Formal Charge vs Oxidation Number Flashcards: Electron Accounting

Practice formal charge vs oxidation number with 48 original English flashcards: bond allocation, single-atom calculations, charge checks, and error repairs.

Về bộ thẻ này

Practice formal charge vs oxidation number (oxidation state) with 48 original English flashcards for introductory chemistry learners who can read short Lewis structures.

The cards practice method → electron-allocation rule; a specified structure and target atom → one signed result; a faulty count or label → its repair; atom values and multiplicities → the species total; and a claim or missing input → whether the conclusion is supported. Two selected comparisons ask for formal charge and oxidation number together, for carbon in CO and oxygen in OF2. Other cards distinguish an atom's bookkeeping number from the charge of the whole species or an electron-density-based partial charge.

Six counting anchors come first. Short calculations then mix neutral molecules, ions, single and multiple bonds, and identical-element bonds. Error repairs follow, with related variants separated by at least three unrelated cards. The last part mixes total checks, a specified nitrate resonance form, peroxide and oxygen–fluorine cases, matching numerical answers, and missing-information decisions. Every structure-based calculation supplies its bonds, lone pairs, target, overall charge, and any needed electronegativity order in text; no diagram is required.

This is narrow electron-accounting practice. It excludes number → molecule guessing because one number fits many structures; exhaustive atom and direction permutations; full Lewis-structure drawing, molecule-wide multipart calculations, and redox balancing because they need longer written practice; and radicals, transition-metal coordination chemistry, and complete exam curricula because they exceed this scope. Recall practice should be followed by fresh calculations on paper.

Read the formal charge and oxidation number lesson for a longer explanation, or the chemistry flashcard study guide for practice advice. The accounting conventions are checked against the IUPAC entries for formal charge and oxidation state.

The questions, explanations, selection, and order are independently composed from scientific facts, with AI-assisted drafting and an original AI-generated abstract cover. No textbook, examination, or competitor questions, answers, or figures were copied or adapted. Common knowledge · CC0 1.0 applies to this original expression and media to the extent applicable rights exist; it does not claim ownership of scientific facts or third-party source material. This is an independent study aid with no examination-board or source-publisher affiliation.

Thẻ trong bộ này

  1. Thẻ 1

    Câu hỏi

    Formal charge: how are a bond's electrons assigned?

    Câu trả lời

    Half to each bonded atom, regardless of electronegativity.

  2. Thẻ 2

    Câu hỏi

    Electron accounting: how many nonbonding electrons are in three lone pairs?

    Câu trả lời

    Six electrons. Each lone pair contains two electrons.

  3. Thẻ 3

    Câu hỏi

    Lewis counting: how many bonding electrons are in a single, double, and triple bond?

    Câu trả lời

    Two, four, and six, respectively. Each bond line represents one electron pair.

  4. Thẻ 4

    Câu hỏi

    Oxidation number: how are electrons allocated in an ordinary bond between different elements?

    Câu trả lời

    Assign the entire bonding pair to the more electronegative atom. Nonbonding electrons stay with their atom.

  5. Thẻ 5

    Câu hỏi

    Electron accounting: what must the sum of all atom formal charges, or all atom oxidation numbers, equal?

    Câu trả lời

    The charge of the whole species. For a neutral molecule, each sum is zero.

  6. Thẻ 6

    Câu hỏi

    Formal charge: what's the calculation for a specified Lewis structure?

    Câu trả lời

    Neutral valence electrons − nonbonding electrons − half the bonding electrons. Read the target atom's counts from that particular structure.

  7. Thẻ 7

    Câu hỏi

    NH4+ (charge +1): four N–H single bonds; no lone pairs on N or H. N formal charge? N has 5 valence electrons.

    Câu trả lời

    +1 — N receives 4 bonding electrons: 5 − 0 − 4 = +1.

  8. Thẻ 8

    Câu hỏi

    Neutral H–F: one single bond; F has 3 lone pairs, H none. F is more electronegative than H. H oxidation number? H has 1 valence electron.

    Câu trả lời

    +1 — the bonding pair goes to F, leaving H assigned 0 electrons: 1 − 0 = +1.

  9. Thẻ 9

    Câu hỏi

    Neutral O=C=O: C has no lone pairs; each O has 2. C formal charge? C has 4 valence electrons.

    Câu trả lời

    Zero — two double bonds assign C 4 bonding electrons: 4 − 0 − 4 = 0.

  10. Thẻ 10

    Câu hỏi

    OH− (charge −1): one O–H single bond; O has 3 lone pairs, H none. O formal charge? O has 6 valence electrons.

    Câu trả lời

    −1 — O receives 1 bonding electron: 6 − 6 − 1 = −1.

  11. Thẻ 11

    Câu hỏi

    Neutral N≡N: each N has 1 lone pair. Either N oxidation number? N has 5 valence electrons.

    Câu trả lời

    Zero — each N is assigned 2 nonbonding electrons and 3 bonding electrons: 5 − 2 − 3 = 0.

  12. Thẻ 12

    Câu hỏi

    Neutral CH3Cl: C has 3 C–H single bonds and 1 C–Cl single bond; C and H have no lone pairs, Cl has 3. Cl > C > H in electronegativity. C oxidation number? C has 4 valence electrons.

    Câu trả lời

    −2 — C gets 6 electrons from the C–H bonds and none from C–Cl: 4 − 6 = −2.

  13. Thẻ 13

    Câu hỏi

    Neutral H–C≡N: N has 1 lone pair; C and H have none. N formal charge? N has 5 valence electrons.

    Câu trả lời

    Zero — N receives 3 electrons from the triple bond: 5 − 2 − 3 = 0.

  14. Thẻ 14

    Câu hỏi

    Neutral H2C=CH2: each C has 2 C–H single bonds and the C=C double bond; no atom has lone pairs. C > H in electronegativity. Either C oxidation number? C has 4 valence electrons.

    Câu trả lời

    −2 — C gets 4 electrons from C–H and 2 from half the C=C bond: 4 − 6 = −2.

  15. Thẻ 15

    Câu hỏi

    CN− (charge −1): C≡N; each atom has 1 lone pair. N is more electronegative than C. C oxidation number? C has 4 valence electrons.

    Câu trả lời

    +2 — C keeps only its 2 nonbonding electrons: 4 − 2 = +2.

  16. Thẻ 16

    Câu hỏi

    Neutral CH3F: four single bonds around C; C and H have no lone pairs, F has 3. F formal charge? F has 7 valence electrons.

    Câu trả lời

    Zero — F receives 1 bonding electron: 7 − 6 − 1 = 0.

  17. Thẻ 17

    Câu hỏi

    Neutral CH3–O–H: all bonds are single; O has 2 lone pairs, C and H none. O formal charge? O has 6 valence electrons.

    Câu trả lời

    Zero — O receives 1 electron from each of its two bonds: 6 − 4 − 2 = 0.

  18. Thẻ 18

    Câu hỏi

    Neutral H2C=O: C has 2 C–H single bonds and 1 C=O double bond; O has 2 lone pairs, C and H none. O > C > H in electronegativity. C oxidation number? C has 4 valence electrons.

    Câu trả lời

    Zero — C gets 4 electrons from C–H and none from C=O: 4 − 4 = 0.

  19. Thẻ 19

    Câu hỏi

    NH2− (charge −1): two N–H single bonds; N has 2 lone pairs, H none. N formal charge? N has 5 valence electrons.

    Câu trả lời

    −1 — N receives 2 bonding electrons: 5 − 4 − 2 = −1.

  20. Thẻ 20

    Câu hỏi

    Neutral H–C≡C–H: no atom has lone pairs. C > H in electronegativity. Either C oxidation number? C has 4 valence electrons.

    Câu trả lời

    −1 — C gets 2 electrons from C–H and 3 from half the C≡C bond: 4 − 5 = −1.

  21. Thẻ 21

    Câu hỏi

    H3O+ (charge +1): three O–H single bonds; O has 1 lone pair, H none. O formal charge? O has 6 valence electrons.

    Câu trả lời

    +1 — O receives 3 bonding electrons: 6 − 2 − 3 = +1.

  22. Thẻ 22

    Câu hỏi

    Neutral F–O–F: both bonds are single; O has 2 lone pairs and each F has 3. O formal charge? O has 6 valence electrons.

    Câu trả lời

    Zero — O receives 2 bonding electrons: 6 − 4 − 2 = 0.

  23. Thẻ 23

    Câu hỏi

    Neutral H2N–NH2: all bonds are single; each N has 1 lone pair, H none. N > H in electronegativity. Either N oxidation number? N has 5 valence electrons.

    Câu trả lời

    −2 — N keeps 2 nonbonding electrons, gets 4 from N–H, and 1 from N–N: 5 − 7 = −2.

  24. Thẻ 24

    Câu hỏi

    Neutral F–F: one single bond; each F has 3 lone pairs. Either F oxidation number? F has 7 valence electrons.

    Câu trả lời

    Zero — each F keeps 6 nonbonding electrons and half the bond: 7 − 6 − 1 = 0.

  25. Thẻ 25

    Câu hỏi

    NH4+ (charge +1): four N–H single bonds, no lone pairs. N > H in electronegativity. Giving N all 8 bonding electrons yields −3. Which accounting label belongs on that result?

    Câu trả lời

    Oxidation number. Equal sharing instead gives N formal charge +1: 5 − 4 = +1.

  26. Thẻ 26

    Câu hỏi

    Neutral H–O–H: O has 2 lone pairs; H none. A formal-charge calculation subtracts 2 for O's nonbonding electrons. What count should replace it?

    Câu trả lời

    Four electrons. Two lone pairs contain 4 electrons, so O formal charge is 6 − 4 − 2 = 0.

  27. Thẻ 27

    Câu hỏi

    Neutral O=C=O: C has no lone pairs; each O has 2. A formal-charge calculation assigns C only 2 bonding electrons because it has two neighbors. What count should replace it?

    Câu trả lời

    Four electrons. Each double bond assigns C 2 electrons, giving 4 − 4 = 0.

  28. Thẻ 28

    Câu hỏi

    Neutral F–F: each F has 3 lone pairs. An oxidation-number calculation gives both bonding electrons to the left F. How should this bond be allocated?

    Câu trả lời

    One electron to each F. Identical-element bonds are split equally, so both F oxidation numbers are zero.

  29. Thẻ 29

    Câu hỏi

    OH− (charge −1): O–H single bond; O has 3 lone pairs, H none. A formal-charge calculation gives O both bonding electrons. How many should O receive?

    Câu trả lời

    One bonding electron. Formal charge splits the pair, giving O: 6 − 6 − 1 = −1.

  30. Thẻ 30

    Câu hỏi

    Neutral H2N–NH2: all bonds single; each N has 1 lone pair, H none. N > H in electronegativity. Does the N–N bond contribute +1 or −1 to either N's oxidation number?

    Câu trả lời

    Neither; its contribution is zero. The N–N electrons are shared equally; the two N–H bonds give each N oxidation number −2.

  31. Thẻ 31

    Câu hỏi

    Neutral H–C≡N: N has 1 lone pair; C and H none. A formal-charge calculation assigns N only 1 electron from C≡N. What count should replace it?

    Câu trả lời

    Three electrons. A triple bond contains 6 electrons; N receives half, so 5 − 2 − 3 = 0.

  32. Thẻ 32

    Câu hỏi

    Neutral F–O–F: O has 2 lone pairs, each F has 3. F > O in electronegativity. A learner uses 'oxygen is always −2.' What is O's oxidation number here?

    Câu trả lời

    +2 — both O–F bonding pairs go to F; O keeps 4 electrons: 6 − 4 = +2.

  33. Thẻ 33

    Câu hỏi

    Neutral CH3–O–H: all bonds single; O has 2 lone pairs, C and H none. O > C > H in electronegativity. A learner copies C's formal charge of zero as its oxidation number. What should C's oxidation number be?

    Câu trả lời

    −2 — C gets 6 electrons from its three C–H bonds and none from C–O: 4 − 6 = −2.

  34. Thẻ 34

    Câu hỏi

    CN− (charge −1): C≡N; each atom has 1 lone pair. N > C in electronegativity. Must C have a negative oxidation number because the ion is negative?

    Câu trả lời

    No; C is +2. C keeps 2 nonbonding electrons, while N receives 8 electrons and is −3; +2 − 3 = −1.

  35. Thẻ 35

    Câu hỏi

    A neutral molecule's atom formal charges add to +1. Can that be a complete, consistent charge assignment?

    Câu trả lời

    No. A neutral species requires a total of zero; recheck the counts and any omitted atoms.

  36. Thẻ 36

    Câu hỏi

    Neutral H–O–O–H: all bonds single; each O has 2 lone pairs, H none. O > H in electronegativity. Either O oxidation number? O has 6 valence electrons.

    Câu trả lời

    −1 — O keeps 4 nonbonding electrons, gets 2 from O–H and 1 from O–O: 6 − 7 = −1.

  37. Thẻ 37

    Câu hỏi

    Neutral O=C=O: C has no lone pairs; each O has 2. O > C in electronegativity. Labels C +4 and each O −2 sum to zero. Does that make them correct formal charges?

    Câu trả lời

    No. Those are oxidation numbers; this Lewis structure has formal charge zero on every atom. A correct total doesn't identify the method.

  38. Thẻ 38

    Câu hỏi

    NO3− (charge −1), one specified resonance form: N=Oa, N–Ob, N–Oc; N has no lone pairs, Oa has 2, Ob and Oc have 3 each. Ob formal charge? O has 6 valence electrons.

    Câu trả lời

    −1 — Ob is singly bonded with 6 nonbonding electrons: 6 − 6 − 1 = −1.

  39. Thẻ 39

    Câu hỏi

    H3O+ (charge +1): three O–H single bonds; O has 1 lone pair, H none. O formal charge is +1 and each H is zero. What is their total?

    Câu trả lời

    +1 — add one O and three H: +1 + 3(0) = +1.

  40. Thẻ 40

    Câu hỏi

    Does an atom's formal charge specify its measured local electric charge inside a molecule?

    Câu trả lời

    No. Formal charge follows a counting convention; electron-density-based partial charges require a stated method and are a different quantity.

  41. Thẻ 41

    Câu hỏi

    CO3²− (charge −2): C oxidation number +4, each of the three O atoms −2. What is the oxidation-number total?

    Câu trả lời

    −2 — +4 + 3(−2) = −2, matching the ion charge.

  42. Thẻ 42

    Câu hỏi

    Only the neutral formula C2H4O2 is supplied. Can it uniquely determine the oxidation number of each individual carbon atom?

    Câu trả lời

    No. Connectivity matters: the formula alone doesn't say which bonds belong to each carbon atom.

  43. Thẻ 43

    Câu hỏi

    Neutral C≡O: each atom has 1 lone pair; O > C in electronegativity. C has 4 valence electrons. C formal charge and oxidation number, respectively?

    Câu trả lời

    −1 and +2. Formal charge: 4 − 2 − 3 = −1. Oxidation number: the triple-bond electrons go to O, so 4 − 2 = +2.

  44. Thẻ 44

    Câu hỏi

    NH4+ has overall charge +1. Does that tell you that every atom has formal charge +1?

    Câu trả lời

    No. The +1 is the total for the ion. In its four-single-bond Lewis structure with no lone pairs, N is +1 and each H is zero.

  45. Thẻ 45

    Câu hỏi

    Neutral H–H: one single bond; no lone pairs. Each H has 1 valence electron. Does H having zero for both formal charge and oxidation number make the two methods interchangeable?

    Câu trả lời

    No. Both methods split this identical-element bond equally, but they allocate ordinary different-element bonds differently.

  46. Thẻ 46

    Câu hỏi

    NO3−: in each of its three equivalent Lewis resonance forms, the O formal charges are 0, −1, and −1. Is −2/3 the formal charge of an O atom in one of those forms?

    Câu trả lời

    No. It is the average across equivalent forms; in any specified form an O formal charge is either 0 or −1.

  47. Thẻ 47

    Câu hỏi

    A prompt gives an atom's element and bond orders but omits its nonbonding electrons and overall species charge. Is that enough to determine its formal charge uniquely?

    Câu trả lời

    No. You need the nonbonding-electron count or enough additional information to deduce it.

  48. Thẻ 48

    Câu hỏi

    Neutral F–O–F: O has 2 lone pairs, each F has 3; F > O in electronegativity. O has 6 valence electrons. O formal charge and oxidation number, respectively?

    Câu trả lời

    Zero and +2. Formal charge: 6 − 4 − 2 = 0. Oxidation number: both bonding pairs go to F, so 6 − 4 = +2.

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Formal Charge vs Oxidation Number Flashcards: Electron Accounting

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