClO₃⁻ block; the bare Cl⁻ ion would be chloride, a different compound entirely. One Ca²⁺ needs two ClO₃⁻ ions, kept whole in parentheses: Ca(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "sodium(I) bromide dihydrate",
  "misconception": "you added a Roman numeral to a fixed-charge metal — sodium is always 1+, so no numeral",
  "feedback": "This answer is wrong — you added a Roman numeral to a fixed-charge metal. Sodium is a Group 1 metal and is always 1+, so its name never carries a numeral: sodium bromide. Two waters: dihydrate. NaBr·2H₂O is sodium bromide dihydrate."
 },
 {
  "value": "sodium bromide",
  "misconception": "you dropped the water word — the built-in water must appear in the name",
  "feedback": "This answer is wrong — you dropped the water word. The built-in water must appear in the name: two waters means dihydrate. NaBr·2H₂O is sodium bromide dihydrate."
 }
]

{
 "keyExplanation": "Sodium is a fixed-charge Group 1 metal, so no numeral: sodium bromide. Two waters: dihydrate. NaBr·2H₂O is sodium bromide dihydrate.",
 "anticipatedWrong": [
  {
   "value": "sodium(I) bromide dihydrate",
   "misconception": "you added a Roman numeral to a fixed-charge metal — sodium is always 1+, so no numeral",
   "feedback": "This answer is wrong — you added a Roman numeral to a fixed-charge metal. Sodium is a Group 1 metal and is always 1+, so its name never carries a numeral: sodium bromide. Two waters: dihydrate. NaBr·2H₂O is sodium bromide dihydrate."
  },
  {
   "value": "sodium bromide",
   "misconception": "you dropped the water word — the built-in water must appear in the name",
   "feedback": "This answer is wrong — you dropped the water word. The built-in water must appear in the name: two waters means dihydrate. NaBr·2H₂O is sodium bromide dihydrate."
  }
 ]
}

[
 {
  "value": "CuC₂H₃O₂·H₂O",
  "misconception": "you wrote a 1:1 ratio in the ionic part — Cu²⁺ needs two C₂H₃O₂⁻ ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio in the ionic part. Copper(II) is Cu²⁺ and acetate is C₂H₃O₂⁻, so two acetate ions cancel one 2+ ion, wrapped in parentheses: Cu(C₂H₃O₂)₂. Mono- means 1 water, written with no number: Cu(C₂H₃O₂)₂·H₂O."
 },
 {
  "value": "Cu(C₂H₃O₂)₂",
  "misconception": "you dropped the water — monohydrate still adds ·H₂O",
  "feedback": "This answer is wrong — you dropped the water. Monohydrate still adds ·H₂O after the compound, written with no number: Cu(C₂H₃O₂)₂·H₂O."
 }
]

{
 "keyExplanation": "Copper(II) is Cu²⁺ and acetate is C₂H₃O₂⁻, so the ionic part is Cu(C₂H₃O₂)₂ — check: (2+) + 2 × (1−) = 0. Mono- means 1 water, written with no number: Cu(C₂H₃O₂)₂·H₂O.",
 "anticipatedWrong": [
  {
   "value": "CuC₂H₃O₂·H₂O",
   "misconception": "you wrote a 1:1 ratio in the ionic part — Cu²⁺ needs two C₂H₃O₂⁻ ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio in the ionic part. Copper(II) is Cu²⁺ and acetate is C₂H₃O₂⁻, so two acetate ions cancel one 2+ ion, wrapped in parentheses: Cu(C₂H₃O₂)₂. Mono- means 1 water, written with no number: Cu(C₂H₃O₂)₂·H₂O."
  },
  {
   "value": "Cu(C₂H₃O₂)₂",
   "misconception": "you dropped the water — monohydrate still adds ·H₂O",
   "feedback": "This answer is wrong — you dropped the water. Monohydrate still adds ·H₂O after the compound, written with no number: Cu(C₂H₃O₂)₂·H₂O."
  }
 ]
}

[
 {
  "value": "nitrogen hydride carbonate",
  "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
  "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so (NH₄)₂CO₃ is ammonium carbonate."
 },
 {
  "value": "diammonium carbonate",
  "misconception": "you pulled the subscript into the name — parentheses and subscripts never enter the name",
  "feedback": "This answer is wrong — you pulled the subscript into the name. Parentheses and subscripts never enter the name: (NH₄)₂CO₃ is ammonium carbonate."
 }
]

{
 "keyExplanation": "NH₄ is the ammonium ion and CO₃ is the carbonate ion — both polyatomic, both keeping their own names: (NH₄)₂CO₃ is ammonium carbonate. Parentheses and subscripts never enter the name.",
 "anticipatedWrong": [
  {
   "value": "nitrogen hydride carbonate",
   "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
   "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so (NH₄)₂CO₃ is ammonium carbonate."
  },
  {
   "value": "diammonium carbonate",
   "misconception": "you pulled the subscript into the name — parentheses and subscripts never enter the name",
   "feedback": "This answer is wrong — you pulled the subscript into the name. Parentheses and subscripts never enter the name: (NH₄)₂CO₃ is ammonium carbonate."
  }
 ]
}

[
 {
  "value": "strontium(II) sulfate",
  "misconception": "you added a numeral — strontium, in Group 2, is fixed-charge",
  "feedback": "This answer is wrong — you added a numeral. Strontium, in Group 2, is a fixed-charge metal, so its name never carries a Roman numeral: SrSO₄ is strontium sulfate."
 },
 {
  "value": "strontium sulfide",
  "misconception": "you used the single-atom ion's name — sulfide is bare S²⁻; the block SO₄ is sulfate",
  "feedback": "This answer is wrong — you used the single-atom ion's name. Sulfide is the bare S²⁻ ion; the block SO₄ in SrSO₄ is the polyatomic sulfate ion, and it keeps its own name: strontium sulfate."
 }
]

{
 "keyExplanation": "No dot; strontium is a fixed-charge Group 2 metal, and the block SO₄ is the polyatomic sulfate ion: strontium sulfate.",
 "anticipatedWrong": [
  {
   "value": "strontium(II) sulfate",
   "misconception": "you added a numeral — strontium, in Group 2, is fixed-charge",
   "feedback": "This answer is wrong — you added a numeral. Strontium, in Group 2, is a fixed-charge metal, so its name never carries a Roman numeral: SrSO₄ is strontium sulfate."
  },
  {
   "value": "strontium sulfide",
   "misconception": "you used the single-atom ion's name — sulfide is bare S²⁻; the block SO₄ is sulfate",
   "feedback": "This answer is wrong — you used the single-atom ion's name. Sulfide is the bare S²⁻ ion; the block SO₄ in SrSO₄ is the polyatomic sulfate ion, and it keeps its own name: strontium sulfate."
  }
 ]
}

[
 {
  "value": "Ni₂(PO₄)₃",
  "misconception": "you swapped the counts — nickel takes the 3 and phosphate takes the 2",
  "feedback": "This answer is wrong — you swapped the counts. Nickel takes the 3 and phosphate takes the 2: 3 × (2+) cancels 2 × (3−), so nickel(II) phosphate is Ni₃(PO₄)₂."
 },
 {
  "value": "NiPO₄",
  "misconception": "you wrote a 1:1 ratio — 2+ and 3− do not cancel one for one",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. 2+ and 3− do not cancel one for one: three 2+ ions give 6+ and two 3− ions give 6−, so nickel(II) phosphate is Ni₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
 }
]

{
 "keyExplanation": "Nickel(II) is Ni²⁺ and phosphate is PO₄³⁻. Three 2+ ions give 6+ and two 3− ions give 6−, with phosphate wrapped in parentheses: Ni₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ni₂(PO₄)₃",
   "misconception": "you swapped the counts — nickel takes the 3 and phosphate takes the 2",
   "feedback": "This answer is wrong — you swapped the counts. Nickel takes the 3 and phosphate takes the 2: 3 × (2+) cancels 2 × (3−), so nickel(II) phosphate is Ni₃(PO₄)₂."
  },
  {
   "value": "NiPO₄",
   "misconception": "you wrote a 1:1 ratio — 2+ and 3− do not cancel one for one",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. 2+ and 3− do not cancel one for one: three 2+ ions give 6+ and two 3− ions give 6−, so nickel(II) phosphate is Ni₃(PO₄)₂. Check: 3 × (2+) + 2 × (3−) = 0."
  }
 ]
}

[
 {
  "value": "iron(II) sulfate nonahydrate",
  "misconception": "you gave each iron the charge one sulfate suggests — three SO₄²⁻ total 6−, shared by two iron ions, so each iron is 3+",
  "feedback": "This answer is wrong — you gave each iron the charge one sulfate suggests. Work the whole negative side first: three SO₄²⁻ total 6−, and that 6− is shared by TWO iron ions, so each iron is 3+: Fe₂(SO₄)₃·9H₂O is iron(III) sulfate nonahydrate."
 },
 {
  "value": "iron(III) sulfate",
  "misconception": "you dropped the water word — the raised dot must show up as a prefix plus 'hydrate'",
  "feedback": "This answer is wrong — you dropped the water word. The raised dot must show up as a prefix plus 'hydrate': nine waters make it nonahydrate, so Fe₂(SO₄)₃·9H₂O is iron(III) sulfate nonahydrate."
 }
]

{
 "keyExplanation": "A raised dot: name the part before it first. Three sulfate ions total 6−, shared by two iron ions, so each iron is 3+ — iron(III) sulfate. Nine waters: nonahydrate. Fe₂(SO₄)₃·9H₂O is iron(III) sulfate nonahydrate.",
 "anticipatedWrong": [
  {
   "value": "iron(II) sulfate nonahydrate",
   "misconception": "you gave each iron the charge one sulfate suggests — three SO₄²⁻ total 6−, shared by two iron ions, so each iron is 3+",
   "feedback": "This answer is wrong — you gave each iron the charge one sulfate suggests. Work the whole negative side first: three SO₄²⁻ total 6−, and that 6− is shared by TWO iron ions, so each iron is 3+: Fe₂(SO₄)₃·9H₂O is iron(III) sulfate nonahydrate."
  },
  {
   "value": "iron(III) sulfate",
   "misconception": "you dropped the water word — the raised dot must show up as a prefix plus 'hydrate'",
   "feedback": "This answer is wrong — you dropped the water word. The raised dot must show up as a prefix plus 'hydrate': nine waters make it nonahydrate, so Fe₂(SO₄)₃·9H₂O is iron(III) sulfate nonahydrate."
  }
 ]
}

[
 {
  "value": "strontium difluoride",
  "misconception": "you added a counting word — ionic names never count atoms; the subscript comes back from the charges",
  "feedback": "This answer is wrong — you added a counting word. Ionic names never count atoms — the subscript 2 comes back from the charges whenever the formula is needed — so SrF₂ is just strontium fluoride."
 },
 {
  "value": "strontium fluorine",
  "misconception": "you kept the element's name — in a compound the nonmetal takes -ide: fluoride, not fluorine",
  "feedback": "This answer is wrong — you kept the element's name. In the compound's name the nonmetal takes the -ide ending — F becomes fluoride — so SrF₂ is strontium fluoride."
 }
]

{
 "keyExplanation": "Metal first, unchanged: strontium. The nonmetal takes the -ide ending: fluoride. SrF₂ is strontium fluoride — the subscript never enters the name.",
 "anticipatedWrong": [
  {
   "value": "strontium difluoride",
   "misconception": "you added a counting word — ionic names never count atoms; the subscript comes back from the charges",
   "feedback": "This answer is wrong — you added a counting word. Ionic names never count atoms — the subscript 2 comes back from the charges whenever the formula is needed — so SrF₂ is just strontium fluoride."
  },
  {
   "value": "strontium fluorine",
   "misconception": "you kept the element's name — in a compound the nonmetal takes -ide: fluoride, not fluorine",
   "feedback": "This answer is wrong — you kept the element's name. In the compound's name the nonmetal takes the -ide ending — F becomes fluoride — so SrF₂ is strontium fluoride."
  }
 ]
}

[
 "The 7 after the raised dot counts whole water molecules.",
 "Each water molecule H₂O carries 2 hydrogen atoms: 7 × 2 = 14.",
 "ZnSO₄·7H₂O contains 14 hydrogen atoms."
]

[
 {
  "value": 7,
  "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 7 waters carries 2 hydrogens",
  "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 7 waters carries 2 hydrogens. The 7 after the dot counts whole water molecules: 7 × 2 = 14 hydrogen atoms."
 },
 {
  "value": 2,
  "misconception": "you counted one water molecule's hydrogens — the 7 in front multiplies the whole H₂O",
  "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 7 in front multiplies the whole H₂O. Seven water molecules each carry 2 hydrogen atoms: 7 × 2 = 14 hydrogen atoms."
 },
 {
  "value": 21,
  "misconception": "you counted every atom in the water molecules — each water holds 2 hydrogens and 1 oxygen, but the question asks for hydrogen only",
  "feedback": "This answer is wrong — you counted every atom in the water molecules, but the question asks for hydrogen only. Each of the 7 waters carries 2 hydrogen atoms: 7 × 2 = 14 hydrogen atoms."
 }
]

{
 "keyExplanation": "The 7 after the dot counts whole water molecules, and each water molecule has 2 hydrogen atoms: 7 × 2 = 14 hydrogen atoms.",
 "anticipatedWrong": [
  {
   "value": 7,
   "misconception": "you counted the water molecules, not their hydrogen atoms — each of the 7 waters carries 2 hydrogens",
   "feedback": "This answer is wrong — you counted the water molecules, not their hydrogen atoms; each of the 7 waters carries 2 hydrogens. The 7 after the dot counts whole water molecules: 7 × 2 = 14 hydrogen atoms."
  },
  {
   "value": 2,
   "misconception": "you counted one water molecule's hydrogens — the 7 in front multiplies the whole H₂O",
   "feedback": "This answer is wrong — you counted one water molecule's hydrogens; the 7 in front multiplies the whole H₂O. Seven water molecules each carry 2 hydrogen atoms: 7 × 2 = 14 hydrogen atoms."
  },
  {
   "value": 21,
   "misconception": "you counted every atom in the water molecules — each water holds 2 hydrogens and 1 oxygen, but the question asks for hydrogen only",
   "feedback": "This answer is wrong — you counted every atom in the water molecules, but the question asks for hydrogen only. Each of the 7 waters carries 2 hydrogen atoms: 7 × 2 = 14 hydrogen atoms."
  }
 ]
}

[
 {
  "value": "potassium sulfide",
  "misconception": "you used the single-atom ion's name — sulfide is bare S²⁻; the block SO₄ is sulfate",
  "feedback": "This answer is wrong — you used the single-atom ion's name. Sulfide is the bare S²⁻ ion; the block SO₄ is the sulfate ion, so K₂SO₄ is potassium sulfate."
 },
 {
  "value": "dipotassium sulfate",
  "misconception": "you pulled the subscript into the name — parentheses and subscripts never enter the name",
  "feedback": "This answer is wrong — you pulled the subscript into the name. The subscript 2 comes back from the charges whenever the formula is needed — K₂SO₄ is just potassium sulfate."
 }
]

{
 "keyExplanation": "The block SO₄ is the sulfate ion, so K₂SO₄ is potassium sulfate — the subscript 2 stays out of the name.",
 "anticipatedWrong": [
  {
   "value": "potassium sulfide",
   "misconception": "you used the single-atom ion's name — sulfide is bare S²⁻; the block SO₄ is sulfate",
   "feedback": "This answer is wrong — you used the single-atom ion's name. Sulfide is the bare S²⁻ ion; the block SO₄ is the sulfate ion, so K₂SO₄ is potassium sulfate."
  },
  {
   "value": "dipotassium sulfate",
   "misconception": "you pulled the subscript into the name — parentheses and subscripts never enter the name",
   "feedback": "This answer is wrong — you pulled the subscript into the name. The subscript 2 comes back from the charges whenever the formula is needed — K₂SO₄ is just potassium sulfate."
  }
 ]
}

[
 {
  "value": "Ca₂O₂",
  "misconception": "you skipped the reduce step — criss-cross gives Ca₂O₂, and 2:2 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ca₂O₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: CaO. Check: (2+) + (2−) = 0."
 },
 {
  "value": "CaO₂",
  "misconception": "you gave oxygen a 1− charge — oxide, from Group 16, is O²⁻, so one 2+ cancels one 2−",
  "feedback": "This answer is wrong — you gave oxygen a 1− charge. Oxide comes from Group 16, so it is O²⁻, and one 2+ ion cancels one 2− ion exactly: CaO. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Calcium is Ca²⁺ (Group 2) and oxide is O²⁻ (Group 16). One 2+ cancels one 2−: CaO. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Ca₂O₂",
   "misconception": "you skipped the reduce step — criss-cross gives Ca₂O₂, and 2:2 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Ca₂O₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: CaO. Check: (2+) + (2−) = 0."
  },
  {
   "value": "CaO₂",
   "misconception": "you gave oxygen a 1− charge — oxide, from Group 16, is O²⁻, so one 2+ cancels one 2−",
   "feedback": "This answer is wrong — you gave oxygen a 1− charge. Oxide comes from Group 16, so it is O²⁻, and one 2+ ion cancels one 2− ion exactly: CaO. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "1+",
  "misconception": "you used the unwritten subscript as the charge - subscripts count atoms; the charge comes from the balance",
  "feedback": "This answer is wrong - you read the one-to-one pairing as a 1+ charge. Subscripts count atoms; the charge comes from the balance: the single S²⁻ carries 2−, so the lead ion must contribute 2+, written 2+."
 },
 {
  "value": "2−",
  "misconception": "you flipped the sign - the metal ion is the positive one",
  "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the sulfide ion is 2−, and the lead ion balances it with 2+, written 2+."
 }
]

{
 "keyExplanation": "Sulfide, from Group 16, is S²⁻; one of them totals 2−. One lead ion must contribute 2+. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "1+",
   "misconception": "you used the unwritten subscript as the charge - subscripts count atoms; the charge comes from the balance",
   "feedback": "This answer is wrong - you read the one-to-one pairing as a 1+ charge. Subscripts count atoms; the charge comes from the balance: the single S²⁻ carries 2−, so the lead ion must contribute 2+, written 2+.",
   "check": {
    "expr": "1",
    "expected": 1
   }
  },
  {
   "value": "2−",
   "misconception": "you flipped the sign - the metal ion is the positive one",
   "feedback": "This answer is wrong - you flipped the sign. The metal ion is the positive one: the sulfide ion is 2−, and the lead ion balances it with 2+, written 2+.",
   "check": {
    "expr": "-2",
    "expected": -2
   }
  }
 ]
}

[
 {
  "value": "2+",
  "misconception": "you used iron's subscript as its charge — the 2 counts atoms; the charge comes from the balance",
  "feedback": "This answer is wrong — you used iron's subscript as its charge. The 2 counts atoms; the charge comes from the balance: three S²⁻ total 6−, shared by two iron ions, so each iron ion is 3+."
 },
 {
  "value": "6+",
  "misconception": "you forgot to share the total — 6+ is split between two iron ions",
  "feedback": "This answer is wrong — you forgot to share the total. The 6+ that balances three S²⁻ is split between two iron ions: 6+ ÷ 2 = 3+ each."
 }
]

{
 "keyExplanation": "Sulfide, from Group 16, is S²⁻; three of them total 6−. The total positive charge must be 6+, shared by two iron ions: 6+ ÷ 2 = 3+ each. Check: 2 × (3+) + 3 × (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "2+",
   "misconception": "you used iron's subscript as its charge — the 2 counts atoms; the charge comes from the balance",
   "feedback": "This answer is wrong — you used iron's subscript as its charge. The 2 counts atoms; the charge comes from the balance: three S²⁻ total 6−, shared by two iron ions, so each iron ion is 3+."
  },
  {
   "value": "6+",
   "misconception": "you forgot to share the total — 6+ is split between two iron ions",
   "feedback": "This answer is wrong — you forgot to share the total. The 6+ that balances three S²⁻ is split between two iron ions: 6+ ÷ 2 = 3+ each."
  }
 ]
}

[
 {
  "value": "manganese(I) iodide",
  "misconception": "you counted the single manganese atom for the numeral — the numeral is the ion's 2+ charge, found from the two I⁻",
  "feedback": "This answer is wrong — you counted atoms for the numeral. The numeral is the charge on one manganese ion: two I⁻ ions total 2−, so it is 2+ and MnI₂ is manganese(II) iodide."
 },
 {
  "value": "manganese iodide",
  "misconception": "you left out the numeral — the name must show the metal ion's charge",
  "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two I⁻ ions total 2−, so the manganese ion is 2+ and MnI₂ is manganese(II) iodide."
 }
]

{
 "keyExplanation": "Two I⁻ ions total 2−, so the manganese ion is 2+ — manganese(II). MnI₂ is manganese(II) iodide.",
 "anticipatedWrong": [
  {
   "value": "manganese(I) iodide",
   "misconception": "you counted the single manganese atom for the numeral — the numeral is the ion's 2+ charge, found from the two I⁻",
   "feedback": "This answer is wrong — you counted atoms for the numeral. The numeral is the charge on one manganese ion: two I⁻ ions total 2−, so it is 2+ and MnI₂ is manganese(II) iodide."
  },
  {
   "value": "manganese iodide",
   "misconception": "you left out the numeral — the name must show the metal ion's charge",
   "feedback": "This answer is wrong — you left out the numeral. The name must show the metal ion's charge: two I⁻ ions total 2−, so the manganese ion is 2+ and MnI₂ is manganese(II) iodide."
  }
 ]
}

[
 {
  "value": "chromium(II) oxide",
  "misconception": "you used a subscript as the numeral — the 2 counts oxygen atoms; the numeral is the charge on one metal ion",
  "feedback": "This answer is wrong — you used a subscript as the numeral. The 2 counts oxygen atoms; balancing two O²⁻ (4− in all) makes the single chromium ion 4+, so CrO₂ is chromium(IV) oxide."
 },
 {
  "value": "chromium(III) oxide",
  "misconception": "you gave the wrong charge — the numeral comes from THIS formula's balance, not from chromium's charge in other compounds",
  "feedback": "This answer is wrong — you gave the wrong charge. The numeral comes from this formula's balance, not from chromium's charge in other compounds: two O²⁻ make 4−, so this chromium ion is 4+ and CrO₂ is chromium(IV) oxide."
 }
]

{
 "keyExplanation": "Two O²⁻ ions make 4− in total, and the single chromium ion balances it, so that ion is 4+ — chromium(IV). CrO₂ is chromium(IV) oxide.",
 "anticipatedWrong": [
  {
   "value": "chromium(II) oxide",
   "misconception": "you used a subscript as the numeral — the 2 counts oxygen atoms; the numeral is the charge on one metal ion",
   "feedback": "This answer is wrong — you used a subscript as the numeral. The 2 counts oxygen atoms; balancing two O²⁻ (4− in all) makes the single chromium ion 4+, so CrO₂ is chromium(IV) oxide."
  },
  {
   "value": "chromium(III) oxide",
   "misconception": "you gave the wrong charge — the numeral comes from THIS formula's balance, not from chromium's charge in other compounds",
   "feedback": "This answer is wrong — you gave the wrong charge. The numeral comes from this formula's balance, not from chromium's charge in other compounds: two O²⁻ make 4−, so this chromium ion is 4+ and CrO₂ is chromium(IV) oxide."
  }
 ]
}

[
 {
  "value": "Cr₂F",
  "misconception": "you turned the numeral into chromium's subscript — the II is the charge on one chromium ion",
  "feedback": "This answer is wrong — you turned the numeral into chromium's subscript. The II is the charge on one chromium ion: one 2+ ion needs two 1− ions, so chromium(II) fluoride is CrF₂."
 },
 {
  "value": "CrF",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. Chromium(II) is 2+ and each fluoride is only 1−, so it takes two fluorides to reach zero: CrF₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Chromium(II) is Cr²⁺ and fluoride is F⁻. One 2+ ion needs two 1− ions: CrF₂. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Cr₂F",
   "misconception": "you turned the numeral into chromium's subscript — the II is the charge on one chromium ion",
   "feedback": "This answer is wrong — you turned the numeral into chromium's subscript. The II is the charge on one chromium ion: one 2+ ion needs two 1− ions, so chromium(II) fluoride is CrF₂."
  },
  {
   "value": "CrF",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. Chromium(II) is 2+ and each fluoride is only 1−, so it takes two fluorides to reach zero: CrF₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "Co₂S₂",
  "misconception": "you skipped the reduce step — criss-cross gives Co₂S₂, and 2:2 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Co₂S₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: CoS. Check: (2+) + (2−) = 0."
 },
 {
  "value": "Co₂S",
  "misconception": "you turned the numeral into a subscript — the II is the cobalt ion's 2+ charge",
  "feedback": "This answer is wrong — you turned the numeral into a subscript. The II is the cobalt ion's 2+ charge, and the subscripts come from the balancing: one 2+ ion cancels one 2− sulfide exactly, a 1:1 ratio: CoS. Check: (2+) + (2−) = 0."
 }
]

{
 "keyExplanation": "Cobalt(II) is Co²⁺ and sulfide is S²⁻. One 2+ cancels one 2−: CoS. Check: (2+) + (2−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Co₂S₂",
   "misconception": "you skipped the reduce step — criss-cross gives Co₂S₂, and 2:2 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Co₂S₂, but a formula shows the simplest whole-number ratio of ions, and 2:2 reduces to 1:1: CoS. Check: (2+) + (2−) = 0."
  },
  {
   "value": "Co₂S",
   "misconception": "you turned the numeral into a subscript — the II is the cobalt ion's 2+ charge",
   "feedback": "This answer is wrong — you turned the numeral into a subscript. The II is the cobalt ion's 2+ charge, and the subscripts come from the balancing: one 2+ ion cancels one 2− sulfide exactly, a 1:1 ratio: CoS. Check: (2+) + (2−) = 0."
  }
 ]
}

[
 {
  "value": "Al₃(PO₄)₃",
  "misconception": "you skipped the reduce step — criss-cross gives Al₃(PO₄)₃, and 3:3 reduces to 1:1",
  "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Al₃(PO₄)₃, but a formula shows the simplest whole-number ratio of ions, and 3:3 reduces to 1:1: AlPO₄. Check: (3+) + (3−) = 0."
 },
 {
  "value": "AlP",
  "misconception": "you wrote the bare phosphide ion — phosphate is the whole block PO₄³⁻",
  "feedback": "This answer is wrong — you wrote the bare phosphide ion. Phosphate is the whole block PO₄³⁻, from the polyatomic roster, and one Al³⁺ cancels its 3− exactly: AlPO₄. Check: (3+) + (3−) = 0."
 }
]

{
 "keyExplanation": "Aluminum is Al³⁺ and phosphate is PO₄³⁻. One 3+ cancels one 3− — a 1:1 ratio — and the single phosphate block needs no parentheses: AlPO₄. Check: (3+) + (3−) = 0.",
 "anticipatedWrong": [
  {
   "value": "Al₃(PO₄)₃",
   "misconception": "you skipped the reduce step — criss-cross gives Al₃(PO₄)₃, and 3:3 reduces to 1:1",
   "feedback": "This answer is wrong — you skipped the reduce step. Criss-cross gives Al₃(PO₄)₃, but a formula shows the simplest whole-number ratio of ions, and 3:3 reduces to 1:1: AlPO₄. Check: (3+) + (3−) = 0."
  },
  {
   "value": "AlP",
   "misconception": "you wrote the bare phosphide ion — phosphate is the whole block PO₄³⁻",
   "feedback": "This answer is wrong — you wrote the bare phosphide ion. Phosphate is the whole block PO₄³⁻, from the polyatomic roster, and one Al³⁺ cancels its 3− exactly: AlPO₄. Check: (3+) + (3−) = 0."
  }
 ]
}

[
 {
  "value": "BaC₂H₃O₂",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions. One Ba²⁺ needs two C₂H₃O₂⁻ ions, and the repeated acetate goes in parentheses: Ba(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "BaC₄H₆O₄",
  "misconception": "you multiplied the acetate's atoms through instead of wrapping the repeated ion in parentheses",
  "feedback": "This answer is wrong — you multiplied the acetate's atoms through. The formula must show two whole C₂H₃O₂⁻ ions, not their atoms merged into one block, so the repeated acetate goes in parentheses: Ba(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Barium forms Ba²⁺, and the acetate ion is C₂H₃O₂⁻. One 2+ ion needs two 1− ions, and the acetate ion repeats, so parentheses: Ba(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "BaC₂H₃O₂",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio, but one 2+ ion needs two 1− ions. One Ba²⁺ needs two C₂H₃O₂⁻ ions, and the repeated acetate goes in parentheses: Ba(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "BaC₄H₆O₄",
   "misconception": "you multiplied the acetate's atoms through instead of wrapping the repeated ion in parentheses",
   "feedback": "This answer is wrong — you multiplied the acetate's atoms through. The formula must show two whole C₂H₃O₂⁻ ions, not their atoms merged into one block, so the repeated acetate goes in parentheses: Ba(C₂H₃O₂)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "chromium nitrate nonahydrate",
  "misconception": "you left out the Roman numeral — three NO₃⁻ ions total 3−, so the chromium is 3+",
  "feedback": "This answer is wrong — you left out the Roman numeral. Three NO₃⁻ ions total 3−, so the chromium ion is 3+ and the name must show it: chromium(III) nitrate. Nine waters: nonahydrate. Cr(NO₃)₃·9H₂O is chromium(III) nitrate nonahydrate."
 },
 {
  "value": "chromium(III) nitrate trihydrate",
  "misconception": "you counted the nitrates for the prefix — the prefix counts the nine waters",
  "feedback": "This answer is wrong — you counted the nitrates for the prefix. The prefix counts water molecules only: nine waters means nonahydrate. Cr(NO₃)₃·9H₂O is chromium(III) nitrate nonahydrate."
 }
]

{
 "keyExplanation": "Three nitrate ions total 3−, so the chromium ion is 3+ — chromium(III) nitrate. Nine waters: nonahydrate. Cr(NO₃)₃·9H₂O is chromium(III) nitrate nonahydrate.",
 "anticipatedWrong": [
  {
   "value": "chromium nitrate nonahydrate",
   "misconception": "you left out the Roman numeral — three NO₃⁻ ions total 3−, so the chromium is 3+",
   "feedback": "This answer is wrong — you left out the Roman numeral. Three NO₃⁻ ions total 3−, so the chromium ion is 3+ and the name must show it: chromium(III) nitrate. Nine waters: nonahydrate. Cr(NO₃)₃·9H₂O is chromium(III) nitrate nonahydrate."
  },
  {
   "value": "chromium(III) nitrate trihydrate",
   "misconception": "you counted the nitrates for the prefix — the prefix counts the nine waters",
   "feedback": "This answer is wrong — you counted the nitrates for the prefix. The prefix counts water molecules only: nine waters means nonahydrate. Cr(NO₃)₃·9H₂O is chromium(III) nitrate nonahydrate."
  }
 ]
}

[
 {
  "value": "LiBr₂·H₂O",
  "misconception": "you moved the water count onto the compound — the prefix counts water molecules only",
  "feedback": "This answer is wrong — you moved the water count onto the compound. The prefix counts water molecules only: lithium is Li⁺ and bromide is Br⁻, so the ionic part stays LiBr, and di- adds 2 waters: LiBr·2H₂O."
 },
 {
  "value": "LiBr",
  "misconception": "you dropped the water — dihydrate adds ·2H₂O after the compound",
  "feedback": "This answer is wrong — you dropped the water. Dihydrate adds ·2H₂O after the compound: LiBr·2H₂O."
 }
]

{
 "keyExplanation": "Lithium is Li⁺ and bromide is Br⁻, so the ionic part is LiBr — check: (1+) + (1−) = 0. Di- means 2 waters: LiBr·2H₂O.",
 "anticipatedWrong": [
  {
   "value": "LiBr₂·H₂O",
   "misconception": "you moved the water count onto the compound — the prefix counts water molecules only",
   "feedback": "This answer is wrong — you moved the water count onto the compound. The prefix counts water molecules only: lithium is Li⁺ and bromide is Br⁻, so the ionic part stays LiBr, and di- adds 2 waters: LiBr·2H₂O."
  },
  {
   "value": "LiBr",
   "misconception": "you dropped the water — dihydrate adds ·2H₂O after the compound",
   "feedback": "This answer is wrong — you dropped the water. Dihydrate adds ·2H₂O after the compound: LiBr·2H₂O."
  }
 ]
}

[
 {
  "value": "nitrogen hydride bromide",
  "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
  "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so NH₄Br is ammonium bromide."
 },
 {
  "value": "ammonium bromate",
  "misconception": "you gave the bare ion an -ate name — there is no oxygen block here; a lone bromine is the single-atom Br⁻ ion, bromide",
  "feedback": "This answer is wrong — you gave the bare ion an -ate name. There is no oxygen block here: a lone bromine is the single-atom Br⁻ ion, and the -ide ending is only for single-atom negative ions, so NH₄Br is ammonium bromide."
 }
]

{
 "keyExplanation": "NH₄ is the ammonium ion, and it leads the name. Bromine here is a single-atom negative ion, so it takes the -ide ending: NH₄Br is ammonium bromide.",
 "anticipatedWrong": [
  {
   "value": "nitrogen hydride bromide",
   "misconception": "you broke the ammonium block into element-style names — NH₄⁺ is one intact polyatomic ion whose roster name is ammonium",
   "feedback": "This answer is wrong — you broke the ammonium block into element-style names. NH₄⁺ is one intact polyatomic ion whose roster name is ammonium, so NH₄Br is ammonium bromide."
  },
  {
   "value": "ammonium bromate",
   "misconception": "you gave the bare ion an -ate name — there is no oxygen block here; a lone bromine is the single-atom Br⁻ ion, bromide",
   "feedback": "This answer is wrong — you gave the bare ion an -ate name. There is no oxygen block here: a lone bromine is the single-atom Br⁻ ion, and the -ide ending is only for single-atom negative ions, so NH₄Br is ammonium bromide."
  }
 ]
}

[
 {
  "value": "nickel carbonate",
  "misconception": "you skipped the numeral — one CO₃²⁻ ion is 2−, so the nickel is 2+",
  "feedback": "This answer is wrong — you skipped the numeral. Nickel is a transition metal, so its name must show the charge: the carbonate ion is CO₃²⁻, a 2− charge, making the nickel ion 2+ — nickel(II) carbonate."
 },
 {
  "value": "nickel(I) carbonate",
  "misconception": "you gave the wrong charge — carbonate carries 2−, so the nickel must be 2+",
  "feedback": "This answer is wrong — you gave the wrong charge. Carbonate is CO₃²⁻ — the whole block carries 2−, and one nickel ion must balance all of it: the nickel is 2+ — nickel(II) carbonate."
 }
]

{
 "keyExplanation": "No dot; nickel is a transition metal, and one CO₃²⁻ ion is 2−, so the nickel ion is 2+ — nickel(II). The polyatomic carbonate keeps its name: nickel(II) carbonate.",
 "anticipatedWrong": [
  {
   "value": "nickel carbonate",
   "misconception": "you skipped the numeral — one CO₃²⁻ ion is 2−, so the nickel is 2+",
   "feedback": "This answer is wrong — you skipped the numeral. Nickel is a transition metal, so its name must show the charge: the carbonate ion is CO₃²⁻, a 2− charge, making the nickel ion 2+ — nickel(II) carbonate."
  },
  {
   "value": "nickel(I) carbonate",
   "misconception": "you gave the wrong charge — carbonate carries 2−, so the nickel must be 2+",
   "feedback": "This answer is wrong — you gave the wrong charge. Carbonate is CO₃²⁻ — the whole block carries 2−, and one nickel ion must balance all of it: the nickel is 2+ — nickel(II) carbonate."
  }
 ]
}

[
 {
  "value": "SrClO₃",
  "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− chlorate ions",
  "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2+ strontium ion needs two 1− chlorate ions, wrapped in parentheses: Sr(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 },
 {
  "value": "SrCl₂",
  "misconception": "you wrote chloride — chlorate is the polyatomic ClO₃⁻, with three oxygens",
  "feedback": "This answer is wrong — you wrote chloride. Chlorate is the polyatomic ClO₃⁻, with three oxygens, and one 2+ ion needs two of them, wrapped in parentheses: Sr(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
 }
]

{
 "keyExplanation": "Strontium is Sr²⁺ (Group 2) and chlorate is ClO₃⁻. One 2+ ion needs two 1− ions, and chlorate repeats, so parentheses: Sr(ClO₃)₂. Check: (2+) + 2 × (1−) = 0.",
 "anticipatedWrong": [
  {
   "value": "SrClO₃",
   "misconception": "you wrote a 1:1 ratio — one 2+ ion needs two 1− chlorate ions",
   "feedback": "This answer is wrong — you wrote a 1:1 ratio. One 2+ strontium ion needs two 1− chlorate ions, wrapped in parentheses: Sr(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  },
  {
   "value": "SrCl₂",
   "misconception": "you wrote chloride — chlorate is the polyatomic ClO₃⁻, with three oxygens",
   "feedback": "This answer is wrong — you wrote chloride. Chlorate is the polyatomic ClO₃⁻, with three oxygens, and one 2+ ion needs two of them, wrapped in parentheses: Sr(ClO₃)₂. Check: (2+) + 2 × (1−) = 0."
  }
 ]
}

[
 {
  "value": "magnesium(II) sulfate monohydrate",
  "misconception": "you added a Roman numeral to a fixed-charge metal — magnesium, in Group 2, is always 2+",
  "feedback": "This answer is wrong — you added a Roman numeral to a fixed-charge metal. Magnesium, in Group 2, is always 2+, so its name never carries a numeral: MgSO₄·H₂O is magnesium sulfate monohydrate."
 },
 {
  "value": "magnesium sulfate",
  "misconception": "you dropped the water word — a dot with no number still means one water: monohydrate",
  "feedback": "This answer is wrong — you dropped the water word. A raised dot with no number still counts one water, and a count of one takes the mono- prefix: MgSO₄·H₂O is magnesium sulfate monohydrate."
 }
]

{
 "keyExplanation": "A raised dot: name the part before it first. Magnesium is a fixed-charge Group 2 metal — no numeral — and the polyatomic sulfate keeps its own name: magnesium sulfate. The dot with no number means one water: monohydrate. MgSO₄·H₂O is magnesium sulfate monohydrate.",
 "anticipatedWrong": [
  {
   "value": "magnesium(II) sulfate monohydrate",
   "misconception": "you added a Roman numeral to a fixed-charge metal — magnesium, in Group 2, is always 2+",
   "feedback": "This answer is wrong — you added a Roman numeral to a fixed-charge metal. Magnesium, in Group 2, is always 2+, so its name never carries a numeral: MgSO₄·H₂O is magnesium sulfate monohydrate."
  },
  {
   "value": "magnesium sulfate",
   "misconception": "you dropped the water word — a dot with no number still means one water: monohydrate",
   "feedback": "This answer is wrong — you dropped the water word. A raised dot with no number still counts one water, and a count of one takes the mono- prefix: MgSO₄·H₂O is magnesium sulfate monohydrate."
  }
 ]
}