130k views
0 votes
In a model of the benzene ring, the hydrogen atoms are ignored and the six carbon atoms are treated as individual particles with angular positions, θj​(t),j=1,2,…,6 and mass m. Each atom is connected to its neighbour by a bond, modelled as a spring with constant k, and the absorption of incoming photons applies a sinusoidal external force with frequency Ω>0. Newton's law for the atoms states that mθˉ1​=k(θ2​−θ1​)−k(θ1​−θ6​+2π)+αsinΩtmθˉ2​=k(θ3​−θ2​)−k(θ2​−θ1​)+αsinΩtmθˉ3​=k(θ4​−θ3​)−k(θ3​−θ2​)+αsinΩtmθˉ4​=k(θ5​−θ4​)−k(θ4​−θ3​)+αsinΩtmθˉ5​=k(θ6​−θ5​)−k(θ5​−θ4​)+αsinΩtmθˉ6​=k(2π+θ1​−θ6​)−k(θ6​−θ5​)+αsinΩt​ where α is the amplitude of the photon forcing. Write these ODEs as the system, x¨=Ax+c+asinΩt identifying the coefficient matrix A and the two constant vectors c and a. Verify (by substitution and performing a matrix multiplication) that there is a time-independent particular solution of θj​=31​(j−1)π for α=0 (corresponding to the atoms sitting evenly spaced at equilibrium). Confirm (again by substitution) that A has the eigenvectors, e1​=⎝⎛​111111​⎠⎞​,e2​=⎝⎛​1−11−11−1​⎠⎞​,e3​=⎝⎛​1−101−10​⎠⎞​,e4​=⎝⎛​01−101−1​⎠⎞​,e5​=⎝⎛​110−1−10​⎠⎞​,e6​=⎝⎛​0110−1−1​⎠⎞​, and find the corresponding eigenvalues, λ, in terms of ω=k/m​. Briefly interpret to what motions of the atoms these eigensolutions correspond. At what forcing frequencies does the benzene ring resonate under the photon irradience? Using various matrices, but without performing any detailed algebra or computing any inverses, find the general solution of the problem when resonance does not occur. Highlight the problem with this formal solution when the forcing frequency is resonant.

User Ehcanadian
by
8.0k points

1 Answer

4 votes

Final answer:

The valence bond theory describes the benzene molecule as a hexagon of sp²-hybridized carbon atoms with perpendicular unhybridized p orbitals forming delocalized π bonds, represented by a circle within the ring structure.

Step-by-step explanation:

Valence bond theory describes the benzene molecule and other planar aromatic hydrocarbon molecules as hexagonal rings of sp²-hybridized carbon atoms. The unhybridized p orbital of each carbon atom is perpendicular to the plane of the ring. Three valence electrons in the sp² hybrid orbitals of each carbon atom and the valence electron of each hydrogen atom form the framework of sigma (σ) bonds in the benzene molecule. The fourth valence electron of each carbon atom is involved in delocalized pi (π) bonds formed with the unhybridized p orbitals of adjacent carbon atoms, contributing to the unique properties of benzene.

User Aaron Swan
by
7.9k points