160 Chapter 11 Coordination Chemistry III: Electronic Spectra CHAPTER 11: COORDINATION CHEMISTRY III: ELECTRONIC SPECTRA 11.1 a. p3 There are (6!)/(3!3!) = 20 microstates: MS –3/2 –1/2 1/2 3/2 + – – + – + +2 1 1 0 1 1 0 + – – + – + 1 1 –1 1 1 –1 +1 – – + + – + 1 0 0 1 0 0 + – – + + – 1 0 –1 1 0 –1 – – – – + – + – + + + + ML 0 1 0 –1 1 0 –1 1 0 –1 1 0 –1 1– 0– –1+ 1– 0+ –1+ + – – + – + –1 –1 1 –1 –1 1 –1 – – + + – + –1 0 0 –1 0 0 + – – + – + –2 –1 –1 0 –1 –1 0 Terms: L = 0, S = 3/2: 4S (ground state) L = 2, S = 1/2: 2D 2 L = 1, S = 1/2: P 6! 10! b. p1d1 There are 60 microstates: 1!5! 1!9! M S –1 0 1 – – – + + – + + 3 1 2 1 2 , 1 2 1 2 – – + – – + + + 1 1 1 1 , 1 1 1 1 2 + + 0– 2– 0– 2+, 0+ 2– 0 2 1– 0– 1+ 0–, 0+ 1– 1+ 0+ + + 1 0– 1– 1– 0+, 0– 1+ 0 1 + + –1– 2– –1– 2+, –1+ 2– –1 2 – – + – – + + + 1 –1 1 –1 , 1 –1 1 –1 – – – + + – + + ML 0 0 0 0 0 , 0 0 0 0 + + –1– 1– –1+ 1–, –1– 1+ –1 1 – – + – + – + + –1 0 –1 0 , 0 –1 –1 0 – – – + – + + + –1 0 –1 –1 0 , 0 –1 0 –1 – – – + + – 1+ –2+ 1 –2 1 –2 , 1 –2 –1– –1– –1+ –1–, –1– –1+ –1– –1+ –2 + – – + + + 0– –2– 0 –2 , 0 –2 0 –2 –3 –1– –2– –1– –2+, –1+ –2– –1+ –2+ Terms: L = 3, S = 1 3F (ground state) L = 2, S = 0 1D L = 3, S = 0 1F L = 1, S = 1 3P 3 1 L = 2, S = 1 D L = 1, S = 0 P The two electrons have quantum numbers that are independent of each other, because the electrons are in different orbitals. Because they have different l values, the electrons can have the same ml and ms values. Copyright © 2014 Pearson Education, Inc. Chapter 11 Coordination Chemistry III: Electronic Spectra 161 11.2 For p3: L = 0, S = 3/2, the term 4S has J = 3/2 only (|L+S| = |L–S|). Therefore, the ground 4 state is S3/2. For p1d1: L = 3, S = 1, the term 3F has J = 4, 3, 2. Since both levels are less than half filled, 3 the state having lowest J has lowest energy, and the ground state is F2. 2! 10! 11.3 a. s1d1 There are 20 microstates: 1!1! 1!9! MS No index entries found. mmmm M –1 0 +1 +2 0– 2– 0– 2+, 0+ 2– 0+ 2+ +1 0– 1– 0– 1+, 0+ 1– 0+ 1+ – – – + + – + + ML 0 0 0 00 , 0 0 00 –1 0– –1– 0– –1+, 0+ –1– 0+ –1+ –2 0– –2– 0– –2+, 0+ –2– 0+ –2+ 3 1 b. Terms: L = 2, S = 1: D; L = 2, S = 0: D 3 c. The D, with the higher spin multiplicity, is the lower energy term. 1 1 10! 14! 11.4 a. d f There are 140 microstates: 1!9! 1!13! MS No index entries found. mmmm M –1 0 +1 +5 2– 3– 2– 3+, 2+ 3– 2+ 3+ 2– 2– 2– 2+, 2+ 2– 2+ 2+ +4 1– 3– 1– 3+, 1+ 3– 1+ 3+ 2– 1– 2– 1+, 2+ 1– 2+ 1+ +3 1– 2– 1– 2+, 1+ 2– 1+ 2+ 0– 3– 0– 3+, 0+ 3– 0+ 3+ 2– 0– 2– 0+, 2+ 0– 2+ 0+ 1– 1– 1– 1+, 1+ 1– 1+ 1+ +2 0– 2– 0– 2+, 0+ 2– 0+ 2+ –1– 3– –1– 3+, –1+ 3– –1+ 3+ 2– –1– 2– –1+, 2+ –1– 2+ –1+ 1– 0– 1– 0+, 1+ 0– 1+ 0+ +1 0– 1– 0– 1+, 0+ 1– 0+ 1+ –1– 2– –1– 2+, –1+ 2– –1+ 2+ –2– 3– –2– 3+, –2– 3– –2– 3+ – – – + + – + + ML 2 –2 2 –2 , 2 –2 2 –2 1– –1– 1– –1+, 1+ –1– 1+ –1+ 0 0– 0– 0– 0+, 0+ 0– 0+ 0+ –1– 1– –1– 1+, –1+ 1– –1+ 1+ –2– 2– –2– 2+ , –2+ 2– –2+ 2+ –2– 1– –2– 1+, –2+ 1– –2+ 1+ –1– 0– –1– 0+, –1+ 0– –1+ 0+ –1 0– –1– 0– –1+, 0+ –1– 0+ –1+ 1– –2– 1– –2+, 1+ –2– 1+ –2+ 2– –3– 2– –3+, 2– –3– 2– –3+ continued Copyright © 2014 Pearson Education, Inc. 162 Chapter 11 Coordination Chemistry III: Electronic Spectra – – – + + – + + –2 0 –2 0 , –2 0 –2 0 –1– –1– –1– –1+, –1+ –1– –1+ –1+ –2 0– –2– 0– –2+, 0+ –2– 0+ –2+ 1– –3– 1– –3+, 1+ –3– 1+ –3+ –2– –1– –2– –1+, –2+ –1– –2+ –1+ –3 –1– –2– –1– –2+, –1+ –2– –1+ –2+ 0– –3– 0– –3+, 0+ –3– 0+ –3+ –2– –2– –2– –2+, –2+ –2– –2+ –2+ –4 –1– –3– –1– –3+, –1+ –3– –1+ –3+ –5 –2– –3– –2– –3+, –2+ –3– –2+ –3+ b. Terms: L = 5, S = 1: 3H; L = 5, S = 0: 1H; L = 4, S = 1: 3G; L = 4, S = 0: 1G; L = 3, S = 1: 3F; L = 3, S = 0: 1F ; L = 2, S = 1: 3D; L = 2, S = 0: 1D; L = 1, S = 1: 3P; L = 1, S = 0: 1P c. The lowest energy term is the 3H. For this term, J has the values 6, 5, and 4. Because the subshells are less than half full, the lowest value of J provides the lowest energy: 3 H4. 11.5 a. From Problem 11.1a, for a p3 configuration there are three terms: 4S, 2D, and 2P. The J values for each of these are determined below. 2 For 4S: L = 0, S = 3/2 ; P 28839.31 28838.92 Because J = L + S, L + S – 1, c 2 …|L – S|, the quantum number J D 19233.18 19224.46 can only be 3/2 and there is a 4 4 single state for S: S3/2 2 e For D: L = 2, S = 1/2. Possible J values are 5/2 and 4S 0 3/2, and the two possible states 2 2 are D5/2 and D3/2. 2 2 2 For P: L = 1, S = 1/2. Possible J values are 3/2 and 1/2, with states P3/2 and P1/2. 4 2 2 The lowest energy state is S3/2 (highest multiplicity). D5/2 and D3/2 are next, at –1 2 2 19233.18 and 19224.46 cm , and P3/2 and P1/2 are the highest energy at 28838.92 and 28839.31 cm–1. 4 2 b. The difference in energy between the S and D states is 2e. From the average of 2 –1 the two nearly degenerate D states, e = 9614.41 cm . The difference in energy between the averages of the 2D and 2P states is –1 c = 28839.12 – 19228.82 = 9610.30 cm . Copyright © 2014 Pearson Education, Inc. Chapter 11 Coordination Chemistry III: Electronic Spectra 163 2! 14! 11.6 a. s1f1 There are 28 microstates: 1!1! 1!13! MS –1 0 1 3 0– 3– 0– 3+, 0+ 3– 0+ 3+ – – – + + – + + 2 0 2 0 2 , 0 2 0 2 1 0– 1– 0– 1+, 0+ 1– 0+ 1+ – – – + + – + + ML 0 0 0 0 0 , 0 0 0 0 – – – + + – + + –1 0 –1 0 –1 , 0 –1 0 –1 – – – + + – 0+ –2+ –2 0 –2 0 –2 , 0 –2 –3 0– –3– 0– –3+, 0+ –3– 0+ –3+ b. Terms: L = 3, S = 1 3F (ground state) L = 3, S = 0 1F 3 c. The F term, with the higher spin multiplicity, has the lower energy. This term has 3 J = 2, 3, 4; the lowest energy term, including J, is F2. 2 11.7 a. D has L = 2 and S = 1/2, so ML = –2, –1, 0, 1, 2 and MS = –1/2, 1/2 3 b. G has L = 4 and S = 1, so ML = –4, –3, –2, –1, 0, 1, 2, 3, 4 and MS = –1, 0, 1 4 c. F has L = 3 and S = 3/2, so ML = –3, –2, –1, 0, 1, 2, 3 and MS = –3/2, –1/2, 1/2, 3/2 2 3 2 11.8 a. D with J = 5/2, 3/2 fits an excited state of d , D3/2 3 4 3 b. G with J = 5, 4, 3 fits an excited state of d , G3 4 7 4 c. F with J = 9/2, 7/2, 5/2, 3/2 fits the ground state of d , F9/2 L 11.9 0.038 A 0.10 l 1.00cm mol cm A 0.10 mol A lc; c 2.6 l L L 0.038 1.00cm mol cm __ 11.10 a. 24,900 cm–1 1 1 4.02 10–5 cm = 402 nm 24,900 cm –1 Copyright © 2014 Pearson Education, Inc. 164 Chapter 11 Coordination Chemistry III: Electronic Spectra 2.998 108 m s–1 100 cm m–1 c 14 –1 7.46 10 s 4.02 10–5cm b. 366 nm 8 –1 9 –1 c 2.998 10 m s 10 nm m 8.19 1014 s–1 366 nm E h (6.62610–34 Js)(8.19 1014 s–1) 5.4310–19 J 11.11 J values are included in these answers: 8 a. d Oh MS = 1 = S Spin multiplicity = 2 + 1 = 3 3 Max ML = 2+2+1+1+0+0–1–2 = 3 = L, so F term.