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Chapter 6: Problem 5

Draw Lewis dot structures for these molecules. Notice that in part \(\mathrm{d}\) and part \(\mathrm{f}\), the formulas are written in a way that emphasizes the structure of the molecule. a. \(\mathrm{CF}_{4}\) b. \(\mathrm{CH}_{3} \mathrm{Cl}\) c. \(\mathrm{SiCl}_{2} \mathrm{H}_{2}\) d. \(\mathrm{CH}_{3} \mathrm{OH}\) e. \(\mathrm{HOCl}\) f. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

Short Answer

Expert verified
Draw central atoms with single bonds to surrounding atoms and lone pairs where necessary to satisfy octets.

Step by step solution

01

Understanding Lewis Structures

Lewis structures represent molecules showing all bonding and non-bonding valence electrons. The total available valence electrons are distributed to complete octets around atoms when possible.
02

Drawing CF4

Carbon has 4 valence electrons and each fluorine has 7. Calculating total valence electrons: \(4 + (7 \times 4) = 32\). Place carbon in the center with single bonds to four fluorines. Each fluorine will have three lone pairs, satisfying the octet rule for all atoms.
03

Drawing CH3Cl

Carbon has 4 valence electrons, each hydrogen has 1, and chlorine has 7, totaling \(4 + (1 \times 3) + 7 = 14\). Carbon is central, with single bonds to three hydrogens and a chlorine. Chlorine completes its octet with three lone pairs.
04

Drawing SiCl2H2

Silicon has 4 valence electrons, each chlorine has 7, and each hydrogen has 1, totaling \(4 + (7 \times 2) + (1 \times 2) = 20\). Silicon is central, bonded to two chlorines and two hydrogens. Chlorines complete their octets with three lone pairs each.
05

Drawing CH3OH

Carbon has 4 valence electrons, each hydrogen has 1, and oxygen has 6, totaling \(4 + (1 \times 4) + 6 = 14\). Carbon is central, bonded to three hydrogens and the oxygen. Oxygen also bonds to one hydrogen and has two lone pairs.
06

Drawing HOCl

Hydrogen has 1 valence electron, oxygen has 6, and chlorine has 7, totaling \(1 + 6 + 7 = 14\). Oxygen is central, bonded to both hydrogen and chlorine. Oxygen and chlorine complete their octets with two and three lone pairs, respectively.
07

Drawing CH3NH2

Carbon has 4 valence electrons, each hydrogen has 1, and nitrogen has 5, totaling \(4 + (1 \times 5) + 5 = 14\). Carbon is bonded to three hydrogens and nitrogen. Nitrogen, bonded to two hydrogens, has one lone pair.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the outermost electrons of an atom and play a vital role in chemical bonding. These electrons are involved in forming bonds with other atoms, whether through sharing or transferring. Understanding the distribution of valence electrons is crucial for drawing accurate Lewis structures.
  • To determine the number of valence electrons, look at the element's group number in the periodic table. For example, carbon has 4 valence electrons because it is in group 14.
  • Count the total valence electrons for a molecule by adding up the valence electrons of each atom involved.
  • These electrons form the backbone for Lewis structures by creating bonds and non-bonding pairs, known as lone pairs.
When drawing Lewis structures, assign electrons to satisfy each atom鈥檚 valence capacity, which typically leads to satisfying the octet rule.
Octet Rule
The octet rule is a fundamental principle in chemistry that states atoms tend to form bonds until they are surrounded by eight valence electrons, similar to the configuration of noble gases. This concept is central to understanding and drawing Lewis structures.
  • Most elements strive to complete their outer electron shell with 8 electrons during bonding, except for hydrogen, which aims for 2 electrons.
  • In a Lewis structure, electrons are showcased as dots around the atom symbols to illustrate the sharing or transfer needed for each atom to reach an octet.
  • For example, carbon atoms, when bonding in molecules like \(\text{CF}_4\), share electrons with fluorine atoms to reach an octet, forming a stable configuration.
There are exceptions to the octet rule, such as molecules with odd numbers of electrons, but it provides a strong basis for many common compounds.
Molecular Geometry
Molecular geometry describes the three-dimensional arrangement of atoms within a molecule. Knowing the molecular geometry is important for understanding how molecules interact with each other, which influences their physical and chemical properties.
  • The shape of a molecule is determined by the number of bonds and lone pairs of electrons around the central atom.
  • Lewis structures help predict molecular geometry by showing how atoms are bonded, but techniques like VSEPR theory (Valence Shell Electron Pair Repulsion) are used to determine the actual shape.
  • For instance, a molecule like \(\text{CH}_4\) is tetrahedral, with bond angles of approximately 109.5掳, because its central carbon forms four equal bonds with hydrogen atoms.
Visualizing these shapes helps explain how molecules pack together, which is important for predicting reactivity and interaction.
Chemistry Education
Chemistry education involves introducing students to the core principles and practices of chemistry, covering a wide range of topics, including atomic theory, chemical bonding, and molecular interactions. Effective chemistry education focuses on building a strong foundation and developing critical thinking.
  • Engaging learning tools, such as Lewis structures, help students visualize and understand chemical bonding and structures.
  • Step-by-step approaches and hands-on exercises lead to better comprehension and retention of complex concepts.
  • Applying these concepts in real-world scenarios, like analyzing how alcohols behave compared to hydrocarbons, deepens understanding and illustrates chemistry鈥檚 relevance.
The goal of chemistry education is to empower students to explore how molecular principles apply in various aspects of science and industry, fostering a deep and practical appreciation of chemistry.

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Most popular questions from this chapter

What are isomers?

Draw Lewis dot structures for these molecules. Notice that in part d and part f, the formulas are written in a way that emphasizes the structure of the molecule. a. \(\mathrm{CF}_{4} \quad\) b. \(\mathrm{CH}_{3} \mathrm{Cl} \quad\) c. \(\mathrm{SiCl}_{2} \mathrm{H}_{2}\) d. \(\mathrm{CH}_{3} \mathrm{OH} \quad\) e. HOCl \(\quad\) f. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

Explain how you can convert acetic acid into an ester. Be specific about how the molecule needs to change.

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