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 statements is correct regarding the Lewis dot structure of NO_2 (with N as the central atom) in its neutral form?

The nitrogen atom has one lone pair and forms one double bond and one single bond with the two oxygen atoms.

The nitrogen atom has no lone pairs and forms two double bonds with the oxygen atoms.

The nitrogen atom has two lone pairs and forms two single bonds with the oxygen atoms.

The nitrogen atom has one lone pair and forms two single bonds with the oxygen atoms.

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Verified step by step guidance

Step 1: Determine the total number of valence electrons available for NO_2. Nitrogen (N) has 5 valence electrons, and each oxygen (O) has 6 valence electrons. Since there are two oxygen atoms, total valence electrons = 5 + 2 × 6 = 17 electrons.

Step 2: Recognize that NO_2 is a neutral molecule with an odd number of valence electrons, which means it is a radical and will have an unpaired electron in its Lewis structure.

Step 3: Place nitrogen as the central atom and connect it to the two oxygen atoms with single bonds initially. This uses 4 electrons (2 bonds × 2 electrons each).

Step 4: Distribute the remaining electrons to satisfy the octet rule for the oxygen atoms first, then place any leftover electrons on nitrogen. Because of the odd number of electrons, one electron will remain unpaired on nitrogen, resulting in one lone electron (not a lone pair).

Step 5: To minimize formal charges, form one double bond between nitrogen and one oxygen atom, and one single bond with the other oxygen atom. This arrangement results in nitrogen having one lone electron (not a full lone pair), one double bond, and one single bond with the oxygen atoms.