Chemistry 362 Spring 2013 Dr. Jean M. Standard April 24, 2013 Sample Questions for Non-Cumulative Portion of Final Exam The non-cumulative portion of the Final Exam is worth 75 points. It will consist of questions covering the material we discussed in Chapters 9, 10 and 13, along with Problem Sets 10, 11, and 12 (and assigned textbook problems). Please note: A molecular orbital correlation diagram will be provided for this portion of the final exam along with the usual equation sheet and physical constants. The cumulative portion of the Final Exam is worth 75 points. It will consist of questions taken from Exams 1-3. For the cumulative portion of the final exam, all the equation sheets, integral tables, physical constants, and conversion factors previously included for Exams 1-3 will be provided. 1.) Consider the diatomic anion N2 – . a.) b.) Determine the possible molecular term symbols of the ground electronic state of N2 – . 2 2 2 2 Excited electronic states of N2 – have the following term symbols: Π g , Π u , Σ g , and Σ u . What are the allowed electronic transitions from the ground electronic state to the listed excited states? € 2 a.) b.) € € € Determine all the atomic term symbols arising from a p1 d1 electron configuration. Arrange the terms from part (a) in order of increasing energy according to Hund's rules. 3.) Give the explicit form of the Hamiltonian operator (in atomic units) for the lithium atom. Your expression should not include any summations. Classify the various terms that are included in the Hamiltonian operator. 4 a.) b.) Give a brief description of the Franck-Condon Principle. In order for a transition to be observed from the v"=0 level of the ground electronic state to the v'=0 level of an excited electronic state of a diatomic molecule, what has to be true about the potential energy curves? 5 a.) b.) Draw an orbital energy diagram for the ground state of the Cl atom. Determine the multiplicity of the ground state of the Cl atom. 6 a.) b.) Explain what is meant by LCAO-MO. Show, using sketches, how two p-type atomic orbitals may be combined to form σ bonding and σ* antibonding orbitals. 2 PHYSICAL CONSTANTS AND CONVERSION FACTORS Constant h  c kB e me ε0 N (Avogadro’s No.) SI 6.62607×10–34 J s 1.05457×10–34 J s 2.99793×108 m/s 1.38066×10–23 J/K 1.60217×10–19 C 9.10938×10–31 kg 8.85419×10–12 C2 N– 1m– 2 6.02203×1023 1 amu 1Å 1 nm 1 bohr 1.66057×10–27 kg 1 hartree 1 hartree 1 hartree 4.35975×10–18 J 10–10 m 10–9 m 0.52917706 Å 219474.6 cm– 1 27.2114 eV ISOTOPIC MASSES (in amu) 1H 2D 7Li 12C 14N 16O 1.0078 2.0140 7.0160 12.0000 14.0031 15.9949 19F 23Na 35Cl 37Cl 79Br 81Br 18.9984 22.9898 34.9689 36.9659 78.9183 80.9163 3 Final Exam Equation List Hˆ ψ = Eψ € € ∫ψ * ψ dτ Sˆ 2α = 3 4 Sˆzα = 1 2 € = 1  2α N = 1 [∫ Sˆ 2 β = € Sz = S € € € 3 4 ψ * ψ dτ ] 2β ms s( s + 1) € 1 Hˆ = − 2 n n ∑ ˆ2 ∇ i i=1 − ∑ i=1 Z + ri n n ∑∑ i=1 j =i+1 1 rij 2 2 2 ˆ2 = ∂ + ∂ + ∂ ∇ i 2 2 ∂x i ∂y i ∂z i2 LJ € K ψ MO = ∑c i χ i,AO i=1 Bond Order = 2S +1 € € = A = € 2S +1 € A Multiplicity = 2S + 1 € € € 1/ 2 Sˆz β = − 12 β α cos θ = Probability = ψ * ψ dτ Hˆ = Tˆ + Vˆ Λ g,u # Bonding electrons − # Antibonding electrons 2 ∫ψ * Aˆ ψ dτ 4