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 correctly describes the Lewis dot structure of the neutral compound N4O?

It features a cyclic arrangement of four nitrogen atoms with the oxygen atom double-bonded to one nitrogen, and all atoms have complete octets.

It consists of a chain of four nitrogen atoms, with the oxygen atom double-bonded to a terminal nitrogen, and at least one nitrogen atom has an odd number of electrons (violating the octet rule).

It has a central oxygen atom bonded to all four nitrogen atoms, with each nitrogen atom having a lone pair.

It contains a linear chain of four nitrogen atoms with a terminal oxygen atom, and all atoms obey the octet rule.

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Verified step by step guidance

Step 1: Determine the total number of valence electrons available for the molecule N4O. Nitrogen has 5 valence electrons and oxygen has 6 valence electrons, so calculate the total as $4 \times 5 + 6$.

Step 2: Consider possible structural arrangements of the atoms. Since the formula is N4O, think about whether the oxygen atom is terminal or central, and how the nitrogen atoms might be connected (chain or cyclic).

Step 3: Draw a plausible Lewis structure starting with a chain of four nitrogen atoms and place the oxygen atom at one end, forming a double bond with the terminal nitrogen. Assign single bonds between the nitrogen atoms initially.

Step 4: Distribute the remaining valence electrons as lone pairs to satisfy the octet rule for each atom. Check if all atoms have eight electrons around them or if any atom has an odd number of electrons, indicating a violation of the octet rule.

Step 5: Analyze the resulting structure to see if it matches any of the given descriptions, focusing on whether the structure is cyclic or linear, the bonding of oxygen, and whether all atoms obey the octet rule or if there is an odd electron count on any nitrogen.