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Chapter 11: Problem 79

A conjugated hydrocarbon has an alternation of double and single bonds. Draw the molecular orbitals of the \(\pi\) system of 1,3,5 -hexatriene. If the energy required to excite an electron from the HOMO to the LUMO corresponds to a wavelength of \(256 \mathrm{nm},\) do you expect the wavelength for the corresponding excitation in 1,3,5,7 -octatetraene to be a longer or shorter wavelength?

Short Answer

Expert verified
The wavelength for the corresponding excitation in 1,3,5,7-octatetraene will be longer.

Step by step solution

01

Draw the Pi Molecular Orbitals

First, draw six atoms in a line to represent 1,3,5-hexatriene. Then, draw the \(\pi\) molecular orbitals. There will be six \(pi\) orbitals corresponding to six \(\pi\) electrons. The lowest energy orbital will have no node (area where the electron density is zero), the next will have one node, and so on up to the highest energy orbital, which will have five nodes.
02

Identify the HOMO and LUMO

The highest occupied molecular orbital (HOMO) is the orbital with the highest energy that contains an electron. The lowest unoccupied molecular orbital (LUMO) is the orbital with the lowest energy that does not contain an electron. For 1,3,5-hexatriene, the HOMO is the 3rd orbital (counting from the lowest energy) and the LUMO is the 4th.
03

Apply the particle-in-a-box concept

To calculate the energy difference between the HOMO and LUMO, use the particle-in-a-box concept. In this concept, \(E = h^2n^2/8mL^2\) where E is energy, n is an integer (orbital number), h is Planck's constant, m is the mass of the particle (electron), and L is the length of the box (distance between atoms). For conjugated systems, the 'box length' increases as the number of double bonds increase.
04

Analyse the effect of increased conjugation

Comparing hexatriene (with 3 double bonds) to octatetraene (with 4 double bonds), the increase in conjugation (and therefore box length) will decrease the energy required to excite an electron from the HOMO to the LUMO (since energy and length are inversely proportional in the particle-in-a-box equation).
05

Relate Energy Difference to Light Wavelength

Because the energy difference decreases, the light wavelength required to excite an electron from the HOMO to the LUMO will increase. This is because energy and wavelength are inversely proportional according to Planck's equation \(E = hc/\lambda\), where E is energy, h is Planck's constant, c is the speed of light, and \(\lambda\) is wavelength.

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Most popular questions from this chapter

Represent bonding in the carbon dioxide molecule, \(\mathrm{CO}_{2},\) by \((\mathrm{a})\) a Lewis structure and \((\mathrm{b})\) the valencebond method. Identify \(\sigma\) and \(\pi\) bonds, the necessary hybridization scheme, and orbital overlap.

Explain how it is possible to avoid the concept of resonance by using molecular orbital theory.

Propose a hybridization scheme to account for bonds formed by the central carbon atom in each of the following molecules: (a) hydrogen cyanide, HCN; (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH} ;\) (c) acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) (d) carbamic acid,

The ion \(\mathrm{F}_{2} \mathrm{Cl}^{-}\) is linear, but the ion \(\mathrm{F}_{2} \mathrm{Cl}^{+}\) is bent. Describe hybridization schemes for the central \(\mathrm{Cl}\) atom consistent with this difference in structure.

Describe the molecular geometry of \(\mathrm{H}_{2} \mathrm{O}\) suggested by each of the following methods: (a) Lewis theory; (b) valence-bond method using simple atomic orbitals; (c) VSEPR theory; (d) valence-bond method using hybridized atomic orbitals.
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