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 energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of porbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the $\pi_{2 p}^{\star}$ molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the C-Cbond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(C-C\) bond in ethylene easier to twist in the ground state or in the excited state?

Short Answer

Expert verified
In ethylene, the HOMO is \(\pi_{2 p}\) and the LUMO is \(\pi_{2 p}^{\star}\). The C-C bond is weaker in the excited state than in the ground state due to an electron being promoted to an antibonding orbital. The C=C bond is easier to twist in the excited state as compared to the ground state.

Step by step solution

01

(a) Find the HOMO in ethylene

HOMO stands for the Highest Occupied Molecular Orbital, which means it is the highest energy orbital that has electrons. In the ground state of ethylene, the \(\pi_{2 p}\) orbital is filled with a pair of electrons, making it the HOMO. So, the HOMO in ethylene is \(\pi_{2 p}\).
02

(b) Find the LUMO in ethylene

LUMO stands for the Lowest Unoccupied Molecular Orbital, which means it is the lowest energy orbital which does not have electrons currently. In the ground state of ethylene, the \(\pi_{2 p}^{\star}\) orbital is empty, making it the LUMO. So, the LUMO in ethylene is \(\pi_{2 p}^{\star}\).
03

(c) Compare C-C bond strength in the excited state and ground state

When a photon is absorbed, an electron from the HOMO (\(\pi_{2 p}\)) gets promoted to the LUMO (\(\pi_{2 p}^{\star}\)), this creates an excited state of ethylene. In this excited state, the electrons are in an antibonding orbital instead of a bonding orbital. This in turn reduces the overall stability and increases the energy. Due to this reason, the C-C bond in ethylene is weaker in the excited state as compared to the ground state.
04

(d) Compare the difficulty of twisting the C=C bond in ground state and excited state

In the ground state, the C=C bond has a pair of electrons in the bonding \(\pi_{2 p}\) orbital. This double bond creates a stronger bond between the two carbons which makes it harder to twist the C=C bond. However, in the excited state, an electron from the bonding \(\pi_{2 p}\) orbital is promoted to the antibonding \(\pi_{2 p}^{\star}\) orbital, weakening the bond between the two carbons. As a result, the C=C bond becomes easier to twist in the excited state as compared to the ground state.

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

The orbital diagram that follows presents the final step in the formation of hybrid orbitals by a silicon atom. (a) Which of the following best describes what took place before the step pictured in the diagram: (i) Two 3 p electrons became unpaired, (ii) An electron was promoted from the \(2 p\) orbital to the \(3 s\) orbital, or (iii) An electron was promoted from the \(3 s\) orbital to the \(3 p\) orbital? (b) What type of hybrid orbital is produced in this hybridization? [Section 9.5]

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

Indicate the hybridization of the central atom in (a) \(\mathrm{BCl}_{3}\), (b) \(\mathrm{AlCl}_{4}^{-}\), (c) \(\mathrm{CS}_{2}\), (d) \(\mathrm{GeH}_{4}\) -

Consider the bonding in an \(\mathrm{MgH}_{2}\) molecule. (a) Draw a Lewis structure for the molecule, and predict its molecular geometry. (b) What hybridization scheme is used in \(\mathrm{MgH}_{2}\) ? (c) Sketch one of the two-electron bonds between an \(\mathrm{Mg}\) hybrid orbital and an \(\mathrm{H} 1 \mathrm{~s}\) atomic orbital.

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$ \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \rightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6}\), you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) The molecule IF h has a pentagonal-bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\).

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