3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
The diagram shows two equivalent Lewis structures of the nitrite ion, NO₂⁻, joined by a double-headed arrow. Which statement best describes what the pair of structures represents?
Answer and reasoning
AOne ion in which the double bond moves rapidly back and forth between the two N–O positions A student who thinks a species flips between its resonance structures picks this. The double-headed arrow does not show a change over time; the ion has one unchanging structure with two identical N–O bonds.
BOne ion in which the two N–O bonds are identical, each between a single and a double bondCorrect The two structures are equivalent, so neither is correct alone; together they describe one ion. Its two N–O bonds are identical, each intermediate in length and strength between an N–O single bond and an N=O double bond, and they stay that way at all times.
CA sample in which half of the ions have each of the two bonding arrangements drawn A student who thinks each resonance structure pictures some of the particles in a sample picks this. Every nitrite ion is the same, with two identical N–O bonds; no ion has either arrangement alone.
DOne ion with a shorter N=O bond and a longer N–O bond, drawn as seen from two opposite sides A student who reads one Lewis structure as a literal picture of fixed bonds picks this. The N–O bonds in the nitrite ion are identical, which is why one structure is not enough and resonance must be included.
The diagram shows one Lewis structure of ozone, O₃. What is the formal charge on the central O atom in this structure?
Answer and reasoning
A−2 A student who counts all the electrons in an atom's bonds as its own picks this: 6 − 2 − 6 = −2. Only one electron from each bond is assigned to the atom.
B−1 A student who gives a negative sign to an atom that is assigned fewer electrons than its valence number picks this. The central atom is assigned 5 electrons, one fewer than the 6 of a free O atom, so its formal charge is +1.
C+1Correct The central O atom is assigned its 2 nonbonding electrons and one electron from each of its three bonds, 5 electrons in all. A free O atom has 6 valence electrons, so the formal charge is 6 − 2 − 3 = +1.
D+2 A student who counts a lone pair as one unit picks this: 6 − 1 − 3 = +2. The lone pair is two nonbonding electrons, and both are subtracted.
Working Formal charge = valence electrons − nonbonding electrons − number of bonds. A free O atom has 6 valence electrons. The central O atom has one lone pair (2 nonbonding electrons) and three bonds (one double bond and one single bond). Formal charge = 6 − 2 − 3 = +1.
Nitrogen monoxide, NO, is a gas whose molecules have a total of 11 valence electrons. Which statement best describes how well the Lewis structure model, with its octet rule, represents NO?
Answer and reasoning
AIt is adequate: a triple bond and one lone pair on each atom give both atoms an octet. A student who thinks an octet on every atom can always be reached by adding multiple bonds picks this. That structure contains only 10 electrons; NO has 11, so it is not a structure of NO.
BIt is limited: it leaves N without an octet, yet each atom in a real molecule must have one. A student who thinks atoms must have octets for a molecule to exist picks this. The octet rule is a guideline within the model, not a requirement: with 11 valence electrons, the two atoms of a real NO molecule cannot both have eight.
CIt is adequate: with the unpaired electron on N, both formal charges are zero, so it is complete. A student who thinks zero formal charges make a structure a complete description picks this. That is the best Lewis structure for NO, but N still has only seven electrons in it, so the octet-based model does not fully describe the molecule.
DIt is limited: an odd number of electrons cannot all be paired, so one atom lacks an octet.Correct Lewis structures place electrons in pairs, as bonds or lone pairs. With 11 valence electrons one electron must be unpaired, and the two atoms cannot both have eight electrons in any structure. The model describes such odd-electron molecules only approximately.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.6.A.1 Resonance Fix
Resonance
A refinement of the Lewis structure model that is needed when two or more equivalent Lewis structures can be drawn for the same arrangement of atoms. The species is described by all of the structures taken together, not by any one of them.
Resonance structures (equivalent Lewis structures)
Lewis structures of one species that have the atoms in the same positions and the same total number of valence electrons but differ in where the electrons are drawn, for example in which N–O bond of NO₂⁻ is shown as the double bond. They are joined by a double-headed arrow (↔).
Structure described by resonance
The single, unchanging structure that the resonance structures describe together. In NO₂⁻, O₃, CO₃²⁻ and benzene, the bonds that are drawn as single in one structure and double in another are all identical, with a length between those of a single and a double bond between the same atoms.
Average bond order in a species with resonance
The number of bonding pairs shared between the atoms, averaged over the equivalent positions. In NO₂⁻, three bonding pairs over two N–O positions give one and a half; in CO₃²⁻, four bonding pairs over three C–O positions give one and one-third. A higher average bond order corresponds to a shorter bond between the same two elements.
Students often think A species with resonance flips rapidly back and forth between its resonance structures, so at any instant it looks like one of them. In fact No. The species has one unchanging structure. In NO₂⁻ the two N–O bonds are identical at all times; the separate Lewis structures are drawings that, taken together, describe that one structure.
Students often think A sample contains a mixture of molecules, some with each of the resonance structures, for example half with the double bond on the left and half with it on the right. In fact No. Every molecule or ion in the sample has the same structure, in which the bonds involved in resonance are identical.
2.6.A.2 Formal charge Fix
Formal charge
A bookkeeping charge assigned to an atom in a Lewis structure by counting all of its nonbonding electrons and half of its bonding electrons as its own: formal charge = (valence electrons of the free atom) − (nonbonding electrons) − (number of bonds). The formal charges in a structure add up to the overall charge of the molecule or ion. They are not the actual charges on the atoms.
Nonequivalent Lewis structures
Valid Lewis structures for the same formula that are not equivalent, because they have different arrangements of atoms or different distributions of bonds and formal charges, such as N≡N–O and N=O=N for N₂O. They are not equally good models, and criteria are needed to choose between them.
Criteria for the best Lewis structure
The octet rule and formal charge. A structure in which second-period atoms have no more than eight valence electrons, and in which the formal charges are zero or as small in magnitude as possible, is the better model for predicting the structure and properties of the species.
Students often think All of the electrons around an atom, including both electrons of every bond, are counted as the atom's own when its formal charge is calculated. In fact No. The atom is assigned all of its nonbonding electrons but only half of its bonding electrons, that is, one electron for each bond.
Students often think Lone pairs are counted as single units, like bonds, so the formal charge is the valence electrons minus the number of lone pairs minus the number of bonds. In fact No. Each lone pair is two nonbonding electrons, and both are subtracted. Each bond, which is also a pair, contributes one electron to the atom.
2.6.A.3 Odd-electron species Fix
Odd-electron species
A molecule with an odd total number of valence electrons, such as NO (11) or NO₂ (17). Its electrons cannot all be paired, so every Lewis structure has an unpaired electron and at least one atom without an octet.
Limitations of the Lewis structure model
A Lewis structure is a simplified model that places valence electrons in pairs, as bonds between two atoms or as lone pairs on one atom. It needs resonance when equivalent structures exist, and it describes species with an odd number of valence electrons only approximately.
Students often think A structure with an octet on every atom can always be found by adding multiple bonds, so a structure such as a triple bond with one lone pair on each atom is accepted for NO. In fact No. A molecule with an odd number of valence electrons, such as NO with 11, cannot have all of its electrons in pairs, so at least one atom has fewer than eight electrons in every structure.
Students often think Atoms must have octets, so a molecule in which an atom lacks an octet cannot exist. In fact Yes. The octet rule is a guideline within a model, not a law. NO and NO₂ are well-known gases whose molecules have an odd number of valence electrons and an atom with fewer than eight.
4 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 4
The C–C single bond in ethane, C₂H₆, is 154 pm long, and the C=C double bond in ethene, C₂H₄, is 134 pm long. Two equivalent Lewis structures can be drawn for benzene, C₆H₆, each with alternating C–C and C=C bonds around a ring of six carbon atoms. In an experiment, the six carbon-carbon bond lengths in benzene are measured. Which result does the resonance model of benzene predict?
Answer and reasoning
ASix equal lengths, each between 134 pm and 154 pmCorrect Each carbon-carbon position is drawn as a single bond in one structure and a double bond in the other, so the model predicts six identical bonds, shorter than a single bond and longer than a double bond. The measured length of every carbon-carbon bond in benzene is 139 pm.
BThree lengths of 134 pm and three of 154 pm in each molecule A student who reads one Lewis structure as a literal picture of fixed single and double bonds picks this. That is the prediction of a single structure without resonance, and it is not what is observed.
CLengths that keep switching between 134 pm and 154 pm A student who thinks a molecule flips back and forth between its resonance structures picks this. The resonance model describes one unchanging structure; the bond lengths do not alternate in time.
DTwo sets of lengths, one for each separable form of benzene A student who thinks resonance structures are different forms of a substance, like isomers, picks this. The two structures have the atoms in the same positions and describe a single kind of molecule, so there are no separate forms to measure.
Working Two equivalent structures exist, so resonance must be included: each carbon-carbon position is single in one structure and double in the other. All six bonds are therefore identical and intermediate between a single bond (154 pm) and a double bond (134 pm). (The measured value is 139 pm.)
One Lewis structure, with two C=O bonds, is drawn for CO₂. Three equivalent Lewis structures, each with one C=O bond and two C–O bonds, are drawn for the carbonate ion, CO₃²⁻. Which prediction about the carbon-oxygen bonds in CO₃²⁻, compared with those in CO₂, is correct and is supported by valid reasoning from these structures?
Answer and reasoning
AAll of them are longer than in CO₂, since any bond in resonance is one and a half bonds. A student who thinks every bond that takes part in resonance is one and a half bonds picks this. The bonds are longer than in CO₂, but four bonding pairs over three positions give one and one-third bonds each, not one and a half.
BAll of them are shorter than in CO₂, since resonance makes each bond stronger than a double bond. A student who thinks resonance strengthens bonds beyond a double bond picks this. Each bond in CO₃²⁻ is intermediate between a single and a double bond, so it is longer and weaker than the double bonds in CO₂.
COne of them matches those in CO₂, and the other two are longer, since they are single bonds. A student who reads one Lewis structure as a literal picture of fixed bonds picks this. Because three equivalent structures can be drawn, the three carbon-oxygen bonds in CO₃²⁻ are identical; none is a full double bond.
DAll of them are longer than in CO₂, since four bonding pairs are shared over three positions.Correct In CO₃²⁻ the three structures share four bonding pairs equally over three carbon-oxygen positions, an average of one and one-third bonds each. Each C=O bond in CO₂ is a full double bond. A lower bond order means a longer bond, so all three bonds in CO₃²⁻ are longer than those in CO₂ (129 pm compared with 116 pm).
The diagram shows two Lewis structures, X and Y, drawn for N₂O with the atoms bonded in different orders. Which claim about the better model of N₂O is supported by evidence from the structures?
Answer and reasoning
AX is the better model, as its triple bond gives it more bonding electrons than Y has. A student who thinks the structure with the highest-order bond has the most bonding and is the best picks this. Both structures have four bonding pairs (8 bonding electrons): a triple and a single bond in X, two double bonds in Y. Bond order is not the criterion.
BX is the better model, as its formal charges are smaller in magnitude than those in Y.Correct In X the formal charges are 0 (end N), +1 (central N) and −1 (O). In Y they are −1 on each N and +2 on the central O. Every atom has an octet in both, so formal charge decides: X, with the smaller formal charges, is the better model. The atoms in N₂O are bonded in the order N–N–O.
CY is the better model, as the atom of the different element belongs in the center. A student who thinks the atom of the element that appears once always goes in the center picks this. That rule of thumb has exceptions; with O in the center the formal charges are −1, +2 and −1, larger than in X.
DX and Y are equally good models, as every atom in each of them has a full octet. A student who thinks the octet rule is the only test of a Lewis structure picks this. Every atom does have an octet in both, but the structures are not equivalent, and their formal charges differ: those in X are smaller.
The usual Lewis structures of the nitrate ion, NO₃⁻, have one N=O bond and two N–O bonds, with formal charges of +1 on N and −1 on each singly bonded O. A student proposes a structure with two N=O bonds and one N–O bond instead, in which the formal charge on N is 0. Which evaluation of the student's structure is correct?
Answer and reasoning
ANot valid: N would have ten electrons, while a second-period atom is limited to eight.Correct Two double bonds and a single bond are five bonding pairs, or ten electrons, around N. Nitrogen is in the second period and is not given more than eight valence electrons, so the octet rule excludes this structure even though its formal charges are smaller.
BValid and better: its formal charges are closer to zero than those in the usual structures. A student who thinks zero formal charges matter more than the octet rule picks this. The formal charges are smaller, but the structure puts ten electrons around a second-period atom, so it is not a valid Lewis structure.
CValid and better: a second N=O bond adds bonding electrons, which makes the ion stronger. A student who thinks the structure showing the most bonding is the best picks this. The extra bond can be drawn only by giving N ten electrons, which the octet rule does not allow for a second-period atom.
DNot valid: its formal charges add up to −1, and they are required to add up to zero. A student who thinks formal charges must always add up to zero picks this. The formal charges add up to the charge of the species, which is −1 for NO₃⁻; the structure is not valid for a different reason, the ten electrons around N.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
