30Periodic2

Lithium tends to form predominantly lithium oxide

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Unformatted text preview: nds to form predominantly lithium oxide because this compound has a greater stability (larger lattice energy) than lithium peroxide. The formation of other alkali metal oxides can be similarly explained. Reaction of Group 1A Elements with O2 M(s) + O2(g) 16 electrons M(s) + O2–(s) 17 electrons 2M(s) + O22–(s) 18 electrons M+(s) + O2–(s) 17 electrons M+(s) + O22–(s) 18 electrons 2M+(s) + 2O2–(s) 20 electrons Reaction of Group 1A Elements with O2 M(s) + O2(g) 16 electrons M+(s) + O2–(s) 17 electrons stops here when M = K Reaction of Group 1A Elements with O2 M(s) + O2(g) 16 electrons M(s) + O2–(s) 17 electrons M+(s) + O2–(s) 17 electrons M+(s) + O22–(s) 18 electrons stops here when M = Na Reaction of Group 1A Elements with O2 M(s) + O2(g) 16 electrons M(s) + O2–(s) 17 electrons 2M(s) + O22–(s) 18 electrons M+(s) + O2–(s) 17 electrons M+(s) + O22–(s) 18 electrons 2M+(s) + 2O2–(s) 20 electrons stops here when M = Li Reaction of Group 1A Elements with O2 Li appears to lose electrons to oxygen better than Na and K. Why? It can’t be ionization energy. Lithium has a higher ionization energy than Na and Na has a higher ionization energy than K. The answer lies in the size of Li+ versus Na+ versus K+. Strength of attractive force: proportional to products of charge (Q+Q–) inversely proportionaly to distance between charges Q+Q– E=k r small size of Li+ makes attractive force in LiO2 very large...
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This note was uploaded on 02/13/2009 for the course CHEM 101 taught by Professor Taracarpenter during the Spring '08 term at UMBC.

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