VSEPR and Hybridization - AP Chemistry

In $\mathrm{H_2O}$ , oxygen is the central atom bonded to two hydrogen atoms and has two lone pairs. What are the electron-domain geometry and the molecular geometry around oxygen?
Answer: Electron-domain geometry: tetrahedral; molecular geometry: bent 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.
For $\mathrm{PCl_5}$ , phosphorus is the central atom bonded to five chlorine atoms with no lone pairs on phosphorus. What is the electron-domain geometry around phosphorus?
Answer: Trigonal bipyramidal 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.
In the polyatomic ion $\mathrm{NO_3^-}$ , nitrogen is the central atom bonded to three oxygen atoms and has no lone pairs on nitrogen (treat each N–O bond as one electron domain). What is the molecular geometry around nitrogen?
Answer: Trigonal planar 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.
In $\mathrm{PCl_5}$ , phosphorus is the central atom bonded to five chlorine atoms and has no lone pairs. What is the hybridization of the phosphorus atom in $\mathrm{PCl_5}$ as predicted by VSEPR-based hybridization models?
Answer: $sp^3d$ 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.
In $\mathrm{BF_3}$ , boron is the central atom bonded to three fluorine atoms and has no lone pairs on boron. What is the hybridization of boron?
Answer: $sp^2$ 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.
In sulfur tetrafluoride, $\mathrm{SF_4}$ , sulfur is the central atom with four S–F single bonds and one lone pair on S (five electron domains total). What is the molecular geometry around sulfur?
Answer: Seesaw 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.
In $\mathrm{SF_4}$ , sulfur is the central atom bonded to four fluorine atoms and has one lone pair on sulfur. What is the molecular geometry around sulfur?
Answer: Seesaw This question tests your understanding of how lone pairs affect molecular geometry in trigonal bipyramidal electron-domain arrangements. In SF₄, sulfur has 6 valence electrons, forms 4 bonds with fluorine atoms, and has 1 lone pair, giving 5 electron domains total. Five electron domains create a trigonal bipyramidal electron-domain geometry. The lone pair occupies an equatorial position (to minimize repulsions), leaving 4 fluorine atoms in a seesaw molecular geometry. Students often incorrectly choose tetrahedral (choice B) by counting only the 4 bonds, forgetting that the lone pair affects the overall geometry. When dealing with 5 electron domains and lone pairs, remember that lone pairs preferentially occupy equatorial positions in trigonal bipyramidal arrangements.
In the molecule $\mathrm{SO_2}$ , sulfur is the central atom and is bonded to two oxygen atoms. In a Lewis structure, sulfur has one lone pair (assume each S–O bond counts as one electron domain in VSEPR). What are the electron-domain geometry and the molecular geometry around sulfur?
Answer: Electron-domain geometry: trigonal planar; molecular geometry: bent This question tests your ability to determine electron-domain and molecular geometries using VSEPR theory. In SO₂, sulfur has 2 S-O bonds and 1 lone pair, giving it 3 electron domains total, which results in a trigonal planar electron-domain geometry. However, the molecular geometry only considers the positions of atoms (not lone pairs), so with 2 bonded atoms and 1 lone pair, the molecular shape is bent. Students often confuse this with tetrahedral geometry (choice E), forgetting that SO₂ has only 3 electron domains, not 4. Remember: count all electron domains (bonds + lone pairs) for electron-domain geometry, but only count atoms for molecular geometry.
In $\mathrm{SF_4}$ , sulfur is the central atom bonded to four fluorine atoms and has one lone pair. What is the molecular geometry around sulfur?
Answer: Seesaw This question tests your understanding of molecular geometry for molecules with 5 electron domains. In SF₄, sulfur has 4 S-F bonds and 1 lone pair, giving it 5 electron domains total with a trigonal bipyramidal electron-domain geometry. With one lone pair, the lone pair occupies an equatorial position (to minimize repulsions), leaving the 4 fluorine atoms in a seesaw arrangement. Students often incorrectly choose tetrahedral (choice C), forgetting that SF₄ has 5 electron domains, not 4. Remember: for 5 electron domains with 1 lone pair, the molecular geometry is always seesaw.
For $\mathrm{BrF_5}$ , bromine is the central atom bonded to five fluorine atoms and has one lone pair on bromine. What is the molecular geometry around bromine?
Answer: Square pyramidal This question tests your knowledge of molecular geometry for molecules with six electron domains. In BrF₅, bromine has 7 valence electrons, forms 5 bonds with fluorine atoms, and has 1 lone pair, giving 6 electron domains total. Six electron domains always result in an octahedral electron-domain geometry. With 5 bonding pairs and 1 lone pair, the lone pair occupies one position of the octahedron, leaving 5 fluorine atoms arranged in a square pyramidal molecular geometry (4 atoms forming a square base with 1 atom at the apex). Students often confuse this with trigonal bipyramidal (choice A), which has only 5 electron domains. Remember that 6 electron domains always give octahedral electron-domain geometry, and removing one atom from an octahedron creates a square pyramid.
In $\mathrm{CO_2}$ , carbon is the central atom bonded to two oxygen atoms and has no lone pairs. What is the molecular geometry around carbon?
Answer: Linear This question tests your ability to determine molecular geometry when there are no lone pairs on the central atom. In CO₂, carbon has 4 valence electrons and forms 2 double bonds with oxygen atoms, with no lone pairs on carbon. Each double bond counts as one electron domain, giving 2 electron domains total. With 2 electron domains and no lone pairs, both the electron-domain geometry and molecular geometry are linear, with a 180° bond angle. Students might incorrectly choose bent (choice A) by confusing CO₂ with molecules like H₂O that have lone pairs. When the central atom has no lone pairs, the molecular geometry equals the electron-domain geometry.
Consider $\mathrm{BF_3}$ , where boron is the central atom bonded to three fluorine atoms by single bonds and has no lone pairs on boron. What is the hybridization of the boron atom?
Answer: $sp^2$ This question evaluates knowledge of hybridization based on electron-domain geometry in VSEPR theory. In BF3, boron has three electron domains from single bonds to fluorine and no lone pairs, corresponding to a trigonal planar geometry. This geometry requires sp2 hybridization, where one s and two p orbitals mix to form three equivalent hybrid orbitals arranged at 120 degrees. The absence of a lone pair on boron confirms this hybridization, as it matches the steric number of three. Choice A, sp3, is a tempting distractor stemming from the misconception of assuming all period 2 elements follow the octet rule, ignoring boron's electron deficiency. Always calculate the steric number—the sum of atoms bonded plus lone pairs—to determine the appropriate hybridization.
In $\mathrm{BrF_5}$ , bromine is the central atom bonded to five fluorine atoms and has one lone pair. What are the electron-domain geometry and the molecular geometry around bromine?
Answer: Electron-domain geometry: octahedral; molecular geometry: square pyramidal 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.
In $\mathrm{ClF_3}$ , chlorine is the central atom bonded to three fluorine atoms and has two lone pairs. What is the molecular geometry around chlorine?
Answer: T‑shaped This question tests your ability to determine molecular geometry for molecules with 5 electron domains. In ClF₃, chlorine has 3 Cl-F bonds and 2 lone pairs, giving it 5 electron domains total with a trigonal bipyramidal electron-domain geometry. The 2 lone pairs occupy equatorial positions (to minimize repulsions), leaving the 3 fluorine atoms in a T-shaped arrangement (2 equatorial, 1 axial). Students often incorrectly choose trigonal planar (choice A), confusing the number of bonds with the molecular shape. Remember: for 5 electron domains with 2 lone pairs, the molecular geometry is always T-shaped.
In xenon difluoride, $\mathrm{XeF_2}$ , xenon is the central atom with two Xe–F single bonds and three lone pairs on Xe (five electron domains total). What is the molecular geometry around xenon?
Answer: Linear 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.
In $\mathrm{SO_2}$ , sulfur is the central atom with two S–O bonding regions (each double bond counts as one electron domain) and one lone pair on sulfur. What is the molecular geometry around sulfur?
Answer: Bent This question probes the determination of molecular geometry in molecules with lone pairs using VSEPR theory. In SO2, sulfur has three electron domains: two from S-O double bonds and one lone pair, giving a trigonal planar electron-domain geometry. The molecular geometry is bent because the lone pair occupies one position, distorting the bond angle to about 120 degrees between the oxygen atoms. This bent shape arises from the repulsion between the lone pair and bonding pairs. Choice B, trigonal planar, is a distractor that might appeal if one forgets to account for the lone pair, a misconception of focusing only on bonded atoms. To avoid errors, remember to include lone pairs in electron domain counts and distinguish between electron-domain and molecular geometries.
In $\mathrm{CH_4}$ , carbon is the central atom with four C–H single bonds and no lone pairs on carbon. What are the electron-domain geometry and the hybridization of carbon?
Answer: Tetrahedral; $sp^3$ This question tests the linkage between electron-domain geometry and hybridization in VSEPR theory. In CH4, carbon has four electron domains from single C-H bonds and no lone pairs, resulting in a tetrahedral electron-domain geometry. This geometry corresponds to sp3 hybridization, where one s and three p orbitals hybridize to form four equivalent orbitals at 109.5 degrees. The molecular geometry is also tetrahedral, matching the electron-domain arrangement without lone pairs. Choice D, tetrahedral with sp2, is incorrect and reflects the misconception of associating tetrahedral geometry with trigonal planar hybridization. A transferable approach is to match the number of electron domains to the hybridization type: two for sp, three for sp2, four for sp3.
In $\mathrm{CO_2}$ , carbon is the central atom bonded to two oxygen atoms and has no lone pairs on the central atom. What is the molecular geometry around carbon?
Answer: Linear 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.
In $\mathrm{CO_2}$ , carbon is the central atom with two regions of electron density (two C=O double bonds) and no lone pairs on carbon. What is the molecular geometry around the carbon atom?
Answer: Linear 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.
A molecule has a central atom $X$ bonded to four atoms (four single bonds) and has one lone pair on $X$ , for a total of five electron domains around $X$ . What are the electron-domain geometry and molecular geometry around $X$ ?
Answer: Trigonal bipyramidal; seesaw 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.

Start this quiz