CHEM 2060 Lecture Notes - Lecture 6: Antibonding Molecular Orbital, Bond Order, Paramagnetism

• School: University of Guelph
• Department: Chemistry
• Course: CHEM 2060
• Professor: Dennis Baker

Document Summary

58 (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:9) (cid:9) (cid:9) (cid:9) (cid:9) (cid:12)(cid:13)(cid:14)(cid:15)(cid:16)(cid:17)(cid:18)(cid:11)(cid:18) (cid:19)(cid:15)(cid:4)(cid:8)(cid:20)(cid:21)(cid:2)(cid:8)(cid:7)(cid:5)(cid:9)(cid:22)(cid:5)(cid:23)(cid:6)(cid:3)(cid:7)(cid:8)(cid:24)(cid:25)(cid:9) (cid:26)(cid:22)(cid:27)(cid:28)(cid:1)(cid:29)(cid:22)(cid:30)(cid:26)(cid:9) (cid:9) Case iii is of highest energy (and is totally antibonding). For the f2 cation the bond order for f2 is 1. + the bond order is 1 . (electron comes out of an. 59 (a) different for c2 and f2 . In order to answer this question you must realise that the mo schemes are. For f2 the homo is a * and the lumo is a *. For c2 the homo is u and the lumo is p g. For nitrogen there is s+p mixing (+enough to push above ). In this question we see how the energies of bonding and antibonding orbitals are calculated. First we need to write down the true normalized form of the wavefunction. Now we are not allowed to neglect s the overlap integral. (see lecture notes). We can simplify matters by setting c1 = c2. 2(1 s) (see dekock and gray p. 190+191)