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## N 2 , O 2 , and F 2

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**A basis for N2:**N2, O2, and F2 - No 1s orbitals, because these are core orbitals. - The dimension of the problem has to be reduced by symmetry.**N2, O2, and F2 (continued)**Intermezzo: The 2p orbitals.**N2, O2, and F2 (continued)**Intermezzo: The 2p orbitals.**N2, O2, and F2 (continued)**A B Symmetry elements:**N2, O2, and F2 (continued)**A B Make plus and minus combination from orbitals that transform into each other.**N2, O2, and F2 (continued)**combination group I II III IV V VI II I**N2, O2, and F2 (continued)**group I: group II: group III: group IV: group V: group VI: - There is no interaction between orbitals of different groups. - The eight-dimensional problem is reduced to six smaller problems. - The one-dimensional problems directly give solutions of the Fock-equation. - Groups III and V, and groups IV and VI give degenerate solutions.**N2, O2, and F2 (continued)**N N2 N**N2, O2, and F2 (continued)**N N2 N**N2, O2, and F2 (continued)**Bond-order = (# bonding electrons - # antibonding electrons)/2 = (8 - 2) / 2 = 3 N N2 N**N2, O2, and F2 (continued)**MO of N2:**N2, O2, and F2 (continued)**MO of N2:**N2, O2, and F2 (continued)**MO of N2:**N2, O2, and F2 (continued)**MO of N2:**N2, O2, and F2 (continued)**MO of N2:**N2, O2, and F2 (continued)**MO of N2:**N2, O2, and F2 (continued)**O O2 O Bond-order = (# bonding electrons - # antibonding electrons)/2 = (8 - 4) / 2 = 2**N2, O2, and F2 (continued)**Hund’s second rule: The Slater-determinant has the lowest energy if electrons are distributed as much as possible over degenerated orbitals. Hund’s third rule: The Slater-determinant has the lowest energy if electrons are placed in degenerated orbitals with the same spin.**N2, O2, and F2 (continued)**F F2 F Bond-order = (# bonding electrons - # antibonding electrons)/2 = (8 - 6) / 2 = 1