Covalent Bonding and Lewis Structures (5B)
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MCAT Chemical and Physical Foundations of Biological Systems › Covalent Bonding and Lewis Structures (5B)
In an atmospheric chemistry assay, ozone (O$_3$) is generated and its reactivity is correlated with its electronic structure. A valid Lewis description uses O as the central atom with one O=O double bond, one O–O single bond, and one lone pair on the central O; resonance interconverts the positions of the single and double bonds. Based on the Lewis structure, which statement best describes the formal charges in a major resonance contributor?
The central oxygen has formal charge $-1$ and the doubly bonded terminal oxygen has formal charge $+1$.
The central oxygen has formal charge $+1$ and the singly bonded terminal oxygen has formal charge $-1$.
One terminal oxygen has formal charge $-2$ and the other has formal charge $+2$ due to complete electron transfer.
All three oxygen atoms have formal charge 0 because each has an octet and identical bonding.
Explanation
This question tests understanding of covalent bonding and Lewis structures, specifically calculating formal charges. Lewis structures illustrate electron distribution and formal charges help identify the most stable resonance contributors. In O₃, the Lewis structure shows a central oxygen with one O=O double bond, one O-O single bond, and one lone pair, with resonance between equivalent forms. Choice B is correct because formal charge calculations show: central O has 6 valence electrons - 2 lone pair electrons - 3 bonds = +1; singly bonded terminal O has 6 valence electrons - 6 lone pair electrons - 1 bond = -1; doubly bonded O has formal charge 0. Choice A is incorrect because the asymmetric bonding pattern (one single, one double bond) necessarily creates non-zero formal charges. When calculating formal charges, use the formula: valence electrons - lone pair electrons - number of bonds, remembering that each bond counts as one regardless of bond order.
A synthetic chemistry report notes that boron trifluoride (BF$_3$) readily accepts an electron pair from Lewis bases. The Lewis structure places B in the center with three B–F single bonds and no lone pairs on B (6 electrons around B). Based on the Lewis structure, which property is best explained by BF$_3$?
BF$_3$ is a Lewis acid because boron has an incomplete octet and can accept an electron pair.
BF$_3$ is strongly polar because its trigonal planar molecular shape forces all three B–F bond dipoles to add.
BF$_3$ is tetrahedral because boron must expand its octet to accommodate three bonds and one lone pair.
BF$_3$ is a Lewis base because boron has a lone pair that can be donated to electrophiles.
Explanation
This question tests understanding of covalent bonding and Lewis structures to predict chemical reactivity. Lewis structures illustrate electron distribution, revealing electron-deficient sites that can act as Lewis acids. In BF₃, the Lewis structure shows boron with only three B-F single bonds and no lone pairs, giving boron only 6 electrons instead of a complete octet. Choice A is correct because boron's incomplete octet creates an empty p orbital that can accept an electron pair from a Lewis base, making BF₃ a Lewis acid. Choice B is incorrect because boron has no lone pairs - it's electron-deficient, not electron-rich. When identifying Lewis acids and bases from structures, look for incomplete octets (Lewis acids) or available lone pairs (Lewis bases), remembering that second-period elements like boron cannot expand their octets.
A formulation scientist compares boiling points of small covalent hydrides for a volatile delivery system. Consider ammonia (NH$_3$) and methane (CH$_4$), both with central atoms from the second period. Lewis structures show N bonded to three H atoms with one lone pair, while C is bonded to four H atoms with no lone pairs. Based on the Lewis structure, which interaction is most likely to explain why NH$_3$ has a higher boiling point than CH$_4$?
CH$_4$ has a lower boiling point because its Lewis structure contains resonance that reduces intermolecular attraction.
CH$_4$ has a lower boiling point because carbon is more electronegative than nitrogen, making CH$_4$ more polar.
NH$_3$ has a higher boiling point because it is linear, increasing surface area relative to tetrahedral CH$_4$.
NH$_3$ exhibits hydrogen bonding because N has a lone pair and N–H bonds, whereas CH$_4$ cannot hydrogen bond.
Explanation
This question tests understanding of covalent bonding and Lewis structures to predict intermolecular forces and physical properties. Lewis structures illustrate electron distribution, revealing molecular features that determine intermolecular interactions and boiling points. In NH₃, the Lewis structure shows nitrogen with three N-H bonds and one lone pair, while CH₄ shows carbon with four C-H bonds and no lone pairs. Choice A is correct because NH₃ can form hydrogen bonds - it has both N-H bonds (hydrogen bond donors) and a lone pair on nitrogen (hydrogen bond acceptor), whereas CH₄ has neither highly polar X-H bonds nor lone pairs, limiting it to weaker dispersion forces. Choice B is incorrect because NH₃ is trigonal pyramidal (not linear) due to its lone pair. When comparing boiling points of small molecules, identify whether hydrogen bonding is possible by checking for both polar X-H bonds (X = N, O, F) and lone pairs on electronegative atoms.
In a corrosion-inhibition study, thiocyanate (SCN$^-$) is evaluated as a ligand. Valid Lewis structures include resonance forms that can be written as $^-!$S–C≡N and S=C=N$^-$. Based on the Lewis structure, what prediction can be made regarding the carbon atom in SCN$^-$?
Carbon is sp$^3$-hybridized because it forms four electron domains when resonance is considered.
Carbon is sp-hybridized because the dominant resonance contributors keep the S–C–N arrangement linear.
Carbon is sp$^2$-hybridized because it must be trigonal planar to allow resonance between S and N.
Carbon is not hybridized because formal charges reside only on S and N, leaving C uninvolved in bonding.
Explanation
This question tests understanding of covalent bonding and Lewis structures to predict hybridization states. Lewis structures illustrate electron distribution and bonding patterns that determine atomic hybridization. In SCN⁻, the resonance structures show carbon forming either a triple bond with N and single bond with S, or double bonds with both S and N, but in both cases carbon has only two electron domains. Choice B is correct because two electron domains around carbon require sp hybridization, which produces a linear geometry consistent with the observed S-C-N arrangement. Choice C is incorrect because sp² hybridization requires three electron domains, but carbon in SCN⁻ has only two bonding domains regardless of resonance form. When determining hybridization, count the number of electron domains (sigma bonds plus lone pairs) around the atom - two domains mean sp, three mean sp², and four mean sp³.
A bioinorganic study examines carbon monoxide (CO) binding to heme iron. The best Lewis structure for CO satisfies octets and yields a triple bond between C and O. Based on that Lewis structure, what prediction can be made about the formal charges on C and O in CO?
Both C and O are 0, because a triple bond always implies zero formal charge on both atoms.
C is $-2$ and O is $+2$, because the electronegativity difference forces complete charge separation.
C is $+1$ and O is $-1$, because oxygen must carry the negative charge in any C–O bond.
C is $-1$ and O is $+1$, consistent with a triple bond and one lone pair on each atom.
Explanation
This question tests understanding of covalent bonding and Lewis structures, specifically formal charge calculations in molecules with multiple bonds. Lewis structures illustrate electron distribution, and formal charges help predict electron density and reactivity. In CO with a triple bond, the Lewis structure shows C≡O with each atom having one lone pair to complete their octets. Choice A is correct because formal charge calculations yield: C has 4 valence electrons - 2 lone pair electrons - 3 bonds = -1; O has 6 valence electrons - 2 lone pair electrons - 3 bonds = +1. Choice B is incorrect because it assumes electronegativity determines formal charge, but formal charge is based solely on electron bookkeeping, not electronegativity. When calculating formal charges in multiply-bonded molecules, remember that each bond (single, double, or triple) counts as one bond in the formal charge formula, and the more electronegative atom doesn't automatically carry negative formal charge.
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.
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.
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 gas-phase kinetics experiment compares the reactivity of phosgene (COCl$_2$) with nucleophiles. The most stable Lewis structure places C as the central atom with a C=O double bond and two C–Cl single bonds; the carbon has no lone pairs. Assuming ideal VSEPR behavior and that bond dipoles add vectorially, which property is best explained by the Lewis structure of COCl$_2$?
COCl$_2$ is polar because the central carbon is sp$^3$-hybridized, creating a trigonal pyramidal molecular shape.
COCl$_2$ is polar because its trigonal planar molecular shape has unequal substituents, preventing full cancellation of bond dipoles.
COCl$_2$ is nonpolar because its tetrahedral electron geometry ensures complete cancellation of all bond dipoles.
COCl$_2$ is nonpolar because resonance makes the C–Cl bonds identical to the C=O bond, producing a symmetric charge distribution.
Explanation
This question tests understanding of covalent bonding and Lewis structures to predict molecular polarity. Lewis structures illustrate electron distribution, which determines molecular geometry and dipole moments through VSEPR theory. In COCl₂, the Lewis structure shows carbon with three bonding domains (one C=O double bond and two C-Cl single bonds) and no lone pairs, resulting in three electron domains total. Choice B is correct because three electron domains around carbon create a trigonal planar molecular shape, and since the three substituents are not identical (one O, two Cl), the bond dipoles cannot completely cancel, making the molecule polar. Choice D is incorrect because carbon with three electron domains is sp²-hybridized (not sp³), and the molecular shape is trigonal planar (not pyramidal). When evaluating polarity, consider both the molecular geometry and whether all substituents are identical - even symmetric shapes can be polar if substituents differ.
In a solvent-selection study, sulfur dioxide (SO$_2$) is used as a polar aprotic solvent under pressure. The dominant Lewis representation shows S bonded to two O atoms with resonance between two equivalent S=O placements; S also has one lone pair (total of three electron domains around S). Based on the Lewis structure, which interaction is most likely between SO$_2$ molecules in the condensed phase?
Strong hydrogen bonding, because SO$_2$ contains an O–H bond that can donate hydrogen bonds.
Dipole–dipole attraction, because the bent molecular shape yields a net molecular dipole.
Ion–ion attraction, because the best Lewis structure assigns full $+2$ and $-1$ formal charges to create an ionic lattice.
No intermolecular attraction beyond dispersion, because resonance forces a linear geometry with zero net dipole.
Explanation
This question tests understanding of covalent bonding and Lewis structures to predict intermolecular forces. Lewis structures illustrate electron distribution, which determines molecular shape and polarity, thereby predicting intermolecular interactions. In SO₂, the Lewis structure shows sulfur with two bonding domains (resonating S=O bonds) and one lone pair, creating three electron domains total. Choice B is correct because three electron domains with one lone pair yield a bent molecular shape, and with polar S-O bonds that don't cancel due to the bent geometry, SO₂ has a net dipole moment enabling dipole-dipole attractions. Choice A is incorrect because SO₂ contains no O-H bonds, only S-O bonds, so hydrogen bonding is impossible. When predicting intermolecular forces, first determine molecular polarity from the Lewis structure and geometry, then identify the strongest applicable interaction type.
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 longer because the best Lewis structure places the negative charge on nitrogen, weakening only one N–O bond.
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.
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 laboratory investigates sulfur dioxide (SO$_2$) as a reducing agent in aqueous microreactors. The Lewis structure is constructed with sulfur central and two oxygens terminal; total valence electrons are $6 + 2(6) = 18$. Spectroscopic data suggest resonance with two equivalent S–O bonds. Based on the Lewis structure (including resonance) and VSEPR, what prediction can be made from the Lewis structure regarding the molecular shape and polarity of SO$_2$?
Bent and polar because sulfur has three electron domains (two bonds and one lone pair).
Linear and nonpolar because resonance forces 180° bond angles.
Trigonal planar and nonpolar because all atoms lie in one plane.
Tetrahedral and nonpolar because sulfur has four lone pairs in the Lewis structure.
Explanation
This question tests understanding of covalent bonding and Lewis structures in predicting molecular shape and polarity. Lewis structures illustrate electron distribution, with resonance showing delocalized electrons and VSEPR theory predicting geometry based on electron domains. In SO₂, the Lewis structure shows sulfur with two double bonds to oxygen atoms and one lone pair on sulfur, giving three electron domains total. Choice C is correct because three electron domains with one lone pair result in bent molecular geometry, and the asymmetrical shape prevents dipole cancellation, making SO₂ polar. Choice A is incorrect because three electron domains do not produce linear geometry, and choice B fails because the lone pair prevents trigonal planar molecular shape. When analyzing Lewis structures with resonance, count all electron domains including lone pairs to predict the correct molecular geometry.