MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5b Molecular Geometry Hybridization

What this quiz covers

This quiz focuses on 5b Molecular Geometry Hybridization, 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.

How to use this quiz

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.

Question 1

1. Linear, because oxygen has two bonds
2. Trigonal planar, because oxygen has three electron domains (two bonds and one lone pair)
3. Bent, because oxygen has four electron domains (two bonds and two lone pairs) (correct answer)
4. Tetrahedral, because molecular geometry equals electron-domain geometry when lone pairs are present

Explanation: This question tests VSEPR theory for determining molecular geometry around oxygen atoms. In both ethanol and dimethyl ether, the oxygen atom has two bonding pairs (C-O and O-H in ethanol; two C-O bonds in ether) and two lone pairs, giving four total electron domains. Four electron domains arrange tetrahedrally, but with two positions occupied by lone pairs, the molecular geometry is bent (angular). The bond angle is approximately 104.5°, slightly compressed from the tetrahedral angle due to lone pair repulsion. Choice A incorrectly counts only three electron domains, while choice D confuses electron-domain geometry with molecular geometry. Remember that for VSEPR, molecular geometry describes only the arrangement of atoms, not lone pairs, so four domains with two lone pairs always gives a bent shape.

Question 2

1. sp$^2$, because three electron domains at carbon give trigonal planar electron geometry (correct answer)
2. sp$^3$, because carbon is single-bonded to two nitrogens and must be tetrahedral
3. sp, because planarity requires linear hybrid orbitals
4. sp$^3$d, because resonance requires five equivalent electron domains

Explanation: This question tests hybridization in molecules with resonance structures. In urea CO(NH₂)₂, the carbonyl carbon is bonded to one oxygen and two nitrogen atoms, giving three electron domains. Three electron domains require sp² hybridization, resulting in a trigonal planar arrangement that allows all four atoms (C, O, N, N) to lie in the same plane. The sp² hybridization provides an unhybridized p orbital that participates in resonance between the C=O double bond and partial double bond character in the C-N bonds. Choice B incorrectly assumes sp³ hybridization based on single bonds, ignoring the resonance that creates partial double bond character. When analyzing molecules with resonance, count electron domains based on the resonance hybrid structure, not individual resonance forms.

Question 3

1. Tetrahedral, because resonance implies four electron domains around carbon
2. Trigonal planar, because carbon has three electron domains and no lone pairs (correct answer)
3. Trigonal pyramidal, because carbon has one lone pair in the anion
4. Linear, because a $\pi$ bond forces 180° bond angles

Explanation: This question tests VSEPR geometry prediction for the bicarbonate ion's carbon center. In HCO₃⁻, the carbon atom is bonded to three oxygen atoms with resonance structures showing one C=O double bond and two C-O single bonds (though all three bonds are equivalent due to resonance). Regardless of resonance, the carbon has three electron domains (three sigma bonds) and no lone pairs, resulting in a trigonal planar geometry according to VSEPR theory. This corresponds to sp² hybridization at carbon, with bond angles of 120°. Choice A incorrectly interprets resonance as creating four electron domains. Choice C wrongly assigns a lone pair to carbon, which would violate its valence. Choice D suggests linear geometry, impossible with three substituents. The transferable principle is that resonance affects bond order and electron distribution but doesn't change the number of electron domains for geometry determination.

Question 4

1. sp$^3$, because four electron domains surround the carbonyl carbon
2. sp, because two electron domains create a linear carbonyl carbon
3. sp$^2$, because three electron domains form a trigonal planar arrangement with one unhybridized p orbital for the $\pi$ bond (correct answer)
4. dsp$^2$, because a double bond requires d-orbital participation in second-row elements

Explanation: This question tests understanding of hybridization states and molecular geometry around a carbonyl carbon. In carbonyl compounds, the carbon atom forms three sigma bonds (one C=O double bond consists of one sigma and one pi bond, plus two C-C single bonds) and has no lone pairs, giving it three electron domains. According to VSEPR theory, three electron domains arrange in a trigonal planar geometry with 120° bond angles, which requires sp² hybridization. The sp² hybridization uses three hybrid orbitals for sigma bonding, leaving one unhybridized p orbital perpendicular to the plane for pi bond formation with oxygen. Choice A incorrectly counts four electron domains by double-counting the pi bond, while choice B suggests only two domains, ignoring the methyl groups. Choice D incorrectly invokes d-orbital participation, which is not possible for carbon. To verify hybridization, count sigma bonds plus lone pairs on the atom of interest - for carbonyl carbon, this gives 3 + 0 = 3, confirming sp² hybridization.

Question 5

1. Electron-domain: trigonal planar; molecular: trigonal planar
2. Electron-domain: tetrahedral; molecular: trigonal pyramidal (correct answer)
3. Electron-domain: tetrahedral; molecular: tetrahedral
4. Electron-domain: trigonal bipyramidal; molecular: seesaw

Explanation: This question tests the distinction between electron-domain geometry and molecular geometry in VSEPR theory. For ammonia (NH₃), the nitrogen atom has four electron domains: three N-H bonds and one lone pair. Four electron domains arrange in a tetrahedral electron-domain geometry to minimize repulsion. However, the molecular geometry only considers the positions of atoms, not lone pairs, resulting in a trigonal pyramidal shape with bond angles slightly less than 109.5° due to lone pair repulsion. This corresponds to sp³ hybridization at nitrogen. Choice A incorrectly assigns trigonal planar electron-domain geometry, which would require only three electron domains. Choice C fails to distinguish between electron-domain and molecular geometry. Choice D suggests five electron domains, which is impossible for nitrogen. The key principle is that electron-domain geometry includes all electron domains (bonds + lone pairs), while molecular geometry only describes atom positions.

Question 6

1. Trigonal planar, consistent with sp$^2$ hybridization at the carbonyl carbon (correct answer)
2. Tetrahedral, consistent with sp$^3$ hybridization at the carbonyl carbon
3. Trigonal pyramidal, consistent with sp$^3$ hybridization and one lone pair on carbon
4. Linear, consistent with sp hybridization at the carbonyl carbon

Explanation: This question tests understanding of carbonyl carbon geometry before nucleophilic attack. In aldehydes and ketones, the carbonyl carbon forms three electron domains: one double bond to oxygen (counting as one domain) and two single bonds to other groups. With three electron domains and no lone pairs on carbon, VSEPR theory predicts a trigonal planar geometry with bond angles of 120°. This geometry corresponds to sp² hybridization, where three sp² hybrid orbitals form sigma bonds in a plane, and the unhybridized p orbital forms the pi bond with oxygen. Choice B incorrectly suggests tetrahedral geometry, which would require four electron domains. Choice C impossibly assigns a lone pair to carbon in a neutral carbonyl. Choice D suggests linear geometry, which requires only two electron domains. The key insight is that the pi bond in C=O doesn't create a separate electron domain - the entire double bond counts as one domain for geometry determination.

Question 7

1. Trigonal pyramidal, because a lone pair still occupies one vertex
2. Tetrahedral, because four bonding domains surround nitrogen (correct answer)
3. Trigonal planar, because protonation converts nitrogen to sp$^2$
4. Linear, because nitrogen forms four equivalent bonds by sp hybridization

Explanation: This question tests molecular geometry determination using VSEPR theory for a protonated amine. When a tertiary amine (NR₃) becomes protonated to form an ammonium ion (NR₃H⁺), the nitrogen atom transitions from having three bonds plus one lone pair to having four bonds with no lone pairs. According to VSEPR theory, four electron domains (all bonding) around a central atom arrange themselves in a tetrahedral geometry with bond angles of approximately 109.5°. The molecular geometry equals the electron-domain geometry when no lone pairs are present, so the protonated nitrogen exhibits tetrahedral geometry. Choice A incorrectly suggests a lone pair remains after protonation, while choice C wrongly claims sp² hybridization for a four-coordinate nitrogen. Choice D impossibly suggests linear geometry for four substituents. A key check is recognizing that protonation adds a fourth bond without changing the total electron count around nitrogen - it simply converts the lone pair into a bonding pair.

Question 8

1. Bent, because two double bonds create unequal electron repulsions
2. Linear, because two electron domains around carbon are arranged 180° apart (correct answer)
3. Trigonal planar, because carbon uses sp$^2$ hybridization in CO$_2$
4. Tetrahedral, because carbon has four total bonding pairs in two double bonds

Explanation: This question tests molecular geometry and its relationship to molecular polarity. Carbon dioxide (CO₂) has a central carbon atom bonded to two oxygen atoms through double bonds. The carbon has two electron domains (each double bond counts as one domain) and no lone pairs, resulting in a linear geometry with a 180° O-C-O bond angle according to VSEPR theory. This linear arrangement causes the two C=O bond dipoles to point in opposite directions and cancel out, resulting in no net dipole moment for the molecule. Choice A incorrectly suggests bent geometry, which would require lone pairs on carbon. Choice C suggests trigonal planar geometry, requiring three electron domains. Choice D misinterprets double bonds as creating four domains. The contrast with SO₂ (which is bent due to a lone pair on sulfur) reinforces that molecular symmetry determines overall polarity. A key check is that linear geometry with identical terminal atoms always produces a nonpolar molecule.

Question 9

1. sp3 → sp2 (correct answer)
2. sp2 → sp3
3. sp3 → sp
4. sp → sp2

Explanation: This question tests understanding of hybridization changes during oxidation reactions. In a primary alcohol (R–CH2OH), the carbon bearing the OH group has four single bonds (two to H, one to R, one to O), giving it four electron domains and sp3 hybridization with tetrahedral geometry. When oxidized to an aldehyde (R–CHO), this same carbon now has three electron domains (one double bond to O, one bond to R, one to H), adopting sp2 hybridization with trigonal planar geometry. The change from sp3 to sp2 reflects the loss of one electron domain as two C-H bonds are replaced by one C=O double bond. A common error is reversing the hybridization change (choice B) or thinking the carbon becomes sp hybridized (choice C). To track hybridization changes, count electron domains before and after the reaction: alcohols have sp3 carbons while carbonyl carbons are sp2.

Question 10

1. Linear molecular geometry at O because there are two sigma bonds and no lone pairs
2. Trigonal planar molecular geometry at O because there are three electron domains around O
3. Bent molecular geometry at O because there are four electron domains around O (correct answer)
4. Tetrahedral molecular geometry at O because O forms four sigma bonds

Explanation: This question tests understanding of molecular geometry around oxygen atoms using VSEPR theory. In both ethanol (CH3CH2OH) and dimethyl ether (CH3OCH3), the oxygen atom has two sigma bonds and two lone pairs, giving it four electron domains total. According to VSEPR theory, four electron domains arrange tetrahedrally in space to minimize electron-pair repulsion. However, when determining molecular geometry, we only consider the positions of atoms, not lone pairs, resulting in a bent molecular shape around oxygen. A common misconception is thinking that two bonds automatically mean linear geometry (choice A), but this ignores the lone pairs that also occupy space around the oxygen. To verify molecular geometry, always count all electron domains (bonds + lone pairs) first for electron-domain geometry, then consider only bonded atoms for molecular geometry.

Question 11

1. Trigonal planar molecular geometry; trigonal planar electron-domain geometry
2. Tetrahedral molecular geometry; tetrahedral electron-domain geometry
3. Trigonal pyramidal molecular geometry; tetrahedral electron-domain geometry (correct answer)
4. Bent molecular geometry; trigonal planar electron-domain geometry

Explanation: This question tests the distinction between electron-domain geometry and molecular geometry for ammonia. The nitrogen atom in NH3 has three N-H bonds and one lone pair, giving it four electron domains total. Four electron domains arrange tetrahedrally to minimize repulsion, so the electron-domain geometry is tetrahedral. However, molecular geometry only considers atom positions, not lone pairs, so with three atoms bonded to nitrogen in a tetrahedral arrangement with one position occupied by a lone pair, the molecular geometry is trigonal pyramidal. A common error is confusing electron-domain geometry with molecular geometry (choice A), forgetting that lone pairs affect shape but aren't included in molecular geometry descriptions. The lone pair on nitrogen also makes NH3 polar and enables its function as a base. Always distinguish between total electron domains (for electron-domain geometry) and bonded atoms only (for molecular geometry).

Question 12

1. Linear
2. Trigonal planar
3. Bent (correct answer)
4. Tetrahedral

Explanation: This question tests VSEPR prediction of bent geometry in molecules with lone pairs. VSEPR arranges three electron domains trigonal planar, but one lone pair yields bent molecular geometry. In SO₂, sulfur has two bonding domains to oxygen and one lone pair, with resonance. Bent geometry is consistent because the lone pair occupies space, bending the O–S–O angle to about 120°. Choice A fails by assuming linear, ignoring the lone pair's effect on shape. Include lone pairs in electron domain counts for accurate molecular geometry. Verify by noting bond angles less than 120° due to lone pair repulsion.

Question 13

1. Trigonal planar (correct answer)
2. Trigonal pyramidal
3. Tetrahedral
4. Linear

Explanation: This question tests geometry in electron-deficient compounds like BF₃. VSEPR predicts trigonal planar for three electron domains with no lone pairs. In BF₃, boron has three bonds to fluorine and an empty p orbital, but no lone pairs. Trigonal planar geometry is consistent because it minimizes repulsion among the three bonding domains. Choice B fails by suggesting pyramidal, which requires a lone pair not present on boron. Account for exceptions in electron-deficient atoms when applying VSEPR. Verify by noting 120° bond angles in spectroscopic data.

Question 14

1. Bent, causing dipoles to add
2. Linear, causing dipoles to cancel (correct answer)
3. Trigonal planar, causing partial cancellation
4. Tetrahedral, because carbon has four electron domains

Explanation: This question evaluates how geometry affects molecular polarity in linear molecules. VSEPR predicts linear geometry for two electron domains, allowing symmetric dipole cancellation. In CO₂, carbon has two double bonds, counting as two domains with no lone pairs. Linear geometry is consistent because it places polar C=O bonds in opposition, canceling the net dipole. Choice A fails by suggesting bent, which would make CO₂ polar like SO₂, but CO₂ is nonpolar. Assess polarity by considering geometry and vector sum of bond dipoles. Verify by comparing to bent molecules where dipoles add constructively.

Question 15

1. sp$^3$
2. sp$^2$ (correct answer)
3. sp
4. sp$^3$d

Explanation: This question evaluates hybridization in carboxylate ions with resonance. Hybridization for three electron domains is sp², supporting trigonal planar geometry. In –COO⁻, the carbon has three domains: two to oxygen (resonance-averaged) and one to R. Sp² hybridization is consistent because it enables pi delocalization, making C–O bonds equivalent and planar. Choice A fails by assuming sp³, which would imply non-equivalent bonds without resonance. Examine resonance for effective domain count and hybridization. Verify by noting bond lengths intermediate between single and double.

Question 16

1. Trigonal pyramidal to tetrahedral (correct answer)
2. Tetrahedral to trigonal pyramidal
3. Trigonal planar to trigonal pyramidal
4. Linear to trigonal planar

Explanation: This question evaluates changes in molecular geometry upon protonation using VSEPR theory. VSEPR predicts geometry based on electron domains, where four domains with one lone pair give trigonal pyramidal, and four bonding pairs give tetrahedral. In trimethylamine (N(CH₃)₃), nitrogen has three bonds and one lone pair, becoming four bonds in the protonated form (N(CH₃)₃H⁺). The change from trigonal pyramidal to tetrahedral is consistent because protonation converts the lone pair to a bonding pair, altering the molecular shape. Choice B fails due to reversing the order, mistakenly assuming deprotonation instead of protonation. To check geometry changes, compare electron domain counts before and after reaction. Verify by noting that pyramidal shapes invert rapidly, while tetrahedral are more rigid.

Question 17

1. Trigonal planar
2. Tetrahedral (correct answer)
3. Square planar
4. Seesaw

Explanation: This question assesses molecular geometry in tetrahedral ions like phosphate. VSEPR theory predicts tetrahedral geometry for four electron domains with no lone pairs. In PO₄³⁻, phosphorus has four bonds to oxygen, with resonance making them equivalent, resulting in four domains. Tetrahedral geometry is consistent because it minimizes repulsion among the four bonding domains, matching observed symmetry. Choice C fails by suggesting square planar, which requires dsp² hybridization and is uncommon for phosphorus. Always count all bonding domains, considering resonance for equivalence. Verify by comparing to similar species like sulfate, which also exhibit tetrahedral geometry.

Question 18

1. sp to sp$^2$
2. sp$^2$ to sp$^3$ (correct answer)
3. sp$^3$ to sp$^2$
4. sp$^2$ to sp

Explanation: This question tests hybridization changes in addition reactions to alkenes. Hybridization depends on electron domains, shifting from sp² (three domains) to sp³ (four domains) upon saturation. In the fluorination, each alkene carbon goes from three domains (two single bonds, one double) to four single bonds. The sp² to sp³ change is consistent because addition breaks the pi bond, adding two sigma bonds per carbon. Choice A fails by reversing the order, as if describing elimination instead. Track domain count changes in reactions to predict hybridization shifts. Verify by noting product stereochemistry, often anti addition for alkenes.

Question 19

1. Tetrahedral, with symmetric bond dipole cancellation (correct answer)
2. Trigonal pyramidal, with one lone pair on carbon
3. Square planar, with dipoles opposing in a plane
4. Bent, with two lone pairs on carbon

Explanation: This question examines how tetrahedral geometry leads to dipole cancellation in symmetric molecules. VSEPR predicts tetrahedral molecular geometry for four electron domains with no lone pairs. In CCl₄, carbon has four equivalent C–Cl bonds, arranged tetrahedrally. Tetrahedral geometry with symmetry is consistent because it allows complete cancellation of polar bond dipoles, resulting in zero net dipole. Choice B fails by suggesting pyramidal, which requires a lone pair absent in CCl₄. Consider symmetry when evaluating net polarity from geometry. Verify by comparing to CH₃Cl, where asymmetry produces a dipole.

Question 20

1. Tetrahedral
2. Seesaw (correct answer)
3. Square planar
4. Trigonal planar

Explanation: This question assesses seesaw geometry in SF₄-like molecules. VSEPR predicts seesaw for five electron domains with one lone pair. In SF₄, sulfur has four bonds and one lone pair, in trigonal bipyramidal electron geometry. Seesaw shape is consistent because the lone pair occupies an equatorial position, distorting the axial bonds. Choice A fails by assuming tetrahedral, ignoring the extra domain from the lone pair. Include lone pairs to differentiate from tetrahedral. Verify by noting bond angles deviating from ideal due to lone pair repulsion.