Physical Chemistry I Fall 2016 Second Hour Exam (100 points) Name: (20 points) 1. Quantum calculations suggest that the molecule U 2 H 2 is planar and has symmetry D 2h. D 2h E C 2 (z) C 2 (y) C 2 (x) i σ xy σ xz σ yz A g 1 1 1 1 1 1 1 1 x 2, y 2, z 2 B 1g 1 1-1 -1 1 1-1 -1 R z xy B 2g 1-1 1-1 1-1 1-1 R y xz B 3g 1-1 -1 1 1-1 -1 1 R x yz A u 1 1 1 1-1 -1-1 -1 B 1u 1 1-1 -1-1 -1 1 1 z B 2u 1-1 1-1 -1 1-1 1 y B 3u 1-1 -1 1-1 1 1-1 x a) Create a reducible representation for all 3N degrees of freedom in U 2 H 2 and reduce that representation to decide the symmetry representations of all of the vibrational modes. *HINT: contributions per unshifted atom: E(3), C 2 (-1), σ(1), i(-3).
1. (continued) b) Two of the in-plane vibrations of the bridging hydrogens in U 2 H 2 are shown. Designate the symmetry representation of each vibration and decide if each vibration shown is infraredactive, Raman-active, both, or neither.
1. (continued) c) The frontier orbitals (HOMO and LUMO) orbitals for U 2 H 2 are shown below. Designate i. the D 2h symmetry representation for each orbital, ii. whether a HOMO > LUMO electronic transition is dipole-allowed. (Note the coordinate system.) EXPLAIN. HOMO LUMO
(40 points) 13.6 ev 2. a) The energy of a hydrogen atom we know exactly to be E H = n 2 = 109691cm 1 n 2. Hydrogen, H 2, in its ground electronic state (X state) dissociates into two ground state (n = 1) hydrogen atoms. H 2 in its first excited electronic state (B state) dissociates into one ground state (n = 1) hydrogen atom and one excited (n = 2) hydrogen atom. Calculate the energy difference (E*) in cm -1 between electronically excited H 2 that has dissociated and ground state H 2 that has dissociated. b) The rotational constant B in the ground electronic state of H 2 is 60.80 cm -1, while the B value of the excited electronic state is 20.016 cm -1. Calculate the equilibrium bond distance for H 2 in its ground (R e ) and first excited electronic states (R e ). *HINT: units. c) The vibrational term (cm -1 ) for H 2 in its ground electronic state is G"(v) = 4395.2 v + 1 2 107.2 v + 1 2 2 where 4395.2 cm -1 is the fundamental vibrational frequency ν e and 107.2 cm -1 is the first anharmonicity constant ν e x e. Calculate the dissociation energy D e "in cm -1 for the ground electronic X state of H 2.
2. (continued) d) The electronic separation of the X and the B states in H 2 (T e ) is 91689 cm -1. Sketch the potential energy curves for the ground and first excited electronic states vs. R H H, the internuclear separation. Clearly indicate on this sketch E*, R e, R e, D e, D e and T e, e) Using the sketch in part (d) calculate the dissociation energy D e ' of the first excited B electronic state of H 2.
(30 points) 3. (i) For transitions within the same electronic state of a diatomic molecule the infrared transition moment is given by µ fi = µ 0 Ψ' vib Ψ" vib R 2 dr R=0 π θ=0 µ + Ψ' vib R Ψ" vib R 2 dr R=0 Ψ' rot cosθ Ψ" rot sinθ dθ π θ=0 2π φ=0 dφ Ψ' rot cosθ Ψ" rot sinθ dθ EXPLAIN the types of transitions allowed by the integral before the + and after the +. 2π φ=0 dφ
3. (ii) Of the following molecules check the appropriate boxes to indicate whether each will absorb light via the transition moment term before the +, after the +, both, or neither. Molecule Absorbs via term before + Absorbs via term after + SO 2 CO 2 CO O 2 3. (iii) By analogy to the hydrogen atom the rotational part of the nuclear wavefunction, Ψ rot, in diatomic molecule is a spherical harmonic. The first two spherical harmonics are 1/2 1 J = 0 Ψ rot = 4π and J = 1 Ψ rot = 3 4π 1/2 cosθ SHOW that the integrals over θ in (i) vanish if J" = J' but are non-zero when J" = 0 and J' = 1. *HINT: (sin ax)(cos m ax)dx = cosm +1 ax. (m +1)a
(10 points) Answer either (i) or (ii). 4. (i) Define the Born-Oppenheimer approximation and identify how it applies to (a) simplification of the Schrodinger equation for molecules, (b) to transition moments and (c) to electronic transitions via the Franck-Condon principle.
4. (ii) The molecular orbital in ground state LiH is best described as a linear combination of a H(1s) orbital with an sp hybrid orbital on Li. Ψ = H(1s) ± (Li(2s) ± Li(2p z )) = H(1s) ± Li(sp hybrid) Sketch the combinations of the two orbitals on Li to create two sp hybrid orbitals. Then, sketch the linear combinations of each of the sp hybrids with the H(1s) orbital to create four molecular orbitals. Clearly identify which of the resultant molecular orbitals is bonding and which is antibonding.
Useful Information h B = 8π 2 cµr 2 rotational constant E rot = hcbj ( J +1) ) rotational energy (rigid rotor) h = 6.6256 x 10-34 J sec h = 6.6256 x 10-27 erg sec c = 2.9927 x 10 8 m sec 1 Planck s constant (mks units) Planck s constant (cgs units) speed of light (mks units) c = 2.9927 x 10 10 cm sec 1 speed of light (cgs units) m H = 1.0079 g / mol molar mass of hydrogen atom µ = m 1 m 2 m 1 + m 2 reduced mass N o = 6.022 x 10 23 molecules/mol Avogadro's number