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.
Tarjetas de este mazo
Tarjeta 1
Pregunta
Formal charge: how are a bond's electrons assigned?
Respuesta
Half to each bonded atom, regardless of electronegativity.
Tarjeta 2
Pregunta
Electron accounting: how many nonbonding electrons are in three lone pairs?
Respuesta
Six electrons. Each lone pair contains two electrons.
Tarjeta 3
Pregunta
Lewis counting: how many bonding electrons are in a single, double, and triple bond?
Respuesta
Two, four, and six, respectively. Each bond line represents one electron pair.
Tarjeta 4
Pregunta
Oxidation number: how are electrons allocated in an ordinary bond between different elements?
Respuesta
Assign the entire bonding pair to the more electronegative atom. Nonbonding electrons stay with their atom.
Tarjeta 5
Pregunta
Electron accounting: what must the sum of all atom formal charges, or all atom oxidation numbers, equal?
Respuesta
The charge of the whole species. For a neutral molecule, each sum is zero.
Tarjeta 6
Pregunta
Formal charge: what's the calculation for a specified Lewis structure?
Respuesta
Neutral valence electrons − nonbonding electrons − half the bonding electrons. Read the target atom's counts from that particular structure.
Tarjeta 7
Pregunta
NH4+ (charge +1): four N–H single bonds; no lone pairs on N or H. N formal charge? N has 5 valence electrons.
Respuesta
+1 — N receives 4 bonding electrons: 5 − 0 − 4 = +1.
Tarjeta 8
Pregunta
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.
Respuesta
+1 — the bonding pair goes to F, leaving H assigned 0 electrons: 1 − 0 = +1.
Tarjeta 9
Pregunta
Neutral O=C=O: C has no lone pairs; each O has 2. C formal charge? C has 4 valence electrons.
Respuesta
Zero — two double bonds assign C 4 bonding electrons: 4 − 0 − 4 = 0.
Tarjeta 10
Pregunta
OH− (charge −1): one O–H single bond; O has 3 lone pairs, H none. O formal charge? O has 6 valence electrons.
Respuesta
−1 — O receives 1 bonding electron: 6 − 6 − 1 = −1.
Tarjeta 11
Pregunta
Neutral N≡N: each N has 1 lone pair. Either N oxidation number? N has 5 valence electrons.
Respuesta
Zero — each N is assigned 2 nonbonding electrons and 3 bonding electrons: 5 − 2 − 3 = 0.
Tarjeta 12
Pregunta
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.
Respuesta
−2 — C gets 6 electrons from the C–H bonds and none from C–Cl: 4 − 6 = −2.
Tarjeta 13
Pregunta
Neutral H–C≡N: N has 1 lone pair; C and H have none. N formal charge? N has 5 valence electrons.
Respuesta
Zero — N receives 3 electrons from the triple bond: 5 − 2 − 3 = 0.
Tarjeta 14
Pregunta
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.
Respuesta
−2 — C gets 4 electrons from C–H and 2 from half the C=C bond: 4 − 6 = −2.
Tarjeta 15
Pregunta
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.
Respuesta
+2 — C keeps only its 2 nonbonding electrons: 4 − 2 = +2.
Tarjeta 16
Pregunta
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.
Respuesta
Zero — F receives 1 bonding electron: 7 − 6 − 1 = 0.
Tarjeta 17
Pregunta
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.
Respuesta
Zero — O receives 1 electron from each of its two bonds: 6 − 4 − 2 = 0.
Tarjeta 18
Pregunta
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.
Respuesta
Zero — C gets 4 electrons from C–H and none from C=O: 4 − 4 = 0.
Tarjeta 19
Pregunta
NH2− (charge −1): two N–H single bonds; N has 2 lone pairs, H none. N formal charge? N has 5 valence electrons.
Respuesta
−1 — N receives 2 bonding electrons: 5 − 4 − 2 = −1.
Tarjeta 20
Pregunta
Neutral H–C≡C–H: no atom has lone pairs. C > H in electronegativity. Either C oxidation number? C has 4 valence electrons.
Respuesta
−1 — C gets 2 electrons from C–H and 3 from half the C≡C bond: 4 − 5 = −1.
Tarjeta 21
Pregunta
H3O+ (charge +1): three O–H single bonds; O has 1 lone pair, H none. O formal charge? O has 6 valence electrons.
Respuesta
+1 — O receives 3 bonding electrons: 6 − 2 − 3 = +1.
Tarjeta 22
Pregunta
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.
Respuesta
Zero — O receives 2 bonding electrons: 6 − 4 − 2 = 0.
Tarjeta 23
Pregunta
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.
Respuesta
−2 — N keeps 2 nonbonding electrons, gets 4 from N–H, and 1 from N–N: 5 − 7 = −2.
Tarjeta 24
Pregunta
Neutral F–F: one single bond; each F has 3 lone pairs. Either F oxidation number? F has 7 valence electrons.
Respuesta
Zero — each F keeps 6 nonbonding electrons and half the bond: 7 − 6 − 1 = 0.
Tarjeta 25
Pregunta
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?
Respuesta
Oxidation number. Equal sharing instead gives N formal charge +1: 5 − 4 = +1.
Tarjeta 26
Pregunta
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?
Respuesta
Four electrons. Two lone pairs contain 4 electrons, so O formal charge is 6 − 4 − 2 = 0.
Tarjeta 27
Pregunta
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?
Respuesta
Four electrons. Each double bond assigns C 2 electrons, giving 4 − 4 = 0.
Tarjeta 28
Pregunta
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?
Respuesta
One electron to each F. Identical-element bonds are split equally, so both F oxidation numbers are zero.
Tarjeta 29
Pregunta
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?
Respuesta
One bonding electron. Formal charge splits the pair, giving O: 6 − 6 − 1 = −1.
Tarjeta 30
Pregunta
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?
Respuesta
Neither; its contribution is zero. The N–N electrons are shared equally; the two N–H bonds give each N oxidation number −2.
Tarjeta 31
Pregunta
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?
Respuesta
Three electrons. A triple bond contains 6 electrons; N receives half, so 5 − 2 − 3 = 0.
Tarjeta 32
Pregunta
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?
Respuesta
+2 — both O–F bonding pairs go to F; O keeps 4 electrons: 6 − 4 = +2.
Tarjeta 33
Pregunta
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?
Respuesta
−2 — C gets 6 electrons from its three C–H bonds and none from C–O: 4 − 6 = −2.
Tarjeta 34
Pregunta
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?
Respuesta
No; C is +2. C keeps 2 nonbonding electrons, while N receives 8 electrons and is −3; +2 − 3 = −1.
Tarjeta 35
Pregunta
A neutral molecule's atom formal charges add to +1. Can that be a complete, consistent charge assignment?
Respuesta
No. A neutral species requires a total of zero; recheck the counts and any omitted atoms.
Tarjeta 36
Pregunta
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.
Respuesta
−1 — O keeps 4 nonbonding electrons, gets 2 from O–H and 1 from O–O: 6 − 7 = −1.
Tarjeta 37
Pregunta
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?
Respuesta
No. Those are oxidation numbers; this Lewis structure has formal charge zero on every atom. A correct total doesn't identify the method.
Tarjeta 38
Pregunta
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.
Respuesta
−1 — Ob is singly bonded with 6 nonbonding electrons: 6 − 6 − 1 = −1.
Tarjeta 39
Pregunta
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?
Respuesta
+1 — add one O and three H: +1 + 3(0) = +1.
Tarjeta 40
Pregunta
Does an atom's formal charge specify its measured local electric charge inside a molecule?
Respuesta
No. Formal charge follows a counting convention; electron-density-based partial charges require a stated method and are a different quantity.
Tarjeta 41
Pregunta
CO3²− (charge −2): C oxidation number +4, each of the three O atoms −2. What is the oxidation-number total?
Respuesta
−2 — +4 + 3(−2) = −2, matching the ion charge.
Tarjeta 42
Pregunta
Only the neutral formula C2H4O2 is supplied. Can it uniquely determine the oxidation number of each individual carbon atom?
Respuesta
No. Connectivity matters: the formula alone doesn't say which bonds belong to each carbon atom.
Tarjeta 43
Pregunta
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?
Respuesta
−1 and +2. Formal charge: 4 − 2 − 3 = −1. Oxidation number: the triple-bond electrons go to O, so 4 − 2 = +2.
Tarjeta 44
Pregunta
NH4+ has overall charge +1. Does that tell you that every atom has formal charge +1?
Respuesta
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.
Tarjeta 45
Pregunta
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?
Respuesta
No. Both methods split this identical-element bond equally, but they allocate ordinary different-element bonds differently.
Tarjeta 46
Pregunta
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?
Respuesta
No. It is the average across equivalent forms; in any specified form an O formal charge is either 0 or −1.
Tarjeta 47
Pregunta
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?
Respuesta
No. You need the nonbonding-electron count or enough additional information to deduce it.
Tarjeta 48
Pregunta
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?
Respuesta
Zero and +2. Formal charge: 6 − 4 − 2 = 0. Oxidation number: both bonding pairs go to F, so 6 − 4 = +2.
48 tarjetas
Formal Charge vs Oxidation Number Flashcards: Electron Accounting
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