Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

11. Bonding & Molecular Structure

Lewis Dot Structures: Neutral Compounds

General Chemistry

11. Bonding & Molecular Structure

Lewis Dot Structures: Neutral Compounds

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Multiple Choice

Which of the following best describes the electron domain geometry and molecular geometry of the neutral compound IF₃ based on its Lewis structure?

Electron domain geometry: trigonal bipyramidal; Molecular geometry: T-shaped

Electron domain geometry: tetrahedral; Molecular geometry: trigonal planar

Electron domain geometry: trigonal planar; Molecular geometry: linear

Electron domain geometry: octahedral; Molecular geometry: square planar

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Verified step by step guidance

Draw the Lewis structure of IF₃ by first counting the total valence electrons: Iodine (I) has 7 valence electrons and each Fluorine (F) has 7 valence electrons, so total electrons = 7 + 3 × 7 = 28 electrons.

Place the iodine atom in the center and arrange the three fluorine atoms around it, then form single bonds between iodine and each fluorine, using 6 electrons (3 bonds × 2 electrons each).

Distribute the remaining electrons as lone pairs to satisfy the octet rule, placing lone pairs on the fluorine atoms first, then placing any leftover electrons as lone pairs on the iodine atom.

Determine the electron domain geometry by counting the total regions of electron density (bonding pairs + lone pairs) around the central iodine atom. For IF₃, this total is 5 regions, which corresponds to a trigonal bipyramidal electron domain geometry.

Determine the molecular geometry by considering only the positions of atoms (ignoring lone pairs). With 3 bonding pairs and 2 lone pairs on iodine, the molecular geometry is T-shaped.