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