self_test_answers

Self_test_answers - Shriver Atkins Inorganic Chemistry 5e Answers to self-tests and exercises CHAPTER 1 1.7 Take the summation of the rest masses

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Answers to self-tests and exercises CHAPTER 1 Self-tests S1.1 80 35 Br + 1 0 n 81 35 + γ S1.2 d orbitals, 5 orbitals S1.3 4 S1.4 3 p S1.5 The added p electron is in a different ( p ) orbital, so it is less shielded. S1.6 82 Ni:[Ar]3 4 ds , 28 N i: [ A r ] 3 d + S1.7 Period 4, Group 2, s block S1.8 Going down a group the atomic radius increases and the first ionization energy generally decreases. S1.9 Group 16. The first four electrons are removed with gradually increasing values. Removing the fifth electron requires a large increase in energy, indicating breaking into a complete subshell. S1.10 Adding another electron to C would result in the stable half filled p subshell. S1.11 Cs + Exercises 1.1 (a) 7 14 N + 2 4 He 8 17 O + 1 1 p+ γ (b) 6 12 C + 1 1 p 7 13 N + (c) 7 14 N + 0 1 n 1 3 H + 6 12 C 1.2 96 246 Cm + 6 12 C 112 257 Uub + 0 1 n 1.3 The higher value of I 2 for Cr relative to Mn is a consequence of the special stability of half- filled subshell configurations and the higher Z eff of a 3 d electron verses a 4 s electron. 1.4 22 4 25 1 10 2 12 0 Ne He Mg n +→ + 1.5 4 9 Be + 4 9 Be 6 12 C + 2 4 He +2 0 1 n 1.6 Since helium-4 is the basic building block, most additional fusion processes will produce nuclei with even atomic numbers. 1.7 Take the summation of the rest masses of all the nuclei of the products minus the masses of the nuclei of the reactants. If you get a negative number, energy will be released. But what you have calculated is the mass difference, which in the case of a nuclear reaction is converted to energy. 1.8 0.25 1.9 –13.2 eV 1.10 1524nm, 1.524 X 10 4 cm -1 1.11 1 λ = R 1 1 2 1 2 = 1.0974 X 10 7 m 1 1 = R 1 1 2 1 4 2 = 1.0288 X 10 7 m 1 1 = R 1 1 2 1 3 2 = 9.7547 X 10 6 m 1 1 = R 1 1 2 1 2 2 = 8.2305 X 10 6 m 1 1.12 0 up to n- X 1.13 n 2 1.14 N l m l Orbital designation Number of orbitals 2 1 +1, 0, 1 2 p 3 3 2 +2, +1, …, 2 3 d 5 4 0 0 4 s 1 4 3 +3, +2, …, 3 4 f 7 1.15 n=5, l = 3, and m l = -3,-2,-1,0,1,2,3 1.16 Li: σ = Z – Z eff ; σ = 3-1.28 = 1.72 Be: σ = Z – Z eff ; σ = 4-1.19 = 2.09 B: σ = Z – Z eff ; σ = 5-2.42 = 2.58 C: σ = Z – Z eff ; σ = 6-3.14 = 2.86 N: σ = Z – Z eff ; σ = 7-3.83 = 3.17 O: σ = Z – Z eff ; σ = 8-4.45 = 3.55 F: σ = Z – Z eff ; σ = 9-5.10 = 3.90 1.17 The 1s electrons shield the positive charge form the 2s electrons. 1 Shriver & Atkins: Inorganic Chemistry 5e
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ANSWERS TO SELF-TESTS AND EXERCISES 2 1.18 See Figs 1.11 through 1.16 1.19 See Table 1.6 and discussion. 1.20 Table 1.6 shows SR > Ba < Ra. Ra is anomalous because of higher Z eff due to lanthanide contraction. 1.21 Anomalously high value for Cr is associated with the stability of a half filled d shell. 1.22 (a) [He]2 s 2 2 p 2 (b) [He]2 s 2 2 p 5 (c) [Ar]4 s 2 (d) [Ar]3 d 10 (e) [Xe]4 f 14 5 d 10 6 s 2 6 p 3 (f) [Xe]4 f 14 5 d 10 6 s 2 1.23 (a) [Ar]3 d 1 4 s 2 (b) [Ar]3 d 2 (c) [Ar] 3d 5 (d) [Ar] 3d 4 (e) [Ar] 3d 6 (f) [Ar] (g) [Ar]3 d 10 4 s 1 (h) [Xe]4 f 7 1.24 (a) Xe]4 f 14 5 d 4 6 s 2 (b) [Kr]4 d 6 (c) [Xe]4 f 6 (d) [Xe]4 f 7 (e) [Ar] (f) [Kr]4 d 2 1.25 (a) S (b) Sr (c) V (d) Tc (e) In (f) Sm 1.26 See Figure 1.4.
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This note was uploaded on 02/08/2012 for the course CHEM 161 taught by Professor Shaklovich during the Spring '10 term at Harvard.

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Self_test_answers - Shriver Atkins Inorganic Chemistry 5e Answers to self-tests and exercises CHAPTER 1 1.7 Take the summation of the rest masses

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