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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 51m
7. Gases3h 52m
8. Thermochemistry2h 38m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 10m
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 34m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

11. Bonding & Molecular Structure

Lewis Dot Structures: Exceptions

General Chemistry

11. Bonding & Molecular Structure

Lewis Dot Structures: Exceptions

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Multiple Choice

Draw the Lewis Dot Structure for the radical hydroxide, OH.

Lewis Dot Structure for hydroxide radical, showing O and H with unpaired electron.

Alternative Lewis Dot Structure for hydroxide radical, depicting O and H with unpaired electron.

Another representation of Lewis Dot Structure for hydroxide radical, illustrating O and H with unpaired electron.

Different version of Lewis Dot Structure for hydroxide radical, showing O and H with unpaired electron.

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Verified step by step guidance

Identify the total number of valence electrons available. Oxygen has 6 valence electrons and hydrogen has 1 valence electron, making a total of 7 valence electrons for the hydroxide radical (OH).

Determine the central atom. In the hydroxide radical, oxygen is the central atom because it is more electronegative than hydrogen.

Draw a single bond between the oxygen and hydrogen atoms. This bond will use 2 of the 7 available valence electrons.

Distribute the remaining valence electrons around the oxygen atom to satisfy its octet. Place the remaining 5 electrons as lone pairs around the oxygen atom.

Since this is a radical, one electron will remain unpaired. Place this unpaired electron on the oxygen atom, indicating the radical nature of the hydroxide ion.