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Student Solutions Manual for
DESCRIPTIVE INORGANIC CHEMISTRY 6th edition
Correlation Guide from 5th edition Solutions Manual
Chapter 1 Exercises 1.19 [Xe]4f7 1.21 was 1.19 1.23 was 1.21 1.25 was 1.23 1.27 Helium is assigned as a member of the alkaline earth metals when in fact it is a noble gas. Orbital filling begins with the alkali metals so, from that perspective, the arrangement is unhelpful. 1.29 Of course, one can argue that orbitals are human constructs only! This question is a good topic for debate, but these authors veer towards the view that an orbital actually exists only when it is populated. Empty orbitals only potentially exist. Chapter 3 Exercises 3.25 Figure 3.32 not figure 3.46 3.37 Figure 3.24 not figure 3.42 3.37 Figure 3.25 not figure 3.43 3.41 [New] We need to look up values of Allred‐Rochow electronegativities of the elements, calculate the average and difference and see where the values are plotted on the bond triangle. (a) EN(Co) = 1.70, EN(Zn) = 1.66, average EN = 1.68, difference in EN = 0.04. The values fall in the metallic/covalent region. (b) EN(B) = 1.83, EN(F) = 4.43, average EN = 3.13, difference in EN = 2.60. The values fall in the covalent region. Beyond the basics Each question number should be +2. Chapter 5 Exercises 5.29 was 5.35 5.31 was 5.37 5.33 was 5.39 5.35 was 5.41 5.37 was 5.43 5.39 [should read zirconium ions] Zr3+ Chapter 7 Beyond the basics 7.58 Figure 7.4 should be 7.5. Chapter 15 Exercises 15.42 Page 421 not 402‐403 Chapter 18 Exercises 18.21 Page 517 not page 496 Chapter 19 Exercises 19.19 Table 19.3 not Table 19.4 19.21 The mixed metal oxide NiFe2O4 will adopt the inverse spinel structure, (Fe3+)t(Ni2+,Fe3+)oO4 as the Fe3+ ion has a zero CFSE, hence the optimum situation energetically is for the Ni2+ ion to occupy the octahedral sites with Δoct rather than the tetrahedral sites with 4/9Δoct. 19.23 Six nitrogen‐donor atoms will give a stronger field (larger splitting) than four nitrogen‐ and two oxygen‐donor atoms. The greater splitting for the former is enough to overcome the pairing energy. Beyond the basics 19.25 The large cation will tend to stabilize the larger pentachloro‐complex. In addition, the cation and anion charges will match, enabling the formation of a simple alternating anion‐cation lattice such as the sodium chloride lattice. The name will be hexaamminecobalt(III) pentachlorocuprate(II). 19.27 If the nickel complex is paramagnetic, then it has tetrahedral stereochemistry and only the one form. The diamagnetic palladium analog would be square‐planar and have two geometric isomers—cis and trans. 19.29 The diiodoaurate(II) ion should be the most stable as it involves a soft acid/soft base combination. 19.31 These ligands will complex any trace toxic metal ions that leach from the can walls. As firmly complexed ions, their toxicity will be substantially diminished. The complexed ions will be excreted rather than absorbed. 19.33 In its normal oxidation state of +1, silver has a complete d10 configuration; thus it is acting as a main group metal. Only in its rare higher oxidation states of +2 and +3 does it have less than a filled d‐set. For “normal” chemistry it is probably more useful to consider silver as a main group metal. 19.35 [New] Cl
Cl
Pt
Cl
Cl
NH3
H3N
Cl
Pt
Cl
Cl
PPh3
H3N
PPh3
Pt
Cl
Cl
PPh3
Ph3P
Cl
Pt
Cl
Cl
NH3
Ph3P
Cl
Pt
Cl
NH3
19.37 was 19.35 Chapter 20 Exercises 20.29 Pages 601‐602 not 572‐573