solut20 - CHAPTER TWENTY TRANSITION METALS AND COORDINATION...

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560 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 b. Re: [Xe]6s 2 4f 14 5d 5 c. Ir: [Xe]6s 2 4f 14 5d 7 Ti 2+ : [Ar]3d 2 Re 2+ : [Xe]4f 14 5d 5 Ir 2+ : [Xe]4f 14 5d 7 Ti 4+ : [Ar] or [Ne]3s 2 3p 6 Re 3+ : [Xe]4f 14 5d 4 Ir 3+ : [Xe]4f 14 5d 6 7. Cr and Cu are exceptions to the normal filling order of electrons. a. Cr: [Ar]4s 1 3d 5 b. Cu: [Ar]4s 1 3d 10 c. V: [Ar]4s 2 3d 3 Cr 2+ : [Ar]3d 4 Cu + : [Ar]3d 10 V 2+ : [Ar]3d 3 Cr 3+ : [Ar]3d 3 Cu 2+ : [Ar]3d 9 V 3+ : [Ar]3d 2 8. Since transition metals form bonds to species which donate lone pairs of electrons, then transition metals are Lewis acids (electron pair acceptors). 9. Most transition metals have unfilled d orbitals, which creates a large number of valence electrons that can be removed. Stable ions of the representative metals are determined by how many s and p valence electrons can be removed. In general, representative metals lose all of the s and p valence electrons to form their stable ions. Transition metals generally lose the s electron(s) to form +1 and +2 ions, but they can also lose some (or all) of the d electrons to form other oxidation states as well. 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 12.85). This leads to a greater similarity in the chemistry of the 4d and 5d elements in a given vertical group. 11. a. molybdenum(IV) sulfide; molybdenum(VI) oxide
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561 b. MoS 2 , +4; MoO 3 , +6; (NH 4 ) 2 Mo 2 O 7 , +6; (NH 4 ) 6 Mo 7 O 24 4 H 2 O, +6 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. 13. TiF 4 : ionic compound containing Ti 4+ ions and F - ions. TiCl 4 , TiBr 4 , and TiI 4 : covalent compounds containing discrete, tetrahedral TiX 4 molecules. As these covalent molecules get larger, the bp and mp increase because the London dispersion forces increase. TiF 4 has the highest bp since the interparticle forces are usually stronger in ionic compounds as compared to covalent compounds. Coordination Compounds 14. 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.
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This homework help was uploaded on 04/06/2008 for the course CHEM 142 taught by Professor Zoller,williamh during the Fall '07 term at University of Washington.

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solut20 - CHAPTER TWENTY TRANSITION METALS AND COORDINATION...

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