Unit 5 — End of Topic Tests (reviewer copy: all keys and model answers visible)

[
 {
  "value": 1,
  "misconception": "you read the first element's count — the subscript 4 sits after Br, so it counts bromine",
  "feedback": "This answer is wrong — you read the first element's count — the subscript 4 sits after Br, so it counts bromine."
 },
 {
  "value": 5,
  "misconception": "you counted every atom in the molecule — the question asks for the bromine atoms alone",
  "feedback": "This answer is wrong — you counted every atom in the molecule — the question asks for the bromine atoms alone."
 }
]

{
 "keyExplanation": "The subscript after Br is 4, so one SeBr₄ molecule contains 4 bromine atoms. The selenium, with no subscript, counts as one atom — but it is not bromine.",
 "anticipatedWrong": [
  {
   "value": 1,
   "misconception": "you read the first element's count — the subscript 4 sits after Br, so it counts bromine",
   "feedback": "This answer is wrong — you read the first element's count — the subscript 4 sits after Br, so it counts bromine."
  },
  {
   "value": 5,
   "misconception": "you counted every atom in the molecule — the question asks for the bromine atoms alone",
   "feedback": "This answer is wrong — you counted every atom in the molecule — the question asks for the bromine atoms alone."
  }
 ]
}

[
 {
  "value": 4,
  "misconception": "you counted the pairs, not the electrons — four pairs touch the selenium, and each pair holds two electrons",
  "feedback": "This answer is wrong — you counted the pairs, not the electrons — four pairs touch the selenium, and each pair holds two electrons."
 },
 {
  "value": 6,
  "misconception": "you counted the double bond as one shared pair — a double bond is two shared pairs, four electrons",
  "feedback": "This answer is wrong — you counted the double bond as one shared pair — a double bond is two shared pairs, four electrons."
 },
 {
  "value": 18,
  "misconception": "you counted every electron in the diagram — the oxygens' lone pairs belong to the oxygens, not the selenium",
  "feedback": "This answer is wrong — you counted every electron in the diagram — the oxygens' lone pairs belong to the oxygens, not the selenium."
 }
]

{
 "keyExplanation": "Count what is drawn around the selenium: the double bond is two shared pairs (4 electrons), the single bond is one shared pair (2), and the selenium's own lone pair adds 2. 4 + 2 + 2 = 8 electrons.",
 "anticipatedWrong": [
  {
   "value": 4,
   "misconception": "you counted the pairs, not the electrons — four pairs touch the selenium, and each pair holds two electrons",
   "feedback": "This answer is wrong — you counted the pairs, not the electrons — four pairs touch the selenium, and each pair holds two electrons."
  },
  {
   "value": 6,
   "misconception": "you counted the double bond as one shared pair — a double bond is two shared pairs, four electrons",
   "feedback": "This answer is wrong — you counted the double bond as one shared pair — a double bond is two shared pairs, four electrons."
  },
  {
   "value": 18,
   "misconception": "you counted every electron in the diagram — the oxygens' lone pairs belong to the oxygens, not the selenium",
   "feedback": "This answer is wrong — you counted every electron in the diagram — the oxygens' lone pairs belong to the oxygens, not the selenium."
  }
 ]
}

[
 "Write down the values in the question:",
 "valence electrons = 6",
 "Write down the relationship:",
 "typical bonds = 8 − valence electrons",
 "Substitute in the values, and calculate:",
 "typical bonds = 8 − 6",
 "typical bonds = 2"
]

[
 {
  "value": 16,
  "misconception": "you used the group number as the bond count — Group 16 atoms have six valence electrons and need just two more",
  "feedback": "This answer is wrong — you used the group number as the bond count — Group 16 atoms have six valence electrons and need just two more."
 },
 {
  "value": 6,
  "misconception": "you found the valence count and stopped — six is the valence count, and the bonds equal 8 − 6",
  "feedback": "This answer is wrong — you found the valence count and stopped — six is the valence count, and the bonds equal 8 − 6."
 },
 {
  "value": 8,
  "misconception": "you answered with the octet target — eight is the electron goal, not the number of bonds",
  "feedback": "This answer is wrong — you answered with the octet target — eight is the electron goal, not the number of bonds."
 }
]

{
 "keyExplanation": "Write down the values in the question:\nvalence electrons = 6\nWrite down the relationship:\ntypical bonds = 8 − valence electrons\nSubstitute in the values, and calculate:\ntypical bonds = 8 − 6\ntypical bonds = 2",
 "anticipatedWrong": [
  {
   "value": 16,
   "misconception": "you used the group number as the bond count — Group 16 atoms have six valence electrons and need just two more",
   "feedback": "This answer is wrong — you used the group number as the bond count — Group 16 atoms have six valence electrons and need just two more."
  },
  {
   "value": 6,
   "misconception": "you found the valence count and stopped — six is the valence count, and the bonds equal 8 − 6",
   "feedback": "This answer is wrong — you found the valence count and stopped — six is the valence count, and the bonds equal 8 − 6."
  },
  {
   "value": 8,
   "misconception": "you answered with the octet target — eight is the electron goal, not the number of bonds",
   "feedback": "This answer is wrong — you answered with the octet target — eight is the electron goal, not the number of bonds."
  }
 ]
}

[
 "Write down the values in the question:",
 "tin: 1 atom × 4 valence electrons",
 "bromine: 4 atoms × 7 valence electrons",
 "Write down the relationship:",
 "total valence electrons = the sum over every element of (atoms × valence electrons)",
 "Substitute in the values, and calculate:",
 "total valence electrons = (1 × 4) + (4 × 7) = 4 + 28 = 32"
]

[
 {
  "value": 11,
  "misconception": "you counted one bromine — the subscript 4 means four bromine atoms",
  "feedback": "This answer is wrong — you counted one bromine — the subscript 4 means four bromine atoms."
 },
 {
  "value": 82,
  "misconception": "you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven",
  "feedback": "This answer is wrong — you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven."
 },
 {
  "value": 5,
  "misconception": "you counted the atoms in the formula, not their valence electrons",
  "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
 }
]

{
 "keyExplanation": "tin: 1 atom × 4 valence electrons\nbromine: 4 atoms × 7 valence electrons\ntotal valence electrons = (1 × 4) + (4 × 7) = 4 + 28 = 32",
 "anticipatedWrong": [
  {
   "value": 11,
   "misconception": "you counted one bromine — the subscript 4 means four bromine atoms",
   "feedback": "This answer is wrong — you counted one bromine — the subscript 4 means four bromine atoms."
  },
  {
   "value": 82,
   "misconception": "you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven",
   "feedback": "This answer is wrong — you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven."
  },
  {
   "value": 5,
   "misconception": "you counted the atoms in the formula, not their valence electrons",
   "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
  }
 ]
}

[
 {
  "value": "phosphorus hexoxide",
  "misconception": "you dropped tetra- — only mono- is dropped, and only on the first element",
  "feedback": "This answer is wrong — you dropped tetra- — only mono- is dropped, and only on the first element."
 },
 {
  "value": "tetraphosphorus oxide",
  "misconception": "you left the six oxygens uncounted — the second element needs hex(a)-",
  "feedback": "This answer is wrong — you left the six oxygens uncounted — the second element needs hex(a)-."
 }
]

{
 "keyExplanation": "Four phosphorus atoms keep tetra-, six oxygens take hexa-, and hexa- + oxide elides to hexoxide. P₄O₆ is tetraphosphorus hexoxide.",
 "anticipatedWrong": [
  {
   "value": "phosphorus hexoxide",
   "misconception": "you dropped tetra- — only mono- is dropped, and only on the first element",
   "feedback": "This answer is wrong — you dropped tetra- — only mono- is dropped, and only on the first element."
  },
  {
   "value": "tetraphosphorus oxide",
   "misconception": "you left the six oxygens uncounted — the second element needs hex(a)-",
   "feedback": "This answer is wrong — you left the six oxygens uncounted — the second element needs hex(a)-."
  }
 ]
}

[
 {
  "value": "Br₅F",
  "misconception": "you attached the count to the wrong element — penta- belongs to fluoride, so the 5 follows F",
  "feedback": "This answer is wrong — you attached the count to the wrong element — penta- belongs to fluoride, so the 5 follows F."
 },
 {
  "value": "BrF",
  "misconception": "you dropped the count — penta- means the formula needs a 5 after F",
  "feedback": "This answer is wrong — you dropped the count — penta- means the formula needs a 5 after F."
 }
]

{
 "keyExplanation": "Bromine has no prefix — one Br, no subscript. penta- means five, so F takes the subscript 5. Bromine pentafluoride is BrF₅.",
 "anticipatedWrong": [
  {
   "value": "Br₅F",
   "misconception": "you attached the count to the wrong element — penta- belongs to fluoride, so the 5 follows F",
   "feedback": "This answer is wrong — you attached the count to the wrong element — penta- belongs to fluoride, so the 5 follows F."
  },
  {
   "value": "BrF",
   "misconception": "you dropped the count — penta- means the formula needs a 5 after F",
   "feedback": "This answer is wrong — you dropped the count — penta- means the formula needs a 5 after F."
  }
 ]
}

[
 {
  "value": "trimagnesium dinitride",
  "misconception": "you counted with prefixes — magnesium is a metal, so Mg₃N₂ is ionic, and ionic names never use prefixes",
  "feedback": "This answer is wrong — you counted with prefixes — magnesium is a metal, so Mg₃N₂ is ionic, and ionic names never use prefixes."
 },
 {
  "value": "magnesium(II) nitride",
  "misconception": "you added a Roman numeral — magnesium forms only Mg²⁺, so no numeral is needed and none is written",
  "feedback": "This answer is wrong — you added a Roman numeral — magnesium forms only Mg²⁺, so no numeral is needed and none is written."
 }
]

{
 "keyExplanation": "First element: magnesium, a metal — ionic naming, no prefixes. The subscripts come from the charges (three Mg²⁺ balance two N³⁻), so the name is simply magnesium nitride.",
 "anticipatedWrong": [
  {
   "value": "trimagnesium dinitride",
   "misconception": "you counted with prefixes — magnesium is a metal, so Mg₃N₂ is ionic, and ionic names never use prefixes",
   "feedback": "This answer is wrong — you counted with prefixes — magnesium is a metal, so Mg₃N₂ is ionic, and ionic names never use prefixes."
  },
  {
   "value": "magnesium(II) nitride",
   "misconception": "you added a Roman numeral — magnesium forms only Mg²⁺, so no numeral is needed and none is written",
   "feedback": "This answer is wrong — you added a Roman numeral — magnesium forms only Mg²⁺, so no numeral is needed and none is written."
  }
 ]
}

[
 {
  "value": 2,
  "misconception": "you counted the elements, not the atoms — add the two subscripts: 4 + 10",
  "feedback": "This answer is wrong — you counted the elements, not the atoms — add the two subscripts: 4 + 10."
 },
 {
  "value": 10,
  "misconception": "you took the largest subscript as the total — every element's count joins the sum",
  "feedback": "This answer is wrong — you took the largest subscript as the total — every element's count joins the sum."
 },
 {
  "value": 4,
  "misconception": "you read only the first subscript — the 10 hydrogens count too",
  "feedback": "This answer is wrong — you read only the first subscript — the 10 hydrogens count too."
 }
]

{
 "keyExplanation": "Each subscript counts its own element: 4 carbon and 10 hydrogen. The total is 4 + 10 = 14 atoms in one butane molecule.",
 "anticipatedWrong": [
  {
   "value": 2,
   "misconception": "you counted the elements, not the atoms — add the two subscripts: 4 + 10",
   "feedback": "This answer is wrong — you counted the elements, not the atoms — add the two subscripts: 4 + 10."
  },
  {
   "value": 10,
   "misconception": "you took the largest subscript as the total — every element's count joins the sum",
   "feedback": "This answer is wrong — you took the largest subscript as the total — every element's count joins the sum."
  },
  {
   "value": 4,
   "misconception": "you read only the first subscript — the 10 hydrogens count too",
   "feedback": "This answer is wrong — you read only the first subscript — the 10 hydrogens count too."
  }
 ]
}

[
 {
  "value": 4,
  "misconception": "you counted the lines, not the electrons — each of the four lines is a pair of two electrons",
  "feedback": "This answer is wrong — you counted the lines, not the electrons — each of the four lines is a pair of two electrons."
 },
 {
  "value": 6,
  "misconception": "you skipped the silicon–silicon bond — its shared pair surrounds the left silicon too",
  "feedback": "This answer is wrong — you skipped the silicon–silicon bond — its shared pair surrounds the left silicon too."
 },
 {
  "value": 14,
  "misconception": "you counted every electron in the diagram — count only the pairs touching the left silicon",
  "feedback": "This answer is wrong — you counted every electron in the diagram — count only the pairs touching the left silicon."
 }
]

{
 "keyExplanation": "Four lines touch the left silicon: three Si–H bonds and the Si–Si bond. Each line is one shared pair of two electrons: 4 × 2 = 8 electrons surround the left silicon.",
 "anticipatedWrong": [
  {
   "value": 4,
   "misconception": "you counted the lines, not the electrons — each of the four lines is a pair of two electrons",
   "feedback": "This answer is wrong — you counted the lines, not the electrons — each of the four lines is a pair of two electrons."
  },
  {
   "value": 6,
   "misconception": "you skipped the silicon–silicon bond — its shared pair surrounds the left silicon too",
   "feedback": "This answer is wrong — you skipped the silicon–silicon bond — its shared pair surrounds the left silicon too."
  },
  {
   "value": 14,
   "misconception": "you counted every electron in the diagram — count only the pairs touching the left silicon",
   "feedback": "This answer is wrong — you counted every electron in the diagram — count only the pairs touching the left silicon."
  }
 ]
}

[
 "Write down the values in the question:",
 "valence electrons of E = 5",
 "Write down the relationship:",
 "typical bonds = 8 − valence electrons",
 "Substitute in the values, and calculate:",
 "typical bonds = 8 − 5 = 3",
 "Each J atom forms one bond, so three bonds means 3 J atoms."
]

[
 {
  "value": 5,
  "misconception": "you used E's valence count as the number of partners — the partners equal the bonds, which is 8 − 5",
  "feedback": "This answer is wrong — you used E's valence count as the number of partners — the partners equal the bonds, which is 8 − 5."
 },
 {
  "value": 8,
  "misconception": "you used the full octet count as the number of partners — eight is the electron target, not the bond count",
  "feedback": "This answer is wrong — you used the full octet count as the number of partners — eight is the electron target, not the bond count."
 },
 {
  "value": 1,
  "misconception": "you assumed E, like its partner, forms just one bond — E needs three more electrons, so it forms three bonds",
  "feedback": "This answer is wrong — you assumed E, like its partner, forms just one bond — E needs three more electrons, so it forms three bonds."
 }
]

{
 "keyExplanation": "E typically forms 8 − 5 = 3 bonds. Each J atom forms one bond, so each bond uses one J atom. Three bonds means three J atoms.",
 "anticipatedWrong": [
  {
   "value": 5,
   "misconception": "you used E's valence count as the number of partners — the partners equal the bonds, which is 8 − 5",
   "feedback": "This answer is wrong — you used E's valence count as the number of partners — the partners equal the bonds, which is 8 − 5."
  },
  {
   "value": 8,
   "misconception": "you used the full octet count as the number of partners — eight is the electron target, not the bond count",
   "feedback": "This answer is wrong — you used the full octet count as the number of partners — eight is the electron target, not the bond count."
  },
  {
   "value": 1,
   "misconception": "you assumed E, like its partner, forms just one bond — E needs three more electrons, so it forms three bonds",
   "feedback": "This answer is wrong — you assumed E, like its partner, forms just one bond — E needs three more electrons, so it forms three bonds."
  }
 ]
}

[
 "Write down the values in the question:",
 "boron: 1 atom × 3 valence electrons",
 "iodine: 3 atoms × 7 valence electrons",
 "Write down the relationship:",
 "total valence electrons = the sum over every element of (atoms × valence electrons)",
 "Substitute in the values, and calculate:",
 "total valence electrons = 3 + 21 = 24"
]

[
 {
  "value": 10,
  "misconception": "you counted one iodine — the subscript 3 means three iodine atoms",
  "feedback": "This answer is wrong — you counted one iodine — the subscript 3 means three iodine atoms."
 },
 {
  "value": 54,
  "misconception": "you used iodine's group number 17 as its valence count — Group 17 atoms have seven valence electrons",
  "feedback": "This answer is wrong — you used iodine's group number 17 as its valence count — Group 17 atoms have seven valence electrons."
 },
 {
  "value": 4,
  "misconception": "you counted the atoms in the formula, not their valence electrons",
  "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
 }
]

{
 "keyExplanation": "boron: 1 atom × 3 valence electrons\niodine: 3 atoms × 7 valence electrons\ntotal valence electrons = (1 × 3) + (3 × 7) = 3 + 21 = 24",
 "anticipatedWrong": [
  {
   "value": 10,
   "misconception": "you counted one iodine — the subscript 3 means three iodine atoms",
   "feedback": "This answer is wrong — you counted one iodine — the subscript 3 means three iodine atoms."
  },
  {
   "value": 54,
   "misconception": "you used iodine's group number 17 as its valence count — Group 17 atoms have seven valence electrons",
   "feedback": "This answer is wrong — you used iodine's group number 17 as its valence count — Group 17 atoms have seven valence electrons."
  },
  {
   "value": 4,
   "misconception": "you counted the atoms in the formula, not their valence electrons",
   "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
  }
 ]
}

[
 {
  "value": "sulfur dichloride",
  "misconception": "you dropped di- from the first element — only mono- is ever dropped; two sulfurs need di-",
  "feedback": "This answer is wrong — you dropped di- from the first element — only mono- is ever dropped; two sulfurs need di-."
 },
 {
  "value": "disulfur chloride",
  "misconception": "you left the two chlorines uncounted — the second element needs di- too",
  "feedback": "This answer is wrong — you left the two chlorines uncounted — the second element needs di- too."
 },
 {
  "value": "disulfur dichlorine",
  "misconception": "you skipped the -ide ending — the second element's ending changes to -ide",
  "feedback": "This answer is wrong — you skipped the -ide ending — the second element's ending changes to -ide."
 }
]

{
 "keyExplanation": "Two sulfurs take di- (a prefix other than mono- is never dropped, even on the first element), and two chlorines take di- with the -ide ending. S₂Cl₂ is disulfur dichloride.",
 "anticipatedWrong": [
  {
   "value": "sulfur dichloride",
   "misconception": "you dropped di- from the first element — only mono- is ever dropped; two sulfurs need di-",
   "feedback": "This answer is wrong — you dropped di- from the first element — only mono- is ever dropped; two sulfurs need di-."
  },
  {
   "value": "disulfur chloride",
   "misconception": "you left the two chlorines uncounted — the second element needs di- too",
   "feedback": "This answer is wrong — you left the two chlorines uncounted — the second element needs di- too."
  },
  {
   "value": "disulfur dichlorine",
   "misconception": "you skipped the -ide ending — the second element's ending changes to -ide",
   "feedback": "This answer is wrong — you skipped the -ide ending — the second element's ending changes to -ide."
  }
 ]
}

[
 {
  "value": "I₇F",
  "misconception": "you attached the count to the wrong element — hepta- belongs to fluoride, so the 7 follows F",
  "feedback": "This answer is wrong — you attached the count to the wrong element — hepta- belongs to fluoride, so the 7 follows F."
 },
 {
  "value": "IF₆",
  "misconception": "you slid one prefix down — hexa- is 6; hepta- is 7",
  "feedback": "This answer is wrong — you slid one prefix down — hexa- is 6; hepta- is 7."
 }
]

{
 "keyExplanation": "Iodine has no prefix — one I, no subscript. hepta- means seven, so F takes the subscript 7. Iodine heptafluoride is IF₇.",
 "anticipatedWrong": [
  {
   "value": "I₇F",
   "misconception": "you attached the count to the wrong element — hepta- belongs to fluoride, so the 7 follows F",
   "feedback": "This answer is wrong — you attached the count to the wrong element — hepta- belongs to fluoride, so the 7 follows F."
  },
  {
   "value": "IF₆",
   "misconception": "you slid one prefix down — hexa- is 6; hepta- is 7",
   "feedback": "This answer is wrong — you slid one prefix down — hexa- is 6; hepta- is 7."
  }
 ]
}

[
 {
  "value": "sulfur oxide",
  "misconception": "you skipped the count — two oxygens need di-",
  "feedback": "This answer is wrong — you skipped the count — two oxygens need di-."
 },
 {
  "value": "monosulfur dioxide",
  "misconception": "you kept mono- on the first element — mono- is always dropped there",
  "feedback": "This answer is wrong — you kept mono- on the first element — mono- is always dropped there."
 },
 {
  "value": "disulfur oxide",
  "misconception": "you attached the prefix to the wrong element — each prefix counts the element it sits in front of",
  "feedback": "This answer is wrong — you attached the prefix to the wrong element — each prefix counts the element it sits in front of."
 }
]

{
 "keyExplanation": "Sulfur and oxygen are both nonmetals, so SO₂ takes prefix naming: one sulfur keeps no prefix (mono- is dropped on the first element), and two oxygens take di- with the -ide ending — sulfur dioxide.",
 "anticipatedWrong": [
  {
   "value": "sulfur oxide",
   "misconception": "you skipped the count — two oxygens need di-",
   "feedback": "This answer is wrong — you skipped the count — two oxygens need di-."
  },
  {
   "value": "monosulfur dioxide",
   "misconception": "you kept mono- on the first element — mono- is always dropped there",
   "feedback": "This answer is wrong — you kept mono- on the first element — mono- is always dropped there."
  },
  {
   "value": "disulfur oxide",
   "misconception": "you attached the prefix to the wrong element — each prefix counts the element it sits in front of",
   "feedback": "This answer is wrong — you attached the prefix to the wrong element — each prefix counts the element it sits in front of."
  }
 ]
}

[
 {
  "value": 1,
  "misconception": "you read the first element's count — the subscript 3 sits after Br, so it counts bromine",
  "feedback": "This answer is wrong — you read the first element's count — the subscript 3 sits after Br, so it counts bromine."
 },
 {
  "value": 4,
  "misconception": "you counted every atom in the molecule — the question asks for the bromine atoms alone",
  "feedback": "This answer is wrong — you counted every atom in the molecule — the question asks for the bromine atoms alone."
 }
]

{
 "keyExplanation": "The subscript after Br is 3, so one BBr₃ molecule contains 3 bromine atoms. The boron, with no subscript, counts as one atom — but it is not bromine.",
 "anticipatedWrong": [
  {
   "value": 1,
   "misconception": "you read the first element's count — the subscript 3 sits after Br, so it counts bromine",
   "feedback": "This answer is wrong — you read the first element's count — the subscript 3 sits after Br, so it counts bromine."
  },
  {
   "value": 4,
   "misconception": "you counted every atom in the molecule — the question asks for the bromine atoms alone",
   "feedback": "This answer is wrong — you counted every atom in the molecule — the question asks for the bromine atoms alone."
  }
 ]
}

[
 {
  "value": 4,
  "misconception": "you counted the triple bond as one shared pair — a triple bond is three shared pairs, six electrons",
  "feedback": "This answer is wrong — you counted the triple bond as one shared pair — a triple bond is three shared pairs, six electrons."
 },
 {
  "value": 6,
  "misconception": "you stopped at the triple bond — the single bond's shared pair also surrounds the middle carbon",
  "feedback": "This answer is wrong — you stopped at the triple bond — the single bond's shared pair also surrounds the middle carbon."
 },
 {
  "value": 16,
  "misconception": "you counted every electron in the diagram — count only the pairs touching the middle carbon",
  "feedback": "This answer is wrong — you counted every electron in the diagram — count only the pairs touching the middle carbon."
 }
]

{
 "keyExplanation": "The triple bond is three shared pairs: 6 electrons. The single bond to the neighboring carbon adds one more pair: 2 electrons. 6 + 2 = 8 electrons surround the middle carbon.",
 "anticipatedWrong": [
  {
   "value": 4,
   "misconception": "you counted the triple bond as one shared pair — a triple bond is three shared pairs, six electrons",
   "feedback": "This answer is wrong — you counted the triple bond as one shared pair — a triple bond is three shared pairs, six electrons."
  },
  {
   "value": 6,
   "misconception": "you stopped at the triple bond — the single bond's shared pair also surrounds the middle carbon",
   "feedback": "This answer is wrong — you stopped at the triple bond — the single bond's shared pair also surrounds the middle carbon."
  },
  {
   "value": 16,
   "misconception": "you counted every electron in the diagram — count only the pairs touching the middle carbon",
   "feedback": "This answer is wrong — you counted every electron in the diagram — count only the pairs touching the middle carbon."
  }
 ]
}

[
 "Write down the values in the question:",
 "valence electrons = 4",
 "Write down the relationship:",
 "typical bonds = 8 − valence electrons",
 "Substitute in the values, and calculate:",
 "typical bonds = 8 − 4",
 "typical bonds = 4"
]

[
 {
  "value": 8,
  "misconception": "you answered with the octet target — eight is the electron goal, not the number of bonds",
  "feedback": "This answer is wrong — you answered with the octet target — eight is the electron goal, not the number of bonds."
 },
 {
  "value": 12,
  "misconception": "you added the valence count to eight — the relationship subtracts it: bonds = 8 − 4",
  "feedback": "This answer is wrong — you added the valence count to eight — the relationship subtracts it: bonds = 8 − 4."
 }
]

{
 "keyExplanation": "Write down the values in the question:\nvalence electrons = 4\nWrite down the relationship:\ntypical bonds = 8 − valence electrons\nSubstitute in the values, and calculate:\ntypical bonds = 8 − 4\ntypical bonds = 4",
 "anticipatedWrong": [
  {
   "value": 8,
   "misconception": "you answered with the octet target — eight is the electron goal, not the number of bonds",
   "feedback": "This answer is wrong — you answered with the octet target — eight is the electron goal, not the number of bonds."
  },
  {
   "value": 12,
   "misconception": "you added the valence count to eight — the relationship subtracts it: bonds = 8 − 4",
   "feedback": "This answer is wrong — you added the valence count to eight — the relationship subtracts it: bonds = 8 − 4."
  }
 ]
}

[
 "Write down the values in the question:",
 "germanium: 1 atom × 4 valence electrons",
 "hydrogen: 4 atoms × 1 valence electron",
 "Write down the relationship:",
 "total valence electrons = the sum over every element of (atoms × valence electrons)",
 "Substitute in the values, and calculate:",
 "total valence electrons = 4 + 4 = 8"
]

[
 {
  "value": 5,
  "misconception": "you counted one hydrogen — the subscript 4 means four hydrogen atoms",
  "feedback": "This answer is wrong — you counted one hydrogen — the subscript 4 means four hydrogen atoms."
 },
 {
  "value": 18,
  "misconception": "you used the group number 14 as germanium's valence count — Group 14 atoms have four valence electrons",
  "feedback": "This answer is wrong — you used the group number 14 as germanium's valence count — Group 14 atoms have four valence electrons."
 },
 {
  "value": 40,
  "misconception": "you gave every atom eight electrons — count each atom's own valence electrons instead",
  "feedback": "This answer is wrong — you gave every atom eight electrons — count each atom's own valence electrons instead."
 }
]

{
 "keyExplanation": "germanium: 1 atom × 4 valence electrons\nhydrogen: 4 atoms × 1 valence electron\ntotal valence electrons = 4 + 4 = 8",
 "anticipatedWrong": [
  {
   "value": 5,
   "misconception": "you counted one hydrogen — the subscript 4 means four hydrogen atoms",
   "feedback": "This answer is wrong — you counted one hydrogen — the subscript 4 means four hydrogen atoms."
  },
  {
   "value": 18,
   "misconception": "you used the group number 14 as germanium's valence count — Group 14 atoms have four valence electrons",
   "feedback": "This answer is wrong — you used the group number 14 as germanium's valence count — Group 14 atoms have four valence electrons."
  },
  {
   "value": 40,
   "misconception": "you gave every atom eight electrons — count each atom's own valence electrons instead",
   "feedback": "This answer is wrong — you gave every atom eight electrons — count each atom's own valence electrons instead."
  }
 ]
}

[
 {
  "value": "monoselenium tetrafluoride",
  "misconception": "you added mono- to the first element — mono- is always dropped on the first element",
  "feedback": "This answer is wrong — you added mono- to the first element — mono- is always dropped on the first element."
 },
 {
  "value": "selenium fluoride",
  "misconception": "you dropped the tetra- count from the second element — the prefix must carry the four",
  "feedback": "This answer is wrong — you dropped the tetra- count from the second element — the prefix must carry the four."
 },
 {
  "value": "tetraselenium monofluoride",
  "misconception": "you crossed the prefixes over — SeF₄ has ONE Se and FOUR F, so tetra- must count the F atoms, not the Se",
  "feedback": "This answer is wrong — you crossed the prefixes over — SeF₄ has ONE Se and FOUR F, so tetra- must count the F atoms, not the Se."
 }
]

{
 "keyExplanation": "One Se atom drops mono- (mono- is dropped only on the FIRST element, never the second), and the four F atoms take tetra- with the -ide ending: SeF₄ is selenium tetrafluoride.",
 "anticipatedWrong": [
  {
   "value": "monoselenium tetrafluoride",
   "misconception": "you added mono- to the first element — mono- is always dropped on the first element",
   "feedback": "This answer is wrong — you added mono- to the first element — mono- is always dropped on the first element."
  },
  {
   "value": "selenium fluoride",
   "misconception": "you dropped the tetra- count from the second element — the prefix must carry the four",
   "feedback": "This answer is wrong — you dropped the tetra- count from the second element — the prefix must carry the four."
  },
  {
   "value": "tetraselenium monofluoride",
   "misconception": "you crossed the prefixes over — SeF₄ has ONE Se and FOUR F, so tetra- must count the F atoms, not the Se",
   "feedback": "This answer is wrong — you crossed the prefixes over — SeF₄ has ONE Se and FOUR F, so tetra- must count the F atoms, not the Se."
  }
 ]
}

[
 {
  "value": "Te₄Cl",
  "misconception": "you attached the count to the wrong element — tetra- belongs to chloride, so the 4 follows Cl",
  "feedback": "This answer is wrong — you attached the count to the wrong element — tetra- belongs to chloride, so the 4 follows Cl."
 },
 {
  "value": "TeCl",
  "misconception": "you dropped the count — tetra- means the formula needs a 4 after Cl",
  "feedback": "This answer is wrong — you dropped the count — tetra- means the formula needs a 4 after Cl."
 }
]

{
 "keyExplanation": "Tellurium has no prefix — one Te, no subscript. tetra- means four, so Cl takes the subscript 4. Tellurium tetrachloride is TeCl₄.",
 "anticipatedWrong": [
  {
   "value": "Te₄Cl",
   "misconception": "you attached the count to the wrong element — tetra- belongs to chloride, so the 4 follows Cl",
   "feedback": "This answer is wrong — you attached the count to the wrong element — tetra- belongs to chloride, so the 4 follows Cl."
  },
  {
   "value": "TeCl",
   "misconception": "you dropped the count — tetra- means the formula needs a 4 after Cl",
   "feedback": "This answer is wrong — you dropped the count — tetra- means the formula needs a 4 after Cl."
  }
 ]
}

[
 {
  "value": "strontium dibromide",
  "misconception": "you counted with prefixes — strontium is a metal, so SrBr₂ is ionic, and ionic names never use prefixes",
  "feedback": "This answer is wrong — you counted with prefixes — strontium is a metal, so SrBr₂ is ionic, and ionic names never use prefixes."
 },
 {
  "value": "strontium(II) bromide",
  "misconception": "you added a Roman numeral — strontium forms only Sr²⁺, so no numeral is needed and none is written",
  "feedback": "This answer is wrong — you added a Roman numeral — strontium forms only Sr²⁺, so no numeral is needed and none is written."
 }
]

{
 "keyExplanation": "First element: strontium, a metal — ionic naming, no prefixes. The two bromines come from the charges (two Br⁻ balance one Sr²⁺), so the name is simply strontium bromide.",
 "anticipatedWrong": [
  {
   "value": "strontium dibromide",
   "misconception": "you counted with prefixes — strontium is a metal, so SrBr₂ is ionic, and ionic names never use prefixes",
   "feedback": "This answer is wrong — you counted with prefixes — strontium is a metal, so SrBr₂ is ionic, and ionic names never use prefixes."
  },
  {
   "value": "strontium(II) bromide",
   "misconception": "you added a Roman numeral — strontium forms only Sr²⁺, so no numeral is needed and none is written",
   "feedback": "This answer is wrong — you added a Roman numeral — strontium forms only Sr²⁺, so no numeral is needed and none is written."
  }
 ]
}

[
 {
  "value": 2,
  "misconception": "you counted the elements, not the atoms — add the two subscripts: 5 + 12",
  "feedback": "This answer is wrong — you counted the elements, not the atoms — add the two subscripts: 5 + 12."
 },
 {
  "value": 12,
  "misconception": "you took the largest subscript as the total — every element's count joins the sum",
  "feedback": "This answer is wrong — you took the largest subscript as the total — every element's count joins the sum."
 },
 {
  "value": 5,
  "misconception": "you read only the first subscript — the 12 hydrogens count too",
  "feedback": "This answer is wrong — you read only the first subscript — the 12 hydrogens count too."
 }
]

{
 "keyExplanation": "Each subscript counts its own element: 5 carbon and 12 hydrogen. The total is 5 + 12 = 17 atoms in one pentane molecule.",
 "anticipatedWrong": [
  {
   "value": 2,
   "misconception": "you counted the elements, not the atoms — add the two subscripts: 5 + 12",
   "feedback": "This answer is wrong — you counted the elements, not the atoms — add the two subscripts: 5 + 12."
  },
  {
   "value": 12,
   "misconception": "you took the largest subscript as the total — every element's count joins the sum",
   "feedback": "This answer is wrong — you took the largest subscript as the total — every element's count joins the sum."
  },
  {
   "value": 5,
   "misconception": "you read only the first subscript — the 12 hydrogens count too",
   "feedback": "This answer is wrong — you read only the first subscript — the 12 hydrogens count too."
  }
 ]
}

[
 {
  "value": 8,
  "misconception": "you answered from the octet pattern instead of counting — this diagram is unfinished, so count what is actually drawn",
  "feedback": "This answer is wrong — you answered from the octet pattern instead of counting — this diagram is unfinished, so count what is actually drawn."
 },
 {
  "value": 4,
  "misconception": "you counted only the bonding electrons — the lone pair already placed also surrounds the selenium",
  "feedback": "This answer is wrong — you counted only the bonding electrons — the lone pair already placed also surrounds the selenium."
 },
 {
  "value": 2,
  "misconception": "you counted only the lone pair — the two shared pairs also surround the selenium",
  "feedback": "This answer is wrong — you counted only the lone pair — the two shared pairs also surround the selenium."
 }
]

{
 "keyExplanation": "Count what is drawn, not what the finished diagram would show. The two Se–H shared pairs give 4 electrons, and the one lone pair placed so far gives 2 more. 4 + 2 = 6 electrons currently surround the selenium.",
 "anticipatedWrong": [
  {
   "value": 8,
   "misconception": "you answered from the octet pattern instead of counting — this diagram is unfinished, so count what is actually drawn",
   "feedback": "This answer is wrong — you answered from the octet pattern instead of counting — this diagram is unfinished, so count what is actually drawn."
  },
  {
   "value": 4,
   "misconception": "you counted only the bonding electrons — the lone pair already placed also surrounds the selenium",
   "feedback": "This answer is wrong — you counted only the bonding electrons — the lone pair already placed also surrounds the selenium."
  },
  {
   "value": 2,
   "misconception": "you counted only the lone pair — the two shared pairs also surround the selenium",
   "feedback": "This answer is wrong — you counted only the lone pair — the two shared pairs also surround the selenium."
  }
 ]
}

[
 "Write down the values in the question:",
 "valence electrons of M = 6",
 "Write down the relationship:",
 "typical bonds = 8 − valence electrons",
 "Substitute in the values, and calculate:",
 "typical bonds = 8 − 6 = 2",
 "Each R atom forms one bond, so two bonds means 2 R atoms."
]

[
 {
  "value": 6,
  "misconception": "you used M's valence count as the number of partners — the partners equal the bonds, which is 8 − 6",
  "feedback": "This answer is wrong — you used M's valence count as the number of partners — the partners equal the bonds, which is 8 − 6."
 },
 {
  "value": 8,
  "misconception": "you used the full octet count as the number of partners — eight is the electron target, not the bond count",
  "feedback": "This answer is wrong — you used the full octet count as the number of partners — eight is the electron target, not the bond count."
 },
 {
  "value": 1,
  "misconception": "you assumed M, like R, forms just one bond — M needs two more electrons, so it forms two bonds",
  "feedback": "This answer is wrong — you assumed M, like R, forms just one bond — M needs two more electrons, so it forms two bonds."
 }
]

{
 "keyExplanation": "M typically forms 8 − 6 = 2 bonds. Each R atom forms one bond, so each bond uses one R atom. Two bonds means two R atoms.",
 "anticipatedWrong": [
  {
   "value": 6,
   "misconception": "you used M's valence count as the number of partners — the partners equal the bonds, which is 8 − 6",
   "feedback": "This answer is wrong — you used M's valence count as the number of partners — the partners equal the bonds, which is 8 − 6."
  },
  {
   "value": 8,
   "misconception": "you used the full octet count as the number of partners — eight is the electron target, not the bond count",
   "feedback": "This answer is wrong — you used the full octet count as the number of partners — eight is the electron target, not the bond count."
  },
  {
   "value": 1,
   "misconception": "you assumed M, like R, forms just one bond — M needs two more electrons, so it forms two bonds",
   "feedback": "This answer is wrong — you assumed M, like R, forms just one bond — M needs two more electrons, so it forms two bonds."
  }
 ]
}

[
 "Write down the values in the question:",
 "carbon: 2 atoms × 4 valence electrons",
 "fluorine: 6 atoms × 7 valence electrons",
 "Write down the relationship:",
 "total valence electrons = the sum over every element of (atoms × valence electrons)",
 "Substitute in the values, and calculate:",
 "total valence electrons = 8 + 42 = 50"
]

[
 {
  "value": 11,
  "misconception": "you counted one atom of each element — the subscripts mean two carbons and six fluorines",
  "feedback": "This answer is wrong — you counted one atom of each element — the subscripts mean two carbons and six fluorines."
 },
 {
  "value": 130,
  "misconception": "you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven",
  "feedback": "This answer is wrong — you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven."
 },
 {
  "value": 8,
  "misconception": "you counted the atoms in the formula, not their valence electrons",
  "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
 }
]

{
 "keyExplanation": "carbon: 2 atoms × 4 valence electrons\nfluorine: 6 atoms × 7 valence electrons\ntotal valence electrons = (2 × 4) + (6 × 7) = 8 + 42 = 50",
 "anticipatedWrong": [
  {
   "value": 11,
   "misconception": "you counted one atom of each element — the subscripts mean two carbons and six fluorines",
   "feedback": "This answer is wrong — you counted one atom of each element — the subscripts mean two carbons and six fluorines."
  },
  {
   "value": 130,
   "misconception": "you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven",
   "feedback": "This answer is wrong — you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven."
  },
  {
   "value": 8,
   "misconception": "you counted the atoms in the formula, not their valence electrons",
   "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
  }
 ]
}

[
 {
  "value": "iodine pentoxide",
  "misconception": "you dropped di- from the first element — only mono- is ever dropped",
  "feedback": "This answer is wrong — you dropped di- from the first element — only mono- is ever dropped."
 },
 {
  "value": "diiodine oxide",
  "misconception": "you dropped the penta- count from the second element — the prefix must carry the five",
  "feedback": "This answer is wrong — you dropped the penta- count from the second element — the prefix must carry the five."
 },
 {
  "value": "pentaiodine dioxide",
  "misconception": "you swapped the two prefixes — each prefix belongs to the element it sits in front of",
  "feedback": "This answer is wrong — you swapped the two prefixes — each prefix belongs to the element it sits in front of."
 }
]

{
 "keyExplanation": "Two I atoms take di- (di- is never dropped — only mono- is), and five O atoms take penta-: penta- + oxide becomes pentoxide. I₂O₅ is diiodine pentoxide.",
 "anticipatedWrong": [
  {
   "value": "iodine pentoxide",
   "misconception": "you dropped di- from the first element — only mono- is ever dropped",
   "feedback": "This answer is wrong — you dropped di- from the first element — only mono- is ever dropped."
  },
  {
   "value": "diiodine oxide",
   "misconception": "you dropped the penta- count from the second element — the prefix must carry the five",
   "feedback": "This answer is wrong — you dropped the penta- count from the second element — the prefix must carry the five."
  },
  {
   "value": "pentaiodine dioxide",
   "misconception": "you swapped the two prefixes — each prefix belongs to the element it sits in front of",
   "feedback": "This answer is wrong — you swapped the two prefixes — each prefix belongs to the element it sits in front of."
  }
 ]
}

[
 {
  "value": "P₁₀S₄",
  "misconception": "you swapped the two counts — tetra- belongs to phosphorus and deca- to sulfide",
  "feedback": "This answer is wrong — you swapped the two counts — tetra- belongs to phosphorus and deca- to sulfide."
 },
 {
  "value": "PS₁₀",
  "misconception": "you dropped tetra- — four phosphorus atoms need the subscript 4",
  "feedback": "This answer is wrong — you dropped tetra- — four phosphorus atoms need the subscript 4."
 }
]

{
 "keyExplanation": "tetra- means four, so P takes the subscript 4; deca- means ten, so S takes the subscript 10. Tetraphosphorus decasulfide is P₄S₁₀.",
 "anticipatedWrong": [
  {
   "value": "P₁₀S₄",
   "misconception": "you swapped the two counts — tetra- belongs to phosphorus and deca- to sulfide",
   "feedback": "This answer is wrong — you swapped the two counts — tetra- belongs to phosphorus and deca- to sulfide."
  },
  {
   "value": "PS₁₀",
   "misconception": "you dropped tetra- — four phosphorus atoms need the subscript 4",
   "feedback": "This answer is wrong — you dropped tetra- — four phosphorus atoms need the subscript 4."
  }
 ]
}

[
 {
  "value": "selenium fluoride",
  "misconception": "you skipped the count — selenium and fluorine are both nonmetals, so prefixes are required: six fluorines need hexa-",
  "feedback": "This answer is wrong — you skipped the count — selenium and fluorine are both nonmetals, so prefixes are required: six fluorines need hexa-."
 },
 {
  "value": "selenium(VI) fluoride",
  "misconception": "you used a Roman numeral — numerals mark metals with more than one common ion charge; molecular compounds count with prefixes",
  "feedback": "This answer is wrong — you used a Roman numeral — numerals mark metals with more than one common ion charge; molecular compounds count with prefixes."
 }
]

{
 "keyExplanation": "Both elements are nonmetals, so SeF₆ is molecular and takes prefix naming: one selenium (mono- dropped), six fluorines — selenium hexafluoride.",
 "anticipatedWrong": [
  {
   "value": "selenium fluoride",
   "misconception": "you skipped the count — selenium and fluorine are both nonmetals, so prefixes are required: six fluorines need hexa-",
   "feedback": "This answer is wrong — you skipped the count — selenium and fluorine are both nonmetals, so prefixes are required: six fluorines need hexa-."
  },
  {
   "value": "selenium(VI) fluoride",
   "misconception": "you used a Roman numeral — numerals mark metals with more than one common ion charge; molecular compounds count with prefixes",
   "feedback": "This answer is wrong — you used a Roman numeral — numerals mark metals with more than one common ion charge; molecular compounds count with prefixes."
  }
 ]
}

[
 {
  "value": 5,
  "misconception": "you read only the bromine subscript — the phosphorus with no subscript still counts as one atom",
  "feedback": "This answer is wrong — you read only the bromine subscript — the phosphorus with no subscript still counts as one atom."
 },
 {
  "value": 2,
  "misconception": "you counted the elements, not the atoms — add the counts: 1 + 5",
  "feedback": "This answer is wrong — you counted the elements, not the atoms — add the counts: 1 + 5."
 }
]

{
 "keyExplanation": "The P with no subscript counts as one atom, and the subscript after Br counts five more. The total is 1 + 5 = 6 atoms in one PBr₅ molecule.",
 "anticipatedWrong": [
  {
   "value": 5,
   "misconception": "you read only the bromine subscript — the phosphorus with no subscript still counts as one atom",
   "feedback": "This answer is wrong — you read only the bromine subscript — the phosphorus with no subscript still counts as one atom."
  },
  {
   "value": 2,
   "misconception": "you counted the elements, not the atoms — add the counts: 1 + 5",
   "feedback": "This answer is wrong — you counted the elements, not the atoms — add the counts: 1 + 5."
  }
 ]
}

[
 {
  "value": 6,
  "misconception": "you skipped the phosphorus's own lone pair — its two electrons surround the phosphorus too",
  "feedback": "This answer is wrong — you skipped the phosphorus's own lone pair — its two electrons surround the phosphorus too."
 },
 {
  "value": 4,
  "misconception": "you counted the pairs, not the electrons — four pairs touch the phosphorus, each holding two electrons",
  "feedback": "This answer is wrong — you counted the pairs, not the electrons — four pairs touch the phosphorus, each holding two electrons."
 },
 {
  "value": 26,
  "misconception": "you counted every electron in the diagram — the bromines' lone pairs belong to the bromines",
  "feedback": "This answer is wrong — you counted every electron in the diagram — the bromines' lone pairs belong to the bromines."
 }
]

{
 "keyExplanation": "Three shared pairs touch the phosphorus (6 electrons), and its own lone pair adds 2 more. 6 + 2 = 8 electrons surround the phosphorus.",
 "anticipatedWrong": [
  {
   "value": 6,
   "misconception": "you skipped the phosphorus's own lone pair — its two electrons surround the phosphorus too",
   "feedback": "This answer is wrong — you skipped the phosphorus's own lone pair — its two electrons surround the phosphorus too."
  },
  {
   "value": 4,
   "misconception": "you counted the pairs, not the electrons — four pairs touch the phosphorus, each holding two electrons",
   "feedback": "This answer is wrong — you counted the pairs, not the electrons — four pairs touch the phosphorus, each holding two electrons."
  },
  {
   "value": 26,
   "misconception": "you counted every electron in the diagram — the bromines' lone pairs belong to the bromines",
   "feedback": "This answer is wrong — you counted every electron in the diagram — the bromines' lone pairs belong to the bromines."
  }
 ]
}

[
 "Write down the values in the question:",
 "valence electrons = 4",
 "Write down the relationship:",
 "typical bonds = 8 − valence electrons",
 "Substitute in the values, and calculate:",
 "typical bonds = 8 − 4",
 "typical bonds = 4"
]

[
 {
  "value": 14,
  "misconception": "you used the group number as the bond count — Group 14 atoms have four valence electrons and need four more",
  "feedback": "This answer is wrong — you used the group number as the bond count — Group 14 atoms have four valence electrons and need four more."
 },
 {
  "value": 8,
  "misconception": "you answered with the octet target — eight is the electron goal, not the number of bonds",
  "feedback": "This answer is wrong — you answered with the octet target — eight is the electron goal, not the number of bonds."
 }
]

{
 "keyExplanation": "Write down the values in the question:\nvalence electrons = 4\nWrite down the relationship:\ntypical bonds = 8 − valence electrons\nSubstitute in the values, and calculate:\ntypical bonds = 8 − 4\ntypical bonds = 4",
 "anticipatedWrong": [
  {
   "value": 14,
   "misconception": "you used the group number as the bond count — Group 14 atoms have four valence electrons and need four more",
   "feedback": "This answer is wrong — you used the group number as the bond count — Group 14 atoms have four valence electrons and need four more."
  },
  {
   "value": 8,
   "misconception": "you answered with the octet target — eight is the electron goal, not the number of bonds",
   "feedback": "This answer is wrong — you answered with the octet target — eight is the electron goal, not the number of bonds."
  }
 ]
}

[
 "Write down the values in the question:",
 "carbon: 2 atoms × 4 valence electrons",
 "chlorine: 4 atoms × 7 valence electrons",
 "Write down the relationship:",
 "total valence electrons = the sum over every element of (atoms × valence electrons)",
 "Substitute in the values, and calculate:",
 "total valence electrons = 8 + 28 = 36"
]

[
 {
  "value": 11,
  "misconception": "you counted one atom of each element — the subscripts mean two carbons and four chlorines",
  "feedback": "This answer is wrong — you counted one atom of each element — the subscripts mean two carbons and four chlorines."
 },
 {
  "value": 96,
  "misconception": "you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven",
  "feedback": "This answer is wrong — you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven."
 },
 {
  "value": 6,
  "misconception": "you counted the atoms in the formula, not their valence electrons",
  "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
 }
]

{
 "keyExplanation": "carbon: 2 atoms × 4 valence electrons\nchlorine: 4 atoms × 7 valence electrons\ntotal valence electrons = (2 × 4) + (4 × 7) = 8 + 28 = 36",
 "anticipatedWrong": [
  {
   "value": 11,
   "misconception": "you counted one atom of each element — the subscripts mean two carbons and four chlorines",
   "feedback": "This answer is wrong — you counted one atom of each element — the subscripts mean two carbons and four chlorines."
  },
  {
   "value": 96,
   "misconception": "you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven",
   "feedback": "This answer is wrong — you used the group numbers 14 and 17 as valence counts — the valence counts are four and seven."
  },
  {
   "value": 6,
   "misconception": "you counted the atoms in the formula, not their valence electrons",
   "feedback": "This answer is wrong — you counted the atoms in the formula, not their valence electrons."
  }
 ]
}

[
 {
  "value": "nitrogen tetrafluoride",
  "misconception": "you dropped di- from the first element — only mono- is ever dropped; two nitrogens need di-",
  "feedback": "This answer is wrong — you dropped di- from the first element — only mono- is ever dropped; two nitrogens need di-."
 },
 {
  "value": "dinitrogen fluoride",
  "misconception": "you left the four fluorines uncounted — the second element needs tetra-",
  "feedback": "This answer is wrong — you left the four fluorines uncounted — the second element needs tetra-."
 }
]

{
 "keyExplanation": "Two nitrogens keep di-, four fluorines take tetra- with the -ide ending: fluorine becomes fluoride. N₂F₄ is dinitrogen tetrafluoride.",
 "anticipatedWrong": [
  {
   "value": "nitrogen tetrafluoride",
   "misconception": "you dropped di- from the first element — only mono- is ever dropped; two nitrogens need di-",
   "feedback": "This answer is wrong — you dropped di- from the first element — only mono- is ever dropped; two nitrogens need di-."
  },
  {
   "value": "dinitrogen fluoride",
   "misconception": "you left the four fluorines uncounted — the second element needs tetra-",
   "feedback": "This answer is wrong — you left the four fluorines uncounted — the second element needs tetra-."
  }
 ]
}

[
 {
  "value": "Si₄I",
  "misconception": "you attached the count to the wrong element — tetra- belongs to iodide, so the 4 follows I",
  "feedback": "This answer is wrong — you attached the count to the wrong element — tetra- belongs to iodide, so the 4 follows I."
 },
 {
  "value": "SiI",
  "misconception": "you dropped the count — tetra- means the formula needs a 4 after I",
  "feedback": "This answer is wrong — you dropped the count — tetra- means the formula needs a 4 after I."
 }
]

{
 "keyExplanation": "Silicon has no prefix — one Si, no subscript. tetra- means four, so I takes the subscript 4. Silicon tetraiodide is SiI₄.",
 "anticipatedWrong": [
  {
   "value": "Si₄I",
   "misconception": "you attached the count to the wrong element — tetra- belongs to iodide, so the 4 follows I",
   "feedback": "This answer is wrong — you attached the count to the wrong element — tetra- belongs to iodide, so the 4 follows I."
  },
  {
   "value": "SiI",
   "misconception": "you dropped the count — tetra- means the formula needs a 4 after I",
   "feedback": "This answer is wrong — you dropped the count — tetra- means the formula needs a 4 after I."
  }
 ]
}

[
 {
  "value": "dilithium sulfide",
  "misconception": "you counted with prefixes — lithium is a metal, so Li₂S is ionic, and ionic names never use prefixes",
  "feedback": "This answer is wrong — you counted with prefixes — lithium is a metal, so Li₂S is ionic, and ionic names never use prefixes."
 },
 {
  "value": "lithium(I) sulfide",
  "misconception": "you added a Roman numeral — lithium forms only Li⁺, so no numeral is needed and none is written",
  "feedback": "This answer is wrong — you added a Roman numeral — lithium forms only Li⁺, so no numeral is needed and none is written."
 }
]

{
 "keyExplanation": "First element: lithium, a metal — ionic naming, no prefixes. The two lithiums come from the charges (two Li⁺ balance one S²⁻), so the name is simply lithium sulfide.",
 "anticipatedWrong": [
  {
   "value": "dilithium sulfide",
   "misconception": "you counted with prefixes — lithium is a metal, so Li₂S is ionic, and ionic names never use prefixes",
   "feedback": "This answer is wrong — you counted with prefixes — lithium is a metal, so Li₂S is ionic, and ionic names never use prefixes."
  },
  {
   "value": "lithium(I) sulfide",
   "misconception": "you added a Roman numeral — lithium forms only Li⁺, so no numeral is needed and none is written",
   "feedback": "This answer is wrong — you added a Roman numeral — lithium forms only Li⁺, so no numeral is needed and none is written."
  }
 ]
}

[
 {
  "value": 4,
  "misconception": "you counted each double bond as two groups — a bond of any multiplicity counts once",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you added the sulfurs' lone pairs — outer-atom lone pairs never count at the central atom",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted electrons around the carbon instead of groups",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the carbon: two C=S double bonds = 2 groups. Each double bond counts once, and the sulfurs' lone pairs belong to the sulfurs. 2 + 0 = 2 electron groups."
}

[
 {
  "value": 2,
  "misconception": "you counted only the bonds — the sulfur's two lone pairs each count as one group",
  "feedback": null
 },
 {
  "value": 10,
  "misconception": "you added the chlorines' lone pairs — outer-atom lone pairs never count at the central atom",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted electrons around the sulfur instead of groups",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the sulfur: two S–Cl single bonds = 2 groups. Lone pairs on the sulfur: two = 2 groups; the chlorines' pairs belong to the chlorines. 2 + 2 = 4 electron groups."
}

[
 {
  "value": 4,
  "misconception": "you counted the double bond as two groups — a bond of any multiplicity counts once",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you counted the groups around both carbons — count at one atom at a time",
  "feedback": null
 },
 {
  "value": 2,
  "misconception": "you skipped the double bond and counted only the C–H bonds",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the left carbon: two C–H single bonds and one C=C double bond = 3 groups. The double bond counts once, and the other carbon's bonds are its own. 3 + 0 = 3 electron groups."
}

[
 {
  "value": 2,
  "misconception": "you counted only the bonds — the two lone pairs on the atom each count as one group",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you added the right O atom's lone pairs — count at one atom at a time",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted electrons around the atom instead of groups",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the left O atom: one O–H single bond and one O–O single bond = 2 groups. Lone pairs on that atom: two = 2 groups; the right O atom's lone pairs are its own. 2 + 2 = 4 electron groups."
}

[
 {
  "value": 4,
  "misconception": "you counted the double bond as two groups — a bond of any multiplicity counts once",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you counted the groups around both carbons — count at one atom at a time",
  "feedback": null
 },
 {
  "value": 2,
  "misconception": "you skipped the double bond and counted only the C–F bonds",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the left carbon: two C–F single bonds and one C=C double bond = 3 groups. The double bond counts once, and the other carbon's bonds are its own. 3 + 0 = 3 electron groups."
}

[
 {
  "value": 3,
  "misconception": "you skipped the lone pair on the As atom — a central-atom lone pair counts as one group",
  "feedback": null
 },
 {
  "value": 13,
  "misconception": "you added the Br atoms' lone pairs — outer-atom lone pairs never count at the central atom",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you counted the bonding electrons instead of the groups",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the As atom: three As–Br single bonds = 3 groups. Lone pairs on the As atom: one = 1 group; the Br atoms' lone pairs are their own. 3 + 1 = 4 electron groups."
}

[
 {
  "value": 4,
  "misconception": "you counted the triple bond as three groups — a bond of any multiplicity counts once",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you counted the groups around both carbons — count at one atom at a time",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you added the N atom's lone pair — outer-atom lone pairs never count at the central atom",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at that carbon: one C–C single bond and one C≡N triple bond = 2 groups. A triple bond counts once, and the N atom's lone pair belongs to the N atom. 2 + 0 = 2 electron groups."
}

[
 {
  "value": 4,
  "misconception": "you counted the double bond as two groups — a bond of any multiplicity counts once",
  "feedback": null
 },
 {
  "value": 2,
  "misconception": "you skipped the lone pair on the N atom — a central-atom lone pair counts as one group",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you added the outer atoms' lone pairs — they never count at the central atom",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the N atom: one N=O double bond (counts once) and one N–Cl single bond = 2 groups. Lone pairs on the N atom: one = 1 group. 2 + 1 = 3 electron groups."
}

[
 {
  "value": 4,
  "misconception": "you counted the triple bond as three groups — a bond of any multiplicity counts once",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted the shared electrons around the middle carbon instead of the groups",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you gave the middle carbon a lone pair it does not have",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the middle carbon: one single bond to the CH₃ carbon and one triple bond to the CH carbon = 2 groups. A triple bond counts once. 2 + 0 = 2 electron groups."
}

[
 {
  "value": 2,
  "misconception": "you counted only the bonds — the two lone pairs on the O atom each count as one group",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted electrons around the O atom instead of groups",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you counted only one of the O atom's two lone pairs",
  "feedback": null
 }
]

{
 "keyExplanation": "Bonds at the O atom: one O–C single bond and one O–H single bond = 2 groups. Lone pairs on the O atom: two = 2 groups. 2 + 2 = 4 electron groups."
}

[
 {
  "value": 3,
  "misconception": "you counted only the hydrogens — the C–C bond belongs to each carbon's count as well",
  "feedback": null
 },
 {
  "value": 7,
  "misconception": "you added both carbons' bonds together — the question asks for the count at EACH carbon, which is four",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted every atom in the molecule — the count is the bonds at one carbon: three to H plus one to C",
  "feedback": null
 }
]

{
 "keyExplanation": "A carbon atom forms four covalent bonds. Each carbon in ethane has three bonds to hydrogen atoms and one bond to the other carbon: 3 + 1 = 4."
}

[
 {
  "value": "C₃H₈",
  "misconception": "you gave the double-bonded carbons a full set of hydrogens — the double bond uses two of each carbon's four lines, leaving fewer for hydrogen: the count is 2 + 1 + 3 = 6",
  "feedback": "This answer is wrong — you gave the double-bonded carbons a full set of hydrogens — the double bond uses two of each carbon's four lines, leaving fewer for hydrogen: the count is 2 + 1 + 3 = 6."
 },
 {
  "value": "C₄H₆",
  "misconception": "you counted the double bond as an extra carbon — two parallel lines are one double bond between the SAME two carbons",
  "feedback": "This answer is wrong — you counted the double bond as an extra carbon — two parallel lines are one double bond between the SAME two carbons."
 },
 {
  "value": "C₃H₃",
  "misconception": "you dropped the CH₃ end's hydrogens — the third carbon carries three: 2 + 1 + 3 = 6",
  "feedback": "This answer is wrong — you dropped the CH₃ end's hydrogens — the third carbon carries three: 2 + 1 + 3 = 6."
 }
]

{
 "keyExplanation": "Count the atoms from the structure: three carbons, and hydrogens 2 + 1 + 3 = 6, so propene is C₃H₆. The two parallel lines are one C=C double bond — two shared pairs — which is why those carbons carry fewer hydrogens than a fully single-bonded chain.",
 "anticipatedWrong": [
  {
   "value": "C₃H₈",
   "misconception": "you gave the double-bonded carbons a full set of hydrogens — the double bond uses two of each carbon's four lines, leaving fewer for hydrogen: the count is 2 + 1 + 3 = 6",
   "feedback": "This answer is wrong — you gave the double-bonded carbons a full set of hydrogens — the double bond uses two of each carbon's four lines, leaving fewer for hydrogen: the count is 2 + 1 + 3 = 6."
  },
  {
   "value": "C₄H₆",
   "misconception": "you counted the double bond as an extra carbon — two parallel lines are one double bond between the SAME two carbons",
   "feedback": "This answer is wrong — you counted the double bond as an extra carbon — two parallel lines are one double bond between the SAME two carbons."
  },
  {
   "value": "C₃H₃",
   "misconception": "you dropped the CH₃ end's hydrogens — the third carbon carries three: 2 + 1 + 3 = 6",
   "feedback": "This answer is wrong — you dropped the CH₃ end's hydrogens — the third carbon carries three: 2 + 1 + 3 = 6."
  }
 ]
}

[
 {
  "value": 7,
  "misconception": "you counted every bond in the molecule — the question asks for C–H bonds only, and the C–C and C–Cl lines are not C–H bonds",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you counted the C–Cl bond as a C–H bond — chlorine is not hydrogen: the C–H count is 3 + 2 = 5",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you counted only one carbon's hydrogens — both carbons carry hydrogens: 3 on one, 2 on the other",
  "feedback": null
 }
]

{
 "keyExplanation": "Trace each line from a carbon to an H symbol: the first carbon has three C–H bonds and the second has two, so the structure contains 3 + 2 = 5 C–H bonds. The C–C line and the C–Cl line are bonds too, but not C–H bonds."
}

[
 {
  "value": 3,
  "misconception": "you counted the double bond as one — it counts as two of the carbon's bonds: 2 + 2 = 4",
  "feedback": null
 },
 {
  "value": 2,
  "misconception": "you counted only the hydrogens — the double bond to the neighboring carbon adds two more: 2 + 2 = 4",
  "feedback": null
 },
 {
  "value": 6,
  "misconception": "you doubled everything — only the double bond counts twice: 2 + 2 = 4",
  "feedback": null
 }
]

{
 "keyExplanation": "Count lines, not neighbors. Two single bonds to hydrogens count 2, and a double bond counts as two more: 2 + 2 = 4 — a carbon atom forms four covalent bonds."
}

[
 {
  "value": "C₄H₁₀",
  "misconception": "you gave the double-bonded carbons full sets of hydrogens — the double bond uses two of each middle carbon's four lines: the count is 3 + 1 + 1 + 3 = 8",
  "feedback": "This answer is wrong — you gave the double-bonded carbons full sets of hydrogens — the double bond uses two of each middle carbon's four lines: the count is 3 + 1 + 1 + 3 = 8."
 },
 {
  "value": "C₅H₈",
  "misconception": "you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons",
  "feedback": "This answer is wrong — you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons."
 },
 {
  "value": "C₄H₆",
  "misconception": "you shorted the end carbons — each end carbon carries three hydrogens: 3 + 1 + 1 + 3 = 8",
  "feedback": "This answer is wrong — you shorted the end carbons — each end carbon carries three hydrogens: 3 + 1 + 1 + 3 = 8."
 }
]

{
 "keyExplanation": "Count the atoms: four carbons, and hydrogens 3 + 1 + 1 + 3 = 8, so the formula is C₄H₈. The two parallel lines between the middle carbons are one double bond, which is why those two carry one hydrogen each.",
 "anticipatedWrong": [
  {
   "value": "C₄H₁₀",
   "misconception": "you gave the double-bonded carbons full sets of hydrogens — the double bond uses two of each middle carbon's four lines: the count is 3 + 1 + 1 + 3 = 8",
   "feedback": "This answer is wrong — you gave the double-bonded carbons full sets of hydrogens — the double bond uses two of each middle carbon's four lines: the count is 3 + 1 + 1 + 3 = 8."
  },
  {
   "value": "C₅H₈",
   "misconception": "you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons",
   "feedback": "This answer is wrong — you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons."
  },
  {
   "value": "C₄H₆",
   "misconception": "you shorted the end carbons — each end carbon carries three hydrogens: 3 + 1 + 1 + 3 = 8",
   "feedback": "This answer is wrong — you shorted the end carbons — each end carbon carries three hydrogens: 3 + 1 + 1 + 3 = 8."
  }
 ]
}

[
 {
  "value": 10,
  "misconception": "you counted every bond in the molecule — the two C–C lines and the C–Cl line are not C–H bonds",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted the C–Cl bond as a C–H bond — chlorine is not hydrogen: the count is 3 + 2 + 2 = 7",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you counted only the first carbon's hydrogens — all three carbons carry hydrogens",
  "feedback": null
 }
]

{
 "keyExplanation": "Trace each line from a carbon to an H symbol: 3 on the first carbon, 2 on the second, and 2 on the third give 3 + 2 + 2 = 7 C–H bonds. The C–C and C–Cl lines are bonds, but not C–H bonds."
}

[
 {
  "value": 2,
  "misconception": "you counted neighbors instead of bonds — the triple bond is worth three of the carbon's four",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you dropped the C–H bond — the single bond to the hydrogen completes the count: 1 + 3 = 4",
  "feedback": null
 },
 {
  "value": 5,
  "misconception": "you gave the triple bond four — a triple bond counts as exactly three",
  "feedback": null
 }
]

{
 "keyExplanation": "Count lines, not neighbors. The single bond to the hydrogen counts 1 and the triple bond counts 3: 1 + 3 = 4 — a carbon atom forms four covalent bonds."
}

[
 {
  "value": "CH₄",
  "misconception": "you read the two Cl symbols as hydrogens — two of the four attachments are chlorines",
  "feedback": "This answer is wrong — you read the two Cl symbols as hydrogens — two of the four attachments are chlorines."
 },
 {
  "value": "CH₂Cl",
  "misconception": "you counted only one of the chlorines — the carbon carries two Cl symbols",
  "feedback": "This answer is wrong — you counted only one of the chlorines — the carbon carries two Cl symbols."
 },
 {
  "value": "CH₃Cl",
  "misconception": "you traded a chlorine for a hydrogen — the structure shows two of each",
  "feedback": "This answer is wrong — you traded a chlorine for a hydrogen — the structure shows two of each."
 }
]

{
 "keyExplanation": "Count the atoms: one carbon, two hydrogens, and two chlorines, so the formula is CH₂Cl₂. All four of the carbon's lines are single bonds — two to H and two to Cl.",
 "anticipatedWrong": [
  {
   "value": "CH₄",
   "misconception": "you read the two Cl symbols as hydrogens — two of the four attachments are chlorines",
   "feedback": "This answer is wrong — you read the two Cl symbols as hydrogens — two of the four attachments are chlorines."
  },
  {
   "value": "CH₂Cl",
   "misconception": "you counted only one of the chlorines — the carbon carries two Cl symbols",
   "feedback": "This answer is wrong — you counted only one of the chlorines — the carbon carries two Cl symbols."
  },
  {
   "value": "CH₃Cl",
   "misconception": "you traded a chlorine for a hydrogen — the structure shows two of each",
   "feedback": "This answer is wrong — you traded a chlorine for a hydrogen — the structure shows two of each."
  }
 ]
}

[
 {
  "value": 3,
  "misconception": "you counted the loop as an open row — the bond that closes the ring counts too",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted the C–H bonds instead — the question asks for C–C bonds only",
  "feedback": null
 },
 {
  "value": 12,
  "misconception": "you counted every bond in the structure — only the four ring lines are C–C bonds",
  "feedback": null
 }
]

{
 "keyExplanation": "Trace the ring: each of the four carbons bonds to the next, and the fourth bonds back to the first — four C–C bonds in all. The eight C–H bonds are bonds too, but not C–C bonds."
}

[
 {
  "value": 2,
  "misconception": "you counted only the ring neighbors — the two C–H bonds belong to the count as well",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you dropped one of the four connections — two ring bonds plus two C–H bonds make 2 + 2 = 4",
  "feedback": null
 },
 {
  "value": 8,
  "misconception": "you counted every bond twice — doubling this carbon's four real connections gives 8, but each bond counts once: 2 + 2 = 4",
  "feedback": null
 }
]

{
 "keyExplanation": "Ring or chain, the count is the same. Two bonds to ring carbons plus two bonds to hydrogens give 2 + 2 = 4 — a carbon atom forms four covalent bonds."
}

[
 {
  "value": "C₂H₆",
  "misconception": "you read the F symbol as another hydrogen — one attachment is a fluorine",
  "feedback": "This answer is wrong — you read the F symbol as another hydrogen — one attachment is a fluorine."
 },
 {
  "value": "C₂H₅",
  "misconception": "you dropped the fluorine — every symbol in the structure goes into the formula",
  "feedback": "This answer is wrong — you dropped the fluorine — every symbol in the structure goes into the formula."
 },
 {
  "value": "C₂H₆F",
  "misconception": "you gave the fluorine's carbon a full set of hydrogens — the F occupies one of its four lines, so the hydrogens count 3 + 2 = 5",
  "feedback": "This answer is wrong — you gave the fluorine's carbon a full set of hydrogens — the F occupies one of its four lines, so the hydrogens count 3 + 2 = 5."
 }
]

{
 "keyExplanation": "Count the atoms: two carbons, hydrogens 3 + 2 = 5, and one fluorine, so the formula is C₂H₅F. The C–F line is a bond like any other and its atom must appear in the formula.",
 "anticipatedWrong": [
  {
   "value": "C₂H₆",
   "misconception": "you read the F symbol as another hydrogen — one attachment is a fluorine",
   "feedback": "This answer is wrong — you read the F symbol as another hydrogen — one attachment is a fluorine."
  },
  {
   "value": "C₂H₅",
   "misconception": "you dropped the fluorine — every symbol in the structure goes into the formula",
   "feedback": "This answer is wrong — you dropped the fluorine — every symbol in the structure goes into the formula."
  },
  {
   "value": "C₂H₆F",
   "misconception": "you gave the fluorine's carbon a full set of hydrogens — the F occupies one of its four lines, so the hydrogens count 3 + 2 = 5",
   "feedback": "This answer is wrong — you gave the fluorine's carbon a full set of hydrogens — the F occupies one of its four lines, so the hydrogens count 3 + 2 = 5."
  }
 ]
}

[
 {
  "value": 11,
  "misconception": "you counted every line in the drawing — the triple bond's three lines and the C–C singles are not C–H bonds",
  "feedback": null
 },
 {
  "value": 5,
  "misconception": "you dropped the triple-bond carbon's single hydrogen — it counts: 1 + 2 + 3 = 6",
  "feedback": null
 },
 {
  "value": 10,
  "misconception": "you ignored the triple bond and gave every carbon a full set — the true count is 1 + 0 + 2 + 3 = 6",
  "feedback": null
 }
]

{
 "keyExplanation": "Trace each line from a carbon to an H symbol: 1 on the first carbon, none on the second, 2 on the third, and 3 on the fourth give 1 + 2 + 3 = 6 C–H bonds. The triple bond and the C–C singles are bonds, but not C–H bonds."
}

[
 {
  "value": 3,
  "misconception": "you counted only the carbon neighbors — the C–H bond belongs to the count as well",
  "feedback": null
 },
 {
  "value": 2,
  "misconception": "you counted connection kinds, not bonds — carbons and hydrogen make two kinds, but the three C–C bonds count separately: 3 + 1 = 4",
  "feedback": null
 },
 {
  "value": 5,
  "misconception": "you added an extra bond for the branch point — branching carbons still form exactly four",
  "feedback": null
 }
]

{
 "keyExplanation": "A branch point changes nothing about the count. Three bonds to carbons plus one bond to a hydrogen give 3 + 1 = 4 — a carbon atom forms four covalent bonds."
}

[
 {
  "value": "C₅H₁₂",
  "misconception": "you gave the double-bonded carbons full sets of hydrogens — the double bond uses two lines at each of its carbons: the count is 2 + 1 + 2 + 2 + 3 = 10",
  "feedback": "This answer is wrong — you gave the double-bonded carbons full sets of hydrogens — the double bond uses two lines at each of its carbons: the count is 2 + 1 + 2 + 2 + 3 = 10."
 },
 {
  "value": "C₆H₁₀",
  "misconception": "you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons",
  "feedback": "This answer is wrong — you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons."
 },
 {
  "value": "C₅H₈",
  "misconception": "you charged the double bond twice at each carbon — it costs the pair just one hydrogen each, so the count is 2 + 1 + 2 + 2 + 3 = 10",
  "feedback": "This answer is wrong — you charged the double bond twice at each carbon — it costs the pair just one hydrogen each, so the count is 2 + 1 + 2 + 2 + 3 = 10."
 }
]

{
 "keyExplanation": "Count the atoms: five carbons, and hydrogens 2 + 1 + 2 + 2 + 3 = 10, so the formula is C₅H₁₀. The two parallel lines at the chain's end are one C=C double bond.",
 "anticipatedWrong": [
  {
   "value": "C₅H₁₂",
   "misconception": "you gave the double-bonded carbons full sets of hydrogens — the double bond uses two lines at each of its carbons: the count is 2 + 1 + 2 + 2 + 3 = 10",
   "feedback": "This answer is wrong — you gave the double-bonded carbons full sets of hydrogens — the double bond uses two lines at each of its carbons: the count is 2 + 1 + 2 + 2 + 3 = 10."
  },
  {
   "value": "C₆H₁₀",
   "misconception": "you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons",
   "feedback": "This answer is wrong — you counted the double bond as an extra carbon — two parallel lines are one bond between the SAME two carbons."
  },
  {
   "value": "C₅H₈",
   "misconception": "you charged the double bond twice at each carbon — it costs the pair just one hydrogen each, so the count is 2 + 1 + 2 + 2 + 3 = 10",
   "feedback": "This answer is wrong — you charged the double bond twice at each carbon — it costs the pair just one hydrogen each, so the count is 2 + 1 + 2 + 2 + 3 = 10."
  }
 ]
}

[
 {
  "value": 13,
  "misconception": "you counted every bond in the molecule — the three C–C lines and the C–Br line are not C–H bonds",
  "feedback": null
 },
 {
  "value": 10,
  "misconception": "you counted the C–Br bond as a C–H bond — bromine is not hydrogen: the count is 3 + 2 + 2 + 2 = 9",
  "feedback": null
 },
 {
  "value": 3,
  "misconception": "you counted only the first carbon's hydrogens — all four carbons carry hydrogens",
  "feedback": null
 }
]

{
 "keyExplanation": "Trace each line from a carbon to an H symbol: 3 + 2 + 2 + 2 = 9 C–H bonds. The three C–C lines and the C–Br line are bonds too, but not C–H bonds."
}