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.
دربارهٔ این دسته
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.
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Formal charge: how are a bond's electrons assigned?
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Half to each bonded atom, regardless of electronegativity.
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Electron accounting: how many nonbonding electrons are in three lone pairs?
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Six electrons. Each lone pair contains two electrons.
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Lewis counting: how many bonding electrons are in a single, double, and triple bond?
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Two, four, and six, respectively. Each bond line represents one electron pair.
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Oxidation number: how are electrons allocated in an ordinary bond between different elements?
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Assign the entire bonding pair to the more electronegative atom. Nonbonding electrons stay with their atom.
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Electron accounting: what must the sum of all atom formal charges, or all atom oxidation numbers, equal?
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The charge of the whole species. For a neutral molecule, each sum is zero.
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Formal charge: what's the calculation for a specified Lewis structure?
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Neutral valence electrons − nonbonding electrons − half the bonding electrons. Read the target atom's counts from that particular structure.
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NH4+ (charge +1): four N–H single bonds; no lone pairs on N or H. N formal charge? N has 5 valence electrons.
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+1 — N receives 4 bonding electrons: 5 − 0 − 4 = +1.
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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.
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+1 — the bonding pair goes to F, leaving H assigned 0 electrons: 1 − 0 = +1.
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Neutral O=C=O: C has no lone pairs; each O has 2. C formal charge? C has 4 valence electrons.
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Zero — two double bonds assign C 4 bonding electrons: 4 − 0 − 4 = 0.
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OH− (charge −1): one O–H single bond; O has 3 lone pairs, H none. O formal charge? O has 6 valence electrons.
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−1 — O receives 1 bonding electron: 6 − 6 − 1 = −1.
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Neutral N≡N: each N has 1 lone pair. Either N oxidation number? N has 5 valence electrons.
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Zero — each N is assigned 2 nonbonding electrons and 3 bonding electrons: 5 − 2 − 3 = 0.
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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.
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−2 — C gets 6 electrons from the C–H bonds and none from C–Cl: 4 − 6 = −2.
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Neutral H–C≡N: N has 1 lone pair; C and H have none. N formal charge? N has 5 valence electrons.
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Zero — N receives 3 electrons from the triple bond: 5 − 2 − 3 = 0.
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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.
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−2 — C gets 4 electrons from C–H and 2 from half the C=C bond: 4 − 6 = −2.
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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.
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+2 — C keeps only its 2 nonbonding electrons: 4 − 2 = +2.
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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.
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Zero — F receives 1 bonding electron: 7 − 6 − 1 = 0.
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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.
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Zero — O receives 1 electron from each of its two bonds: 6 − 4 − 2 = 0.
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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.
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Zero — C gets 4 electrons from C–H and none from C=O: 4 − 4 = 0.
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NH2− (charge −1): two N–H single bonds; N has 2 lone pairs, H none. N formal charge? N has 5 valence electrons.
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−1 — N receives 2 bonding electrons: 5 − 4 − 2 = −1.
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Neutral H–C≡C–H: no atom has lone pairs. C > H in electronegativity. Either C oxidation number? C has 4 valence electrons.
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−1 — C gets 2 electrons from C–H and 3 from half the C≡C bond: 4 − 5 = −1.
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H3O+ (charge +1): three O–H single bonds; O has 1 lone pair, H none. O formal charge? O has 6 valence electrons.
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+1 — O receives 3 bonding electrons: 6 − 2 − 3 = +1.
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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.
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Zero — O receives 2 bonding electrons: 6 − 4 − 2 = 0.
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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.
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−2 — N keeps 2 nonbonding electrons, gets 4 from N–H, and 1 from N–N: 5 − 7 = −2.
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Neutral F–F: one single bond; each F has 3 lone pairs. Either F oxidation number? F has 7 valence electrons.
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Zero — each F keeps 6 nonbonding electrons and half the bond: 7 − 6 − 1 = 0.
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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?
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Oxidation number. Equal sharing instead gives N formal charge +1: 5 − 4 = +1.
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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?
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Four electrons. Two lone pairs contain 4 electrons, so O formal charge is 6 − 4 − 2 = 0.
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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?
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Four electrons. Each double bond assigns C 2 electrons, giving 4 − 4 = 0.
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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?
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One electron to each F. Identical-element bonds are split equally, so both F oxidation numbers are zero.
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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?
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One bonding electron. Formal charge splits the pair, giving O: 6 − 6 − 1 = −1.
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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?
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Neither; its contribution is zero. The N–N electrons are shared equally; the two N–H bonds give each N oxidation number −2.
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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?
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Three electrons. A triple bond contains 6 electrons; N receives half, so 5 − 2 − 3 = 0.
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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?
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+2 — both O–F bonding pairs go to F; O keeps 4 electrons: 6 − 4 = +2.
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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?
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−2 — C gets 6 electrons from its three C–H bonds and none from C–O: 4 − 6 = −2.
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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?
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No; C is +2. C keeps 2 nonbonding electrons, while N receives 8 electrons and is −3; +2 − 3 = −1.
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A neutral molecule's atom formal charges add to +1. Can that be a complete, consistent charge assignment?
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No. A neutral species requires a total of zero; recheck the counts and any omitted atoms.
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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.
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−1 — O keeps 4 nonbonding electrons, gets 2 from O–H and 1 from O–O: 6 − 7 = −1.
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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?
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No. Those are oxidation numbers; this Lewis structure has formal charge zero on every atom. A correct total doesn't identify the method.
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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.
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−1 — Ob is singly bonded with 6 nonbonding electrons: 6 − 6 − 1 = −1.
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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?
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+1 — add one O and three H: +1 + 3(0) = +1.
کارت ۴۰
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Does an atom's formal charge specify its measured local electric charge inside a molecule?
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No. Formal charge follows a counting convention; electron-density-based partial charges require a stated method and are a different quantity.
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CO3²− (charge −2): C oxidation number +4, each of the three O atoms −2. What is the oxidation-number total?
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−2 — +4 + 3(−2) = −2, matching the ion charge.
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Only the neutral formula C2H4O2 is supplied. Can it uniquely determine the oxidation number of each individual carbon atom?
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No. Connectivity matters: the formula alone doesn't say which bonds belong to each carbon atom.
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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?
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−1 and +2. Formal charge: 4 − 2 − 3 = −1. Oxidation number: the triple-bond electrons go to O, so 4 − 2 = +2.
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NH4+ has overall charge +1. Does that tell you that every atom has formal charge +1?
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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.
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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?
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No. Both methods split this identical-element bond equally, but they allocate ordinary different-element bonds differently.
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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?
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No. It is the average across equivalent forms; in any specified form an O formal charge is either 0 or −1.
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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?
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No. You need the nonbonding-electron count or enough additional information to deduce it.
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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?
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Zero and +2. Formal charge: 6 − 4 − 2 = 0. Oxidation number: both bonding pairs go to F, so 6 − 4 = +2.
۴۸ کارت
Formal Charge vs Oxidation Number Flashcards: Electron Accounting
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