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.

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.

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.

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.

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.

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.

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.

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.

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.

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.