Molecular Spectra and Molecular Structure - 1976 PS3 Sol

Problem Set #3 ANSWERS 1. (a) Write out the electron configurations for the molecules O+2 , O2 , O−2 , and O22− . ANSWER: O+2 → KK(2sσg )2 (2sσ⋆u )2 (2pσg )2 (2pπu )4 (2pπ⋆g ) O2 → KK(2sσg )2 (2sσ⋆u )2 (2pσg )2 (2pπu )4 (2pπ⋆g )2 O−2 → KK(2sσg )2 (2sσ⋆u )2 (2pσg )2 (2pπu )4 (2pπ⋆g )3 O22− → KK(2sσg )2 (2sσ⋆u )2 (2pσg )2 (2pπu )4 (2pπ⋆g )4 (b) Determine the ground-state term symbols � MΛ g,u ± � for O+2 , O2 , O2− , and O22− . If there are two or more low-lying states, select one as that of the ground state and justify your selection. ANSWER: O+2 → 2 Πg (only term arising) 1 1 + O2 → Δg , Σg , 3 Σ−g By Hund’s rules 3 Σ−g ground state [largest S and largest Λ] O−2 → 2 Π (only term arising) O−2 2 → 1 Σ+g (only term arising) 2. (a) Using symbols appropriate to the separated-atom approximation to a molecular orbital, write the elec­ tronic configuration of lowest energy for the diatomic species C2 , NO+ , and S2 . ANSWER: C2 : KK(2sσg )2 (2sσ⋆u )2 (2pπu )4 NO+ : KK(2sσg )2 (2sσ⋆u )2 (2pσg )2 (2pπu )4 S2 : KK(2sσg )2 (2sσ⋆u )2 (2pσg )2 (2pπu )4 (2pπ⋆g )4 (2pσ⋆u )2 (3sσg )2 (3sσ⋆u )2 (3pσg )2 (3pπu )4 (3pπ⋆g )2 (b) Write the term symbols � MΛ g,u ± � for all the electronic states derivable from the configuration of lowest energy for all three species. Which of these states will have non-zero magnetic moments? ANSWER: C2 from configuration in (a) only term 1 Σ+g ← ground state. There will be no magnetic moment from this configuration. NO+ only term arising is 1 Σ+ there will be no magnetic moment. S 2 : 1 Δg , 3 Σ−g , 1 Σ+g The ground state is 3 Σ−g by Hund’s Rules and it will have a magnetic moment. The 1 Δg state will also have a non-zero magnetic moment. page 2 3. Gaseous HCl is normally a 3 : 1 mixture of H35 Cl and H37 Cl. To a high approximation, the rotational energy levels of such diatomic rotators are E(J)(in cm−1 ) = J(J + 1)B − J 2 (J + 1)2 D where B, the rotational constant, is larger by a factor 1.0015 for H35 Cl than for H37 Cl, and the centrifugal distortion constant D is the same for both molecules within the error of its measurement. For H35 Cl and H39 Cl E(J) = J(J + 1)B − J 2 (J + 1)2 D where DH35 Cl = DH37 Cl and BH35 Cl = 1.0015BH37 Cl (a) Derive an expression for the separation of the pure rotational absorption lines of H35 Cl and H37 Cl as a function of J′ , the J–value for the upper state (ΔJ = J′ − J′′ = +1). ANSWER: Find ΔE as a function of J ′ ΔJ = +1 ⇒ J ′ − J ′′ = 1 ⇒ J ′′ = J ′ − 1 ΔE(J ′′ , J ′ ) = J ′ (J ′ + 1)B − J ′ 2 (J ′ + 1)2 D − J ′′ (J ′′ + 1)B + J ′′ 2 (J ′′ + 1)2 D ΔE(J ′ ) = [J ′ (J ′ + 1) − (J ′ − 1)J ′ ]B + [(J ′ − 1)2 J ′ 2 − J ′ 2 (J ′ + 1)2 ]D = [J ′ (J ′ + 1 − J ′ + 1)]B + [J ′ 2 ((J ′ − 1)2 − (J ′ + 1)2 )]D � � ΔE(J ′ ) = 2J ′ B − 4J ′ 3 D for H37 Cl ′ B = BH37 Cl δE = 0.003BJ ΔE(J ′ ) = 2.003JB − 4J ′ 3 D for H35 Cl (b) What is the spacing in cm−1 of the two lines for which J′ = 10? ANSWER: Find the spacing for the two lines where J ′ = 10. for J ′ = 10 δE = 0.03 BH37 Cl = 0.0301 BH35 Cl page 3 4. The “transition moment,” or the probability of transition, between two rotational levels in a linear molecule may be assumed to depend only on the permanent electric dipole moment of the molecule and thus to be the same for all allowed pure-rotational transitions. In the pure-rotational emission spectrum of H35 Cl gas, lines at 106.0 cm−1 and 233.2 cm−1 are observed to have equal intensities. What is the temperature of the gas? The rotational constant B for H35 Cl is known to be 10.6 cm−1 , and the ratio hc/k has the value 1.44 cm·deg. ANSWER: Since the transition moment is the same ′ ′ I106.0cm−1 = (2J ′ + 1)e−BJ (J +1)/kT = I233.2cm−1 �′ �′ = (2 J�′ + 1)e−BJ (J +1)/kT ΔE = 2J ′ B since 106.0cm−1 = 2BJ ′ ⇒ J ′ = 5 233.2cm−1 = 2BJ�′ ⇒ J�′ = 11 Thus 11e−318cm −1 /kT −457.5 ⇒ 11e T 457.5 2.398 + T 1557.7 T ⇒T = 23e−1399.2/kT = 23e −2013.2 T = 3.135 + −2013.2 T = 0.737 = 2110 ◦ K 5. What would happen to the Birge-Sponer extrapolation scheme for a molecular potential correlating with ionic states of the separated atoms? ANSWER: For correlation to ions, the potential would be 1r at long distances, hence would have an infinite number of vibrational levels. So a Birge-Sponer plot would approach zero asymptotically: Morse potential ∆ G(v + 1/2) v + 1/2 page 4 6. The ground state and a low-lying excited electronic state of the BeO molecule have the following properties Term symbol Electronic energy, T e /cm−1 ωe /cm−1 ωe xe /cm−1 re /10−8 cm 1 Σ+ 1Π 0 1487.3 11.8 1.33 9405.6 1144.2 8.4 1.46 (Note that the electronic energy T e is the energy from the minimum of one curve to the minimum of the other; this is not equal to the vibrational origin of the 0 − 0 band.) (a) Construct a Deslandres table of the vibrational band origins of the 1 Π − 1 Σ+ system, for v′′ = 0 through 3 and v′ = 0 through 5. Which of these vibrational bands would you expect to be the most intense, when the system is observed in absorption? Comment on the relative intensities that you would expect for the other bands in your table. ANSWER v′′ \v′ 0 1 2 3 0 9234.9 7771.3 6331.1 4914.6 1 10362.3 8898.6 7458.5 6042.0 2 11472.8 10009.2 8569.1 7152.6 3 12566.6 11103.0 9662.9 8246.4 4 13643.7 12180.1 10739.9 9323.4 5 14703.9 13240.2 11800.1 10383.6 The most intense bands in absorption would be those from v′′ = 0 because of the Boltzmann distribution. (b) In the rotational structure of the individual vibronic bands in this system, what branches would you expect to observe? In which branch would you expect to observe a band head? Calculate the transition in J that will give rise to a line at the band head, and the distance in cm−1 from the band head to the vibrational band origin. ANSWER P, Q, & R branches since this is a 1 Π − 1 Σ+ transition. reΠ > reΣ ⇒ BeΠ < BeΣ thus you would expect a band head in the R branch. Be = 8π2hcµr2 e B′e = 1.37 cm−1 −1 B′′ e = 1.65 cm ′ ′′ B′′ B −3B′ ′′ e −3Be Jhead = 2(B ′ −B′′ ) ≈ 2( B′ −B′′ ) � 4 e e ′ =5 Jhead The head occurs +8 cm−1 from the band origin. page 5 (c) What would you guess about the MO configurations corresponding to these two states? (HINT: Note that BeO is isoelectronic with C2 , so that the MO’s may be expected to be somewhat similar, except that the g-u property will be lost, and the orbitals will be distorted toward the higher nuclear charge of the O-atom.) Would you suspect the presence of any other excited electronic states below the 1 Π state? If so, what would its term symbol be? ANSWER 1 Σ+ : (1σ)2 (2σ)2 (3σ2 )(4σ)2 (1π)4 1 Π, 3 Π: (1σ)2 (2σ)2 (3σ2 )(4σ)2 (1π)3 (5σ) The 3 Π may be lie below the 1 Π state. page 6 7. The following bands are observed in the second positive system of nitrogen (units are reciprocal centimeters corrected to vacuum): 35,522 cm−1 35,453 33,852 33,751 33,583 32,207 32,076 31,878 31,643 30,590 30,438 30,212 29,940 cm−1 29,654 29,010 28,819 28,559 28,267 27,949 27,451 27,226 26,942 26,621 26,274 25,913 cm−1 25,669 25,354 25,003 24,627 24,414 24,137 23,800 23,414 23,016 ′′ ′ ′ Arrange these in a Deslandres table, and find values for ω′′ e , ωe xe , ωe , and ωe xe . Important Suggestion: Look at the pattern of bands first, before doing anything else. Do any natural groupings seem to suggest themselves? It may help to draw a “stick spectrum” of the band origins, to scale, in order to pick out these patterns. Remember that bands having the same Δv fall along diagonals on the Deslandres table.) Is there any suggestion of a cubic term in either state? If so, derive an expression for the third difference, � �3 Δ3 Gv+1/2 , including terms in ωe ye v + 12 in Gv+1/2 , and estimate ωe ye . ANSWER v′ \v′′ 0 1 2 3 4 for the lower state for the upper state 0 29654 31643 33583 35453 — 1 27949 29940 31878 33751 35522 2 26274 28267 30212 32076 33852 3 24627 26621 28559 30438 32207 4 23016 25003 26942 28819 30590 5 6 7 8 23414 25354 27226 29010 23800 25669 27451 25913 24414 Δ2 Gv+1/2 = −29.3 ⇒ ωe xe′′ = 14.6 cm−1 ′′ ω′′ e = ΔG v+1/2 + 2(v + 1)ωe xe −1 ′′ ωe = 1733 cm 2 Δ Gv+1/2 = −71.8 cm−1 ⇒ ωe xe′ = 35.9 cm−1 ω′e = 2074 cm−1 page 7 8. The first strong electronic band system of carbon monoxide (the ground-state vibrational frequency of which is observed at 2140 cm−1 in the infrared) appears in absorption at room temperature at about 1550 Å in the vacuum ultraviolet. The system shows a progression with a spacing of 1480 cm−1 . The vibronic bands show a single set of unperturbed P–, Q–, and R–branches degraded to the red. Analysis by combination differences −1 of these branches gives B′e = 1.61 cm−1 , B′′ e = 1.93 cm . In each band, the lines nearest the origin are P(2), Q(1), and R(0). (a) Deduce all you can about the two electronic states involved in the transition from these data and your general knowledge of the properties of carbon monoxide. ANSWER The ground state of CO is 1 Σ+ and since P, Q, and R branches are observed, the upper state has Λ � 0. � h = 2 8π cµB′e R′′ e = 1.13Å R′e �1/2 = 1.24Å (b) Sketch the lower portions of the potential curves in cm−1 for CO, roughly, to scale, from these data [find the harmonic force constant, and use the potential U(r) = 21 k(r − re )2 ]. Use the Franck-Condon principle to find the strongest vibronic bands in the spectrum. ANSWER 4 3 2 0 1 E v″ = 0 0 r 0 From the Franck-Condon principle it appears that the (1, 0), (2, 0), (3, 0), (4, 0), and (5, 0) bands will be the strongest. The strongest of them will probably be the (2, 0) band since it has the largest FC factor (from the graph). page 8 (c) Interpret the electronic terms of the two states in terms of the most likely MO configurations of each. ANSWER Ground State: (1σ)2 (2σ)2 (3σ)2 (4σ)2 (1π)4 (5σ)2 − 1 Σ+ Excited State: (1σ)2 (2σ)2 (3σ)2 (4σ)2 (1π)4 (5σ)(2π) − 1 Π, 3 Π One might think its a 1 Π − 1 Σ+ transition (actually these are the “forbidden” Cameron bands due to the a3 Π − X 1 Σ+ transition of CO). (d) The dipole moment of the ground state of CO is about 0.1 Debye (1 Debye = 10−18 esu-cm). Show how an optical Stark effect experiment can be used to find the dipole moment of the excited states; estimate the magnitudes of the splittings for the P(2), Q(1), and R(0) lines, for an applied field of 10,000 V/cm (1 statvolt = 300 ordinary volts), and an assumed excited-state moment of 1.0 Debye. ANSWER ????? 9. (a) What states arise from the σ2 σ∗, σ2 π, and σπ2 configurations? (There should be a total of 6 states.) ANSWER σ 2 σ ∗ − 2 Σ+ σ2 π − 2 Π σπ2 − 2 Σ+ , 2 Δ, 4 Σ− , 2 Σ− (b) Write the linear combinations of Slater determinants which correspond to the e–parity components of each of these states. (There should be a total of 9 component states, hence 9 linear combinations.) ANSWER J−1/2 ϕ e–parity states transform as σ��v ϕe = (−1) e � � ��2 + ��2 + �� 2 ∗ 2 + −1/2 σ σ Σ ϕe = 2 +� Σ = 2−1/2 [|0+ 0− 0+∗ | + |0+ 0− 0−∗ | � Σ σπ2 2 Σ+ 2 Σ− 4 Σ− 1/2 σ− π 2Δ 5/2 2Δ 3/2 4 Σ− 3/2 2Π 1/2 2Π 3/2 1/2 −1/2 −1 + − + 2 [|0 1 − 1 | − |0+ 1+ ϕe = − 1− | + |0− 1− − 1+ | − |0− 1+ − 1− |] −1/2 − + + + ϕe = 12 [2|0 1 − 1 | − |0 1− − 1+ | − |0+ 1+ − 1− | − 2|0+ 1− − 1− | +|0− 1+ − 1− | + |0− 1− − 1+ |] ϕe = 6−1/2 [|0− 1+ − 1+ | + |0+ 1− − 1+ | + |0+ 1+ − 1− | + |0+ 1− − 1− | +|0− 1+ − 1− | + |0− 1− − 1+ | ϕe = 2−1/2 [|0+ 1+ 1− | + |0− − 1+ − 1− |] ϕe = 2−1/2 [|0− 1+ 1− | + |0+ − 1+ − 1− |] ϕe = 2−1/ [|0+ 1+ − 1− | + |0− 1− − 1− |] ϕe = 2−1/2 [|0+ 0− 1− | + |0+ 0− − 1+ |] ϕe = 2−1/2 [|0+ 0− 1+ | + |0+ 0− − 1− |] page 9 (c) Which of these 9 substates can perturb each other? Make a table specifying the terms in the Hamiltonian � � HElect , HSO , HCoriolis which cause these perturbations. ANSWER: σ2 σ∗ 2 Σ+ σ2 π 2Π 2 Σ+ 2Π 1/2 HSO + Hcor HSO Hcor 2Π 3/2 2 Σ+ 3/2 cor H Hcor + Hcor Hel 1/2 3Π Hcor 2 Σ− 4 Σ− 1/2 4 Σ− 3/2 2Δ 3/2 2Δ 5/2 Hcor HSO +Hcor HSO Hcor Hcor 4 Σ− 3/2 2Δ 3/2 2Δ 5/2 HSO Hcor HSO +Hcor Hcor Hel HSO +Hcor Hcor HSO +Hcor 2 Σ− 2 Σ+ σπ2 4 Σ− 1/2 HSO +Hcor Hcor HSO HSO HSO HSO HSO HSO HSO +Hcor Hcor Hcor Hcor Hcor Hcor (d) Calculate all possible nonzero spin-orbit perturbation matrix elements between |ΛS Σi |ΩJM|ie states in terms of the one-electron parameters � � a1 ≡ σ∗ |ξℓ− |π a2 ≡ hσ|ξℓ− |πi a3 ≡ hπ|ξℓz |πi . (There will be at least 7 nonzero matrix elements.) ANSWER � � 1 2 Π |HSO |4 Σ− 3/2 3/2 = 2√a2 � � 3 2 Π |HSO |4 Σ− 1/2 1/2 = √6 a2 � � 2 Π |HSO |2 Σ− = − 6 a 1/2 2 � �12 √ 2 Π |HSO |2 Σ+ , σπ2 = 2 a 4 2 � 1/2 � 1 2 Π |HSO |2 Δ 3/2 3/2 = − 2 a2 2 σπ ����� 4 Σ− |HSO | 2 Σ+ = 1 1/2 � � 2Π � 2 1/2 √ − 6 3 a2 |HSO |2 Σ+ , σ2 σ⋆ 1 2 2 Σ+ , σπ2 |HSO |2 Σ− � = � = a1 √ 3 3 a2 page 10 (e) In the absence of substate mixing, between which of these 9 e–parity substates can there exist electric dipole allowed R or P branch transitions? ANSWER Allowed R or P branch transitions: 2 Σ+ , σ 2 σ ⋆ → 2 Π1/2,3/2 2 Σ+,− , σπ2 → 2 Π1/2,3/2 2Π 1/2 → 2 Δ3/2 2Π 3/2 → 2 Δ5/2 (f) Express the relative intensities � � of the following electric dipole transitions in terms of the one-electron matrix element µ⊥ ≡ σ|µ− |π : σ2 π 2 Π1/2 → σπ2 2 Σ+ σ2 π 2 Π3/2 → σπ2 2 Σ+ σ2 π 2 Π1/2 → σπ2 2 Σ− σ2 π 2 Π1/2 → σπ2 2 Δ3/2 σ2 π 2 Π3/2 → σπ2 2 Δ3/2 So there are some semi-empirical relationships among isoconfigurational transition intensities! ANSWER � � I ∝ |h1|µ|2i|2 where µ = i µi = i µ xi + µyi + µzi In this case, all transitions have ΔΛ = ±1 ⇒ hµzi i = 0 � so µ = i 12 (µ+i + µ−i ) + 2i (µ+i − µ−i ) (the µ±i ’s are one electron operators). The relative intensities are derived using the linear combination of Slaters from part (b) of this problem. σ2 π 2 Π1/2 → σπ2 2 Σ+ : I ∝ µ2⊥ 8 σ2 π 2 Π3/2 → σπ2 2 Σ+ : I ∝ µ2⊥ 8 σ2 π 2 Π1/2 → σπ2 2 Σ− : I ∝ 32 µ2⊥ σ2 π 2 Π1/2 → σπ2 2 Δ3/2 : I ∝ µ2⊥ σ2 π 2 Π3/2 → σπ2 2 Δ3/2 : I ∝ 0 page 11 10. (a) Construct the Hamiltonian matrix for a 5 Δ state by evaluating the matrix elements of HROT = B(J−L−S)2 and HSO = AL · S in the Hund’s case ‘a’ e/ f (parity) basis. ANSWER 5 Δ State: �� � � �� �� � �5 Δ4 e = √1 ��5 Δ4 ± ��5 Δ−4 2 f �� � � �� �� � �5 Δ3 e = √1 ��5 Δ3 ± ��5 Δ−3 2 f �� � � �� �� � �5 Δ2 e = √1 ��5 Δ2 ± ��5 Δ−2 2 f �� � � �� �� � �5 Δ1 e = √1 ��5 Δ1 ± ��5 Δ−1 2 f �� � � �� �� � �5 Δ0 e = √1 ��5 Δ0 , Λ = 2 ± ��5 Δ0 , Λ = −2 2 f 5 Δ4 5 Δ3 5 Δ4 T e + G(v) + Bv (J(J + 1) − 14) + 4Av −2Bv [(J + 4)(J − 3)]1/2 5 Δ3 −2Bv [(J + 4)(J − 3)]1/2 Δ2 5 Δ1 0 0 T e + G(v) + Bv (J(J + 1) − 4) + 2Av −Bv [6(J − 2)(J + 3)]1/2 0 5 0 0 5 Δ0 5 5 Δ2 5 Δ1 Δ0 0 0 0 −Bv [6(J − 2)(J + 3)]1/2 0 0 T e +G(v)+ Bv (J(J +1)+2) −Bv [6(J + 2)(J − 1)]1/2 −Bv [6(J + 2)(J − 1)]1/2 T e + G(v) + Bv (J(J + 1) + 4) − 2Av -2Bv [J(J + 1)]1/2 0 −2Bv [J(J + 1)]1/2 0 T e + G(v) + Bv [J(J + 1) + 2] − 4Av page 12 (b) Construct the matrix of second-order Van Vleck Transformation corrections to HROT +HSO which results from the r–dependence of the “constants” Bv and Av . Define centrifugal distortion parameters Dv ≡ − A Dv = 2 � hv| B(r)|v′ i hv′ | B(r)|vi Ev − Ev′ v′ �v � hv|A(r)|v′ i hv′ |B(r)|vi v′ �v A◦v = Ev − Ev′ � hv|A(r)|v′ i hv′ |A(r)|vi v′ �v Ev − Ev′ and express the matrix elements in terms of these three parameters. ANSWER Second Order Corrections Diagonal Elements: 5Δ 4 5Δ 3 5Δ 2 5Δ 1 5Δ 0 → −[(J(J + 1) − 14)2 + 4(J + 4)(J − 3)]Dv + 16A◦v + 8ADv [J(J + 1) − 14] → −[(J(J + 1) − 4)2 + 4(J + 4)(J − 3) + 6(J − 2)(J + 3)]Dv + 4A◦v + 4ADv [J(J + 1) − 4] → −[(J(J + 1) + 2)2 + 6(J − 2)(J + 3) + 6(J − 1)(J + 2)]Dv → −[(J(J + 1) + 4)2 + 6(J − 1)(J + 2) + 4J(J + 1)]Dv + 4A◦v − 4ADv [J(J + 1) + 4] → −[(J(J + 1) + 2)2 + 4J(J + 1)]Dv + 16A◦v − 8ADv [J(J + 1) + 2] Off Diagonal Elements x = J(J + 1): � � 5 Δ , 5 Δ → +2 {(x − 4) + (x − 14)}[(J + 4)(J − 3)]1/2 D − 12A [(J + 4)(J − 3)]1/2 4 3 Dv � v � 5 Δ , 5 Δ → + {(x − 4) + (x + 2)}[6(J − 2)(J + 3)]1/2 D − 2A [6(J − 2)(J + 3)]1/2 3 2 v Dv � � 5 Δ , 5 Δ → +2 {(x + 4) + (x + 2)}[J(J + 1)]1/2 D + 12A [J(J + 1)]1/2 1 0 v Dv � � 5 Δ , 5 Δ → + {(x + 2) + (x + 4)}[6(J − 1)(J + 2)]1/2 D + 2A [6(J − 1)(J + 2)]1/2 2 1 v Dv