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.
About this deck
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.
Cards in this deck
Card 1
Question
Formal charge: how are a bond's electrons assigned?
Answer
Half to each bonded atom, regardless of electronegativity.
Card 2
Question
Electron accounting: how many nonbonding electrons are in three lone pairs?
Answer
Six electrons. Each lone pair contains two electrons.
Card 3
Question
Lewis counting: how many bonding electrons are in a single, double, and triple bond?
Answer
Two, four, and six, respectively. Each bond line represents one electron pair.
Card 4
Question
Oxidation number: how are electrons allocated in an ordinary bond between different elements?
Answer
Assign the entire bonding pair to the more electronegative atom. Nonbonding electrons stay with their atom.
Card 5
Question
Electron accounting: what must the sum of all atom formal charges, or all atom oxidation numbers, equal?
Answer
The charge of the whole species. For a neutral molecule, each sum is zero.
Card 6
Question
Formal charge: what's the calculation for a specified Lewis structure?
Answer
Neutral valence electrons − nonbonding electrons − half the bonding electrons. Read the target atom's counts from that particular structure.
Card 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.
Answer
+1 — N receives 4 bonding electrons: 5 − 0 − 4 = +1.
Card 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.
Answer
+1 — the bonding pair goes to F, leaving H assigned 0 electrons: 1 − 0 = +1.
Card 9
Question
Neutral O=C=O: C has no lone pairs; each O has 2. C formal charge? C has 4 valence electrons.
Answer
Zero — two double bonds assign C 4 bonding electrons: 4 − 0 − 4 = 0.
Card 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.
Answer
−1 — O receives 1 bonding electron: 6 − 6 − 1 = −1.
Card 11
Question
Neutral N≡N: each N has 1 lone pair. Either N oxidation number? N has 5 valence electrons.
Answer
Zero — each N is assigned 2 nonbonding electrons and 3 bonding electrons: 5 − 2 − 3 = 0.
Card 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.
Answer
−2 — C gets 6 electrons from the C–H bonds and none from C–Cl: 4 − 6 = −2.
Card 13
Question
Neutral H–C≡N: N has 1 lone pair; C and H have none. N formal charge? N has 5 valence electrons.
Answer
Zero — N receives 3 electrons from the triple bond: 5 − 2 − 3 = 0.
Card 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.
Answer
−2 — C gets 4 electrons from C–H and 2 from half the C=C bond: 4 − 6 = −2.
Card 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.
Answer
+2 — C keeps only its 2 nonbonding electrons: 4 − 2 = +2.
Card 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.
Answer
Zero — F receives 1 bonding electron: 7 − 6 − 1 = 0.
Card 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.
Answer
Zero — O receives 1 electron from each of its two bonds: 6 − 4 − 2 = 0.
Card 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.
Answer
Zero — C gets 4 electrons from C–H and none from C=O: 4 − 4 = 0.
Card 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.
Answer
−1 — N receives 2 bonding electrons: 5 − 4 − 2 = −1.
Card 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.
Answer
−1 — C gets 2 electrons from C–H and 3 from half the C≡C bond: 4 − 5 = −1.
Card 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.
Answer
+1 — O receives 3 bonding electrons: 6 − 2 − 3 = +1.
Card 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.
Answer
Zero — O receives 2 bonding electrons: 6 − 4 − 2 = 0.
Card 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.
Answer
−2 — N keeps 2 nonbonding electrons, gets 4 from N–H, and 1 from N–N: 5 − 7 = −2.
Card 24
Question
Neutral F–F: one single bond; each F has 3 lone pairs. Either F oxidation number? F has 7 valence electrons.
Answer
Zero — each F keeps 6 nonbonding electrons and half the bond: 7 − 6 − 1 = 0.
Card 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?
Answer
Oxidation number. Equal sharing instead gives N formal charge +1: 5 − 4 = +1.
Card 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?
Answer
Four electrons. Two lone pairs contain 4 electrons, so O formal charge is 6 − 4 − 2 = 0.
Card 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?
Answer
Four electrons. Each double bond assigns C 2 electrons, giving 4 − 4 = 0.
Card 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?
Answer
One electron to each F. Identical-element bonds are split equally, so both F oxidation numbers are zero.
Card 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?
Answer
One bonding electron. Formal charge splits the pair, giving O: 6 − 6 − 1 = −1.
Card 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?
Answer
Neither; its contribution is zero. The N–N electrons are shared equally; the two N–H bonds give each N oxidation number −2.
Card 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?
Answer
Three electrons. A triple bond contains 6 electrons; N receives half, so 5 − 2 − 3 = 0.
Card 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?
Answer
+2 — both O–F bonding pairs go to F; O keeps 4 electrons: 6 − 4 = +2.
Card 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?
Answer
−2 — C gets 6 electrons from its three C–H bonds and none from C–O: 4 − 6 = −2.
Card 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?
Answer
No; C is +2. C keeps 2 nonbonding electrons, while N receives 8 electrons and is −3; +2 − 3 = −1.
Card 35
Question
A neutral molecule's atom formal charges add to +1. Can that be a complete, consistent charge assignment?
Answer
No. A neutral species requires a total of zero; recheck the counts and any omitted atoms.
Card 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.
Answer
−1 — O keeps 4 nonbonding electrons, gets 2 from O–H and 1 from O–O: 6 − 7 = −1.
Card 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?
Answer
No. Those are oxidation numbers; this Lewis structure has formal charge zero on every atom. A correct total doesn't identify the method.
Card 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.
Answer
−1 — Ob is singly bonded with 6 nonbonding electrons: 6 − 6 − 1 = −1.
Card 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?
Answer
+1 — add one O and three H: +1 + 3(0) = +1.
Card 40
Question
Does an atom's formal charge specify its measured local electric charge inside a molecule?
Answer
No. Formal charge follows a counting convention; electron-density-based partial charges require a stated method and are a different quantity.
Card 41
Question
CO3²− (charge −2): C oxidation number +4, each of the three O atoms −2. What is the oxidation-number total?
Answer
−2 — +4 + 3(−2) = −2, matching the ion charge.
Card 42
Question
Only the neutral formula C2H4O2 is supplied. Can it uniquely determine the oxidation number of each individual carbon atom?
Answer
No. Connectivity matters: the formula alone doesn't say which bonds belong to each carbon atom.
Card 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?
Answer
−1 and +2. Formal charge: 4 − 2 − 3 = −1. Oxidation number: the triple-bond electrons go to O, so 4 − 2 = +2.
Card 44
Question
NH4+ has overall charge +1. Does that tell you that every atom has formal charge +1?
Answer
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.
Card 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?
Answer
No. Both methods split this identical-element bond equally, but they allocate ordinary different-element bonds differently.
Card 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?
Answer
No. It is the average across equivalent forms; in any specified form an O formal charge is either 0 or −1.
Card 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?
Answer
No. You need the nonbonding-electron count or enough additional information to deduce it.
Card 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?
Answer
Zero and +2. Formal charge: 6 − 4 − 2 = 0. Oxidation number: both bonding pairs go to F, so 6 − 4 = +2.
48 cards
Formal Charge vs Oxidation Number Flashcards: Electron Accounting
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