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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.

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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.

Cartes de ce paquet

  1. Carte 1

    Question

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

    Réponse

    Half to each bonded atom, regardless of electronegativity.

  2. Carte 2

    Question

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

    Réponse

    Six electrons. Each lone pair contains two electrons.

  3. Carte 3

    Question

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

    Réponse

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

  4. Carte 4

    Question

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

    Réponse

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

  5. Carte 5

    Question

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

    Réponse

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

  6. Carte 6

    Question

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

    Réponse

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

  7. Carte 7

    Question

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

    Réponse

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

  8. Carte 8

    Question

    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.

    Réponse

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

  9. Carte 9

    Question

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

    Réponse

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

  10. Carte 10

    Question

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

    Réponse

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

  11. Carte 11

    Question

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

    Réponse

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

  12. Carte 12

    Question

    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.

    Réponse

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

  13. Carte 13

    Question

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

    Réponse

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

  14. Carte 14

    Question

    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.

    Réponse

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

  15. Carte 15

    Question

    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.

    Réponse

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

  16. Carte 16

    Question

    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.

    Réponse

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

  17. Carte 17

    Question

    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.

    Réponse

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

  18. Carte 18

    Question

    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.

    Réponse

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

  19. Carte 19

    Question

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

    Réponse

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

  20. Carte 20

    Question

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

    Réponse

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

  21. Carte 21

    Question

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

    Réponse

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

  22. Carte 22

    Question

    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.

    Réponse

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

  23. Carte 23

    Question

    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.

    Réponse

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

  24. Carte 24

    Question

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

    Réponse

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

  25. Carte 25

    Question

    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?

    Réponse

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

  26. Carte 26

    Question

    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?

    Réponse

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

  27. Carte 27

    Question

    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?

    Réponse

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

  28. Carte 28

    Question

    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?

    Réponse

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

  29. Carte 29

    Question

    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?

    Réponse

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

  30. Carte 30

    Question

    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?

    Réponse

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

  31. Carte 31

    Question

    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?

    Réponse

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

  32. Carte 32

    Question

    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?

    Réponse

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

  33. Carte 33

    Question

    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?

    Réponse

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

  34. Carte 34

    Question

    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?

    Réponse

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

  35. Carte 35

    Question

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

    Réponse

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

  36. Carte 36

    Question

    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.

    Réponse

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

  37. Carte 37

    Question

    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?

    Réponse

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

  38. Carte 38

    Question

    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.

    Réponse

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

  39. Carte 39

    Question

    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?

    Réponse

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

  40. Carte 40

    Question

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

    Réponse

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

  41. Carte 41

    Question

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

    Réponse

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

  42. Carte 42

    Question

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

    Réponse

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

  43. Carte 43

    Question

    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?

    Réponse

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

  44. Carte 44

    Question

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

    Réponse

    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. Carte 45

    Question

    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?

    Réponse

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

  46. Carte 46

    Question

    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?

    Réponse

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

  47. Carte 47

    Question

    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?

    Réponse

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

  48. Carte 48

    Question

    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?

    Réponse

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

Overlapping amber and teal glass counting trays with floating spherical counters on an ivory background.

48 cartes

Formal Charge vs Oxidation Number Flashcards: Electron Accounting

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