MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5b Covalent Bonding Lewis Structures

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5b Covalent Bonding Lewis StructuresQuestion 1 of 20

A laboratory evaluates phosphorus pentachloride (PCl $_5$ ) as a chlorinating agent under anhydrous conditions. A Lewis structure with P central and five P–Cl single bonds (no lone pairs on P) is consistent with experimental stoichiometry; phosphorus is a third-row element and can have an expanded valence shell. Based on the Lewis structure, what prediction can be made regarding the molecular shape of PCl $_5$ in the gas phase?

Trigonal bipyramidal, because five bonding electron domains around P minimize repulsion in a

90^\circ/120^\circ

arrangement.

Square planar, because five electron domains arrange as a square with one bond above the plane to minimize repulsion.

Tetrahedral, because phosphorus must obey the octet rule and can form at most four single bonds.

Trigonal pyramidal, because five bonds imply four bonding pairs and one lone pair on phosphorus.

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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5b Covalent Bonding Lewis Structures

Practice 5b Covalent Bonding Lewis Structures in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on 5b Covalent Bonding Lewis Structures, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Trigonal bipyramidal, because five bonding electron domains around P minimize repulsion in a $90^\circ/120^\circ$ arrangement. (correct answer)
Square planar, because five electron domains arrange as a square with one bond above the plane to minimize repulsion.
Tetrahedral, because phosphorus must obey the octet rule and can form at most four single bonds.
Trigonal pyramidal, because five bonds imply four bonding pairs and one lone pair on phosphorus.

Explanation: This question tests understanding of covalent bonding and Lewis structures to predict molecular geometry using VSEPR theory. Lewis structures illustrate electron distribution, and VSEPR theory uses electron domain arrangements to predict three-dimensional molecular shapes. In PCl₅, the Lewis structure shows phosphorus with five P-Cl single bonds and no lone pairs, creating five bonding electron domains. Choice A is correct because five electron domains arrange in a trigonal bipyramidal geometry to minimize electron-electron repulsion, with three equatorial positions at 120° angles and two axial positions at 90° to the equatorial plane. Choice C is incorrect because phosphorus, being a third-row element, can expand its valence shell beyond eight electrons using empty d orbitals. When predicting shapes for molecules with expanded octets, remember that five electron domains always adopt trigonal bipyramidal geometry, while six domains form octahedral geometry.

Question 2

A research group models the geometry of nitrite (NO $_2^-$ ) in aqueous solution to predict its interactions with metal centers. The Lewis structure that minimizes formal charges places N in the center with one N=O double bond, one N–O single bond, and one lone pair on N; two equivalent resonance forms exist. Based on the Lewis structure, what prediction can be made regarding the N–O bond lengths in NO $_2^-$ ?

One N–O bond is significantly shorter than the other because the double bond is fixed in a single resonance form.
Both N–O bonds are the same length because resonance gives each N–O bond partial double-bond character (bond order between 1 and 2). (correct answer)
Both N–O bonds are the same length because the ion is linear, forcing identical bond angles and therefore identical bond lengths.
One N–O bond is longer because the best Lewis structure places the negative charge on nitrogen, weakening only one N–O bond.

Explanation: This question tests understanding of covalent bonding and Lewis structures, specifically how resonance affects bond properties. Lewis structures illustrate electron distribution, and resonance structures show electron delocalization that affects bond lengths and strengths. In NO₂⁻, the Lewis structure shows nitrogen with two N-O bonds that resonate between single and double bond character, plus one lone pair on nitrogen. Choice B is correct because resonance delocalizes the pi electrons equally between both N-O bonds, giving each bond a bond order of 1.5 (between single and double), resulting in identical bond lengths. Choice A is incorrect because resonance prevents the double bond from being fixed in one position - the electrons are delocalized across both bonds. When analyzing resonance structures, remember that the actual molecule is a hybrid of all resonance forms, leading to averaged bond properties rather than distinct single and double bonds.

Question 3

A synthetic protocol uses nitrate (NO $_3^-$ ) as a counterion, and its symmetry is used to rationalize weak ion-pairing in solution. Total valence electrons are $5 + 3(6) + 1 = 24$ . The Lewis structure is represented by three resonance forms with one N=O and two N–O bonds in each form, distributing negative charge over the oxygens. Based on the Lewis structure, what prediction can be made regarding the molecular shape around nitrogen and the equivalence of the N–O bonds?

Trigonal planar with three equivalent N–O bonds due to resonance delocalization. (correct answer)
Tetrahedral with three equivalent N–O bonds because nitrogen has four lone pairs.
Bent with two short and one long N–O bond because resonance localizes the double bond.
Linear with three equivalent N–O bonds because nitrogen forms three $sp$ hybrids.

Explanation: This question tests understanding of covalent bonding and Lewis structures in predicting molecular geometry with resonance. Lewis structures illustrate electron distribution, and resonance in polyatomic ions shows charge delocalization across equivalent positions. In NO₃⁻, the Lewis structure has three resonance forms with the double bond rotating among the three N-O positions, while nitrogen maintains three electron domains (three bonds, no lone pairs). Choice A is correct because three electron domains around nitrogen produce trigonal planar geometry, and resonance makes all three N-O bonds equivalent with bond order 1.33 each. Choice B is incorrect because nitrogen has no lone pairs in NO₃⁻, preventing tetrahedral geometry. When analyzing resonance structures, recognize that rapid interconversion makes all equivalent positions identical in the actual molecule, affecting both bond lengths and molecular symmetry.

Question 4

A gas-phase electron-diffraction experiment compares formal charge assignments for isoelectronic species. For carbon monoxide (CO), total valence electrons are $4 + 6 = 10$ . A common Lewis structure uses a triple bond between C and O with one lone pair on each atom. Based on this Lewis structure, which formal charge assignment is most consistent with the octet rule and electron counting?

C has +1 and O has −1.
C has −1 and O has +1. (correct answer)
C has 0 and O has 0.
C has +2 and O has −2.

Explanation: This question tests understanding of covalent bonding and Lewis structures, specifically formal charge calculations. Lewis structures illustrate electron distribution, and formal charge equals valence electrons minus (lone pair electrons + ½ bonding electrons). In CO with a triple bond (C≡O) and one lone pair on each atom, carbon's formal charge = 4 - (2 + 3) = -1, while oxygen's formal charge = 6 - (2 + 3) = +1. Choice B is correct because these formal charges satisfy the octet rule while accounting for the unusual electron distribution in CO, where the less electronegative carbon carries negative charge. Choice A is incorrect because it reverses the formal charges, ignoring the electron counting rules. When calculating formal charges, carefully count all electrons assigned to each atom, remembering that bonding electrons are shared equally regardless of electronegativity differences.

Question 5

In a kinetic study of acid–base behavior in aprotic solvent, ammonia (NH $_3$ ) is used as a nucleophile. The Lewis structure places nitrogen central with three N–H single bonds and one lone pair; total valence electrons are $5 + 3(1) = 8$ . Assume VSEPR applies and that lone pairs repel more strongly than bonding pairs. What prediction can be made from the Lewis structure regarding the approximate H–N–H bond angle in NH $_3$ ?

Approximately 180° because three bonds maximize separation in a line.
Approximately 120° because nitrogen is $sp^2$ -hybridized with no lone pairs.
Approximately 109.5° because the molecular shape is tetrahedral.
Slightly less than 109.5° because the electron geometry is tetrahedral with one lone pair. (correct answer)

Explanation: This question tests understanding of covalent bonding and Lewis structures in predicting bond angles. Lewis structures illustrate electron distribution, and VSEPR theory predicts that electron domains arrange to minimize repulsion, with lone pairs exerting stronger repulsion than bonding pairs. In NH₃, the Lewis structure shows nitrogen with three N-H bonds and one lone pair, creating four electron domains in a tetrahedral arrangement. Choice D is correct because while the electron geometry is tetrahedral (109.5°), the molecular geometry is trigonal pyramidal, and the lone pair's stronger repulsion compresses the H-N-H angles to slightly less than 109.5° (approximately 107°). Choice C is incorrect because it ignores the lone pair's effect on bond angles. When evaluating bond angles, consider both the electron domain geometry and the differential repulsion of lone pairs versus bonding pairs.

Question 6

A photochemistry lab compared formaldehyde (H2CO) and methanol (CH3OH) as carbonyl-containing versus alcohol-containing quenchers in a fluorescence assay. The team used Lewis structures to infer whether the carbon–oxygen bond is best represented as a single or double bond and how that influences bond length and polarity. Based on the Lewis structure of formaldehyde, which prediction can be made regarding the C–O bond?

Constants (for reference): electronegativity values—H 2.1, C 2.5, O 3.5.

The C–O bond in formaldehyde is a single bond with bond order 1 because oxygen must have three lone pairs to complete its octet.
The C–O bond in formaldehyde has bond order 2, consistent with a shorter, stronger bond than a C–O single bond. (correct answer)
The C–O bond in formaldehyde is nonpolar because carbon and oxygen have similar electronegativities.
The C–O bond in formaldehyde must be a triple bond to satisfy the octet on carbon.

Explanation: This question tests understanding of covalent bonding and Lewis structures to determine bond order and properties. Lewis structures illustrate electron distribution, predicting molecular shape and properties, including bond multiplicity. In formaldehyde (H2CO), the Lewis structure shows carbon forming a double bond with oxygen (C=O) and two single bonds with hydrogen atoms. Choice B is correct because it accurately identifies the C=O double bond with bond order 2, which results in a shorter, stronger bond compared to single C-O bonds. Choice A is incorrect because oxygen only needs two lone pairs (not three) when forming a double bond, satisfying its octet with four shared electrons. In similar cases, ensure to count shared electron pairs between atoms to determine bond order: single bond = 1, double bond = 2, triple bond = 3.

Question 7

In a gas-phase kinetics experiment, nitrosyl chloride (NOCl) was monitored as a transient intermediate. Researchers used Lewis structures to anticipate molecular geometry and whether a net dipole moment should be expected, which can influence collisional energy transfer. Consider NOCl with nitrogen as the central atom. Based on the Lewis structure, what prediction can be made regarding the molecular shape about nitrogen?

Constants (for reference): electronegativity values—N 3.0, O 3.5, Cl 3.0.

Linear, because nitrogen forms two double bonds and has no lone pairs.
Bent, because nitrogen has three electron domains (two bonds and one lone pair) giving a non-linear molecular shape. (correct answer)
Trigonal planar, because nitrogen has three sigma bonds to O, Cl, and a second oxygen from resonance.
Tetrahedral, because nitrogen must have four electron domains to satisfy the octet in NOCl.

Explanation: This question tests understanding of covalent bonding and Lewis structures to predict molecular geometry. Lewis structures illustrate electron distribution, predicting molecular shape and properties through VSEPR theory. In NOCl, the Lewis structure shows nitrogen as the central atom with a double bond to oxygen, a single bond to chlorine, and one lone pair, creating three electron domains. Choice B is correct because three electron domains around nitrogen result in a bent molecular shape due to the lone pair repelling the bonding pairs. Choice A is incorrect because it fails to account for the lone pair on nitrogen; NOCl has three electron domains, not two, preventing a linear geometry. In similar cases, ensure to count all electron domains (single bonds, multiple bonds count as one domain, and lone pairs) to predict molecular geometry using VSEPR theory.

Question 8

A pharmaceutical lab compares the basicity of pyridine (C $_5$ H $_5$ N) and pyrrole (C $_4$ H $_5$ N) using their Lewis structures. In pyridine, the nitrogen is part of an aromatic ring and has a lone pair not used in the aromatic sextet; in pyrrole, the nitrogen lone pair contributes to aromaticity. Based on these Lewis-structure considerations, which interaction is most likely when each is exposed to H $^+$ in water?

Both protonate equally because aromatic rings cannot accept protons.
Pyridine protonates more readily because its nitrogen lone pair is available to form an N–H bond. (correct answer)
Pyrrole protonates more readily because its nitrogen has a positive formal charge in the neutral Lewis structure.
Neither protonates because nitrogen cannot exceed an octet in any Lewis structure.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like basicity from lone pair availability. In pyridine, the Lewis structure shows nitrogen's lone pair in an sp2 orbital, available for protonation, unlike pyrrole's delocalized pair. Choice B is correct because pyridine's accessible lone pair makes it more basic. Choice C is incorrect due to misassigning formal charges; pyrrole's nitrogen is neutral. In similar cases, ensure to determine if lone pairs are delocalized when evaluating Lewis structures for reactivity. Consider aromaticity's impact on electron availability.

Question 9

In a headspace analysis of volatile anesthetics, a lab compares nitrous oxide (N $_2$ O) to carbon dioxide (CO $_2$ ) as calibration gases. For N $_2$ O, the dominant Lewis structure is typically drawn as N–N–O with a total of 16 valence electrons; for CO $_2$ , the dominant Lewis structure is O=C=O with 16 valence electrons. Both gases are linear under the conditions used. The instrument response is sensitive to molecular polarity because polar analytes interact more strongly with a polar stationary phase. Based on the Lewis structure of N $_2$ O, what prediction can be made regarding its polarity relative to CO $_2$ ? (Electronegativity: N = 3.0, O = 3.5, C = 2.5.)

N $_2$ O is nonpolar like CO $_2$ because any bond dipoles cancel in a linear molecule.
N $_2$ O is polar because the terminal atoms are different, giving a net dipole even if the molecule is linear. (correct answer)
N $_2$ O is polar only if it is bent; linear geometry implies zero dipole moment.
N $_2$ O is more nonpolar than CO $_2$ because N–N bonds dominate the electron distribution.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties such as polarity based on bond dipoles and geometry. In N2O, the Lewis structure shows a linear arrangement with N≡N-O or resonance forms, but unequal terminal atoms (N and O) create asymmetric electron distribution. Choice B is correct because the different terminal atoms result in a net dipole despite linearity, unlike symmetric CO2. Choice A is incorrect due to overlooking the asymmetry in N2O, assuming all linear molecules are nonpolar. In similar cases, ensure to consider molecular symmetry and electronegativity differences when evaluating Lewis structures for polarity. Always verify if bond dipoles cancel completely in the overall geometry.

Question 10

A formulation study evaluates carbon tetrachloride (CCl $_4$ ) as a nonpolar solvent. The Lewis structure places carbon central with four single bonds to chlorine; total valence electrons are $4 + 4(7) = 32$ , with each Cl bearing three lone pairs. Assume ideal VSEPR geometry. Which property is best explained by the Lewis structure of CCl $_4$ ?

CCl $_4$ is strongly polar because each C–Cl bond is polar.
CCl $_4$ is nonpolar because a symmetric tetrahedral shape cancels bond dipoles. (correct answer)
CCl $_4$ is nonpolar because carbon has no valence electrons in the Lewis structure.
CCl $_4$ is polar because the Cl atoms force a square-planar geometry around carbon.

Explanation: This question tests understanding of covalent bonding and Lewis structures in predicting molecular polarity. Lewis structures illustrate electron distribution, and molecular geometry determines whether bond dipoles cancel to produce nonpolar molecules. In CCl₄, the Lewis structure shows carbon with four equivalent C-Cl bonds arranged tetrahedrally, with each chlorine having three lone pairs. Choice B is correct because the tetrahedral geometry places the four C-Cl bonds symmetrically in three dimensions, causing all bond dipoles to cancel perfectly despite each C-Cl bond being polar, resulting in a nonpolar molecule. Choice A is incorrect because it ignores the symmetry that cancels individual bond dipoles. When evaluating molecular polarity, consider both individual bond polarities and the three-dimensional arrangement - symmetric molecules with identical substituents are typically nonpolar regardless of bond polarity.

Question 11

A formulation chemist compares ammonia (NH $_3$ ) and ammonium (NH $_4^+$ ) in aqueous solution to rationalize differences in geometry and hydrogen-bonding patterns. The Lewis structure of NH $_3$ has three N–H bonds and one lone pair on N; NH $_4^+$ has four N–H bonds and no lone pairs on N. Based on the Lewis structure, what prediction can be made about the molecular shape of NH $_4^+$ relative to NH $_3$ ? (Assume VSEPR applies.)

NH $_4^+$ is trigonal pyramidal like NH $_3$ because both have four electron domains.
NH $_4^+$ is tetrahedral because it has four bonding pairs and no lone pairs on nitrogen. (correct answer)
NH $_4^+$ is bent because positive charge increases lone-pair repulsion.
NH $_4^+$ is square planar because four N–H bonds maximize separation in a plane.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties via VSEPR and electron domains. In NH4+, the Lewis structure shows four N-H bonds and no lone pairs on nitrogen, yielding four electron domains. Choice B is correct because four bonding domains without lone pairs result in tetrahedral geometry, unlike pyramidal NH3. Choice A is incorrect due to confusing NH4+ with NH3, ignoring the extra bond and charge. In similar cases, ensure to count bonds and lone pairs precisely when evaluating Lewis structures for ions. Compare charged and neutral species to highlight geometry differences.

Question 12

A membrane-permeability assay compares urea (NH $_2$ CONH $_2$ ) to ethane (C $_2$ H $_6$ ). Urea's Lewis structure contains a carbonyl (C=O) and two amide nitrogens, each bearing a lone pair; ethane contains only C–C and C–H single bonds. The assay shows urea has much higher aqueous solubility. Based on the Lewis structure, which property best explains urea's higher solubility in water? (Water can both donate and accept H-bonds.)

Urea is nonpolar because resonance cancels all dipoles, increasing solubility.
Urea can participate in multiple hydrogen bonds via the carbonyl oxygen and N–H groups. (correct answer)
Urea has a smaller molar mass than ethane, so it must be more soluble.
Urea has only single bonds, which are more soluble in water than double bonds.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like solubility through intermolecular interactions. In urea, the Lewis structure shows N-H bonds and a C=O, enabling multiple hydrogen bonds with water. Choice B is correct because these groups allow urea to form H-bonds, enhancing solubility unlike nonpolar ethane. Choice A is incorrect due to misstating urea as nonpolar; resonance doesn't cancel polarity here. In similar cases, ensure to identify H-bond donors and acceptors when evaluating Lewis structures for solubility. Compare polar functional groups to nonpolar ones for predictions.

Question 13

A laboratory evaluates the dipole moments of isoelectronic species CO and N $_2$ . Both have 10 valence electrons total. A common Lewis structure for CO includes a triple bond with one lone pair on each atom and formal charges C $^-$ and O $^+$ . N $_2$ is drawn as Nc3N with one lone pair on each N and no formal charges. Based on the Lewis structure, which statement best explains why CO has a nonzero dipole moment while N $_2$ does not?

CO is nonpolar because triple bonds always eliminate dipole moments.
CO is polar because its Lewis structure assigns unequal formal charges to different atoms, unlike homonuclear N $_2$ . (correct answer)
CO is nonpolar because carbon and oxygen have identical electronegativities.
N $_2$ is polar because each nitrogen has a lone pair that creates a permanent dipole.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like dipole moments from bond polarity and formal charges. In CO, the Lewis structure with a triple bond and formal charges (C- and O+) creates a dipole, unlike symmetric N2. Choice B is correct because unequal atoms and charges yield a net dipole in CO. Choice A is incorrect due to assuming all triple bonds eliminate dipoles, ignoring atomic differences. In similar cases, ensure to consider formal charges and electronegativity when evaluating Lewis structures for polarity. Compare isoelectronic species to highlight differences.

Question 14

A spectroscopic study tracks the geometry of water (H $_2$ O) and hydrogen sulfide (H $_2$ S) in the gas phase. Both are drawn with two single bonds to the central atom and two lone pairs on the central atom. Based on the Lewis structure and VSEPR, what prediction can be made about the electron geometry and molecular shape of H $_2$ O?

Electron geometry tetrahedral; molecular shape bent, due to two lone pairs on oxygen. (correct answer)
Electron geometry trigonal planar; molecular shape bent, due to one lone pair on oxygen.
Electron geometry linear; molecular shape linear, because oxygen forms two bonds.
Electron geometry tetrahedral; molecular shape tetrahedral, because lone pairs do not affect shape.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties via VSEPR and electron domains. In H2O, the Lewis structure shows two O-H bonds and two lone pairs, yielding four electron domains. Choice A is correct because tetrahedral electron geometry with two lone pairs results in bent molecular shape. Choice B is incorrect due to miscounting domains; oxygen has four, not three. In similar cases, ensure to include lone pairs in domain counts when evaluating Lewis structures for geometry. Apply VSEPR rules consistently for accurate predictions.

Question 15

A catalysis paper describes hydrogen cyanide (HCN) as a linear ligand in metal complexes. HCN has 10 valence electrons total (H = 1, C = 4, N = 5) and is commonly drawn as H–Cc3N with one lone pair on N. Based on the Lewis structure, what prediction can be made about the bond angle H–C–N and the hybridization of carbon?

$\approx 109.5^\circ$ ; carbon is sp $^3$ hybridized due to four electron domains.
$\approx 120^\circ$ ; carbon is sp $^2$ hybridized due to three electron domains.
$\approx 180^\circ$ ; carbon is sp hybridized due to two electron domains. (correct answer)
$\approx 90^\circ$ ; carbon is d $^2$ sp $^3$ hybridized due to octahedral geometry.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like bond angles and hybridization from electron domains. In HCN, the Lewis structure shows H-C≡N, with carbon having two domains (one single, one triple). Choice C is correct because two domains yield linear geometry and sp hybridization. Choice B is incorrect due to miscounting domains; triple bonds count as one domain. In similar cases, ensure to treat multiple bonds as single domains when evaluating Lewis structures. Correlate domain count with hybridization for accuracy.

Question 16

A polymer lab evaluates the reactivity of epoxides toward nucleophilic ring opening. Ethylene oxide (C $_2$ H $_4$ O) is represented by a three-membered ring with oxygen bonded to two carbons and bearing two lone pairs. Based on the Lewis structure, which statement best predicts why the C–O bonds in the epoxide are susceptible to nucleophilic attack under acidic conditions?

Protonation of oxygen increases the electrophilicity of adjacent carbons by polarizing the C–O bonds. (correct answer)
Protonation of oxygen removes ring strain by converting the ring to a linear molecule without bond breaking.
Acidic conditions create a carbanion at carbon by adding a lone pair to carbon in the Lewis structure.
Nucleophiles attack oxygen because oxygen is always positively charged in the neutral epoxide Lewis structure.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like reactivity from bond polarization and strain. In ethylene oxide, the Lewis structure shows strained C-O bonds with oxygen's lone pairs. Choice A is correct because protonation makes oxygen positive, polarizing C-O bonds and enhancing carbon electrophilicity. Choice B is incorrect due to misunderstanding; protonation doesn't linearize without breaking bonds. In similar cases, ensure to consider charge effects on bond polarity when evaluating Lewis structures for reactions. Analyze acidic vs. basic conditions for mechanism differences.

Question 17

A biochemistry lab analyzes the peptide bond and notes restricted rotation about the C–N bond in amides. In a simplified Lewis description of an amide, resonance contributors place a C=O double bond in one form and a C=N double bond with O $^-$ in another. Based on the Lewis structure, which prediction best explains the restricted rotation around the amide C–N bond?

The C–N bond has partial double-bond character due to resonance, increasing bond order and rigidity. (correct answer)
The C–N bond is purely ionic, so rotation is prevented by electrostatic attraction.
The nitrogen is sp $^3$ hybridized with a pyramidal geometry, locking the bond in place.
Rotation is restricted because oxygen cannot have more than six electrons in its valence shell.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like bond rotation from resonance and bond order. In amides, the Lewis structure shows resonance giving C-N partial double-bond character. Choice A is correct because this partial double bond restricts rotation, increasing rigidity. Choice B is incorrect due to overstating ionicity; amides are covalent. In similar cases, ensure to evaluate resonance for bond order when assessing Lewis structures for conformational properties. Compare to single bonds for rotation freedom.

Question 18

A solvent-screening experiment compares dichloromethane (CH $_2$ Cl $_2$ ) and carbon tetrachloride (CCl $_4$ ) for extracting a moderately polar analyte. Both are tetrahedral at carbon in a Lewis description. Based on the Lewis structure and symmetry, what prediction can be made about the net molecular polarity of CH $_2$ Cl $_2$ relative to CCl $_4$ ?

CH $_2$ Cl $_2$ is nonpolar like CCl $_4$ because tetrahedral geometry always cancels dipoles.
CH $_2$ Cl $_2$ is polar because substituents differ, so bond dipoles do not fully cancel. (correct answer)
CCl $_4$ is polar because C–Cl bonds are polar and add constructively in a tetrahedron.
Both are equally polar because they have the same number of lone pairs on carbon.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like polarity from symmetry and bond dipoles. In CH2Cl2, the Lewis structure shows tetrahedral geometry with two H and two Cl, lacking full symmetry. Choice B is correct because differing substituents prevent dipole cancellation, making it polar unlike symmetric CCl4. Choice A is incorrect due to assuming all tetrahedrals are nonpolar. In similar cases, ensure to assess substituent symmetry when evaluating Lewis structures for net dipole. Visualize vector addition of bond dipoles.

Question 19

A reaction screen examines nucleophilic substitution on methyl chloride (CH $_3$ Cl) versus methyl fluoride (CH $_3$ F). Both can be drawn with a single C–X bond and three C–H bonds. Based on Lewis structures and electronegativity (F = 4.0, Cl = 3.0, C = 2.5), which prediction is most consistent with the relative polarization of the C–X bond and the partial charges on carbon?

CH $_3$ F has a more polarized C–X bond, making carbon more $\delta^+$ than in CH $_3$ Cl. (correct answer)
CH $_3$ Cl has a more polarized C–X bond, making carbon more $\delta^+$ than in CH $_3$ F.
Both have identical polarization because halogens always form purely covalent bonds to carbon.
Carbon is $\delta^-$ in both because halogens donate electron density through lone pairs in a single bond.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like bond polarity from electronegativity differences. In CH3F and CH3Cl, the Lewis structures show C-X bonds, with F more electronegative than Cl, polarizing C-F more. Choice A is correct because greater polarization in CH3F makes carbon more δ+ susceptible to nucleophiles. Choice B is incorrect due to reversing electronegativity order. In similar cases, ensure to use electronegativity values when evaluating Lewis structures for partial charges. Predict reactivity based on δ+ sites.

Question 20

A clinical lab monitors nitrite (NO $_2^-$ ) levels. Nitrite has 18 valence electrons and is commonly represented by two resonance structures with one N=O and one N–O bond, with the negative charge on oxygen in each contributor. Based on the Lewis structure, what prediction can be made about the O–N–O bond angle in NO $_2^-$ ? (Assume VSEPR applies.)

Approximately $180^\circ$ , because resonance forces a linear arrangement.
Less than $120^\circ$ , because nitrogen has three electron domains (two bonding regions and one lone pair) giving a bent shape. (correct answer)
Exactly $120^\circ$ , because any molecule with three electron domains must have a $120^\circ$ bond angle.
Approximately $109.5^\circ$ , because nitrogen has four electron domains in nitrite.

Explanation: This question tests understanding of covalent bonding and Lewis structures. Lewis structures illustrate electron distribution, predicting molecular shape and properties like bond angles via VSEPR and resonance. In NO2-, the Lewis structure shows three electron domains on nitrogen (two bonds, one lone pair) due to resonance. Choice B is correct because trigonal planar electron geometry yields bent shape with angle <120° from lone-pair repulsion. Choice C is incorrect due to ignoring lone-pair effects on angles. In similar cases, ensure to average resonance for domain count when evaluating Lewis structures. Adjust ideal angles for repulsion types.