Problem Set 11 Solutions

Transcription

1 Chemistry 62 Dr. Jean M. Standard Problem Set 11 Solutions 1. Use the Pauli and Aufbau principles, along with Hund's Rule of maximum multiplicity, to draw orbital energy diagrams for the ground states of the Si and P atoms. Classify the multiplicity of the ground states of the Si and P atoms as singlet, doublet, triplet, or quartet. Using the Pauli and Aufbau Principles, the electron configuration of Si is 1s 2 2s 2 2 p 6 s 2 p 2. The electron configuration of P is 1s 2 2s 2 2p 6 s 2 p. Orbital energy diagrams for Si and P are shown below. Note that using Hund's rule of maximum multiplicity, the electrons in the p orbitals are placed in different orbitals and have the same electron spin. Atomic Energy Levels of Si Atomic Energy Levels of P p p s s E 2p E 2p 2s 2s 1s 1s For Si, placing two electrons in the p set of orbitals leads to two unpaired electrons. Thus, the total spin of the two unpaired electrons is S=1, and therefore the multiplicity is 2S+1 = =. Therefore, the multiplicity of the ground state of Si is a triplet. For P, placing three electrons in the p set of orbitals leads to three unpaired electrons. Thus, the total spin of the three unpaired electrons is S=/2, and therefore the multiplicity is 2S+1 = 2 (/2)+1 = 4. Therefore, the multiplicity of the ground state of P is a quartet.

2 2 2. A researcher employs the following function to represent the spatial part of the ground state wavefunction of the Li atom, ψ( 1,2,) = χ 1s ( 1) χ 1s ( 2) χ 2s ( ). Here, χ 1s or χ 2s is shorthand notation used to refer to the spatial form of the atomic orbital; for example, χ 1s ( 1) = $ & % 1 πa o ' ) ( 1/ 2 e r 1 / a o. Does the form given above for the Li atom correspond to an acceptable spatial wavefunction? Explain. No, it is not an acceptable function. The electron configuration of the ground state of Li is 1s 2 2s 1, so the atomic orbitals used to construct the function are correct. However, for example, the function specifically places electron in the 2s orbital; thus, the function distinguishes between the electrons. An acceptable function would allow for electron1, 2, or being placed in the 2s orbital. Shown below is an example of an acceptable spatial wavefunction that does not distinguish between the electrons: ψ( 1,2,) = 1 [ χ 1s ( 2) χ 1s ( ) χ 2s ( 1) + χ 1s ( 1) χ 1s ( ) χ 2s ( 2) + χ 1s ( 1) χ 1s ( 2) χ 2s ( ) ]. This function places each of the three electrons in the 2s orbital, so it does not distinguish. This example spatial wavefunction happens to be symmetric with respect to interchange of electrons. When constructing the full wavefunction for Li, including spatial and spin components, this symmetric spatial wavefunction would have to be combined with an antisymmetric spin wavefunction so that the overall wavefunction was antisymmetric and satisfied the Pauli Principle.. Determine the possible atomic term symbols arising from a d 1 electron configuration. Since the d 1 electron configuration only involves one outer electron, the total orbital angular momentum quantum number L and total spin angular momentum quantum number S are identical to the orbital and spin angular momentum quantum numbers of the outer electron. Thus, S = s 1 = 1 2 and L = l 1 = 2. The multiplicity 2S+1 is therefore 2 (a doublet) and the term corresponds to a 2 D term. Then, all that is needed is to determine the total angular momentum quantum number J. The total angular momentum quantum number J ranges from L S to L + S. For the 2 D term, L=2 and S = 1 2 leads to J = 2, 5 2. Therefore, the possible term symbols for the d 1 electron configuration are 2 D / 2 and 2 D 5 / 2.

3 4. The following terms arise from a d 8 electron configuration: 1 S, 1 D, 1 G, P, F. Use the first two of Hund's rules to properly order these terms according to energy. Then, determine the term symbols arising from each term, and use the third Hund's rule to complete the energy ordering. Hund's Rules #1 and 2 allow us to order the terms according to increasing energy: F < P < 1 G < 1 D < 1 S. Hund's Rule #1 tells us that both the triplet terms ( P and F ) will be lower in energy than the singlet terms ( 1 S, 1 D, and 1 G ). Hund's Rule #2 tells us that among the triplets, the one with the highest L value will be lowest energy ( F < P ), and similarly among the singlets, the one with the highest L will have the lowest energy ( 1 G < 1 D < 1 S ). Next, the complete term symbols for each term may be determined: For F : L = and S = 1 leads to J = 2,, 4 and term symbols F 2, F, F 4. For P : L = 1 and S = 1 leads to J = 0, 1, 2 and term symbols P 0, P 1, P 2. For 1 G : L = 4 and S = 0 leads to J = 4 and the term symbol 1 G 4. For 1 D : L = 2 and S = 0 leads to J = 2 and the term symbol 1 D 2. For 1 S : L = 0 and S = 0 leads to J = 0 and the term symbol 1 S 0. Finally, Hund's Rule # may be used to order the term symbols with each term. Since the d 8 electron configuration corresponds to a shell that is greater than 1/2 filled, the term symbol with the highest J value will be lowest in energy. Therefore, for the F term, the energy ordering is F 4 < F < F 2, and for the P term, the energy ordering is P 2 < P 1 < P 0. To summarize, the overall energy ordering is therefore: F 4 < F < F 2 < P 2 < P 1 < P 0 < 1 G 4 < 1 D 2 < 1 S 0.

4 4 5. Determine the possible atomic term symbols arising from a 2s 1 2p 1 electron configuration. Give the degeneracy of each term symbol. Use Hund's rules to order the term symbols with respect to increasing energy. Show an energy diagram that includes splittings due to spin-orbit coupling. The spin angular momentum is calculated using s 1 = 1 2 and s 2 = 1 2. The total spin angular momentum S ranges from s 1 s 2 to s 1 + s 2 in steps of 1. Thus, S = ,, , or S = 0, 1. The multiplicity 2S+1 is therefore 1 or (singlet or triplet). The orbital angular momentum is calculated using l 1 = 0 and l 2 = 1. The total orbital angular momentum L ranges from l 1 l 2 to l 1 + l 2 in steps of 1. Therefore, L = 0 1,, 0 +1, or L=1. The state is therefore a P state, and the possible terms are 1 P and P. To complete the term symbols, the total angular momentum J must be determined. The total angular momentum J ranges from L S to L + S. For 1 P : L = 1 and S = 0 leads to J = 1. For P : L = 1 and S = 1 leads to J = 0, 1, 2. The possible term symbols (and their degeneracies) are therefore Term Symbol 1 P 1 P 0 P 1 P 2 Degeneracy (2J+1) 1 5

5 5 5. continued The terms for the 2s 1 2p 1 electron configuration are 1 P and P. Using Hund's rule 1, the P term is lower in energy because it has a higher multiplicity. Spin-orbit coupling will split the P 0, P 1, and P 2 term symbols by a small amount. A sketch of the energy level diagram is shown below. 1 P 1 P 1 E P P o, P 1, P 2 2s 1 2p 1 elec-elec repulsions spin-orbit coupling

6 6 6. Determine the possible atomic term symbols arising from a s 1 d 1 electron configuration. Give the degeneracy of each term symbol. The spin angular momentum is calculated using s 1 = 1 2 and s 2 = 1 2. The total spin angular momentum S ranges from s 1 s 2 to s 1 + s 2 in steps of 1. Thus, S = ,, , or S = 0, 1. The multiplicity 2S+1 is therefore 1 or (singlet or triplet). The orbital angular momentum is calculated using l 1 = 0 and l 2 = 2. The total orbital angular momentum L ranges from l 1 l 2 to l 1 + l 2 in steps of 1. Therefore, L = 0 2,, 0 + 2, or L = 2. The state is therefore a D state, and the possible terms are 1 D and D. To complete the term symbols, the total angular momentum J must be determined. The total angular momentum J ranges from L S to L + S. For 1 D : L = 2 and S = 0 leads to J = 2. For D : L = 2 and S = 1 leads to J = 1, 2,. The possible term symbols (and their degeneracies) are therefore Term Symbol 1 D 2 D 1 D 2 D Degeneracy (2J+1) 5 5 7

7 7 7. Determine the possible atomic term symbols arising from a p 1 d 1 electron configuration. Give the degeneracy of each term symbol. The spin angular momentum is again calculated using s 1 = 1 2 and s 2 = 1 2. The total spin angular momentum S ranges from s 1 s 2 to s 1 + s 2 in steps of 1. Thus, S = ,, , or S = 0, 1. The multiplicity 2S+1 is therefore 1 or (singlet or triplet). The orbital angular momentum is calculated using l 1 = 1 and l 2 = 2. The total orbital angular momentum L ranges from l 1 l 2 to l 1 + l 2 in steps of 1. Therefore, L = 1 2,, 1+ 2, or L = 1, 2,. The states possible are therefore P, D, and F states, and the possible terms are 1 P, P, 1 D, D, 1 F, and F. To complete the term symbols, the total angular momentum J must be determined. The total angular momentum J ranges from L S to L + S. For 1 P : L = 1 and S = 0 leads to J = 1. For P : L = 1 and S = 1 leads to J = 0, 1, 2. For 1 D: L = 2 and S = 0 leads to J = 2. For D : L = 2 and S = 1 leads to J = 1, 2,. For 1 F : L = and S = 0 leads to J =. For F : L = and S = 1 leads to J = 2,, 4.

8 8 7. continued The possible term symbols (and their degeneracies) are therefore Term Symbol Degeneracy (2J+1) 1 P 1 P 0 1 P 1 P 2 1 D 2 D 1 D 2 D 1 F F 2 F F Classify the following atomic transitions as allowed or not. If a transition is not allowed, indicate which selection rule is violated. a.) P 2 1 D 2 Not allowed. This transition corresponds to ΔS= 1, which violates the ΔS=0 selection rule. Otherwise, it corresponds to ΔL=1 and ΔJ=0, which are both OK. b.) 2 P /2 2 S 1/2 Allowed. This transition has ΔS=0, ΔL= 1, and ΔJ= 1, which are all allowed by the selection rules. c.) D 2 F 2 Allowed. This transition has ΔS=0, ΔL=1, and ΔJ=0, which are all allowed by the selection rules.