CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.Born-Oppenheimer Approximation1. What is the main assumption of the Born-Oppenheimer approximation? The Born-Oppenheimer approximation assumes that the nuclei are stationary, and electron motion can be treated separately. Fixed bond distances and bond angles are assumed, and a Hamiltonian operator is written for electronic motion only (Mortimer, p.648).2. Using the Born-Oppenheimer approximation, write the Hamiltonian for the H2 molecule (2 electrons). 122121022221211111142ˆrrrrrRemHBBAAABRABR12RA2RA1RB2RB1ABe1e2where rA1 is the distance between nucleus A and electron 1; rA2 is the distance between nucleus A and electron 2; rB1 is the distance between nucleus B and electron 1; rB2 is the distance between nucleus B and electron 2; r12 is the distance between electron 1 and electron 2; and RAB is the distance between nucleus A and nucleus B (and remains constant by the Born-Oppenheimer approximation).Variation Method; LCAO-MO3. Determine the energy for 2H in terms of Haa, Hab, and S using the variation method. The secular determinant for this system is 0bbbbbabaababaaaaESHESHESHESHIn this case, 1 , , , bbaabaabbaabbbaaSSSSSHHHH, giving0EHESHESHEHaaababaa               SHHEHHSEHHSEESEHEHESHEHESHEHESHEHabaaabaaabaaabaaabaaabaaabaa1100011111122or1CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.    SHHEHHSEHHSEESEHEHabaaabaaabaaabaa110222222Normalization; Hybrid Orbitals4. Show that the sp2 hybrid orbital  322pssp is normalized if the s and p orbitals are also normalized.     10213122231231222*dsppsdpsdNote: We used  12ds;  12dp; and  0spd for normalized, orthogonal s and p orbitals. Electronic Configuration for Atoms & Molecules; Pauli Exclusion Principle5. What are the electron configurations for H, Li+, O2, F, Na+, and Mg2+?                       6222622622622222221:221:221:221:1:1:pssMgpssNapssFpssOsLisH6. Which of the following transitions are allowed in the normal electronic emission spectrum of an atom(a) 2s to 1s(b) 2p to 1s(c) 3d to 2pFor a single-electron transition, 1l and n = any integer.(a) not allowed, 0l(b) allowed, 1l (c) allowed, 1l 7. Write the electronic configurations for N2, N2+, N2. The configurations are                   1*242*22*22142*22*22242*22*222222211:222211:222211:ppσpsσsσsσsσNpσpsσsσsσsσNpσpsσsσsσsσNgguugugguugugguugugTerm Symbols for Diatomics; Electronic Transitions2CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.8. Determine the complete term symbol for each of the following electronic configurations(a) 1gσ(b) 1uσ(c) 2uσ(d) 1u(e) 3u(a) 1gσ, the single unpaired electron gives a doublet state because S = 1/2 so that2121212 S. The symbol is g2(b) 1uσ, the term symbol is u2(c) 2uσ, the electrons must be paired. Therefore, S = 0 and the state is a singlet. The product of two ungerade functions is gerade. The term symbol is g1(d) 1u, the electron could be in either the  1 or the  1 state. Therefore, there is a degenerate pair of states with 1 and the term symbol isu2.(e) 3u, there are two possible configurations:    12 for which 1111  and    21 for which 1111 . Therefore 1 and the state is . Also, the function is ungerade because it is a product of three ungerade functions. The term symbol is therefore u2. 9. Write the electronic configuration for Li2, and predict the term symbol for the ground level. The electronic configuration is    22*2211 sσsσsσgug. For this configuration, 0LM. Since LM, we know that 0 and S = 0, which results in the symbol 1. Because the wave function for a σ molecular orbital does not change sign upon reflection across the xz plane, the + superscript is used. The parity can be found by multiplying the parities of the orbitals being used, according to the laws of odd (u) and even (g) multiplication:uguguuggg  ; ;. Therefore,    gug 24, and the complete term symbol is g1.10. Which of the following electronic transitions are allowed?(a)gu11(b)gu31(c)ug 113CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.The selection rules for electronic transitions in diatomic molecules are allowednot only; and 01,0ugS(a) ;;0;0  ugSThe transition is allowed.(b) The transition is forbidden because 1S.(c) ugS  ;0;1The transition is allowed.11. The ground-level term for a heteronuclear diatomic molecule is 3. Write the term symbols of the electronic transitions allowed for this molecule. According to the transition rules (see the previous problem), the transition must be; the superscript must be 3; and 0 or 1. The allowed transitions are 33and  33. Note that the notations g and u are not used for heteronuclear diatomic