5.61 F17 Solutions to Problem Set 7

5.61 Fall 2017 Problem Set #7 Solutions 1. Hydrogenic Systems. In each of the following cases, state which of the two quantities is larger. Justify your answers. You do not need to do any integrals here. Some equations on page 333 of McQuarrie might be helpful. A. The average value of r for a 2s electron versus a 2p electron. Solution: The average value of hri is larger for a 2s electron than a 2p electron. B. The average value of 1/r for a 2s electron versus a 2p electron. Solution: 1 r 1 r is 4aZ0 for both the 2s and 2p cases. We can confirm this since we have derived an expression for from the virial theorem in Question 4, and have proved that it is independent of the ` value. C. The average value of r for a 2s electron in He+ versus a 1s electron in H. Solution: hri for 2s He+ is 3a0 and for 1s Hydrogen it’s 32 a0 . It makes sense that hri of the electron in the He+ 2s would be double the distance (on average) than that of the 1s electron. D. The average value of r for a 3d electron in Fe25+ versus a 1s electron in C5+ . Solution: hri for the 3d electron of Fe25+ is ≈ 0.4a0 , whereas for the 1s electron of C5+ it is 0.25a0 . It is expected that, although the nucleus of Fe25+ has a larger positive charge and would therefore pull its electron in the 3d shell closer, it’s still not as close to the nucleus as the 1s electron of C5+ . 5.61 Fall 2017 Problem Set #7 Solutions Page 1 E. The number of radial nodes in an 8g orbital versus the number of angular nodes in an 8g orbital. Solution: The number of radial nodes is given as n − ` + 1 while the number of angular nodes is given by `. Therefore 8g has 3 radial nodes and 4 angular nodes. F. The spacing between the radial nodes for a 14s orbital versus the spacing between the radial nodes for a 16d orbital. Solution: Both have the same number of nodes, but in general since the 16d electron will have a larger hri value, we can expect that the node spacing should be larger for this case. NOTE: Atomic properties such as the relative n` orbital energies, En` , and radii, hrn in` , atomic ionization energy and electron affinity, and the dependence of these properties on electron configuration are explained by the systematic shielding of the +Z nuclear charge experienced by an electron in the n` atomic orbital by all of the electrons in the other occupied n0 `0 atomic orbitals. One computes the effective nuclear charge experienced by an electron in the n` orbital due to eff (configuration). This shielding by all of the other electrons in the electronic configuration, Zn` effective nuclear charge is inserted in the standard hydrogenic formulas for orbital energy, En` , orbital radius, hrin` , and other integer powers of r, hrn in` . The Periodic Table is explained by shielding! A table of these inter-orbital shielding effects, sometimes known as Burns’ rules, is given in a paper by Gerald Burns, “Atomic Shielding Parameters,” J. Chem. Phys. 41, 1521-1522 (1964), https://doi.org/10.1063/1.1726113. 2. Spin 3/2 Periodic Table Consider a universe where the electron has spin 3/2 instead of spin 1/2. A. Draw the periodic table (up to Hafnium) in this alternate universe. Solution: If the electron were to have s = 3/2, then there would be four possible ms values, ms = +3/2, +1/2, −1/2, −3/2, instead of the usual two. This means that each atomic orbital can be occupied by four electrons and still satisfy the Pauli exclusion principle (anti-symmetrization). s–orbitals can hold 4, p–orbitals can hold 12, etc. We build up the periodic table in the same way we are used to, adding electrons to shells as we go. The modified table (up to Hf) is shown in Figure 1. 5.61 Fall 2017 Problem Set #7 Solutions Page 2 orbital can be occupied by four electrons and still satisfy the Pauli exclusion principle / antisymmetrization. s-orbitals can hold 4, p-orbitals can hold 12, etc. We build up the periodic table in the same way we are used to, adding electrons to shells as we go. The modified table (up to Hf) is shown in Figure 1. That the 3d block should come after the 4s block isn’t obvious (and may even not I II III IV s block (III) (IV) (V) (VI) (VII) (VIII) (IX) (X) (XI) (XII) (XIII) (XIV) (XV) (XVI) (I) (II) (III) (IV) V VI VII VIII IX d block X XI XII XIII XIV XV XVI p block 1 H He Li Be 2 B C N O 3 Sc Ti V Cr F Ne Na Mg Al Si 4 Rb Sr Y Zr Nb Mo Te Ru Rh Pd Ag Cd In Sn Sb Te P S Cl Ar K Ca Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr I Xe Cs Ba La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Hf Figure 1: One reasonable ordering for a spin-3/2 periodic table. That the 3dbeblock should the 4sInblock isn’ttable, obvious may not even be true for the case for acome spin after 3/2 table). the real this (and occurs because e-e repulsion is a spin for 4s electrons (which arerather much further the core electrons), 3/2 table). significantly In the realreduced table, this lower energy for 4s than 3dfrom occurs because e-e repulsion thanreduced for 3d electrons. Even though the one-electron orbital from energythe forcore 4s levels is higherthan for is significantly for 4s electrons (which are much further electrons), than 3d levels, this reduction in e-e repulsion in many-electron atoms causes the reorder3d electrons. Even though the one-electron orbital energy for 4s levels is higher than 3d levels, ing. We assume the same trend will exist here. this reduction in e-e repulsion in many-electron atoms causes the reordering. We assume the same trend will exist here. There are 3 radial nodes for 4s and zero radial nodes for 3d. b. The analog of the ”noble gases” would be those with complete n-shells, namely Be, Ca, Kr, and Hf. With regards to the ”alkali earth metals,” we could argue that any element with an incomplete s-shell would count, which includes groups I-III. You might argue B. Which elements would be “noble gases”? Which would be alkaline earth elements? Which that only (ns)1 configurations should count, in which case you consider only group I elements would be inInthe periodthere as carbon? elements. thesame real table, is no ambiguity because the only incomplete s-shells have 1 electron anyway. Carbon is still in the second period, along with elements with Z = 5 − 20 (B through Ca). Solution: The analogc.of the “noble gases” would be those with complete n–shells, namely Be, Ca, Kr, and Hf. Since each MO can also be filled with 4 electrons now, the bond order of He2 would be With regard to the “alkaline earth metals,” we could argue that any element with an incomplete 4(!) O2 , on the other hand, now has a bond order of 0 (the two σ and two σ ∗1 orbitals configurations s–shell would count, which includes groups I-III. You might argue that only (ns) from the O 1s and 2s orbitals are all completely filled with 16 electrons). You should be should count, in which caseimportant you consider only group thechemical real table, there is no realizing just how the electron’s spin Iis elements. in determingInthe structure ambiguity and because the only incompletely filled s–shells have 1 electron anyway. For S = 3/2, patterns of the universe. carbon is in the second period, along with elements with Z = 5 − 20 (B through Ca). 1 2 Page C. What would the bond order of He2 be in this universe? What about O2 ? Solution: Since each MO can also be filled with 4 electrons now, the bond order of He2 would be 4(!). O2 , on the other hand, now has a bond order of 0 (the two σ and two σ ∗ orbitals from the O 1s and 2s orbitals are each completely filled with 16 electrons). You should be realizing just how important the electron’s spin is in determining the chemical structure and patterns of the universe. D. What would the equivalent of the octet rule be in this alternate universe? [NOTE: you could spend a lot of time answering this last question. It is intended to be fun. When it stops being fun, your answer is long enough.] 5.61 Fall 2017 Problem Set #7 Solutions Page 3 Solution: The octet rule is based on having complete n-shells. It should be clear that the simplest analog in the spin 3/2 table is the “hexadectet” rule, i.e. n = 2 and 3 orbitals prefer to have 16 electrons in their outer shells (4 in s and 12 in p). 5.61 Fall 2017 Problem Set #7 Solutions Page 4 3. Rydberg States of a Many-Electron Atom. The subject of this problem is potassium, which has a closed-shell ion-core: K(1s2 2s2 2p6 3s2 3p6 )n∗ ` The ionization energy (in cm−1 units) from the “4s” electronic ground state is 35009.78 cm−1 . The Rydberg constant for K is < = 109737.32cm−1 . A. Why is it reasonable to ignore the anti-symmetrization requirement for Rydberg states of this 19 electron atom? Solution: We approximate the K atom as a K+ core and an electron. As the K+ core is complete (all filled shells), anti-symmetrization between the core and the single valence electron has a negligible effect, so we ignore it. B. Consider three consecutive members of the ns, np, and nd Rydberg series: n 8 9 10 s-series 31764.95 cm−1 32648.17 cm−1 33214.39 cm−1 n 40 41 42 p-series 34934.97 cm−1 34938.72 cm−1 34942.20 cm−1 n 9 10 11 d-series 33572.11 cm−1 33851.76 cm−1 34056.90 cm−1 Compute n∗ -values for all 9 of the tabulated energy levels. Do the n∗ levels increase in steps of ∼1.00? Solution: We are provided with excitation energies from the ground state. To convert to energies under ionization, we use En∗ = Eionization − Estate . We combine this with En∗ = − Thus ∗ r n = Plugging in the tabulated values, ns 8 9 10 5.61 Fall 2017 n* 5.815 6.817 7.818 δs 2.185 2.183 2.182 np 40 41 42 < . n∗2 < . Estate − Eionization n* 38.300 39.297 40.297 δp 1.700 1.703 1.703 ns 9 10 11 Problem Set #7 Solutions n* 8.737 9.735 10.731 δs 0.263 0.265 0.269 Page 5 C. The quantum defects, δ` , are defined as n − n∗ . Compute the approximately n-independent quantum defects for the s, p, and d series of K. Solution: See above table. δ` = n − n∗ . We see that the defect for ns is about 2.2, for np is about 1.7, and for nd is about 0.3. D. Suggest a reason why δs  δp > δd . Solution: s orbitals penetrate further into the inner core than orbitals with higher angular momentum. From this we expect δs  δp  δd ≈ 0. E. The n∗ values you have determined from real spectroscopic data may be considered “experimentally measured.” But the tabulated integer n quantum numbers are not measured. They are inferred from some sort of physical argument. Can you suggest what this argument is? [HINT: the lowest s, p, and d states of K are called 4s, 4p, and 3d.] Solution: The quantum defects above are correct, even down to the 4s, 4p, and 4d states, to within 0.1. We don’t record n∗ because, to an excellent approximation, it increases in steps of 1. 4. Two Electron Wavefunctions: Spin. For two electrons, the total z component of the spin angular momentum for the system is Sbz, total = Sbz1 + Sbz2 while the total spin operator is given by 2  2  2  2 2 b2 b2 bx + Sbx by + Sby bz + Sbz Sbtotal = Sbx, + S + S + S + S + S 1 2 1 2 1 2 total y, total z, total 5.61 Fall 2017 Problem Set #7 Solutions Page 6 A. Show that both 1 1sα(1) 2sβ (1) 1 ψαβ = √ ≡ √ (2sα(1)2sβ(2) − 2sβ(1)1sα(2)) 2 1sα(2) 2sβ (2) 2 and 1 1sβ (1) 2sα(1) 1 ψβα = √ ≡ √ (1sβ(1)2sα(2) − 2sα(1)1sβ(2)) 1 sβ (2) 2 sα (2) 2 2 are antisymmetric. Show also that ψαβ and ψβα are eigenfunctions of Sbz, total . What are the eigenvalues in each case? Solution: 1 Ψαβ = √ [1sα(1)2sβ(2) − 2sβ(1)1sα(2)] 2 switching 1 and 2 by the P̂12 permutation operator 1 P̂12 Ψαβ = √ [1sα(2)2sβ(1) − 2sβ(2)1sα(1)] 2 1 = − √ [2sβ(2)1sα(1) − 1sβ(1)1sα(2)] 2 = −Ψαβ . This indicates that Ψαβ is anti-symmetric: 1 Ψβα = √ [1sβ(1)2sα(2) − 2sα(1)1sβ(2)] 2 switching 1 and 2 by P̂12 1 P̂12 Ψβα = √ [1sβ(2)2sα(1) − 2sα(2)1sβ(1)] 2 1 = − √ [2sα(2)1sβ(1) − 1sβ(2)2sα(1)] 2 = −Ψβα . This indicates that Ψβα is anti-symmetric. Now the eigenvalue for Ψαβ : Sbz, total = Sbz 1 + Sbz 2   Sbz, total Ψαβ = Sbz1 + Sbz2 Ψαβ i 1 h = √ Sbz1 {1sα(1)2sβ(2) − 2sβ(1)1sα(2)} + Sbz2 {1sα(1)2sβ(2) − 2sβ(1)1sα(2)} 2     1 } } =√ 1sα(1)2sβ(2) − − 2sβ(1)1sα(2) 2 2 2      } } + − 1sα(1)2sβ(2) − 2sβ(1)1sα(2) 2 2 =0 5.61 Fall 2017 Problem Set #7 Solutions Page 7 Now the eigenvalue for Ψβα :   Sbz, total Ψβα = Sbz1 + Sbz2 Ψβα i 1 h = √ Sbz1 {1sβ(1)2sα(2) − 2sα(1)1sβ(2)} + Sbz2 {1sβ(1)2sα(2) − 2sα(1)1sβ(2)} 2     } } 1 2sα(1)1sβ(2) − 1sβ(1)2sα(2) − =√ 2 2 2      } } + {1sβ(1)2sα(2) − − 2sα(1)1sβ(2) 2 2 =0 This indicates that both Ψαβ and Ψβα are eigenfunctions of Sbz, total , both with eigenvalues of 0. B. Show that, while ψαβ and ψβα cannot be written in the form ψspace ψspin , the combinations of ψαβ ± ψβα can both be cast in the form ψspace ψspin . Solution: 1 Ψαβ = √ [1sα(1)2sβ(2) − 2sβ(1)1sα(2)] 2 1 = √ [(1s(1)2s(2)α(1)β(2)) − (2s(1)1s(2)β(1)α(2))] 2 = 6 Ψspin Ψspace 1 Ψβα = √ [1sβ(1)2sα(2) − 2sα(1)1sβ(2)] 2 1 = √ [(1s(1)2s(2)β(1)α(2)) − (2s(1)1s(2)α(1)β(2))] 2 = 6 Ψspin Ψspace . However 1 Ψαβ + Ψβα = √ [1sα(1)2sβ(2) − 2sβ(1)1sα(2) + 1sβ(1)2sα(2) − 2sα(1)1sβ(2)] 2 1 = √ [{1s(1)2s(2)α(1)β(2)} − {2s(1)1s(2)β(1)α(2)} 2 + {1s(1)2s(2)β(1)α(2)} − {2s(1)1s(2)α(1)β(2)}] 1 = √ {1s(1)2s(2) − 2s(1)1s(2)}{α(1)β(2) + β(1)α(2)} 2 1 = √ Ψspace · Ψspin . 2 5.61 Fall 2017 Problem Set #7 Solutions Page 8 Similarly, 1 Ψαβ − Ψβα = √ [1sα(1)2sβ(2) − 2sβ(1)1sα(2) − 1sβ(1)2sα(2) + 2sα(1)1sβ(2)] 2 1 = √ [{1s(1)2s(2)α(1)β(2)} − {2s(1)1s(2)β(1)α(2)} 2 − {1s(1)2s(2)β(1)α(2)} + {2s(1)1s(2)α(1)β(2)}] 1 = √ {1s(1)2s(2) + 2s(1)1s(2)}{α(1)β(2) − β(1)α(2)} 2 1 = √ Ψspace · Ψspin . 2 C. Verify that the total spin operator can be re-written in terms of raising and lowering operators:   Sb2 = Sb12 + Sb22 + 2Sb1z Sb2z + Sb1+ Sb2− + Sb1− Sb2+ . total Solution: 2 Sbtotal = (Sb1 + Sb2 )(Sb1 + Sb2 ) = Sb12 + Sb1 Sb2 + Sb2 Sb1 + Sb22 = Sb 1 + Sb 2 + 2(Sb1 Sb2 ) = 1 Sb12 + 2 Sb22 + 2(Sbx1 Sbx2 + Sby1 Sby2 + Sbz1 Sbz2 ) ! !# " Sb+1 + Sb−1 Sb+2 + Sb−2 Sb+1 − Sb−1 Sb+2 − Sb−2 2 2 b b b b = S1 + S2 + 2Sz1 Sz2 + 2 + 2 2 2i 2i h 1 b b S+1 S+2 + Sb+1 Sb−2 + Sb−1 Sb+2 + Sb−1 Sb−2 − Sb+1 Sb+2 + Sb+1 Sb−2 = Sb12 + Sb22 + 2Sbz1 Sbz2 + i 2 +Sb−1 Sb+2 − Sb−1 Sb−2 = Sb12 + Sb22 + 2Sbz1 Sbz2 + (Sb+1 Sb−2 + Sb−1 Sb+2 ) 5.61 Fall 2017 Problem Set #7 Solutions Page 9 2 . That is to say, show that neither D. Show that neither ψαβ nor ψβα is an eigenfunction of Sbtotal of these wavefunctions is a total spin eigenstate. Solution: h i 2 Sbtotal Ψαβ = Sb12 + Sb22 + 2Sbz1 Sbz2 + Sb+1 Sb−2 + Sb−1 Sb+2 Ψαβ 3 3 Sb12 Ψαβ = } 2 Ψαβ Sb22 Ψαβ = } 2 Ψαβ 4 4      2 } } 2sβ(1)1sα(2) 2Sbz1 Sbz2 Ψαβ = √ Sbz1 − 1sα(1)2sβ(2) − 2 2 2      } } } = −√ 2sβ(1)1sα(2) 1sα(1)2sβ(2) + − 2 2 2 }2 = − Ψαβ 2 " # r 1 3 1 Sb+1 Sb−2 Ψαβ = √ Sb+1 0 − } + 2sβ(1)1sβ(2) 4 4 2 r 3 1 } = −√ · } + 2sα(1)1sβ(2) 4 4 2 }2 = − √ 2sα(1)1sβ(2) 2 " r # 1 3 1 Sb−1 Sb+2 Ψαβ = √ Sb−1 } + 1sα(1)2sα(2) + 0 4 4 2 } = √ · }1sβ(1)2sα(2) 2 2 } = √ 1sβ(1)2sα(2) 2 Therefore, summing over all the terms gives }2 2 Sbtotal Ψαβ = }2 Ψαβ + √ [1sβ(1)2sα(2) − 2sα(1)1sβ(2)] 2 2 = } [Ψαβ + Ψβα ] Repeating similar steps for Ψβα gives 2 Sbtotal Ψβα = }2 [Ψαβ + Ψβα ]. 2 . Therefore, neither Ψαβ nor Ψβα is an eigenfunction of Sbtotal 5.61 Fall 2017 Problem Set #7 Solutions Page 10 2 . E. Finally, show that the combinations ψαβ ± ψβα are eigenfunctions of both Sbz, total and Sbtotal Solution: 1 Sbz, total √ (Ψαβ ± Ψβα ) = 2 1 2 √ (Ψαβ + Ψβα ) = Sbtotal 2 1 2 √ (Ψαβ Sbtotal 2 1 √ (0 ± 0) = 0 2 2 } √ (Ψαβ + Ψβα + Ψαβ + Ψβα ) 2   1 = 2}2 √ (Ψαβ + Ψβα ) 2 }2 − Ψβα ) = √ (Ψαβ + Ψβα − Ψαβ − Ψβα ) 2 =0 F. Note that ψαβ , ψβα , and ψαβ ± ψβα are all degenerate states within the non-interacting electron picture. Comment on why your work above shows that ψαβ ± ψβα are more realistic eigenstates of the Hamiltonian. Solution: We note that Sb 2 and Sbz correspond to the magnitude squared of the spin and its projection onto the z–axis, and therefore it would be more realistic if our wavefunctions are eigenfunctions of these operators. Since √12 (Ψαβ + Ψβα ) and √12 (Ψαβ − Ψβα ) are in fact eigenfunctions of our operators, they are more realistic than either Ψαβ or Ψβα (which are not eigenfunctions of Sb 2 and Sbz . 5. Independent Particle Model. The following concern the independent particle mode. You may find the following set of Coulomb and exchange integrals useful (energies in eV): J1s1s = 17.0 Z J1s2p = 6.6 Z J2pi ,2pi = 3.9 Z 5.61 Fall 2017 J1s2s = 4.8 Z K1s2p = 0.5 Z J2pi ,2pk = 3.5 Z K1s2s = 0.9 Z J2s2p = 4.4 Z K2pi ,2pk = 0.2 Z i = 6 k Problem Set #7 Solutions J2s2s = 3.5 Z K2s2p = 0.8 Z Page 11 A. Using the independent particle model discussed in class, what is the energy difference between the 1s2 px2 configuration and the 1s2 2s2 configuration? How do you justify your result? Solution: We are asked to calculate the enregy difference between a 1s2 2p2x and a 1s2 2s2 configuration. Let’s compute the energy for each using the independent particle model E[1s2 2p2x ] = X i Ei + i>j X i,j e ij Jeij − K = 2E1s + 2E2p + Je1sα,1sβ + Je1sα,2px α + Je1sα,2p e 1sα,1sβ −K + Je1sβ,2px α + Je1sβ,2px β + Je2px α,2px β e 2p α,2p β e 1sβ,2p β − K e 1sβ,2p α − K e 1sα,2px α − K e 1sα,2p β − K −K x x x x x xβ = 2E1s + 2E2p + J1s,1s + 4J1s,2p + J2pi ,2pi − 2K1s,2p E[1s2 2s2 ] = 2E1s + 2E2s + J1s,1s + 4J1s,2s + J2s,2s − 2K1s,2s (5.1) (5.2) ⇒ ∆E = 4(J1s,2p − J1s,2s ) + (J2pi ,2pi − J2s,2s ) − 2(K1s,2p − K1s,2s ) = Z[4(6.6 − 4.8) − (3.9 − 3.5) − 2(0.5 − 0.9)] = +7.6Z eV (5.3) B. What is the energy difference between the ground state of Lithium and the spin polarized 1s↑ 2s↑ 2pz ↑ state? Is this energy about the size you expected? Solution: E[1s2 2s1 ] = 2E1s + E2s + J1s1s + 2J1s2s − K1s2s   Z2 2 1 =− + (27.2 eV ) + Z(17.0 + 2 · 4.8 − 0.9) 2 12 22 = −32 (9/8)(27.2) + 3(25.7) = −198.3 eV (5.4) = −32 (3/4)(27.2) + 3(13.6) = −142.8 eV (5.5) E[1s↑ 2s↑ 2p↑x ] = E1s + E2s + E2p + J1s2s + J1s2p + J2s2p − K1s2s − K1s2p − K2s2p   Z2 1 2 =− + (27.2 eV ) + Z(4.8 + 6.6 + 4.4 − 0.9 − 0.5 − 0.8) 2 12 22 ⇒ ∆E = −142.8 + 198.3 = 55.5 eV (5.6) Consulting the NIST Atomic Spectra Database [physics.nist.gov/cgi-bin/ASD], we find that the high-spin 1s2s2p state lies 57.469 eV above the ground state. Our IPM prediction is remarkably close given we’ve only included first order e-e repulsion corrections! 5.61 Fall 2017 Problem Set #7 Solutions Page 12 C. Use the IPM To compute the ionization potential of B (Boron): IP = E(B+ ) − E(B). Compare your boron result with the ionization potential of Lithium within the IPM. Does this agree with periodic trends? Solution: No solution given. D. Finally, compare the ionization potentials computed in part C. with the experimental results [IP(Li) = 5.4eV, IP(B) = 8.3eV] and the answer you would have gotten if you had assumed that the electrons do not interact. Solution: First, we calculate the ionization energy of Be: E[Be → 1s2 2s2 ] = 2E1s + 2E2s + J1s1s + 4J1s2s + J2s2s − 2K1s2s   Z2 2 2 =− + (27.2 eV ) + Z(17.0 + 4 × 4.8 + 3.5 − 2 × 0.9) 2 12 22 + 2 = −42 (5/4)(27.2) + 4(37.9) = −392.4 eV (5.7) = −42 (9/8)(27.2) + 4(25.7) = −386.8 eV (5.8) E[Be → 1s 2s] = 2E1s + E2s + J1s1s + 2J1s2s − K1s2s   2 Z2 2 + (27.2 eV ) + Z(17.0 + 2 × 4.8 − 0.9) =− 2 12 22 ⇒ IE = ∆E = −386.8 + 392.4 = 5.6 eV (5.9) Now, we do the same for Li. Note the ground state for Li is the same as Be+ with Z = 4 → Z = 3. E[Li → 1s2 2s] = −32 (9/8)(27.2) + 3(25.7) = −198.3 eV E[Li+ → 1s2 ] = 2E1s + J1s1s   Z2 2 =− (27.2 eV ) + Z(17.0) = −193.8 eV 2 12 ⇒ IE = ∆E = −193.8 + 198.3 = 4.5 eV (5.10) (5.11) (5.12) The experimental IEs for Be and Li are 9.32 and 5.39 eV, respectively (NIST ASD). Though the IPM values are significantly off in magnitude, they reproduce the correct periodic trend: elements in the same period have increasing IEs from left to right. In the absence of e-e interactions, the IEs would be the energies of the vacated one-electron orbitals. In each case this is a 2s orbital. The energy for Be would be 42 /2(22 ) × 27.2 = 54.4 eV and for Li, 5.61 Fall 2017 Problem Set #7 Solutions Page 13 32 /2(22 ) × 27.2 = 30.6 eV. These values are extremely high! Even though the IPM predictions are not quantitatively correct, we see that they do account for the majority of the effects caused by e-e interactions. E. Within the IPM, what is the energy difference between a closed shell 1s2 2s2 2p2x configuration and a high spin 1s2 2s↑ 2p↑x 2p↑y 2p↑z configuration for carbon? Does this agree with your intuition? Solution: Our final problem is comparing closed-shell and high-spin configurations for carbon, Z = 6. E[1s2 2s2 2p2x ] = 2E1s + 2E2s + 2E2p + J1s1s + 4J1s2s + 4J1s2p + J2s2s + 4J2s2p + J2px 2px − 2K1s2s − 2K1s2p − 2K2s2p   Z2 2 4 =− + (27.2 eV ) + Z(83.2) 2 12 22 = 969.6 eV (5.13) E[1s2 2s↑ 2p↑x 2p↑y 2p↑z ] = 2E1s + 1E2s + 3E2p + J1s1s + 2J1s2s + 6J1s2p + 3J2s2p + 3J2pi 2pk + K1s2s − 3K1s2p − 3K2s2p − 3K2pi 2pk   Z2 2 4 =− + (27.2 eV ) + Z(84.5) 2 12 22 = −961.8 eV ⇒ E[high spin] − E[closed shell] = ∆E = 7.8 eV. (5.14) (5.15) This agrees with our chemical intuition, where we expect closed shell configurations to be the most stable. 5.61 Fall 2017 Problem Set #7 Solutions Page 14