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12/05/18, Chem433 Final Exam answers Last Name or Student ID 1. (2 pts) 12. (3 pts) 2. (6 pts) 13. (3 pts) 3. (3 pts) 14. (2 pts) 4. (3 pts) 15. (3 pts) 5. (4 pts) 16. (3 pts) 6. (2 pts) 17. (15 pts) 7. (9 pts) 18. (6 pts) 8. (2 pts) 19. (14 pts) 9. (3 pts) 20. (2 pts) 10. (2 pts) 21. (3 pts) 11. (2 pts) 22. (8 pts) Honors (10 pts) Total: (100pts) The exam is close book close notes with mostly conceptual problems. Give a brief explanation for maximum credit. You can submit corrections to your answers online before midnight 12/05/18. If answered correctly online, you get half the face value credit for that question. If you do not use a cheat sheet (A4 page only), you get 10% extra. The extra questions for honor students will be counted as extra if you decide to answer them (only in class) 1

1. (2 pts) How many angular and how many radial nodal surfaces does 5g orbital have? Number of angular nodes = l, i.e. 4 in this case Number of radial nodes = n 1 - l, i.e. 5 1 4 = 0 in this case 2. (6 pts) Identify whether a function, f(x), is an eigen function of the corresponding operator, A, and with what eigen value: a) f(x) = (cosx) 2 and A = d 2 /dx 2 d 2 /dx 2 (cosx) 2 = d/dx(-2cosx sinx) = -2cos(2x) not an eigen function b) f(x) = x 4 and A = x d/dx : x(d/dx)(x 4 ) = (x)(4x 3 ) = 4x 4 eigen function with the eigen value 4 c) f(x,y) = exp(ax+by) and A = d 2 /dx dy : d/dx(exp(ax + by)) = (ab) exp(ax+by) = (ab)f eigen function with the eigen value = ab 3. (3 pts) Identify dxz orbital: a) b) c) d) e) 4. (3 pts) Which of the following statements is/are wrong about Hermitian operators: a) the eigenvalues of a Hermitian operator are real numbers b) if two linear Hermitian operators have the same set of eigenfunctions then they commute c) eigenfunctions of a linear Hermitian operator with different eigen values are orthogonal to each other d) a Hermitian operator can be a complex operator e) the eigen functions of a linear Hermitian operator cannot be real 2

5. (4 pts) Calculate the commutator of the operators p 2 x and x 4. Express your answer through powers of x and px. [p 2 x, x 4 ] = (p 2 xx 4 - x 4 p 2 x) = - ħ 2 (d 2 /dx 2 (x 4 + x 4 ħ 2 (d 2 /dx 2 ) = - ħ 2 (4x 3 + x 4 ( ) + x 4 ħ 2 = - ħ 2 (12x 2 + 8x 3 + x 4 ( ) + x 4 ħ 2 = - ħ 2 (12x 2 + 8x 3 ) [ p 2 x, x 4 ] = - ħ 2 (12x 2 + 8x 3 d/dx) = - 12ħ 2 x 2 - i8 ħx 3 px 6. (2 pts) What is the degeneracy of J = 6 rotational state in CH4? For spherical rotors, the degeneracy is gj = (2J+1) 2, which is different from the linear rotors, where gj = (2J+1). There are not only 2J+1 different projections of J (i.e. MJ) on the external axis but also the same number of projections of J on the internal axis. Thus gj = 13 2 = 169 7. (9 pts) Consider an excited state electron configuration of fluorine 1s 2 2s 1 2p 5 3d 1 5 electrons are only one electron away from the totally filled subshell, i.e. for them l2 = 1 and s2 = 1/2; a) (1 pt) Identify all possible values of S for it. s1 = 1/2, s2 = 1/2, s3 = 1/2. Thus possible S = 1/2, 3/2 b) (1 pt) Identify all possible values of L for it. l1 = 0, l2 = 1, l3 = 2. Thus possible L = 1, 2, 3 c) (3 pts) Identify all possible terms that can arise from this electron configuration. Based on L = 1, 2, 3 and S = 1/2, 3/2, the possible terms are 2 P, 2 D, 2 F, 4 P, 4 D, 4 F, d) (2 pts) Based on the Hund s rules, what is the lowest energy term? What is its degeneracy? Hund s rules predict that the lowest energy term is with maximal S and L, i.e. 4 F. The degeneracy is gt = (2S+1)(2L+1) = 4*7=28 e) (2 pts) What are the possible states (levels) for the lowest energy term and their degeneracies? Since L = 3 and S =3/2, the possible J = 3/2, 5/2, 7/2, 9/2. The degeneracy of each is gj = (2J+1). Thus the possible states and their degeneracies are 4 F3/2 (4), 4 F5/2 (6), 4 F7/2 (8), 4 F9/2 (10). 3

8. (2 pts) The lowest energy state of para hydrogen for hydrogen molecule is characterized by: a) zero overall nuclear spin of two hydrogens, I = 0 b) overall nuclear spin of two hydrogens, I = 1 c) zero overall spin of two electrons, S = 0 d) overall spin of two electrons, S = 1 e) zero angular momentum of its rotational state, J = 0 f) angular momentum of its rotational state, J = 1 from parity, this state is not allowed for para hydrogen g) a, c, and e h) b, d, and f i) b, c, and f 9. (3 pts) What is the ground electron configuration of ClO ion? For clarity, sketch the orbital diagram illustrating the AOs mostly contributing to the MOs. Pay attention to the relative energies of AOs and MOs. 1 2 2 2 3 2 4 2 5 2 1 4 2 4 10. (2 pts) What is the bond order for this ground electron configuration of ClO? BO = (n n*)/2 = (6 4)/2 = 1 11. (2 pts) Which atomic orbital(s) contribute the most to the LUMO of ClO? Explain. The LUMO is made by destructive interference of 3s of Cl and 2s of O, where contribution from Cl is more because it is less electronegative Cl, i.e., has a higher energy 12. (3 pts) Which molecule(s) will show microwave absorption? a) CH4 b) CO2 c) CH2Cl2 - nonzero dipole moment d) HC CH e) N C-C N f) none g) a and b h) a and c i) all of them j) there is no correct answer 4

13. (3 pts) Which molecule(s) will show rotational Raman spectrum? a) CH4 b) CO2 c) CH2Cl2 d) HC CH e) N C-C N f) none g) all of them h) c, d, and e i) a and c j) b, c, d, and e all of them have anisotropic polarizability k) there is no correct answer 14. (2 pts) What is the difference between a microcanonical and canonical ensembles? Microcanonical ensemble the ensemble of equi-energetic states, i.e., the same N, V, E Canonical ensemble the ensemble of states with the same N, V, T 15. (3 pts) Calculate the probability of a macrostate {5,2,1}. Assume that the system consists of identical particles, and the particle states have identical energy. (Hint: what is the number of energy levels for each particle? What is the number of particles in the system?) There are N = 8 particles among the 3 levels The multiplicity is W = 8!/5!/2! = 3*7*8 = 168 P = 168/3 8 = 0.0256; 16. (3 pts) What are the probabilities of measuring l = 3 and lz (or m) = 2 for a dx2-y2 orbital? Explain or show derivations. In the basis of Yl m = l,m>, the orbital dx2-y2 = ( 2,2> + 2,-2>)/2 ½. The probability of measuring l =3 is then P(l =1) = 0 The probability of measuring lz = 2 is 1/2 5

17. (15 pts) Consider CO2 molecule in the gas phase. a) (4 pts) How many unique fundamental vibrational frequencies does it have? What are the values for these frequencies in cm -1 (you don t need to be accurate, very crude estimate is enough): There are (3N-5) = 4 vibrational modes in this molecule, 2 stretching (# of bonds),, and the remaining 2 modes are represented by a doubly degenerate bending,. Thus, there are 3 fundamental frequencies. 2 stretching modes: symmetric stretching ( g) at ~ 1300 cm -1 g, antisymmetric stretching ( u) at ~ 2300 cm -1 O C O, 1 bending mode ( u) at ~ 700 cm -1. u u b) (3 pts) How many of them are IR- active (allowed)? You may want to illustrate each mode by displacement vectors. Only u and u are IR allowed, i.e. two fundamental transitions are visible in IR. O C O O C O c) (8 pts) Sketch (or explain) its IR spectrum in the gas phase with as much details as you can. Identify the features one needs to pay attention to. I expect you to identify the allowed and forbidden IR transitions rovibronic structure with P, Q, and R branches, when appropriate, and their relative intensities pattern for each. Neglect anharmoncity and centrifugal distortion. Reminder: B (cm -1 ) = 16.86/I(amuÅ 2 ). Take the bond lengths RCO = 1.16 Å, Antisymmetric stretching, near 2300 cm -1, has only P and R branches where peaks are separated by 2B= 0.3 cm -1 : I = 2*mORCO 2 = 2*16*1.16 2 = 43.06 amu Å 2 => 2B = 0.78 cm -1. The envelope of the intensities of the rotational substructure has a maximum corresponding Jmax ~ (kt/2b) ½ - ½ = (203/0.78) ½ - ½ ~ 15.6, i.e. J =15 or 16 will be the most intense. The doubly degenerate bending mode at ~ 780cm -1 has P, Q, and R branches. The P and R branches have the same pattern as for the stretching modes. ~ 780 cm -1 2B ~ 0.8 cm -1 ~ 2350 cm -1 J max ~ 16 P Q R P R 6

18. (6 pts) Choose the Slater determinants that are correctly representing the states arising from 1s 1 2p 1 electron configuration of He atom. Among those that are, identify the states that are eigen functions of the operator S 2 and label their spins and projections values of Ms. Are these functions also eigen functions of L 2? If they are, what are their eigen values (or values of L)? Hint: calculating the determinant might clarify the choices. (1) 2 p(1) 1 a) 1 s(1)2 p(2) (1) (2) 2 p(1)1 (1) (2) Not an 2 1 2 p(2) (2) 2 antisymmetric function with respect to exchanging two electrons 2 p(1) (1) 1 b) 1 s(1)2 p(2) (1) (2) 2 p(1)1 (1) (2) Not an eigen 2 1 2 p(2) (2) 2 function of S 2 but Ms = 0 and L=1 (1) 2 p(1) 1 c) 1 s(1)2 p(2) (1) (2) 2 p(1)1 (1) (2) Not an eigen 2 1 (2) 2 p(2) 2 function of S 2 but Ms = 0 and L=1 (1) 2 p(1) (1) 1 d) 1s (1)2 p(2) 2 p(1)1 (1) (2) Eigen function of S 2, 2 1 (2) 2 p(2) (2) 2 S=1, Ms = -1, L=1 2 p(1) 1 e) 1s (1)2 p(2) 2 p(1)1 (1) (2) Eigen function of S 2, 2 1 2 p(2) 2 S=1 Ms=+1, L=1 2 p(1) (1) 1 f) 1 s(1)2 p(2) (1) (2) 2 p(1)1 (1) (2) Not an 2 1 (2) 2 p(2) 2 antisymmetric function with respect to exchanging two electrons 19. (14 pts) A wavefunction of a particle of mass m on a harmonic potential with force constant k is given by = 2 0> + 3 1>, where n> are the eigen functions of Hamiltonian, Ĥ n> = ħ (n+1/2) n>. a) (2 pts) Is this an eigen function of the Hamiltonian? It is NOT an eigen function because it is a linear combination of eigen functions with different eigen values. b) (4 pts) Sketch the function, Hint: recall how each of the eigenfunction in the superposition looks like and add them with the corresponding weight. 2 +3 = 7

c) (4 pts) Is it this function normalized, i.e. is < > =1? If not, normalize it. Hint: you do not need to integrate; represent the function as a superposition of the normalized eigen functions, cn n, and evaluate the result by recalling that the eigen functions have to be orthogonal. < > = (2<0 + 3<1 )(2 0> + 3 1>) = (4<0 0> + 9<1 1> + 6<1 0> + 6<0 1>) = 13 - NOT NORMALIZED To normalize, the wavefunction should be divided by the square root of the norm calculated above, i.e. normalized function is: = (2 0> + 3 1>)/13 ½ d) (4 pts) Calculate the expectation value for energy in this state, i.e. <E> = < Ĥ >. <E> = < Ĥ > = (2<0 + 3<1 ) Ĥ (2 0> + 3 1>) /13 = 4/13E0 + 9/13E1 = 4/13ħ (0+1/2) + 9/13ħ (1+1/2) = 31/26ħ e) (Honors, 4 pts) Calculate the expectation value for <x(t)> = < (x,t) x (x,t)>. Hint: Remember that operator x can be expressed through the lowering and raising operators and the time dependence of an eigen function n of energy is given by exp(-ient/ħ), i.e., n (x,t) = n(x) exp(- ient/ħ) x = (a + a + )(ħ/2m ) ½ x (x,t) = (2 0>exp(-i t/2) + 3 1>exp(-3i t/2))/13 ½ x (x,t)>= (ħ/2m ) ½ (a + a + )(2 0>exp(-i t/2) + 3 1>exp(-3i t/2))/13 ½ = 2 1>exp(-i t/2) + 3(2 ½ 2>+ 0>)exp(-3i t/2))/13 ½ <x(t)> = < (x,t) x (x,t)> = (ħ/2m ) ½ (2<0 exp(i t/2) + 3<1 exp(3i t/2)) = (2 1>exp(-i t/2) + 3(2 ½ 2>+ 0>)exp(-3i t/2))/13 = (ħ/2m ) ½ (6 exp(i t)+ 6 exp(-i t)/13 = (ħ/2m ) ½ 12 cos( t)/13 20. (2 pts) Phosphorescence is: a) Any transition from an excited to the ground state b) Transition from an excited to the ground state that is accompanied by emission of a photon c) Any transition from the ground to an excited state d) Transition from the ground to an excited state that is accompanied by absorption of a photon e) A spin allowed transition f) A spin forbidden transition g) d and e h) b and f i) a and e 21. (3 pts) Calculate the partition function for an ideal gas with the thermal de Broglie wavelength of 16 pm at 300K, 1atm. 8

22. (8 pts) Sketch a high resolution proton NMR spectrum of Ethyl acetate (see structure) at low concentration in an aprotic solvent presuming that the coupling constants are identical (J = 7 Hz). Identify the approximate chemical shifts, the total (relative) integrals for each group, and the intensity pattern due to J-coupling (presuming that the latter). 1:3:3:1 1:2:1 9

Honors (6 pts) Consider scattering of an electron on a potential step of the height Vo originating at x = 0 and being higher than the particle s kinetic energy, E < Vo. How fast does the probability for the electron to penetrate at V the depth x into the wall, P = Poexp(- x), decline with E? o Calculate the numeric value of 1/ for E = 1 ev and Vo = 1.03 ev (1eV = 1.6 10-19 J). The mass of electron is me = 9.11 10-31 kg and Planck s constant, h = 6.626 10-34 Js. What is different in the reflected wave when Vo is finite as compared to the infinite one? V=0 The solutions in two regions (x<0) = e ikx + Be -ikx and (x>0) = Ae -k x have to satisfy the boundary conditions of continuity (0) and d (0)/dx, i.e., 1 + B = A and ik(1-b) = -k A which can be solved to give A = 2ik/(ik - k ) and B = (ik + k )/(k - ik), where k = (2mE) ½ /ħ and k = (2m(Vo-E)) ½ /ħ The probability of reflection is R = B 2 = 1, independent of Vo, but the phase in B changes from: 0 o for k >>k, B(Vo ) 1, to 180 o for k <<k, B((Vo E) 0) -1 The probability of penetration, P(x) = (x>0) 2 = A 2 exp(-2k x) = 4k 2 /(k 2 +k 2 )*exp(-2k x) - exponentially declines with distance: P = Poexp( -2x(2m(Vo-E)) ½ /ħ) => = 2(2m(Vo-E)) ½ /ħ = 2(2*0.03*1.6 10-19 J*9.11 10-31 kg) ½ /1.05 10-34 Js = 2(2*0.03*1.6*9.11) ½ /1.05 10-9 m -1 = 1.78 nm -1 or the characteristic length of 0.56 nm. Note that Po also changes with E: Po= 4k 2 /(k 2 +k 2 ) = 4E/Vo. It goes from close to zero at E<<Vo, to almost 4 at Vo-E<<Vo. 10