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$$ \begin{aligned} &\text { Draw Lewis structures for HCFC-123 (CF }_{3} \mathrm{CHCl}_{2} \text { ) and }\\\ &\mathrm{CF}_{3} \mathrm{CFH}_{2} \end{aligned} $$

Short Answer

Expert verified
HCFC-123 has a structure with C-C in the center, three F on one C, two Cl on another C, and one H. CF₃CFH₂ has two C connected, three F on one C, and two H on another C.

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01

Count Total Valence Electrons

For each molecule, calculate the total number of valence electrons. - HCFC-123 (CF₃CHCl₂): Carbon (C) has 4, Fluorine (F) has 7, Chlorine (Cl) has 7, and Hydrogen (H) has 1 valence electron. The total is: (2 C × 4) + (3 F × 7) + (1 Cl × 7) + (2 Cl × 7) + (1 H × 1) = 48 electrons. - CF₃CFH₂: (2 C × 4) + (3 F × 7) + (2 H × 1) = 50 electrons.
02

Determine Central Atom and Sketch Skeleton

For each compound, identify the central atom and connect the peripheral atoms with single bonds. - HCFC-123: Use carbon atoms as central, connect other atoms (F, Cl, H) to the central carbons with single bonds. - CF₃CFH₂: Connect the two carbon atoms and then attach each fluorine and hydrogen.
03

Distribute Remaining Electrons to Fulfill Octets

Fulfill the octet rule by placing the remaining electrons around the atoms. - For HCFC-123: After forming single bonds, distribute additional electrons as lone pairs to ensure F and Cl have complete octets. - For CF₃CFH₂: Similarly, place lone pairs on F atoms to fulfill their octet.
04

Finalize Lewis Structures

Adjust structures if necessary to ensure every atom has a complete valence shell (except for hydrogen, which needs 2 electrons). Ensure that the total number of electrons used matches the total count from Step 1. Check for resonance structures if applicable.

Key Concepts

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

Valence Electrons
In chemistry, valence electrons are the outermost electrons of an atom and are crucial in forming chemical bonds. These electrons determine how atoms interact with each other.
For the molecules HCFC-123 and CF₃CFH₂, calculating the total number of valence electrons is the first step. Here's how we do it:
  • HCFC-123 (CF₃CHCl₂): Each atom contributes its own valence electrons. Carbon has 4, fluorine has 7, chlorine has 7, and hydrogen has 1 valence electron. Add them all up, considering the number of each atom, to get a total of 48 electrons.
  • CF₃CFH₂: Similarly, with two carbon atoms, three fluorine atoms, and two hydrogens, you end up with 50 valence electrons.
This understanding is essential for drawing accurate Lewis structures, where electrons are represented as dots.
Octet Rule
The octet rule is a simple guideline in chemistry, stating that atoms tend to form bonds until they have eight electrons in their valence shell. This gives them the same electron configuration as a noble gas, which is very stable.
When drawing Lewis structures for HCFC-123 and CF₃CFH₂, ensure each atom follows the octet rule:
  • Carbon, fluorine, and chlorine strive for eight electrons in their valence shell.
  • Hydrogen is an exception and only needs two electrons to be stable.
Use remaining valence electrons to complete the octets by placing them around atoms as lone pairs, especially around fluorine and chlorine.
Central Atom
Selecting the central atom is an important step in creating a Lewis structure. Generally, the least electronegative element (except hydrogen) is chosen as the central atom.
For HCFC-123 and CF₃CFH₂:
  • Carbon is typically chosen as the central atom since it can form four bonds, making it ideal for connecting multiple atoms.
  • In HCFC-123, two carbons serve as central atoms, each surrounded by other elements like fluorine, chlorine, and hydrogen.
  • In CF₃CFH₂, the two carbon atoms are central, with fluorine and hydrogen attached.
Properly identifying the central atom helps in arranging other atoms around it effectively.
Lone Pairs
Lone pairs are pairs of valence electrons that are not involved in bonding and belong exclusively to one atom. They play a critical role in the shape and properties of molecules.
After placing single bonds in Lewis structures, the next step is distributing remaining electrons as lone pairs to satisfy the octet rule:
  • Fluorine and chlorine typically hold onto lone pairs because they need to achieve a complete octet.
  • In both HCFC-123 and CF₃CFH₂, once all single bonds are formed, distribute the leftover electrons as lone pairs on the fluorine and chlorine atoms.
  • This not only completes their octet but also helps in predicting molecular geometry and reactivity.
Lone pairs significantly affect molecular shapes by repelling bonded electron pairs, thus determining the molecule's three-dimensional structure.

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Most popular questions from this chapter

A concentration of \(8.00 \times 10^{2}\) ppm by volume of \(\mathrm{CO}\) is considered lethal to humans. Calculate the minimum mass of \(\mathrm{CO}\) (in grams) that would become a lethal concentration in a closed room \(17.6 \mathrm{~m}\) long, \(8.80 \mathrm{~m}\) wide, and \(2.64 \mathrm{~m}\) high. The temperature and pressure are \(20.0^{\circ} \mathrm{C}\) and \(756 \mathrm{mmHg}\), respectively.

The balance between \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) is important in understanding acid rain formation in the troposphere. From the following information at \(25^{\circ} \mathrm{C}\) : $$ \begin{aligned} \mathrm{S}(s)+\mathrm{O}_{2}(g) & \rightleftarrows \mathrm{SO}_{2}(g) & & K_{1}=4.2 \times 10^{52} \\ 2 \mathrm{~S}(s)+3 \mathrm{O}_{2}(g) & \rightleftarrows 2 \mathrm{SO}_{3}(g) & & K_{2}=9.8 \times 10^{128} \end{aligned} $$ calculate the equilibrium constant for the reaction: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) $$

Although the hydroxyl radical (OH) is present only in a trace amount in the troposphere, it plays a central role in its chemistry because it is a strong oxidizing agent and can react with many pollutants as well as some \(\mathrm{CFC}\) substitutes. The hydroxyl radical is formed by the following reactions: $$ \mathrm{O}_{3} \stackrel{\lambda=320 \mathrm{nm}}{\longrightarrow} \mathrm{O}^{*}+\mathrm{O}_{2} $$ where \(\mathrm{O}^{*}\) denotes an electronically excited atom. (a) Explain why the concentration of \(\mathrm{OH}\) is so small even though the concentrations of \(\mathrm{O}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) are quite large in the troposphere. (b) What property makes OH a strong oxidizing agent? (c) The reaction between \(\mathrm{OH}\) and \(\mathrm{NO}_{2}\) contributes to acid rain. Write an equation for this process (d) The hydroxyl radical can oxidize \(\mathrm{SO}_{2}\) to \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The first step is the formation of a neutral \(\mathrm{HSO}_{3}\) species, followed by its reaction with \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) to form \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and the hydroperoxyl radical \(\left(\mathrm{HO}_{2}\right)\). Write equations for these processes.

Water and sulfur dioxide are both polar molecules, and their geometry is similar. Why is \(\mathrm{SO}_{2}\) not considered a major greenhouse gas?

Describe three human activities that generate carbon dioxide. List two major mechanisms for the uptake of carbon dioxide.

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