5.61 F17 Third Hour Exam Solutions

5.61 Exam III ANSWERS Fall, 2017 Page 1 of 13 pages MASSACHUSETTS INSTITUTE OF TECHNOLOGY 5.61 Physical Chemistry Fall, 2017 Professor Robert W. Field FIFTY MINUTE EXAMINATION III ANSWERS Thursday, November 30 I. LCAO-MO A. (20 points) This problem deals with homonuclear diatomic molecules from the Li-Ne row of the periodic table. (i) (5 points) Be2 (40 POINTS) Draw an MO diagram for Be2 and use it to answer the two questions: 4 electrons Be 2p Be 2p πu(2p) σg(2p) σu*(2s) 2s 2s σs(2p) 5.61 Exam III ANSWERS Fall, 2017 Page 2 of 13 pages (a) (3 points) Is Be2 expected to be bound in its electronic ground state? Justify your answer. Key points: σu*(2s) is more anti-bonding than σg(2s) is bonding. The interaction between σg(2p) and σg(2s) will make σg(2s) more bonding. The interaction between σu*(2p) and σu*(2s) will have a small effect on reduction of anti-bonding character because of a much larger energy denominator. The expectation is that the single bond from σg(2s)2 will be almost perfectly cancelled by the single anti-bond from σu*(2s)2. The ground state of Be2 is likely to be unbound. (b) (2 points) Do you expect that the lowest energy excited state of Be2 is bound? Justify your answer. The first excited state will involve promotion of an electron from the σu*(2s) orbital to the σg(2p) orbital. This σ g (2s)2 σ u* (2s)1σ g (2 p)1 3 + ∑u state will certainly be bound. The isoconfigurational 1 ∑+u state is also likely to be bound. Electronic states of diatomic molecules are specified as 2S+1Λ (∑ for Λ = 0, ∆ for Λ = 2). A π2 configuration gives rise to three electronic states: 3∑–, 1∆, and 1∑+. Which one of these three states is predicted by Hund’s first rule to have the lowest energy? Hund's first rule specifies that the lowest energy electronic state from each electronic configuration has maximum multiplicity. For a π2 configuration the lowest energy state would be 3∑–. (ii) (5 points) (iii) (5 points) Draw an MO diagram that explains why O2 has a triplet ground state. Explain why this is a safe prediction. O2 has 12 valence electrons. The MO diagram is 5.61 Exam III ANSWERS 0 Fall, 2017 σu*(2p) Page 3 of 13 pages 0 πg*(2p) 2p 2p πu(2p) σg(2p) x σu(2s) 2s 2s σg(2s) The 12 electrons are filled into the molecular orbitals from the lowest one in energy order: σ g (2s)2 σ u* (2s)2 σ g (2 p)2 π u (2 p)4 π g* (2 p)2 The lowest energy state is 3 ∑−g . There is no other choice. σu*(2p) is slightly pushed up (more antibonding) by interaction with σu*(2s). There is no πg* valence orbital for πg*(2p) to interact with. So the energy gap between the predicted HOMO and predicted LUMO is very large. The prediction is iron-clad! 5.61 Exam III ANSWERS (iv) (5 points) Fall, 2017 Page 4 of 13 pages A similar argument might predict that C2 also has a triplet ground state. The three lowest lying electronic configurations for C2 are: σ g (2s)2 σ u* (2s)2 σ g (2 p)2 π u (2 p)2 : 3 Σ−g , 1∆ g , 1 Σ+g σ g (2s)2 σ u* (2s)2 σ g (2 p)1 π u (2 p)3 : 3 Πu , 1 Πu σ g (2s)2 σ u* (2s)2 σ g (2 p)0 π u (2 p)4 : 1 Σ+g Show, using an MO diagram, why the O2-like argument is unsafe for C2. [HINT: the two σg orbitals derive from the 2s and 2p AOs, which are relatively close in energy for carbon.] MO diagram for C2. C2 has 8 valence electrons. σg(2p) de-stabilized by σg(2s) πu(2p) (no mixing) very small energy difference σu*(2s) negligible stabilization by σu*(2p) σg(2s) extra stabilization It is likely that the ground state of C2 is σg(2s)2σu*(2s)2πu(2p)4 1∑g+. It is possible that the πu(2p)→σg(2p) promotion energy is small relative to the e–– e– repulsion due to 4 electrons in the πu(2p) orbital. So the σg(2s)2σu*(2s)2πu(2p)3σg(2p) 3∏u state is very lowlying. The analogous state to O2, σg(2s)2σu*(2s)2πu(2p)2σg(2p)2 3∑g– state is definitely out of contention for being the electronic ground state. Everything is far less conclusive than for O2. 5.61 Exam III ANSWERS B. (10 points) Fall, 2017 Page 5 of 13 pages Isoelectronic molecules with 8 valence electrons: C2, BN, BeO, LiF. The vibrational frequencies in the electronic ground states of these four molecules are C2: ωe = 1855 cm–1, BN: 1515 cm–1, BeO: 1489 cm–1, LiF: 910 cm–1. Use MO diagrams to explain this trend in bond order. We are looking at hetero-nuclear 8 electron molecules. The ionization energy from the highest energy AO increases as you go from Li to C and also from C to F. C2 C 2p 2s LiF σg(2p) πu(2p) σg*(2s) σg(2s) 1Σ+ F Li C 2p σ*(2p Li) π*(2p Li) 2p 2s σ*(2s Li) 2s π(2p F) σ(2p F) 2p 2s σ(2s F) Ionic bonding. Negligible covalent character. C2 has a covalent bond order of 2. The covalent boning character decreases as the ionization energy difference increases: C2 → BN → BeO → LiF. 5.61 Exam III ANSWERS C. (10 points) Fall, 2017 Page 6 of 13 pages The X 1Σ+g Electronic Ground States of Homologous Molecules: N2 and P2. Re ωe N2 P2 2359 cm–1 781 cm–1 1.10Å 1.89Å extremely stable and non-reactive extremely unstable and reactive These molecular constants suggest that N2 has a triple bond while P2 has a single bond. Given that the electronic ground state configuration in N2 is σ g (2s)2 σ u* (2s)2 σ g (2 p)2 π u (2 p)4 and N(s 2 p 3 ) → N + (sp 3 ) ∆ E2 s = 164,100 cm −1 (hc) P(s 2 p 3 ) → P + (sp 3 ) ∆ E3s = 130, 300 cm −1 (hc) N(s 2 p 3 ) → N + (s 2 p 2 ) ∆ E2 p = 117, 400 cm −1 (hc) P(s 2 p 3 ) → P + (s 2 p 2 ) ∆ E3 p = 84,600 cm −1 (hc) Speculate about the σ(3s) vs. σ(3p) vs. π(3p) dominant character of the single bond in P2 and the reason for the absence or weakness of the bonds associated with the other two normally bonding LCAOMO orbitals. Recall that the orbital ionization energy roughly determines the orbital radius and that the probability density in an orbital is proportional to r–3. Why is the bonding so different between N2 and P2? There is no simple correct answer to this question. Several factors are relevant: 1) The orbital size decreases as ionization energy increases. The orbitals on N are smaller than the corresponding orbitals on P. This suggests that Re will be larger for P2. 2) Orbital density decreases as orbital size increases. This means that bonding (ωe) will be weaker for P2. 3) The largest and most directed orbital is σ(2p). If there is only one bond for P2, it will be from σg(2p)2. 4) The larger size of the valence MOs on P2 suggest that a smaller fraction of the electrons are in the region along the bond axis between the two atoms. The electrons outside the spatial region of overlap will be extremely susceptible to reaction. 5.61 Exam III ANSWERS II. Fall, 2017 Hückel Theory Page 7 of 13 pages (30 POINTS) Consider the butadiene molecule, CH2CHCHCH2 The four Hückel Theory orbital eigen-energies are α + β[3 + 51/2/2]1/2 E0 α + β[3 – 51/2/2]1/2 E1 α – β[3 + 51/2/2]1/2 E2 α – β[3 – 51/2/2]1/2 E3 and the four eigenvectors are ⎛−0.37 ⎞ ⎜ ⎟ ⎜ 0.60 ⎟, ⎜−0.60 ⎟ ⎜ ⎟ ⎝ 0.37 ⎠ C3 ⎛ 0.37 ⎞ ⎜ ⎟ ⎜ 0.60 ⎟, ⎜ 0.60 ⎟ ⎜ ⎟ ⎝ 0.37 ⎠ C0 ⎛ 0.60 ⎞ ⎜ ⎟ ⎜−0.37 ⎟, and ⎜−0.37 ⎟ ⎜ ⎟ ⎝ 0.60 ⎠ C2 ⎛−0.60 ⎞ ⎜ ⎟ ⎜−0.37 ⎟ ⎜ 0.37 ⎟ ⎜ ⎟ ⎝ 0.60 ⎠ C1 A. (5 points) Match each of the four orbital eigen-energies to the corresponding eigenvector. HINT: count the nodes and remember that both α and β are < 0. See above for C0→C3 (counted by number of nodes), and the matching E0→E3. Note that 3 > sqrt(5)/2. B. (13 points) (i) (3 points) How many pz electrons are there in the π-system of butadiene? Each carbon donates one electron, for a total of 4. Also, 2 × 2 pi bonds is 4 5.61 Exam III ANSWERS Fall, 2017 Page 8 of 13 pages (ii) (3 points) Specify the number of electrons in each of the four orbitals. We have 4 electrons, so fill the two lowest orbitals. C0 and C1 then will have two electrons each. (iii) (7 points) For only the Highest energy Occupied Molecular Orbital (HOMO), specify the non-integer number of π-electrons on each of the carbon atoms. C1 is filled and is (–0.6, –0.37, 0.37, 0.6)2 electrons ×C1 squared (element wise) gives 2(0.62, 0.372, 0.372, 0.62) = (0.72, 0.27, 0.27, 0.72) C. (7 points) Calculate the total bond-order between atoms 1 and 2 and between atoms 2 and 3. For a doubly occupied orbital, the bond order between atoms a and b is Oab = 2cacb and the total bond-order between atoms a and b is given by the sum over all of the occupied molecular orbitals. O12: 2*(C1_1*C1_2) + 2*(C0_1*C0_2) = 2*(0.37*0.6+(-0.6)*(-0.37))= 0.89 (or 1.89 if you include the sigma bond). O23: 2*(0.6*0.6 + (0.37)(-0.37)) = 0.45 (or 1.45) D. (5 points) Calculate the bond-order between atoms 1 and 4. Why is this calculated bond-order surprising? O14: 2(0.37 × 0.37 + (-0.6)(0.6)) = –0.45. No sigma bond. Lewis diagrams and chemical intuition tell us that there should be no bond order between non-adjacent atoms. This calculation indicates that there is actually bond order, and that atoms 1 and 4 are antibonding with respect to one another. This is due to the 4 atom orbitals (and thus multiple atom-long bonds) that are the molecular orbitals. 5.61 Exam III ANSWERS Fall, 2017 Page 9 of 13 pages III. Many Electron Atoms and “Shielding” (30 POINTS) The lowest energy electronic configuration for Sc is [Ar]4s23d1. This notation means the electronic configuration of Ar plus 4s23d1. A. (6 points) If we remove one electron from the electronic ground state of Sc, we have Sc+ in one of the two possible configurations: [Ar]4s13d1 [Ar]4s2 A B Express the energy of the two configurations using the symbols ε4s ε3d J4s4s J4s3d J3d3d K4s3d EA = ε4s + ε3d + J4s3d – K4sds EB = 2ε4s + J4s4s B (2 points) Energy of the 4s orbital Energy of the 3d orbital Coulomb integral between 4s and 4s Coulomb integral between 4s and 3d Coulomb integral between 3d and 3d Exchange integral between 4s and 3d (1) (2) The values of several of these integrals are found to be (in atomic units) J4s4s = 0.303 J4s3d = 0.150 K4s3d = 0.002 J4s4s is larger than J4s3d. Why is this consistent with our expectations? Jij is the average coulomb repulsion felt by an electron in orbital i with an electron in orbital j. An electron in the 4s orbital feels greater repulsion with another electron sharing the 4s orbital as opposed to an electron in a different orbital. 5.61 Exam III ANSWERS [Ar]4s13d1 C. (4 points) Fall, 2017 Page 10 of 13 pages [Ar]4s2 2+ For Sc the experimentally measured energy difference between the [Ar]4s1 and [Ar]3d1 configurations is ε4s – ε3d = 0.1164. B A Going from Sc2+ to Sc+, use the cited numerical values of ε, J, and K to determine which of the Sc+ [Ar]4s13d1 or [Ar]4s2 configurations has the lowest energy. Which configuration is more stable? EB – EA = ε4s –ε3d + J4s4s – J4s3d + K4s3d = 0.2714 > 0 This means that Configuration A is more stable. D. (6 points) Next we consider the high-spin and low-spin energy states of Sc+ in the [Ar]4s3d configuration High-Spin Low-Spin D C Calculate the energies of configurations C and D and express your answer in terms of the relevant symbols. EC = ε4s + ε3d + J4s3d — K4s3d ED = e4s + e3d + J4s3d E. (4 points) Which configuration, C (high-spin) or D (low-spin) is more stable? You do not need to do a numerical calculation to answer this question. Why? ED – EC = K4s3d > 0 So Configuration C (high-spin) is more favorable. Hund’s First Rule favors high spin. It can also be explained using the fact that an exchange force keeps the electrons apart in C, so there is less coulomb repulsion. 5.61 Exam III ANSWERS Fall, 2017 Page 11 of 13 pages F. (3 points) Your answers to parts A and B are surprising because the lowest energy state for the iso-electronic atom, Ca, is [Ar]4s2. Suggest a possible explanation for why the lowest energy state for Ca comes from the [Ar]4s2 configuration and the lowest energy state for Sc+ comes from the [Ar]4s13d1 configuration. One acceptable answer is provided. A large positive contribution to EB – EA comes from J4s4s. Sc2+ has the same electrons, but a higher nuclear and effective charge than Ca. The 4s orbital for Ca is farther from the nucleus and thus, larger, which means that J4s4s is smaller for Ca. This may be enough to shift the inequality to EB – EA < 0 Physically, this means that electron repulsion drives the second valence electron out of the 4s orbital and into the 3d orbital in the case of Sc2+. G. (5 points) In parts A and B you were given numerical values for some Coulomb and exchange integrals. Without a computer, you cannot derive values for J and K integrals. However, you should be able to predict which one of a pair of integrals is larger. [All of the integrals express electron-electron repulsion, thus are all positive.] For each of the following pairs of integrals, specify which one is larger and, briefly, why it is larger: J1s,2s vs. J1s,4 s J1s,1s vs. J1s,2s J1s,1s vs. J10s,10s K 2s,2 pz vs. K 2 px ,2 pz J 2s,2s vs. K 2s,2s J1s2s > J1s4s : 4s is more diffuse than 2s, so electrons in the 1s orbitals feel less coulomb repulsion. J1s1s > J1s2s : Electrons in the 1s orbital are, on average, closer to other electrons sharing the 1s orbital than an electron in the 2s orbital, and thus feels more repulsion with other 1s electrons. J1s1s > J10s10s : The 1s orbital is smaller than the 10s orbital, so electrons in the 1s orbital spend more time near each other, resulting in a larger coulomb repulsion. K2s2pz > K2px2pz : Exchange is larger when there is more overlap between the orbitals. 2s and 2pz overlap but 2px and 2pz do not. J2s2s = K2s2s : This is by definition of the integrals. 5.61 Exam III ANSWERS Fall, 2017 Page 12 of 13 pages Some Possibly Useful Constants and Formulas ! = 1.054 × 10–34 J · s ε 0 = 8.854 × 10 −12 Cs 2 kg −1m −3 h = 6.63 × 10–34 J · s c = 3.00 × 108 m/s c = λν me = 9.11 × 10–31 kg mH = 1.67 × 10–27 kg 1 eV = 1.602 x 10-19 J e = 1.602 x 10-19 C E = hν a0 = 5.29 x 10-11 m ν= $1 1' 1 = RH && 2 − 2 )) λ % n1 n2 ( λ = h/p where RH = me4 8ε 02 h3c e±iθ = cosθ ± isinθ = 109,678 cm -1 Free particle: E= 2k 2 2m ψ ( x ) = Acos ( kx ) + Bsin ( kx ) Particle in a box: En = h2 8ma 2 2 n = E1 n 2 # 2 &1 2 # nπ x & ψ (0 ≤ x ≤ a ) = % ( sin % ( $a' $ a ' n = 1, 2, … Harmonic oscillator: ! 1$ En = # n + & ω [units of ω are radians/s] " 2% 14 $α ' 2 ψ0 ( x ) = & ) e−α x 2 , %π ( x̂! ≡ a≡ a† = mω x̂ " 1 x̂! + i!p̂ 2 ( 14 1 $α ' ψ1 ( x ) = & ) 2 %π ( p̂! ≡ ) 1 ( x̂! − i!p̂) 2 14 (2α x) e 12 −α x 2 2 1 $α ' ψ2 ( x ) = & ) 8 %π ( 1 p̂ [units of ω are radians/s] "mω Ĥ 1 1 = aa † − = a †a + !ω 2 2 N̂ = a †a (4α x 2 ) − 2 e−α x 2 2 5.61 Exam III ANSWERS 2π cω! = ω Fall, 2017 Page 13 of 13 pages [units of ω! are cm –1 ] Semi-Classical λ = h/p pclassical(x) = [2m(E – V(x))]1/2 period: τ = 1/ν = 2π/ω For a thin barrier of width ε where ε is very small, located at x0, and height V(x0): (1) H nn =∫ x0 +ε/2 x0 −ε/2 (0) (0) ψ (0)* n V (x)ψ n dx = εV (x0 ) ψ n (x0 ) Perturbation Theory En = En(0) + En(1) + En(2) (1) ψ n = ψ (0) n + ψn (0) (1) ! En(1) = ∫ ψ (0)* n H ψ n dx = H nn (1) ψ (1) n = ∑ m≠n E (2) n =∑ m≠n (1) H nm ψ (0) En(0) − Em(0) m 2 (1) H nm En(0) − Em(0) 2