This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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)...
View Full Document