Physical Chemistry II Test Name: KEY CHEM 464 Spring 18 Chapters 7-11 Average = 1. / 16 6 questions worth a total of 16 points Planck's constant h = 6.63 1-34 J s Speed of light c = 3. 1 8 m/s ħ = h π = 1.55 1 34 J s Electron mass m e =9.19 1-31 kg Gas constant R = 8.314 J mol -1 K -1 Avogadro's number N A = 6. 1 3 mol -1 Boltzmann constant k B = 1.381 1-3 J/K =.695 cm -1 /K Units: 1 bar = 1 5 Pa. 1 GPa = 1 9 Pa 1 amu = 1.665 1-7 kg 1 Joule = 1 Pa m 3. 1 L bar = 1 J. 1L=1-3 m 3. 1 Hartree = 65.5 kj/mol = 4.36 1-18 J Complex exponential: e i k x = cos(k x )+i sin(k x) sin(k x ) = 1 i (ei k x e i k x ) Special values: e πi = e π i =1 e πi = e πi = 1 ^L = ħ [ 1 sin θ θ ( sin θ θ ) + 1 ] sin θ ϕ Integrals: e a x d x = 1 a ea x +C x e a x d x = ( x a 1 e α x d x= π 4α e a x d x = 1 a x n e a x d x = 1 a n+1 x n e a x d x = 1 3 5 (n 1) a ) ea x +C n+1 a n π a x n+1 e a x d x = n! a n+1 1 H 1.1 He 4. 3 Li 6.94 4 Be 9.1 5 B 1.81 6 C 1.1 7 N 14.1 8 O 16. 9 F 19. 1 Ne.18 11 Na.99 1 Mg 4.31 13 Al 6.98 14 Si 8.9 15 P 3.97 16 S 3.6 17 Cl 35.45 18 Ar 39.95 19 K 39.1 Ca 4.8 1 Sc 44.96 Ti 47.88 3 V 5.94 4 Cr 5. 5 Mn 54.94 6 Fe 55.85 7 Co 58.93 8 Ni 58.7 9 Cu 63.55 3 Zn 65.38 31 Ga 69.7 3 Ge 7.59 33 As 74.9 34 Se 78.96 35 Br 79.9 36 Kr 83.8 37 Rb 85.47 38 Sr 87.6 39 Y 88.91 4 Zr 91. 41 Nb 9.91 4 Mo 95.94 43 Tc 98 44 Ru 11.1 45 Rh 1.9 46 Pd 16.4 47 Ag 17.9 48 Cd 11.4 49 In 114.8 5 Sn 118.7 51 Sb 11.8 5 Te 17.6 53 I 16.9 54 Xe 131.3 55 Cs 13.9 56 Ba 137.3 57 La 138.9 7 Hf 178.5 73 Ta 18.9 74 W 183.8 75 Re 186. 76 Os 19. 77 Ir 19. 78 Pt 195.1 79 Au 197. 8 Hg.6 81 Tl 4.4 8 Pb 7. 83 Bi 9. 84 Po 9 85 At 1 86 Rn 87 Fr 3 88 Ra 6. 89 Ac 7. 14 Rf 67 15 Db 68 16 Sg 69 17 Bh 7 18 Hs 69 19 Mt 78 11 Ds 81 111 Rg 8 11 Cn 85 113 Nh 86 114 Fl 89 115Mc 89 116 Lv 93 117 Ts 94 118 Og 94 58 Ce 14.1 59 Pr 14.9 6 Nd 144. 61 Pm 145 6 Sm 15.4 63 Eu 15. 64 Gd 157.3 65 Tb 158.9 66 Dy 16.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 7 Yb 173. 71 Lu 175. 9 Th 3. 91 Pa 31. 9 U 38. 93 Np 37 94 Pu 44 95 Am 43 96 Cm 47 97 Bk 47 98 Cf 51 99 Es 5 1 Fm 57 11 Md 58 1 No 59 13 Lr 6
1. ( points) Hydrogen-atom wave functions are ψ(r,θ, ϕ)=r n,l (r )Y l,ml (θ,ϕ). Four H-atom radial wavefunctions are graphed below. a. Which one is for the 1s orbital? Answer: D. Max at nucleus, no nodes. b. Which one is for the p orbital? Answer: A. Zero at nucleus, no nodes.. ( points) Here is the radial Schrodinger equation for a hydrogen atom. m e r R e A B C ħ m e r d d r [ r d R d r ] + ħ l(l+1) 4 π ϵ r R = E R There are three terms on the left-hand side of the equation. They are labeled A, B and C. Which term represents electron-nucleus attraction? Answer: C Which term is zero for s orbitals? Answer: B
Score: /16 3. ( points) Here is a two-electron wave function: ψ(1,)= ϕ s (1)ϕ p ()α(1)α() a) In terms of spin, does this wave function represent a singlet state or a triplet state? Answer: triplet because M S =1. b) The given wave function does not satisfy the Pauli principle. Show how it fails to satisfy the Pauli principle. Answer: ψ(,1)=ϕ s ()ϕ p (1)[α()α(1)] ψ(1,) 4. ( points) Here is the spherical harmonic function for l= and m l =: Y, = ( 15 3π ) 1/ sin θ e iϕ. a. Prove by operating on Y, with ^L z, that Y, is an eigenfunction of ^L z. Answer: ^L z Y, = i ħ Y, ϕ = ħ Y, b. Y, is an eigenfunction of ^L. What is the eigenvalue? Give the eigenvalue in SI units. Note: It is not necessary to show the operation of ^L on Y,. Answer: The eigenvalue of ^L is l(l+1)ħ =6 ħ =6.7 1 68 J s.
5. (3 points) Consider a phosphorous atom in its ground electron configuration 1s s p 6 3s 3p 3. The lowest-energy term is 4 S 3/. The next-lowest term is D 5/. a) Draw electrons as arrows on the p-orbital diagram at right, showing one (any one) configuration that is consistent with the lowest-energy term. Answer: Any occupation that has M L = and M S =±3/ or ±1/ is correct. b) A light-induced transition between the two terms is forbidden by the usual dipole selection rules. Specifically, write one selection rule that would be violated by 4 S 3/ + photon D 5/. Answer: ΔS=, ΔL= or ±1. However, ΔJ= or ±1 is not violated. c) A Gamess Hartree-Fock calculation that specified a doublet spin state gave this result for total spin: S =.756 ħ. Compare that average to its theoretical value. Answer: A doublet term has S=½ so <S > should be (3/4)ħ. There is 1% error. 6. (5 points) Consider the diatomic molecule AgH. The equilibrium bond length is 1.619 Å. The rotation constant B e = 6.44 cm -1. The harmonic-oscillator frequency is 176 cm -1. Here are two harmonic-oscillator wave functions, the ground state and the first excited state: ψ =( α π ) 1/ 4 e 1 α x ) 1/4 x e 1 α x. The constant α = 56. Å. and ψ 1 =( 4α3 π a. (1 point) Calculate the bond length at which the ground state and the excited state have the same probability density. There are two values. Calculate either one. Note: The limit of infinite bond length is not an acceptable answer. Answer: Probability density is the square of the wave function, so set ψ = ψ 1. ψ = ψ1 ( α π ) 1/ e α x = ( 4α3 π ) 1/ x α x e x = 1 α x =± 1 α = 1 = ±.94 Å 56. Å R=R e +x =1.55 or 1.713 Å
b. ( points) In a rotation-vibration spectrum of AgH, predict ΔE in cm -1 for the transition from (v,j)=(,1) to (1,11). Assume harmonic oscillation and rigid rotation. Answer: ΔE = E(v=1,J=11) - E(v=,J=1) = (3/) 176 + 11(11+1)B - (1/) 176-1(1+1) B ΔE = 176 cm -1 + 11 6.44 cm -1 = 176 + 14 = 19 cm -1. c. ( point) A molecule closely related to AgH is AgD, where D is deuterium. Masses of H, D and Ag are 1.8,.14 and 17.87 amu. Calculate the harmonic-oscillator frequency of AgD, in cm -1. Answer: μ AgD = 17.87.14/17.87+.14 = 1.9771 amu μ AgH = 17.87 1.8/17.87+1.8 =.9987 amu ~ νagd = 1 π μ k ~ ν AgD AgH μ AgH μ = AgD 176.9987 = 176.7118=153 cm 1 1.971 The experimental value is 151.