Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

1. A Review of General Chemistry

Lewis Structure

Organic Chemistry

1. A Review of General Chemistry

Lewis Structure

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Problem 62j

Textbook Question

Draw the Lewis structure for the following molecular formulas.
(j) H₂SO₄

Verified step by step guidance

Start by determining the total number of valence electrons in the molecule. Sulfur (S) has 6 valence electrons, each oxygen (O) has 6 valence electrons, and each hydrogen (H) has 1 valence electron. Add these together: 6 (S) + 4×6 (O) + 2×1 (H).

Place the sulfur atom in the center of the structure because it is less electronegative than oxygen. Arrange the four oxygen atoms around the sulfur atom.

Connect the sulfur atom to each of the four oxygen atoms using single bonds. Each single bond represents 2 electrons, so subtract these electrons from the total valence electrons calculated in step 1.

Assign lone pairs to the oxygen atoms to satisfy their octet rule. Start by placing three lone pairs (6 electrons) on each oxygen atom, then adjust if necessary based on the remaining electrons.

To account for the structure of H₂SO₄, note that two of the oxygen atoms are bonded to hydrogen atoms. Replace one lone pair on each of these two oxygen atoms with a single bond to a hydrogen atom. Additionally, sulfur can expand its octet, so adjust the bonding (e.g., form double bonds) to ensure all atoms have a formal charge as close to zero as possible.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They use dots to represent electrons and lines to represent bonds between atoms. Understanding how to draw Lewis structures is essential for visualizing molecular geometry and predicting reactivity.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial for determining how atoms bond with each other. In the case of H₂SO₄, knowing the number of valence electrons for hydrogen, sulfur, and oxygen helps in accurately constructing the Lewis structure. Each element's valence electrons contribute to the overall electron count needed for bonding.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The shape of a molecule, influenced by the number of bonds and lone pairs, affects its physical and chemical properties. For H₂SO₄, understanding its geometry helps in predicting its behavior in chemical reactions and interactions with other molecules.

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Related Practice

Textbook Question

Given the atoms involved and the number of valence electrons remaining, complete the Lewis structure by placing bonds between atoms such that each has a full octet.

(e)

Textbook Question

Draw the Lewis structure for the following molecular formulas.

(l) SO₄^2-

Textbook Question

Draw Lewis structures for the following free radicals.

a. The ethyl radical, CH₃—ĊH₂

b. The tert-butyl radical, (CH₃)₃C•

Textbook Question

In addition to radicals, anions, and cations, a fourth class of reactive intermediates is carbenes. A neutral species, the simplest carbene has a molecular formula of CH₂.

(a) Draw the Lewis structure of CH₂.

Textbook Question

Draw the Lewis structure for the following molecular formulas.

(v) CH₅N

Textbook Question

Draw a Lewis structure for each of the following:

d. NH₂O⁻

Textbook Question

Give Lewis structures corresponding to the following line–angle structures. Give the molecular formula for each structure.

(a)

(b)

Textbook Question