ch 10 con - Chem. 140B S2008 HW set 9 Due: 4/10/08 Ch. 10:...

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Unformatted text preview: Chem. 140B S2008 HW set 9 Due: 4/10/08 Ch. 10: P10.12, 13, 14, 15, 18, 19, 20, 23, 26 P10.12) List the quantum numbers L and S that are consistent with the terms a) 4S, b) 4G, c) 3P, and d) 2D a) 4S: L = 0, 2S + 1 = 4, S = 3/2 b) 4G: L = 4, 2S + 1 = 4, S = 3/2 c) 3P: L = 1, 2S + 1 = 3, S = 1 d) 2D: L = 2, 2S + 1 = 2, S = ½ P10.13) List the allowed quantum numbers ml and ms for the following subshells and determine the maximum occupancy of the subshells. a) 2p b) 3d c ) 4f d) 5g a) 2p: 1, 0, –1 for ml and ½ and –½ for ms for each of the ml values. Subshell occupancy = 6. b)3d: 2, 1, 0, –1, –2 for ml and ½ and –½ for ms for each of the ml values. Subshell occupancy = 10. c) 4f: 3, 2, 1, 0, –1, –2, –3 for ml and ½ and –½ for ms for each of the ml values. Subshell occupancy = 14. d) 5g: 4, 3, 2, 1, 0, –1, –2, –3, –4 for ml and ½ and –½ for ms for each of the ml values. Subshell occupancy = 18. P10.14) Two angular momenta with quantum numbers j1 = 3/2 and j2 = 5/2 are added together. What are the possible values of J for the resultant angular momentum states? J = J1 + J 2 , J1 + J 2 − 1 , J1 + J 2 − 2 ,..., J1 − J 2 giving possible J values of 4, 3, 2, and 1. P10.15) Calculate the terms that can arise from the configuration np1n′p1 , n ≠ n′. Compare your results with those derived in the text for np2. Which configuration has more terms and why? Because the principle quantum number is different, any combination of ml and ms is allowed. Therefore, S = 0, 1 and L = 0, 1, and 2. Any combination of the two quantum numbers is allowed. This leads to 1S, 1 P, and 1D terms as well as 3S, 3P, and 3D terms. The np1n ′p1 , n ≠ n ′ configuration has more terms because some of the possible terms listed above are not allowed if n = n ′ because of the Pauli principle. P10.18) What atomic terms are possible for the following electron configurations? Which of the possible terms has the lowest energy? a) ns1np1 b) ns1nd1 c) ns2np1 d) ns1np2 1 a) ns1np1 L can only have the value 1, and S can have the values 0 and 1. The possible terms are 1P and 3 P. Hund’s Rules predict that the 3P term will have the lower energy. b) ns1nd1 L can only have the value 2, and S can have the values 0 and 1. The possible terms are 1D and 3 D. Hund’s Rules predict that the 3D term will have the lower energy. c) ns2np1 L can only have the value 1, and S can only have the value 1/2. The only possible term is 2P. d) ns1np2 A table such as the table in the text for the p2 configuration will have three columns, one for each of the electrons, for ML and MS. Each of the fifteen states for the p2 configuration can be combined 1 with ms = ± for the ns electron. This gives a total of 30 states. Working through the table gives 2D, 2 4 P, 2P, and 2S terms. Hund’s Rules predict that the 4P term will have the lowest energy. P10.19) The ground-state level for the phosphorus atom is 4S3/2. List the possible values of L, ML, S, MS, J, and MJ consistent with this level. S = 3/2 and L = 0. ML can range from –L to +L and can only have the value zero in this case. MS can range from –S to +S and can have the values –3/2, –1/2, 1/2, and 3/2 in this case. J lies between L + S and L − S and can only have the value 3/2 for this case. MJ can range from –J to +J and can have the values –3/2, –1/2, 1/2, and 3/2 in this case. P10.20) Derive the ground-state term symbols for the following configurations: a) d2, b) f 9, c) f 14 The method illustrated in Example Problem 10.7 is used for all parts. a) MLmax = 3 and MSmax = 1. Therefore, the ground state term is 3F. b) MLmax = 5 and MSmax = 2.5. Therefore, the ground state term is 6H. c) 2 MLmax = 5 and MSmax = 1. Therefore, the ground state term is 3H. P10.23) Using Table 10.5, which lists the possible terms that arise from a given configuration, and Hund’s rules, write the term symbols for the ground state of the atoms H through F in the form (2S+1) LJ. We use Hund’s rule that the term with the highest multiplicity is the lowest in energy to get the left superscript. For the right subscript, if there are several choices for J, the lowest J value gives the lowest energy if the subshell is less than half full, and the highest J value gives the lowest energy if the subshell is exactly or more than half full. Applying these rules gives rise to the following term symbols. Atom H He Li Be B C N O F Configuration Ground state term symbol 2 1s1 S1/2 2 1 1s S0 21 2 1s 2s S1/2 22 1 1s 2s S0 22 1 2 P1/2 1s 2s 2p 22 2 3 P0 1s 2s 2p 22 3 4 S3/2 1s 2s 2p 22 4 3 P2 1s 2s 2p 22 5 2 1s 2s 2p P3/2 3 P10.26) Derive the ground-state term symbols for the following atoms or ions: a) H, b) F, c) F-, d) Na, e) Na+, f) P, g) Sc We use Hund’s rule that the term with the highest multiplicity is the lowest in energy to get the left superscript. For the right subscript, if there are several choices for J, the lowest J value gives the lowest energy if the subshell is less than half full, and the highest J value gives the lowest energy if the subshell is exactly or more than half full. Applying these rules gives rise to the following term symbols. Atom H F F– Na Na+ P Sc Configuration Ground state term symbol 2 1s1 S1/2 2 5 2 2s 2p P3/2 2 6 1 2s 2p S0 1 2 3s S1/2 2 6 1 2s 2p S0 3 4 3p S3/2 1 2 3d D3/2 4 ...
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This note was uploaded on 06/14/2009 for the course CHEM CHEM 2080 taught by Professor Pathivan during the Spring '05 term at Indian Institute of Technology, Chennai.

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