chem notes 11-7 - [Ne 3s1-Na 1[Ne-alkaline metals-Ca[Ar...

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November 7, 2007 Dr. Stevens Gen. Chem. 1 Chapter 8 Objectives: -understand organization of the periodic table -know how e- configuration relates to periodic properties -determine e- configuration for ions -know trends in radius, ionization E, and e- affinity -know general trends in chemical properties Chapter 7: Stability issues the larger the subshells, the more stability issues are important (esp. d/f) -most stable if subshell is all full -middle stability is ½ full -unstable – partially filled subshell -higher energy for most unstable -very stable: Zn, Cd, Hg -nearby atoms: Cu, Ag, Au -Cu [Ar] 4s2 3d9 = [Ar] 4s1 3d10 -Ag [Kr] 5s2 4d9 = [Kr] 5s1 4d10 -Au [Xe] 6s2 5d9 = [Xe] 6s1 4f14 5d10 -d4 d5 for stability -d5 some stability (Mn, Tc, Re) Chapter 8: Ions -alkali X+1 -Na (neutral)
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Unformatted text preview: -[Ne] 3s1-Na+1 [Ne]-alkaline metals-Ca [Ar] 4s2-Ca+2 [Ar]-isoelectronic: 2 species that have same # of e--ex: Na+ and Ne are isoelectronic (both have 10e-)-anions-O likes to have q = -2-[He] 2s2 2p4-[He] 2s2 2p6 = [Ne] Predicting transition metal changes 1. lose s electrons first 2. lose delectrons to gain stability-Ti = [Ar] 4s2 3d2 [Ar] 3d2 = Ti +2 [Ar] = Ti+4-Ti can become Ti+2 or Ti+4-order of subshell when empty: 1s 2s 3s 4s 3d 4p-when full: 1s 2s 3s 3p 3d 4s-lose s electrons first because they hold higher E than 3d (when full)-Ru = [Kr] 5s2 4d6 [Kr] 4d6 = Ru+2 [Kr] 4d5 = Ru+3 [Kr] = Ru+8-Ru can become Ru+2 or Ru+3 or Ru+8 (least likely)-Cd = [Kr] 5s2 4d10 [Kr] 4d10 = Cd+2-Cd can become Cd+2 (stable and common)...
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