1. Chemistry 11 Review Group Period Family Electronic configuration Orbital Long version Short version Atoms Ions: cations & anions Valence Atomic charge Periodic Trends Atomic radius vs Electronegativity vs Electron affinity Metals Non Metals Metalloids Avogadro # = 6.02 x 10 23 Diagonal rule Balancing Equations Types of Chemical Equations (S), (D), (SD), (DD), (CC), (IC) STP / SATP Mole Calculations Stoechiometry Common Ions (+) & (-) Lewis Diagram Bonding % composition Nomenclature

2. █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ 1 IA 2 IIA 3 IIIB 4 IVB 5 VB 6 VIB 7 VIIB 8 VIIIB 9 VIIIB 10 VIIIB 11 IB 12 IIB 13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA 18 VIIIA Group New Version Old Version Period 1 2 3 4 5 6 7 6 7 Family Alkali metals Alkaline earth metals Metals of transition Boron FamilyCarbon FamilyNitrogen FamilyOxygen FamilyHalogen FamilyNoble Gas Family Orbitals p d f End of the electronic configuration ELECTRONIC CONFIGURATION s Diagonal rule 1s 1 1s 2 2s 1 2s 2 2p 1 2p 2 2p 3 2p 4 2p 5 2p 6 3s 1 3s 2 3p 1 3p 2 3p 3 3p 4 3p 5 3p 6 4s 1 4s 2 Lantanide series Actinides series 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 3d 8 3d 9 3d 10 4p 1 4p 2 4p 3 4p 4 4p 5 4p 6 5s 1 5s 2 4d 1 4d 2 4d 3 4d 4 4d 7 4d 5 4d 6 4d 8 4d 9 4d 10 5p 1 5p 2 5p 3 5p 4 5p 5 5p 6 6s 1 6s 2 5d 1 5d 2 5d 3 5d 4 5d 7 5d 6 5d 8 5d 9 5d 10 6d 1 6d 2 6d 3 6d 4 6d 7 6d 6 6d 8 6d 9 6d 10 5d 5 6d 5 7s 1 7s 2 6p 1 6p 2 6p 3 6p 4 6p 5 6p 6 7p 1 7p 2 7p 3 7p 4 7p 5 7p 6 4f 1 4f 2 4f 3 4f 4 4f 5 4f 6 4f 7 4f 8 4f 9 4f 10 4f 11 4f 12 4f 13 4f 14 5f 1 5f 2 5f 3 5f 4 5f 5 5f 6 5f 7 5f 8 5f 9 5f 10 5f 11 5f 12 5f 13 5f 14 Metals Non-Metals Metalloids

3. Atoms vs Ions Atomic number (Z)(p + ) Mass number (A)(p + +n o ) proton p + neutron n o electron e - 12 24.3u 1224.3u12 10 1735.45u171817 Cat + ions (+) Anions (-) 1735.45u1718

4. Electronic Configuration Long version vs Short version and Ions Long Version 20 Ca: 1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 4s 2 since it has 20 e - and 20 exp. 20 Ca 2+ : 1s 2, 2s 2, 2p 6, 3s 2, 3p 6 since it has 18 e - and not 20 e- 53 I:1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 4s 2, 3d 10, 4p 6, 5s 2, 4d 10, 5p 5 53 I - :1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 4s 2, 3d 10, 4p 6, 5s 2, 4d 10, 5p 6 Short Version Pick the previous noble gas + … 20 Ca: [Ar], 4s 2 20 Ca 2+ : [Ne], 3s 2, 3p 6 53 I: [Kr], 5s 2, 4d 10, 5p 5 53 I-: [Kr], 5s 2, 4d 10, 5p 6

5. █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ █ VALENCE ELECTRON: An electron that occupies the outermost energy level, or shell, of an atom. ATOMIC CHARGE: The electric charge of an ion, equal to the number of electrons the atom has gained or lost. VALENCE CHARGE 11+11+ 22+22+ Most of the metal of transition have a valence of 2 which mean they would have a charge of 2 + 33+33+ 62-62- 53-53- 4 4 +- 71-71- 2 or 8 0

6. Periodic Trends Atomic radius: The distance from the centre of the nucleus to the outermost shell of an atom.

7. Periodic Trends Electronegativity: A relative measure of an atom’s ability to attract the shared electrons in a chemical bond.

8. Periodic Trends Ionization Energy: the energy needed to remove electrons from an atom. Large atoms require low ionization energy while small atoms require high ionization energy.

9. Periodic Trends SUMMARY

10. This means that each human on earth has to eat 35000 eggs per second for 80 years?????? What about the # of chicken to produce that amount of eggs???? Avogadro # = 6.02 x 10 23 How many eggs does each human on earth has to eat every second to add up to 1 mol in 80 years? 80 years x 365 day x 24 hours x 60 min. x 60 sec. = 2 522 880 00 s Current earth population = 6 800 000 000 x 2 522 880 000 s = 1.72 x 10 19 s 1 mole of eggs = 6.02 x 10 23 eggs 6.02 x 10 23 eggs = 35000 eggs / s 1.72 x 10 19 s 1 year1 day 1 hour 1 min.

11. Avogadro # = 6.02 x 10 23 Avogadro’s Number: 1 mole of gas @ … STP – Standard Temperature and Pressure: V= 22.4L, 0ºC(273 K), 101.3 kPa SATP – Standard Ambient Temperature Pressure V= 24.8L, 25ºC(298 K), 101.3kPa

12. % Composition Percent Composition: The percentage, by mass, that each element that makes up a compound, out of 100%. To find the percent composition you must divide the mass of the element by the mass of the compound and multiply the result by 100. H 2 CO 3 H x 2 = 2.0g/mol %H = 2/62 x 100 = 3.2% C x 1 = 12.0g/mol%C = 12/62 x 100 = 19.4% O x 3 = 48.0g/mol%O = 48/62 x 100 = 77.4% Total = 62g/mol

13. 1+ Charge2+ Charge3+ Charge4+ Charge Li + LithiumBe 2+ BerylliumAl 3+ AluminiumSn 4+ Tin (IV) Na + SodiumMg 2+ MagnesiumCr 3+ Chromium (III)Pb 4+ Lead (IV) K+K+ PotassiumCa 2+ CalciumFe 3+ Iron (III) Ag + SilverBa 2+ BariumAu 3+ Gold (III) Cu + Copper (I)Fe 2+ Iron (II)Co 3+ Cobalt (III) NH 4 + AmmoniumPb 2+ Lead (II)Ni 3+ Nickel (III) H+H+ HydrogenNi 2+ Nickel (II) Zn 2+ Zinc Hg 2+ Mercury (I) Sn 2+ Tin (II) Co 2+ Cobalt (II) Cu 2+ Copper (II) Common Ions (+)

14. Common Ions (-)

15. Nomenclature 1. Magnesium Chlorate 2. Mercury II Chlorite 3. Lithium Sulfate 4. Zinc acetate 5. Sodium phosphide 4. Zinc acetate 5. Sodium phosphide 3. Lithium Sulfate 4. Zinc acetate 3. Lithium Sulfate 4. Zinc acetate Mg(ClO 3 ) 2 Hg(ClO) 2 Li 2 SO 2 Zn(CH 3 CH 2 ) 2 Na 3 P Nomenclature Game Dimitry Vinagradov sofware

16. Bonding A covalent bond is a form of chemical bonding that is characterized by the sharing of pairs of electrons between atoms, and other covalent bondschemical bonding You tube video: Covalent Bond An ionic bond is a type of chemical bond formed through electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a metal and a nonmetal ion.chemical bondionsmetalnonmetal You tube video: Ionic Bond

17. Covalent Bonds

18. What kind of bonds are theses? Song

19. Difference in Electronegativity

20. Lewis Symbols or Diagrams Elemental properties and reactions are determined only by electrons in the outer energy levels. Electrons in completely filled energy levels are ignored when considering properties. Simplified Bohr diagrams which only consider electrons in outer energy levels are called Lewis Symbols. A Lewis Symbol consists of the element symbol surrounded by "dots" to represent the number of electrons in the outer energy level as represented by a Bohr Diagram. The number of electrons in the outer energy level is correlated by simply reading the Group number. Lewis symbols for oxygen, fluorine, and sodium are given in the diagram on the left. Lewis Symbols for the elements of the second period. Correlate the number of dots with the group number.

21. Balancing Equations A chemical equation is an expression for a chemical reaction. It is a quantitative statement indicating the number of moles of each reactant and of each product. Reactants  Products In chemical equations, matter must be conserved. The number of atoms of each kind on the reactant side must equal those on the product side. __Fe + __H 2 O → __Fe 2 O 3 + __H 2 2 3 3 __P 4 + __Cl 2 → __PCl 3 46 DONE! __Fe 2 O 3 + __CO → __CO 2 + __Fe 3 3 2 DONE! Σ R = Σ P

22. Types of Chemical Equations SUMMARY Let E= element and C = Compound ReactantsReaction TypeProductsGeneral Equation E 1 + E 2 Combination (C)C R + S  RS CDecomposition (D)E 1 + E 2 RS  R + S E 1 + C 1 Single Displacement (SD) E 2 + C 2 T + RS  TS + R C 1 + C 2 Double Displacement (DD) C 3 + C 4 RS + TU  RU + TS Hydrocarbon + O 2 Complete Combustion (CC) CO 2 + H 2 O CH 4 +2O 2  CO 2 + 2H 2 O Hydrocarbon + O 2 Incomplete Combustion (IC) CO + H 2 O 2CH 4 +3O 2  2CO+ 4H 2 O

23. Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions.chemistryreactantsproducts In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers.

24. Mole Calculations Mass – Mass: 2H 2 + O 2  2H 2 O How many g of O 2 is needed to burn with 10g H 2 ? 10g H 2 x 1 mol H 2 x 1 mol O 2 x 32g O 2 = 2.02g H 2 2 mol H 2 1 mol O 2 80g O 2

25. Mole Calculations Mass – Mol: 2KClO 3  2KCl + 3O 2 With 5g of KClO 3, how many moles of KCl would you get? 5g KClO 3 x 1mol KClO 3 x 2 mol KCl = 122.45g KClO 3 2 mol KClO 3 0.041mol KCl

26. Mole Calculations Mass – Molecules: 2H 2 + O 2  2H 2 O How many molecules of O 2 would you get if you start with 10g H 2 ? 10g H 2 x 1 mol H 2 x 1 mol O 2 x 6.02 x 10 23 molecule O 2 = 2.02g H 2 2 mol H 2 1 mol O 2 1.5 x 10 24 molecule O 2

27. Mole Calculations Mass – Volume: N 2 + 3H 2  2NH 3 With 25g N 2, find the volume of NH 3 at STP 25g N 2 x 1 mol N 2 x 2 mol NH 3 x 22.4L NH ₃ = 28g N 2 1 mol N 2 1 mol NH 3 40L NH 3

28. Limited and Excessive Reactant Limited Reactant (LR) – The substance in the formula that is used up first. Excessive Reactant (ER) – The substance you have more of. NH 3 + HCl  NH 4 Cl What mass of ammonium chloride would you get if you add 1g NH ₃ and 1g HCl? 1g NH 3 = 0.0588 1g HCl = 0.0274 17g NH 3 36.45g HCl 1.47g NH 4 Cl HCl is your limited reactant since you get a smaller ratio. Therefore you get… 0.0274mol HCl x 1mol NH 4 Cl x 53.5g NH 4 Cl = 1 mol HCl 1 mol NH 4 Cl This means that if you started with 1g HCl and it gave you 1.47 g of NH 4 Cl you can conclude that you used 0.47g of NH 3 therefore since we started with 1g of NH 3 we know that we have 0.53 g extra.

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