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

10/28 – 29/2015, Warm-up: Half Life If 20.0 g of a radioactive isotope are present at 1:00 PM and 5.0 g remain at 2:00 PM, what is the half life of the isotope? (Refer to Chap. 4, pp. 125 – 126.) Bombardment of aluminum-27 by alpha particles produces phosphorus-30 and one other particle. Writhe the nuclear equation for this reaction and identify the other particle. LecturePLUS Timberlake2

Chemistry Agenda: 10/28 – 29/2015 Objective: I can determine the number of valence electrons and possible oxidation numbers from an element's electron configuration. Warm-up: If 20.0 g of a radioactive isotope are present at 1:00 PM and 5.0 g remain at 2:00 PM, what is the half life of the isotope? (Refer to Chap. 4, pp. 125 – 126.) –Bombardment of aluminum-27 by alpha particles produces phosphorus-30 and one other particle. Write the nuclear equation for this reaction and identify the other particle. Explore: How does a nuclear reaction relate to half-life? Explain: How does half-life relate to us individually in daily life? Evaluation: Use the handouts Half-Life Review Problems Packet, to solve the following problems to calculate the half-life of various radio-isotopes. –CHECK: Half-life PRACTICE, Nuclear Reactions, Half-Life Review Problems Packet, Lab 17-2: Building an Atom, Warm-ups – 9/24/ /7/2015. –Magic Square Atomic Structure and Theory handout (Refer to your Atomic Timeline organizer & Chap. 5, pp. 146 – 161.) –REVIEW Unit II: Atomic Structure Exam Review –UNIT II EXAM: 1,3,5, Tue., 10/27/2015; 4,6, Mon., 10/26/2015 3

Chemistry Agenda: 10/30, 11/4/2015 Objective: I can relate the position of an element on the periodic table to its electron configuration. CH.2g. Warm-up: What is the difference between the alkali metals and the alkaline earth metals? (Refer to Chap. 6, p. 177.) Explore: What is the difference in the reactivity between the alkali metals and the alkaline earth metals? (Refer to Chap. 6, p. 177.) Explain: What is electron configuration? (Refer to Chap. 5, p. 156.) Evaluation: Use the handouts Atomic Notation Packet, to solve the following problems to write the dot structure. –REVIEW: Naming Atoms, Ions & Isotopes, Section of the Periodic Table. –Bohr Diagram handout (Refer to your Bohr Diagram handout Key, Chap. 5, pp. 146 – 161.) –SNC1P Bohr Diagram handout 4

Chemistry Agenda: 11/20, 23/2015 Objective: I can determine the number of valence electrons and possible oxidation numbers from an element's electron configuration. CH.2g. Warm-up: How many valence electrons does vanadium (V) have from the periodic table of elements? (Refer to Chap. 5, p. 161). Explore: What part of an element’s electron configuration can be used to determine its valence electrons? (Refer to Chap. 5, p. 161.) Explain: How can we decipher the difference in elements’ atomic radius in the periodic table? (Refer to Chap. 6, p. 182.) Evaluation: Use the handouts Unit 3 - Periodic Trends and Electron Configurations Packet, to solve the following problems to write the electron configuration. UNIT III EXAM: 1 st, 3 rd, 5 th – Mon., 11/23/2015; 4 th, 6 th – Tue., 11/24/2015 CW: Valence Electrons handout Unit III Study Guide REVIEW: Periodic Trends (Refer to your Unit III packet, p. 8.) –SNC1P Bohr Diagram handout –Practice: Layout of the Periodic Table –Practice: Electron Configurations –PRACTICE! Periodic Trends, p. 9 –Periodicity of Atomic Radius and Ionization Energy handout 5

LecturePLUS Timberlake6 Chemical Bonds Attraction between two or more atoms Interaction between valence electrons Ionic bonds Covalent bonds

LecturePLUS Timberlake7 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 Timberlake8 Valence Electrons of Transition Metals The 4s orbital is at higher energy than the 3d orbitals (despite the filling order suggesting the opposite). This means that the two 4s electrons are lost first giving a valence of two (which is common to all the first row transition elements). The 3d orbitals is fully occupied so one of these two electrons will be preferentially released due to the repulsion of its pair. This results in a valence of three. These are by far the most common valences of iron. Example: Fe [Ar] 4s 2 3d 6 valence = 6e -

LecturePLUS Timberlake9 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 Timberlake10 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 Timberlake11 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 Timberlake12 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 Timberlake13 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 Timberlake14 Formation of Sodium Ion Sodium atom Sodium ion Na  – e   Na ( = Ne) 11 p + 11 p + 11 e - 10 e

LecturePLUS Timberlake15 Formation of Magnesium Ion Magnesium atom Magnesium ion  Mg  – 2e   Mg (=Ne) 12 p + 12 p + 12 e- 10 e

LecturePLUS Timberlake16 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 Timberlake17 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 Timberlake18 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 Timberlake19 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 Timberlake20 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 Timberlake21 Learning Check A. Why does Ca form a Ca 2+ ion? B. Why does O form O 2- ion?

LecturePLUS Timberlake22 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 Timberlake23 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 Timberlake24 Fluoride Ion unpaired electronoctet     1 - : F  + e  : F :     (= Ne) 9 p+ 9 p + 9 e- 10 e ionic charge

LecturePLUS Timberlake25 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 Timberlake26 Solution 5A 6A 7A N 3  O 2  F  nitride oxide fluoride P 3  S 2  Cl  phosphide sulfide chloride Br  bromide