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zumdahl_chemprin_6e_csm_ch19 - CHAPTER 19 TRANSITION METALS...

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730 CHAPTER 19 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. Chromium ([Ar]4s 0 3d 5 ) and copper ([Ar]4s 1 3d 10 ) have electron configurations that are different from that predicted from the periodic table. Other exceptions to the predicted filling order are transition metal ions. These all lose the s electrons before they lose the d electrons. In neutral atoms, the n s and ( n ! 1)d orbitals are very close in energy, with the n s orbitals slightly lower in energy. However, for transition metal ions, there is an apparent shifting of energies between the n s and ( n ! 1)d orbitals. For transition metal ions, the energy of the ( n ! 1)d orbitals are significantly less than that of the n s electrons. So when transition metal ions form, the highest-energy electrons are removed, which are the n s electrons. For example, Mn 2+ has the electron configuration [Ar]4s 0 3d 5 and not [Ar]4s 2 3d 3 . Most transition metals have unfilled d orbitals, which creates a large number of other 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. 9. 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 the following exercise). This leads to a greater similarity in the chemistry of the 4d and 5d elements in a given vertical group.
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CHAPTER 19 TRANSITION METALS AND COORDINATION CHEMISTRY 731 10. Size also decreases going across a period. Sc and Ti, and Y and Zr are adjacent elements. There are 14 elements (the lanthanides) between La and Hf, making Hf considerably smaller. 11. a. molybdenum(IV) sulfide; molybdenum(VI) oxide b. MoS 2 , +4; MoO 3 , +6; (NH 4 ) 2 Mo 2 O 7 , +6; (NH 4 ) 6 Mo 7 O 24 C 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 .
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