Chapter 20 Even Problem Solutions

# Chapter 20 Even Problem Solutions - CHAPTER TWENTY...

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CHAPTER TWENTY TRANSITION METALS AND COORDINATION CHEMISTRY Transition Metals 6. Transition metal ions lose the s electrons before the d electrons. a. Ti: [Ar]4s 2 3d 2 ; Ti 2+ : [Ar]3d 2 ; Ti 4+ : [Ar] or [Ne]3s 2 3p 6 b. Re: [Xe]6s 2 4f 14 5d 5 ; Re 2+ : [Xe] 4f 14 5d 5 ; Re 3+ : [Xe] 4f 14 5d 4 c. Ir: [Xe]6s 2 4f 14 5d 7 ; Ir 2+ : [Xe]4f 14 5d 7 ; Ir 2+ : [Xe]4f 14 5d 6 8. Since transition metals form bonds that donate lone pairs of electrons, transition metals are Lewis acids (electron pair acceptors). The Lewis bases in coordination compounds are the ligands that must have at least one unshared pair of electrons. The coordinate covalent bond between the ligand and the transition metal just indicates that both electrons in the bond originally came from one of the atoms in the bond. 10. The lanthanide elements are located just before the 5d transition metals. The lanthanide contraction is the steady decrease in the atomic radii of the lanthanide elements when going from left to right across the periodic table. As a result of the lanthanide contraction, the sizes of the 4d and 5d elements are very similar (see Exercise 12.87). This leads to a greater similarity in the chemistry of the 4d and 5d elements in a given vertical group. 12. a. 4 O atoms on faces × 1/2 O/face = 2 O atoms, 2 O atoms inside body, Total: 4 O atoms 8 Ti atoms on corners × 1/8 Ti/corner + 1 Ti atom/body center = 2 Ti atoms Formula of the unit cell is Ti 2 O 4 . The empirical formula is TiO 2 . b. +4 -2 0 0 +4 -1 +4 -2 +2 -2 2 TiO 2 + 3 C + 4 Cl 2 → 2 TiCl 4 + CO 2 + 2 CO; Cl is reduced and C is oxidized. Cl 2 is the oxidizing agent and C is the reducing agent. +4 -1 0 +4 -2 0 TiCl 4 + O 2 → TiO 2 + 2 Cl 2 ; O is reduced and Cl is oxidized. O 2 is the oxidizing agent and TiCl 4 is the reducing agent. 14. If rain were imminent, there would be a lot of water vapor in the air causing the reaction to shift to the right. The indicator would take on the color of CoCl 2 . 6H 2 O, pink. Coordination Compounds

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16. Fe 2 O 3 (s) + 6 H 2 C 2 O 4 (aq) → 2 Fe(C 2 O 4 ) 3 3— (aq) + 3 H 2 O(l) + 6 H + (aq); The oxalate anion forms a soluble complex ion with iron in rust (Fe 2 O 3 ), which allows rust stains to be removed. 18. a. hexaamminecobalt(II) chloride b. hexaaquacobalt(III) iodide c. potassium tetrachloroplatinate(II) d. potassium hexachloroplatinate(II) e. pentaamminechlorocobalt(III) chloride f. triamminetrinitrocobalt(III) 20. a. pentaaquabromochromium(III) bromide b. sodium hexacyanocobaltate(III)
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## This note was uploaded on 04/07/2008 for the course CHEM 027 taught by Professor Mccallum during the Spring '08 term at Pacific.

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Chapter 20 Even Problem Solutions - CHAPTER TWENTY...

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