Covalent Bonding and Lewis Structures (5B)
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MCAT Chemical and Physical Foundations of Biological Systems › Covalent Bonding and Lewis Structures (5B)
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 pyramidal, because five bonds imply four bonding pairs and one lone pair on phosphorus.
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 bipyramidal, because five bonding electron domains around P minimize repulsion in a $90^\circ/120^\circ$ arrangement.
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.
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$_2^-$?
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).
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 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.
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.
Linear with three equivalent N–O bonds because nitrogen forms three $sp$ hybrids.
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.
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.
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 +2 and O has −2.
C has +1 and O has −1.
C has 0 and O has 0.
C has −1 and O has +1.
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.
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 109.5° because the molecular shape is tetrahedral.
Approximately 180° because three bonds maximize separation in a line.
Approximately 120° because nitrogen is $sp^2$-hybridized with no lone pairs.
Slightly less than 109.5° because the electron geometry is tetrahedral with one lone pair.
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.
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 must be a triple bond to satisfy the octet on carbon.
The C–O bond in formaldehyde has bond order 2, consistent with a shorter, stronger bond than a C–O single bond.
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 is nonpolar because carbon and oxygen have similar electronegativities.
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.
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.
Tetrahedral, because nitrogen must have four electron domains to satisfy the octet in NOCl.
Bent, because nitrogen has three electron domains (two bonds and one lone pair) giving a non-linear molecular shape.
Trigonal planar, because nitrogen has three sigma bonds to O, Cl, and a second oxygen from resonance.
Linear, because nitrogen forms two double bonds and has no lone pairs.
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.
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?
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.
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.
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.
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 polar because the terminal atoms are different, giving a net dipole even if the molecule is linear.
N$_2$O is nonpolar like CO$_2$ because any bond dipoles cancel in a linear molecule.
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.
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.
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.