Ethylene is of the point group D2h with symmetry elements

Consider the planar ethylene molecule. Ethylene is of the point group D2h with symmetry elements E, C2(x), C2(y), C2(z), i, sigma(xy), sigma(xz), sigma(yz). You can find a character table for D2h on the web. One location is ethylene molecule below has useful labels as the graph (please see attachment).Let’s assign the molecule to the xy plane with the z-axis perpendicular to the plane and the x-axis along the C-C bond.a. The inversion operator is at the origin of the xyz axes which is at the center of the C-C bond. Now apply the symmetry operations on x, y, and z and confirm x=B3mu, y=B2mu and z = B1mu symmetries.b. Now find Γr, the reducible representation for the four C-H bonds. Remember the characters are the diagonal sums (traces). I will give you the first two: E = 4 since all four bonds remain where they are and C2(z) = 0 since this operation takes Ha → Hd, Hd → Ha, Hb → Hc, and Hc→ Hb and hence none of them remain in their original locations.c. Find the four symmetry components of Γr. There needs to be four because the E character in Γr is 4.d. Now go to the s, Px, Py, and Pz orbitals on the two carbon atoms. Obtain Γs, Γpx, Γpy, and Γpz as you did for Γr. Each of them will be two-dimensional (E character of 2). To get you started C2(x)(Pz)=-2(Pz) and C2(y)(Pz)=0(Pz)e. Find the two symmetry components associated with each reducible representation of part d.f. Now we need to pair up the symmetries of the carbon orbitals with the symmetries found for the four C-H bonds. If you did everything correctly you will find a match between Γr and ΓsΓpxΓpy but not Γpz.g. Now take the C-C σ- bond. Assume it is pointed from C1 → C2. Show that you get a symmetry for Γcc of B3mu. For example i(C − C) = −1 (C − C) since it reverses the direction of the bond. Note this correlates with the symmetries of the Γs and Γpx but not Γpy and Γpz as it should.

Order your essay today and save 30% with the discount code: KIWI20