Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 C-H bond energy between CH4 and CHBr3 can be explained by the electronegativity of the surrounding atoms. Bromine atoms in CHBr3 have a higher electronegativity than hydrogen atoms in CH4, which weakens the C-H bond and results in a lower bond energy in CHBr3. This is consistent with the molecular structures and the effect of surrounding atoms on bond strength.

Step by step solution

01

Understanding bond energy

Bond energy is the amount of energy required to break a chemical bond between two atoms in a molecule. In this case, we are looking at the C-H bond in CH4 and CHBr3.
02

Comparing the molecular structures of CH4 and CHBr3

Methane (CH4) is a simple hydrocarbon molecule with a carbon atom that is bonded to four hydrogen atoms. On the other hand, CHBr3 (bromoform) is a halogenated hydrocarbon, where the carbon atom is bonded to one hydrogen atom and three bromine atoms.
03

Understanding the role of the surrounding atoms in bond energy

The chemical environment has a significant impact on the bond strength. In CHBr3, the carbon atom is bonded to three highly electronegative bromine atoms, which pull electron density away from the carbon atom. This reduces the effective positive charge on the carbon atom, weakening the C-H bond. In CH4, the carbon atom is only bonded to hydrogen atoms, which have lower electronegativity than the bromine atoms, resulting in a stronger C-H bond.
04

Explaining the difference in bond energy

The difference in C-H bond energy between CH4 and CHBr3 can be explained by the electronegativity of the surrounding atoms. Bromine atoms in CHBr3 have a higher electronegativity than hydrogen atoms in CH4, which causes a decrease in the effective positive charge on the carbon atom in CHBr3. This results in a weaker C-H bond and hence lower bond energy in CHBr3 compared to CH4.
05

Understanding whether the lower bond energy in CHBr3 makes sense

The fact that the bond energy is lower in CHBr3 does make sense. It is consistent with the molecular structures and the effect of surrounding atoms on bond strength. In CHBr3, the high electronegativity of bromine atoms weakens the C-H bond compared to the bond in CH4, resulting in a lower bond energy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Order the following species with respect to carbon-oxygen bond length (longest to shortest). $$ \mathrm{CO}, \quad \mathrm{CO}_{2}, \quad \mathrm{CO}_{3}{ }^{2-}, \quad \mathrm{CH}_{3} \mathrm{OH} $$ What is the order from the weakest to the strongest carbonoxygen bond? \(\left(\mathrm{CH}_{3} \mathrm{OH}\right.\) exists as \(\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}\).)

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Li}\) and \(\mathrm{N}\) c. \(\mathrm{Rb}\) and \(\mathrm{Cl}\) b. Ga and \(\mathrm{O}\) d. \(\mathrm{Ba}\) and \(\mathrm{S}\)

Without using Fig. \(8.3\), predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{H}, \mathrm{Si}-\mathrm{H}, \mathrm{Sn}-\mathrm{H}\) b. \(\mathrm{Al}-\mathrm{Br}, \mathrm{Ga}-\mathrm{Br}, \mathrm{In}-\mathrm{Br}, \mathrm{Tl}-\mathrm{Br}\) c. \(\mathrm{C}-\mathrm{O}\) or \(\mathrm{Si}-\mathrm{O}\) d. \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{O}-\mathrm{Cl}\)

Which compound in each of the following pairs of ionic substances has the most exothermic lattice energy? Justify your answers. a. \(\mathrm{NaCl}, \mathrm{KCl}\) b. \(\mathrm{LiF}, \mathrm{LiCl}\) c. \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{MgO}\) d. \(\mathrm{Fe}(\mathrm{OH})_{2}, \mathrm{Fe}(\mathrm{OH})_{3}\) e. \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{O}\) f. \(\mathrm{MgO}, \mathrm{BaS}\)

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{~F}_{2}\left(\mathrm{O}_{2} \mathrm{~F}_{2}\right.\) exists as \(\left.\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F}\right)\). Assign oxidation states and formal charges to the atoms in \(\mathrm{O}_{2} \mathrm{~F}_{2}\). This compound is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{~F}_{2}\) ?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free