LecturePLUS Timberlake1 Chapter 4 Compounds and Their Bonds 4.1 Valence Electrons 4.2 Octet Rule and Ions

LecturePLUS Timberlake2 Chemical Bonds Attraction between two or more atoms Interaction between valence electrons Ionic bonds Covalent bonds

LecturePLUS Timberlake3 Valence Electrons Electrons in the highest (outer) electron level Have most contact with other atoms Known as valence electrons Outer shells of noble gases contain 8 valence electrons (except He = 2) Example: Ne 2, 8 Ar2, 8, 8

LecturePLUS Timberlake4 Electron Dot Structures Symbols of atoms with dots to represent the valence-shell electrons 1A 2A 3A 4A 5A 6A 7A 8A H  He:            Li  Be   B   C   N   O  : F  : Ne :                    Na  Mg   Al   Si   P   S  : Cl  : Ar :        

LecturePLUS Timberlake5 Learning Check A. X would be the electron dot formula for 1) Na2) K3) Al B. X would be the electron dot formula 1) B2) N3) P

LecturePLUS Timberlake6 Solution A. X would be the electron dot formula for 1) Na2) K B. X would be the electron dot formula 2) N3) P

LecturePLUS Timberlake7 Octet Rule An octet in the outer shell makes atoms stable Electrons are lost, gained or shared to form an octet Unpaired valence electrons strongly influence bonding

LecturePLUS Timberlake8 Formation of Ions from Metals Ionic compounds result when metals react with nonmetals Metals lose electrons to match the number of valence electrons of their nearest noble gas Positive ions form when the number of electrons are less than the number of protons Group 1A metals  ion 1+ Group 2A metals  ion 2+ Group 3A metals  ion 3+

LecturePLUS Timberlake9 Formation of Sodium Ion Sodium atom Sodium ion Na  – e   Na ( = Ne) 11 p + 11 p + 11 e - 10 e

LecturePLUS Timberlake10 Formation of Magnesium Ion Magnesium atom Magnesium ion  Mg  – 2e   Mg (=Ne) 12 p + 12 p + 12 e- 10 e

LecturePLUS Timberlake11 Some Typical Ions with Positive Charges (Cations) Group 1AGroup 2AGroup 3A H + Mg 2+ Al 3+ Li + Ca 2+ Na + Sr 2+ K + Ba 2+

LecturePLUS Timberlake12 Learning Check A. Number of valence electrons in aluminum 1) 1 e - 2) 2 e - 3) 3 e - B. Change in electrons for octet 1) lose 3e - 2) gain 3 e - 3) gain 5 e - C.Ionic charge of aluminum 1) 3- 2) 5- 3) 3 +

LecturePLUS Timberlake13 Solution A. Number of valence electrons in aluminum 3) 3 e - B. Change in electrons for octet 1) lose 3e - C.Ionic charge of aluminum 3) 3 +

LecturePLUS Timberlake14 Learning Check Give the ionic charge for each of the following: A. 12 p + and 10 e - 1) 02) 2+3) 2- B. 50p + and 46 e- 1) 2+2) 4+3) 4- C. 15 p + and 18e- 2) 3+ 2) 3-3) 5-

LecturePLUS Timberlake15 Solution Give the ionic charge for each of the following: A. 12 p + and 10 e - 2) 2+ B. 50p + and 46 e- 2) 4+ C. 15 p + and 18e- 2) 3-

LecturePLUS Timberlake16 Learning Check A. Why does Ca form a Ca 2+ ion? B. Why does O form O 2- ion?

LecturePLUS Timberlake17 Solution A. Why does Ca form a Ca 2+ ion? Loses 2 electrons to give octet (like Ar) B. Why does O form O 2- ion? Gains 2 electrons to give octet e (like Ne)

LecturePLUS Timberlake18 Ions from Nonmetal Ions In ionic compounds, nonmetals in 5A, 6A, and 7A gain electrons from metals Nonmetal add electrons to achieve the octet arrangement Nonmetal ionic charge: 3-, 2-, or 1-

LecturePLUS Timberlake19 Fluoride Ion unpaired electronoctet     1 - : F  + e  : F :     (= Ne) 9 p+ 9 p + 9 e- 10 e ionic charge

LecturePLUS Timberlake20 Learning Check Complete the names of the following ions: 5A 6A 7A N 3  O 2  F  nitride __________ fluoride P 3  S 2  Cl  ___________ __________ _________ Br  _________

LecturePLUS Timberlake21 Solution 5A 6A 7A N 3  O 2  F  nitride oxide fluoride P 3  S 2  Cl  phosphide sulfide chloride Br  bromide