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Chapter 8: Problem 4

The bond energy for a \(\mathrm{C}-\mathrm{H}\) bond is about \(413 \mathrm{~kJ} / \mathrm{mol}\) in \(\mathrm{CH}_{4}\) but \(380 \mathrm{~kJ} / \mathrm{mol}\) in \(\mathrm{CHBr}_{3}\). Although these values are relatively close in magnitude, they are different. Explain why they are different. Does the fact that the bond energy is lower in \(\mathrm{CHBr}_{3}\) make any sense? Why?

Short Answer

Expert verified
The difference in bond energies of the C-H bond in CH4 and CHBr3 can be attributed to the surrounding atoms, which impact the bond strength and length. In CH4, the C-H bonds are surrounded by hydrogen atoms with low electronegativity, causing minimal effect on bond energy. In CHBr3, the C-H bond is surrounded by larger and more electronegative bromine atoms, leading to a weaker, elongated C-H bond. The lower bond energy in CHBr3 is expected and makes sense due to these factors.

Step by step solution

01

Compare the structure of CH4 and CHBr3

First, let's look at the structure of the two molecules. CH4 is a symmetrical tetrahedral molecule with 4 C-H bonds with equal strength. On the other hand, CHBr3 is also a tetrahedral molecule but has 3 C-Br bonds and only 1 C-H bond.
02

Consider the impact of the surrounding atoms

In CH4, the C-H bonds are surrounded by other hydrogen atoms which have a low electronegativity, meaning the overall effect on the bond energy is minimal. However, in CHBr3, the C-H bond is surrounded by larger and more electronegative bromine atoms. This results in the electron cloud around the carbon atom being drawn closer to the bromine atoms and away from the hydrogen atom. This change in electron density affects the strength and length of the bond.
03

Discuss the lower bond energy in CHBr3

The fact that the bond energy is lower in CHBr3 compared to CH4 does make sense when considering the impact of the surrounding atoms. The higher electronegativity values of the bromine atoms in CHBr3 result in a weak, elongated C-H bond. Typically, when a bond becomes longer, it becomes weaker, which would explain why the bond energy is lower in CHBr3.
04

Conclusion

In conclusion, the difference in bond energies of the C-H bond in CH4 and CHBr3 can be attributed to the surrounding atoms, which impact the bond strength and length. The lower bond energy in CHBr3 is expected due to the presence of three larger and more electronegative bromine atoms, which lead to a weaker C-H bond.

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Most popular questions from this chapter

Place the species below in order of the shortest to the longest nitrogen- oxygen bond. \(\begin{array}{lllll}\mathrm{H}_{2} \mathrm{NOH}, & \mathrm{N}_{2} \mathrm{O}, & \mathrm{NO}^{+}, & \mathrm{NO}_{2}^{-}, & \mathrm{NO}_{3}^{-}\end{array}\) \(\left(\mathrm{H}_{2} \mathrm{NOH}\right.\) exists as \(\left.\mathrm{H}_{2} \mathrm{~N}-\mathrm{OH} .\right)\)

The compound \(\mathrm{NF}_{3}\) is quite stable, but \(\mathrm{NCl}_{3}\) is very unstable \(\left(\mathrm{NCl}_{3}\right.\) was first synthesized in 1811 by P. L. Dulong, who lost three fingers and an eye studying its properties). The compounds \(\mathrm{NBr}_{3}\) and \(\mathrm{NI}_{3}\) are unknown, although the explosive compound \(\mathrm{NI}_{3} \cdot \mathrm{NH}_{3}\) is known. Account for the instability of these halides of nitrogen.

Identify the five compounds of \(\mathrm{H}, \mathrm{N}\), and \(\mathrm{O}\) described as follows. For each compound, write a Lewis structure that is consistent with the information given. a. All the compounds are electrolytes, although not all of them are strong electrolytes. Compounds \(\mathrm{C}\) and \(\mathrm{D}\) are ionic and compound \(\mathrm{B}\) is covalent. b. Nitrogen occurs in its highest possible oxidation state in compounds A and C; nitrogen occurs in its lowest possible oxidation state in compounds \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E}\). The formal charge on both nitrogens in compound \(\mathrm{C}\) is \(+1\); the formal charge on the only nitrogen in compound \(\mathrm{B}\) is \(0 .\) c. Compounds A and E exist in solution. Both solutions give off gases. Commercially available concentrated solutions of compound A are normally \(16 M\). The commercial, concentrated solution of compound \(\mathrm{E}\) is \(15 \mathrm{M}\). d. Commercial solutions of compound \(\mathrm{E}\) are labeled with a misnomer that implies that a binary, gaseous compound of nitrogen and hydrogen has reacted with water to produce ammonium ions and hydroxide ions. Actually, this reaction occurs to only a slight extent. e. Compound \(\mathrm{D}\) is \(43.7 \% \mathrm{~N}\) and \(50.0 \% \mathrm{O}\) by mass. If compound D were a gas at STP, it would have a density of \(2.86 \mathrm{~g} / \mathrm{L}\). f. A formula unit of compound \(\mathrm{C}\) has one more oxygen than a formula unit of compound D. Compounds \(\mathrm{C}\) and \(\mathrm{A}\) have one ion in common when compound \(\mathrm{A}\) is acting as a strong electrolyte. g. Solutions of compound \(\mathrm{C}\) are weakly acidic; solutions of compound A are strongly acidic; solutions of compounds \(\mathrm{B}\) and \(\mathrm{E}\) are basic. The titration of \(0.726 \mathrm{~g}\) compound \(\mathrm{B}\) requires \(21.98 \mathrm{~mL}\) of \(1.000 M \mathrm{HCl}\) for complete neutralization.

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}\), \(\mathrm{O}-\mathrm{H}, \mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H}\).

Use formal charge arguments to explain why \(\mathrm{CO}\) has a much smaller dipole moment than would be expected on the basis of electronegativity.
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