LecturePLUS Timberlake1 Chemical Bonds Attraction between two or more atoms Interaction between valence electrons Ionic bonds Covalent bonds

LecturePLUS Timberlake2 Valence Electrons Electrons in the highest (outer) electron level Have most contact with other atoms Known as valence electrons Outer shelI of noble gases the outer shell contains 8 valence electrons as s 2 p 6 (except He = 2) Example: Ne 1s 2 2s 2 2p 6 Ar1s 2 2s 2 2p 6 3s 2 3p 6

LecturePLUS Timberlake3 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 Timberlake4 Learning Check B1 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 Timberlake5 Solution B1 A. X would be the electron dot formula for 1) Na2) K B. X would be the electron dot formula 2) N3) P

LecturePLUS Timberlake6 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 Timberlake7 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 Timberlake8 Formation of Sodium Ion Sodium atom Sodium ion Na  – e   Na + 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 ( = Ne) 11 p + 11 p + 11 e - 10 e

LecturePLUS Timberlake9 Formation of Magnesium Ions Magnesium atom Magnesium ion  Mg  – 2e   Mg 2+ 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 (Ne) 12 p + 12 p + 12 e- 10 e

LecturePLUS Timberlake10 Cations of Some Representative Elements Group 1AGroup 2AGroup 3A H + Mg 2+ Al 3+ Li + Ca 2+ Na + Sr 2+ K + Ba 2+

LecturePLUS Timberlake11 Learning Check B 2 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 Timberlake12 Solution B 2 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 Timberlake13 Learning Check B 3 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 Timberlake14 Solution B 3 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 Timberlake15 Learning Check B 4 A. Why does Ca form a Ca 2+ ion? B. Why does O form O 2- ion?

LecturePLUS Timberlake16 Solution B 4 A. Why does Ca form a Ca 2+ ion? Loses 2 electrons to give octet 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 1s 2 2s 2 2p 6 3s 2 3p 6 (like Ar) B. Why does O form O 2- ion? Gains 2 electrons to give octet 1s 2 2s 2 2p 4 + 2e - 1s 2 2s 2 2p (like Ne)

LecturePLUS Timberlake17 Transition Metals Lose s electrons from the highest occupied energy level. May also lose one or more d electrons. Most form 2 or more positive ions

LecturePLUS Timberlake18 Fe 2+ and Fe 3+ Fe atom: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 Loss of 4s 2 electrons (outermost shell) Fe 2+ ion: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 6 Loss of 3d 1 electron Fe 3+ ion: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 5 A 3d 5 (half-filled) subshell is stable

LecturePLUS Timberlake19 Some Transition Metal Ions or or 3+ Ag + Cd 2+ Cu +, Cu 2+ Fe 2+, Fe 3+ silver cadmium copper(I) ion iron(II) ion ion ion copper (II) ion iron(III) ion Zn 2+ zinc ion

LecturePLUS Timberlake20 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 Timberlake21 Fluoride Ions unpaired electronoctet     1 - : F  + e  : F :     1s 2 2s 2 2p 5 1s 2 2s 2 2p 6 = Ne

LecturePLUS Timberlake22 Some Nonmetal Ions 5A 6A 7A N 3  O 2  F  nitride __________ fluoride P 3  S 2  Cl  ___________ __________ _________ Br  _________ I  iodide ion

LecturePLUS Timberlake23 Some Nonmetal Ions 5A 6A 7A N 3  O 2  F  nitride oxide fluoride P 3  S 2  Cl  phosphide sulfide chloride Br  bromide

LecturePLUS Timberlake24 Ionic Compounds Attraction between + ions and - ions Electrons go from metals to nonmetals electron transfer metal nonmetal ion + ion -– Electrons lost = Electrons gain

LecturePLUS Timberlake25 Formulas of Ionic Compounds Formulas of ionic compounds are determined from the charges on the ions atoms ions     – Na  +  F :  Na + : F :  NaF     sodium fluorine sodium fluoride formula Charge balance: = 0

LecturePLUS Timberlake26 Writing a Formula Write the formula for the ionic compound that will form between Ba 2+ and Cl . Solution: 1. Balance charge with + and – ions 2. Write the positive ion of metal first, and the negative ion Ba 2+ Cl  Cl  3. Write the number of ions needed as subscripts BaCl 2

LecturePLUS Timberlake27 Learning Check B 5 A. The formula for the ionic compound of Na + and O 2- is 1) NaO2) Na 2 O3) NaO 2 B. The formula of a compound of aluminum and chlorine is 1) Al 3 Cl2) AlCl 2 3) AlCl 3 C. The formula of Fe 3+ and O 2- is 1) Fe 3 O 2 2) FeO 3 3) Fe 2 O 3

LecturePLUS Timberlake28 Solution B 5 A. The formula for the ionic compound of Na + and O 2- is 2) Na 2 O B. The formula of a compound of aluminum and chlorine is 3) AlCl 3 C. The formula of Fe 3+ and O 2- is 3) Fe 2 O 3