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

Previous problemNext problem

Struggling with General Chemistry?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Given the neutral compound IF₃, what is the hybridization of the central iodine atom in its Lewis structure?

sp³d

sp³d²

sp³

sp²

Previous problemNext problem

Verified step by step guidance

Step 1: Determine the total number of valence electrons for the central atom (iodine) and the surrounding atoms (fluorine). Iodine is in group 17, so it has 7 valence electrons, and each fluorine atom also has 7 valence electrons.

Step 2: Calculate the total valence electrons in the molecule IF₃ by adding iodine's valence electrons to three times the valence electrons of fluorine atoms: $7 + 3 \times 7 = 28$ electrons.

Step 3: Draw the Lewis structure by placing iodine in the center and connecting it to three fluorine atoms with single bonds. Each bond uses 2 electrons, so 6 electrons are used in bonding. Distribute the remaining electrons to satisfy the octet rule for fluorine atoms first, then place any leftover electrons on iodine as lone pairs.

Step 4: Count the regions of electron density (bonding pairs + lone pairs) around the central iodine atom. This total number determines the electron geometry and hybridization. For IF₃, iodine has 3 bonding pairs and 2 lone pairs, making 5 regions of electron density.

Step 5: Use the number of electron density regions to identify the hybridization. Five regions correspond to an $sp^{3}d$ hybridization for the central iodine atom.