Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 36m
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: Ions

General Chemistry

11. Bonding & Molecular Structure

Lewis Dot Structures: Ions

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Multiple Choice

Which of the following is the correct Lewis structure for the ion I₃⁻, including formal charges?

I-I-I with a -1 charge on one of the terminal iodines

I-I-I with a -1 charge on the central iodine

I-I-I with a +1 charge on the central iodine and -1 charges on both terminal iodines

I-I-I with no formal charges

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Verified step by step guidance

Identify the total number of valence electrons for the I₃⁻ ion. Iodine (I) has 7 valence electrons, and there are three iodine atoms, plus an extra electron due to the negative charge, giving a total of 22 valence electrons.

Arrange the iodine atoms in a linear structure: I-I-I. This is the most common arrangement for the triiodide ion.

Distribute the 22 valence electrons. Start by placing a single bond (2 electrons) between each pair of iodine atoms, which uses 4 electrons, leaving 18 electrons to distribute.

Place the remaining 18 electrons as lone pairs around the iodine atoms, starting with the terminal iodines to satisfy the octet rule. Each terminal iodine will have 3 lone pairs (6 electrons each), using up 12 electrons, leaving 6 electrons.

Place the remaining 6 electrons as 3 lone pairs on the central iodine. Calculate the formal charges: the central iodine has 7 valence electrons, 2 bonding electrons, and 6 non-bonding electrons, resulting in a formal charge of -1. The terminal iodines each have a formal charge of 0. Thus, the correct Lewis structure is I-I-I with a -1 charge on the central iodine.