AP Chemistry Quiz: Vsepr And Hybridization

What this quiz covers

This quiz focuses on Vsepr And Hybridization, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

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

Explanation: This question tests your ability to determine molecular geometry using VSEPR theory. In CO₂, carbon has 2 C-O bonds and 0 lone pairs, giving it 2 electron domains total. With only 2 electron domains and no lone pairs, both the electron-domain geometry and molecular geometry are linear, with a bond angle of 180°. Students sometimes incorrectly choose bent (choice A), confusing CO₂ with molecules like H₂O that have lone pairs. Remember: molecules with 2 electron domains and no lone pairs are always linear.

Question 2

1. Seesaw
2. T‑shaped
3. Linear (correct answer)
4. Trigonal planar
5. Square planar

Explanation: This question tests molecular geometry for linear arrangements in five-domain systems. In XeF2, xenon has five electron domains: two bonds and three lone pairs. Electron geometry trigonal bipyramidal, molecular linear with lone pairs equatorial. Opposite bonds align linearly. A tempting distractor is choice B, T-shaped, which is for two lone pairs, miscounting lone pairs. Maximize lone pair separation for correct molecular shape.

Question 3

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

Explanation: This question tests your understanding of molecular geometry for molecules with 6 electron domains. In BrF₅, bromine has 5 Br-F bonds and 1 lone pair, giving it 6 electron domains total with an octahedral electron-domain geometry. With one lone pair, the lone pair occupies one position of the octahedron, leaving the 5 fluorine atoms in a square pyramidal arrangement (4 in a square base, 1 at the apex). Students often incorrectly choose trigonal bipyramidal geometry (choice A), miscounting the electron domains as 5 instead of 6. Remember: 6 electron domains with 1 lone pair always results in a square pyramidal molecular geometry.

Question 4

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

Explanation: This question tests molecular geometry prediction using VSEPR for expanded octets. In SF4, sulfur has five electron domains: four bonds and one lone pair. The electron-domain geometry is trigonal bipyramidal, but the lone pair in an equatorial position yields a seesaw molecular geometry. This minimizes 90-degree repulsions. A tempting distractor is choice A, trigonal bipyramidal, which ignores the lone pair's effect, a misconception of confusing electron and molecular geometries. Always distinguish between electron-domain and molecular geometries by accounting for lone pairs.

Question 5

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

Explanation: This question tests your understanding of electron-domain geometry. In BF₃, boron has 3 B-F bonds and 0 lone pairs, giving it 3 electron domains total. Three electron domains arrange themselves in a trigonal planar geometry with 120° bond angles to minimize electron-electron repulsion. Students sometimes incorrectly choose tetrahedral (choice C), perhaps thinking all molecules follow the octet rule, but boron is an exception and is stable with only 6 valence electrons. Remember: electron-domain geometry depends only on the total number of electron domains, regardless of whether they are bonds or lone pairs.

Question 6

1. Electron-domain geometry: trigonal planar; molecular geometry: linear
2. Electron-domain geometry: trigonal planar; molecular geometry: bent
3. Electron-domain geometry: tetrahedral; molecular geometry: linear
4. Electron-domain geometry: tetrahedral; molecular geometry: bent (correct answer)
5. Electron-domain geometry: linear; molecular geometry: linear

Explanation: This question tests your understanding of both electron-domain and molecular geometries. In H₂O, oxygen has 2 O-H bonds and 2 lone pairs, giving it 4 electron domains total with a tetrahedral electron-domain geometry. However, the molecular geometry only considers the positions of atoms, so with 2 bonded atoms and 2 lone pairs, the molecular shape is bent with a bond angle of approximately 104.5°. Students often incorrectly choose trigonal planar geometry (choice A), forgetting that water has 4 electron domains, not 3. Remember: 4 electron domains with 2 lone pairs always results in a bent molecular geometry.

Question 7

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

Explanation: This question tests your ability to determine hybridization from molecular structure. In PCl₅, phosphorus has 5 P-Cl bonds and 0 lone pairs, giving it 5 electron domains total with a trigonal bipyramidal geometry. Five electron domains require five hybrid orbitals, which means sp³d hybridization (one s orbital + three p orbitals + one d orbital = five sp³d hybrid orbitals). Students often incorrectly choose sp³ (choice A), forgetting that phosphorus can expand its octet and use d orbitals. Remember: the number of hybrid orbitals must equal the number of electron domains.

Question 8

1. Electron-domain geometry: trigonal planar; molecular geometry: linear
2. Electron-domain geometry: trigonal bipyramidal; molecular geometry: linear (correct answer)
3. Electron-domain geometry: tetrahedral; molecular geometry: linear
4. Electron-domain geometry: octahedral; molecular geometry: linear
5. Electron-domain geometry: trigonal bipyramidal; molecular geometry: bent

Explanation: This question tests your ability to apply VSEPR theory to molecules with multiple lone pairs. In XeF₂, xenon has 2 Xe-F bonds and 3 lone pairs, giving it 5 electron domains total with a trigonal bipyramidal electron-domain geometry. The 3 lone pairs occupy the equatorial positions (120° apart) to minimize repulsions, leaving the 2 fluorine atoms in axial positions 180° apart, resulting in a linear molecular geometry. Students often incorrectly choose octahedral geometry (choice D), miscounting the electron domains. Remember: 5 electron domains with 3 lone pairs always gives a linear molecular shape.

Question 9

1. Octahedral
2. Trigonal planar
3. Trigonal bipyramidal (correct answer)
4. Tetrahedral
5. Square planar

Explanation: This question tests your knowledge of electron-domain geometry for molecules with five electron domains. In PCl₅, phosphorus has 5 valence electrons and forms 5 bonds with chlorine atoms, with no lone pairs on phosphorus. Five electron domains always result in a trigonal bipyramidal electron-domain geometry, with three atoms in an equatorial plane and two atoms in axial positions. Since there are no lone pairs, the molecular geometry is also trigonal bipyramidal. A common mistake is to select octahedral (choice A), which requires 6 electron domains, not 5. Remember that electron-domain geometry is determined solely by the number of electron domains: 5 domains always gives trigonal bipyramidal geometry.

Question 10

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

Explanation: This question tests your understanding of molecular geometry in polyatomic ions. In NO₃⁻, nitrogen has 5 valence electrons, forms 3 bonds with oxygen atoms (treating each N-O bond as one electron domain regardless of bond order), and has no lone pairs on nitrogen. Three electron domains with no lone pairs result in both trigonal planar electron-domain geometry and trigonal planar molecular geometry, with 120° bond angles. Students might incorrectly choose trigonal pyramidal (choice C) by confusing this with NH₃, which has a lone pair. Remember that molecular geometry depends on both the number of electron domains and the presence of lone pairs; with no lone pairs, the molecular geometry matches the electron-domain geometry.

Question 11

1. Trigonal bipyramidal (correct answer)
2. Square pyramidal
3. Tetrahedral
4. Octahedral
5. Trigonal planar

Explanation: This question examines VSEPR theory for molecules with expanded octets and five electron domains. In PCl5, phosphorus has five electron domains from bonds to chlorine and no lone pairs, leading to a trigonal bipyramidal electron-domain and molecular geometry. The three equatorial and two axial positions minimize repulsion among the bonding pairs. This shape is typical for AX5 systems in VSEPR notation. Choice D, octahedral, might tempt those confusing five domains with six, a misconception from miscounting bonds. Consistently use the AXE notation, where A is the central atom, X is a bond, and E is a lone pair, to predict geometries accurately.

Question 12

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

Explanation: This question assesses understanding of molecular geometry using VSEPR theory for molecules with multiple bonds. In CO2, the carbon atom has two electron domains from the two C=O double bonds and no lone pairs, resulting in a linear electron-domain geometry. Since there are no lone pairs, the molecular geometry is also linear, with the oxygen atoms positioned 180 degrees apart to minimize repulsion. This configuration is consistent with the VSEPR model for AX2 systems. A common distractor is choice A, bent, which might tempt those miscounting double bonds as separate domains, a misconception that overlooks how each multiple bond counts as a single electron domain. A useful strategy is to draw the Lewis structure first to accurately identify the number of electron domains before predicting geometry.

Question 13

1. Seesaw
2. Linear (correct answer)
3. Bent
4. Trigonal bipyramidal
5. T‑shaped

Explanation: This question evaluates VSEPR theory for hypervalent molecules with multiple lone pairs. In XeF2, xenon has five electron domains: two bonding pairs and three lone pairs, forming a trigonal bipyramidal electron-domain geometry. The molecular geometry is linear because the lone pairs occupy the three equatorial positions, leaving the fluorine atoms in axial positions 180 degrees apart. This arrangement reduces repulsion effectively. Choice D, trigonal bipyramidal, tempts those who forget lone pairs don't contribute to molecular geometry, a misconception of including non-bonding electrons in atom arrangements. A key strategy is to identify the electron-domain geometry first, then remove lone pairs to visualize the molecular shape.

Question 14

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

Explanation: This question tests hybridization determination based on electron domains. In HCN, carbon has two electron domains: one single and one triple bond. Triple bonds count as one domain, leading to sp hybridization. The molecule is linear. A tempting distractor is choice A, sp3, which overcounts the triple bond as three, a multiple bond misconception. Treat multiple bonds as single domains to determine hybridization correctly.

Question 15

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

Explanation: This question assesses VSEPR application to molecules with five electron domains including a lone pair. In SF4, sulfur has five electron domains: four bonding pairs and one lone pair, yielding a trigonal bipyramidal electron-domain geometry. The molecular geometry is seesaw because the lone pair prefers the equatorial position, distorting the shape with axial bonds at about 180 degrees and equatorial at 120 degrees. This minimizes 90-degree repulsions involving the lone pair. Choice B, trigonal bipyramidal, is a distractor for those ignoring the lone pair, a common misconception of equating electron-domain and molecular geometries. When predicting shapes, always position lone pairs to minimize repulsion, preferring equatorial sites in trigonal bipyramidal arrangements.

Question 16

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

Explanation: This question tests your ability to determine hybridization from molecular structure. In BF₃, boron has 3 valence electrons, forms 3 bonds with fluorine atoms, and has no lone pairs, resulting in 3 electron domains total. Three electron domains require three hybrid orbitals, which corresponds to sp² hybridization (one s orbital + two p orbitals = three sp² hybrid orbitals). The trigonal planar geometry of BF₃ with 120° bond angles confirms this sp² hybridization. A common mistake is to select sp³ hybridization (choice A), which would require 4 electron domains. To determine hybridization, count all electron domains around the central atom: 3 domains always indicates sp² hybridization.

Question 17

1. Trigonal bipyramidal; seesaw (correct answer)
2. Tetrahedral; trigonal pyramidal
3. Trigonal planar; trigonal planar
4. Octahedral; square pyramidal
5. Trigonal bipyramidal; trigonal bipyramidal

Explanation: This question tests general VSEPR application to hypothetical molecules. For central X with five electron domains (four bonds, one lone), electron geometry is trigonal bipyramidal, molecular seesaw. Lone pair equatorial minimizes repulsion. Applies to SF4-like. A tempting distractor is choice B, tetrahedral; trigonal pyramidal, which is for four domains, undercounting. Base geometry on total domains, then adjust for lone pairs.

Question 18

1. Octahedral; octahedral
2. Trigonal bipyramidal; seesaw
3. Octahedral; square pyramidal (correct answer)
4. Tetrahedral; trigonal pyramidal
5. Square planar; square planar

Explanation: This question tests VSEPR theory for six-electron-domain molecules with a lone pair. In BrF5, bromine has six electron domains: five bonding pairs and one lone pair, resulting in an octahedral electron-domain geometry. The molecular geometry is square pyramidal, with the lone pair opposite one axial position, creating a pyramid with a square base. This shape minimizes repulsion in an AX5E system. Choice A, octahedral for both, is incorrect due to the misconception of treating the lone pair as equivalent to a bonding pair in molecular geometry. To master such problems, practice distinguishing electron domains from molecular arrangements by using models or diagrams.

Question 19

1. Tetrahedral; trigonal planar
2. Trigonal planar; trigonal planar
3. Tetrahedral; trigonal pyramidal (correct answer)
4. Octahedral; square pyramidal
5. Trigonal planar; trigonal pyramidal

Explanation: This question tests the application of VSEPR theory to determine electron-domain and molecular geometries. In NH3, the nitrogen atom has four electron domains: three bonding pairs from N-H bonds and one lone pair, leading to a tetrahedral electron-domain geometry. The molecular geometry, which considers only the arrangement of atoms, is trigonal pyramidal because the lone pair occupies one position, pushing the three hydrogen atoms into a pyramid shape. This arrangement minimizes electron repulsion according to VSEPR principles. A tempting distractor is choice B, trigonal planar electron-domain geometry and trigonal pyramidal molecular geometry, which is incorrect due to the misconception of counting only bonding domains and ignoring the lone pair. To determine geometries reliably, always count the total electron domains around the central atom, including both bonding pairs and lone pairs, before applying VSEPR rules.

Question 20

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

Explanation: This question examines molecular geometry in polyatomic ions using VSEPR, considering resonance. In CO3^2-, carbon has three electron domains from bonds to oxygen (resonance averages to three equivalent domains) and no lone pairs, giving a trigonal planar electron-domain geometry. The molecular geometry is also trigonal planar, with bond angles of 120 degrees. This is consistent with AX3 notation in VSEPR. Choice B, tetrahedral, might attract those counting individual bonds in one resonance structure, a misconception ignoring resonance delocalization. For ions or resonant molecules, always consider the average structure or equivalent domains to predict geometry correctly.