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Ch7- Ionic Bonds

Ch7- Ionic Bonds. Valence electrons- electrons in the highest energy level (only s & p) - # of valence electrons corresponds to the group. 1 18 H 2 13 14 15 16 17 He

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Ch7- Ionic Bonds

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  1. Ch7- Ionic Bonds Valence electrons- electrons in the highest energy level (only s & p) - # of valence electrons corresponds to the group. 1 18 H 2 13 14 15 16 17He LiBe B C N OF Ne Na Mg Al Si P S ClAr lose electrons to gain electrons to form form positive ions negative ions called called cations. anions. (ANegative ION) All atoms are trying to satisfy the octet rule - get 8 electrons in outer energy level.

  2. Ex1) Write the electron configurations and dot diagrams for: Na Cl Write the e.c. & dot diagram for the ions: Na+ : Cl- : What is the chemical formula for sodium chloride?

  3. Ex1) Write the electron configurations and dot diagrams for: Na Cl [Ne] 3s1 [Ne] 3s2 3p5 Write the e.c. & dot diagram for the ions: Na+ : [Ne] or [Na]+Cl- : [Ne] 3s2 3p6 or [Cl]- What is the chemical formula for sodium chloride? NaCl

  4. Ex 2) Write the electron configure & dot diagram for: Al Br Write the e.c. & d.d. for the ions: Al+3 Br-1 What is the chemical formula for aluminum bromide?

  5. Ex 2) Write the electron configure & dot diagram for: . . . . . . Br . Al . . . [Ne] 3s23p1 [Ar] 4s23d104p5 Write the e.c. & d.d. for the ions: Al+3 = [Ne]+3 Br-1 = [Ar] 4s23d104p6 or [Kr]-1 What is the chemical formula for aluminum bromide? AlBr3

  6. Ex 3) Same stuff for: Atoms: Mg N Ions: Formula:

  7. Ex 3) Same stuff for: Atoms: Ions: Formula: N . . . . . . . Mg [Ne]3s2 [He] 2s22p3 Mg+2 = [Ne]+2N-3= [He] 2s22p6 = [Ne]-3 . Mg . . . Mg3N2 . N . . . Mg . . . . . N Properties of Ionic Compounds - solids at room temp - don’t conduct electricity in solid state - will conduct electricity if 1 molten (liquid state) 2 dissolved in water in both these cases the ions are free to move. . . Mg Ch7 HW#1 .

  8. Ch7 HW#1 1) How many valence electrons for: a) Potassium b) Carbon c) Magnesium d) Oxygen 2) Dot structures & electron configurations: K C Mg O 3) Gain or Lose? K  C  Mg  O  4) E.C. for ions K+1C+4Mg+2 O-2 C-4 5) Why do nonmetals form anions when reacting?

  9. Ch7 HW#1 1) How many valence electrons for: a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6 2) Dot structures & electron configurations: K C Mg O 3) Gain or Lose? K  C  Mg  O  4) E.C. for ions K+1C+4Mg+2 O-2 C-4 5) Why do nonmetals form anions when reacting?

  10. Ch7 HW#1 1) How many valence electrons for: a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6 2) Dot structures & electron configurations: K C Mg O [Ar] 4s1[He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4 3) Gain or Lose? K  C  Mg  O  4) E.C. for ions K+1C+4Mg+2 O-2 C-4 5) Why do nonmetals form anions when reacting?

  11. Ch7 HW#1 1) How many valence electrons for: a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6 2) Dot structures & electron configurations: K C Mg O [Ar] 4s1[He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4 3) Gain or Lose? K  Lose 1 C  gain or lose 4 elec Mg  lose 2 O  gain 2 4) E.C. for ions K+1C+4Mg+2 O-2 C-4 5) Why do nonmetals form anions when reacting?

  12. Ch7 HW#1 1) How many valence electrons for: a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6 2) Dot structures & electron configurations: K C Mg O [Ar] 4s1[He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4 3) Gain or Lose? K  Lose 1 C  gain or lose 4 elec Mg  lose 2 O  gain 2 4) E.C. for ions K+1[Ar] C+4[He] Mg+2[Ne] O-2[Ne]2s22p4 C-4[He] 2s22p6 [Ne] [Ne] 5) Why do nonmetals form anions when reacting?

  13. Ch7 HW#1 1) How many valence electrons for: a) Potassium b) Carbon c) Magnesium d) Oxygen 1 4 2 6 2) Dot structures & electron configurations: K C Mg O [Ar] 4s1[He] 2s22p2 [Ne] 3s2 [Ne] 2s22p4 3) Gain or Lose? K  Lose 1 C  gain or lose 4 elec Mg  lose 2 O  gain 2 4) E.C. for ions K+1[Ar] C+4[He] Mg+2[Ne] O-2[Ne]2s22p4 C-4[He] 2s22p6 [Ne] [Ne] 5) Why do nonmetals form anions when reacting? Have 4 or more electron in outer energy level Easier to gain electrons to satisfy octet rule.

  14. 6) Dot formula to determine formulas: a) potassium & iodine b) Ca & S K I Ca S c) Al & O d) Na & P Al Na Na P Na O O Al O 7) Name:

  15. 6) Dot formula to determine formulas: a) potassium & iodine b) Ca & S K I Ca S KClCaS c) Al & O d) Na & P Al Na Na P Na O O Al O 7) Name:

  16. 6) Dot formula to determine formulas: a) potassium & iodine b) Ca & S K I Ca S KClCaS c) Al & O d) Na & P Al Al2O3 Na Na3P Na P Na O O Al O 7) Name:

  17. 6) Dot formula to determine formulas: a) potassium & iodine b) Ca & S K I Ca S KClCaS c) Al & O d) Na & P Al Al2O3 Na Na3P Na P Na O O Al O 7) Name: Potassium iodide Calcium sulfide Aluminum oxide Sodium phosphate

  18. Ch7.2 – Metallic Compounds + + + + + + + + + + + + + + + + + - + - have free floating valence electrons - good conductors of heat and electricity - if electrons enter one end, others will exit other end. (basis of electric circuits) - ductility and malleability caused by electrons shielding cations from each other, even when the metal is smashed or bent. (ionic solids shatter because like charges get pressed together, then repel.)

  19. Metallic Structures: (common, simple ones) Body-centered cubes (BCC) - every atom has 8 neighbors - Na,K,Fe,Cr,W Face-centered cubic (FCC) - every atom has 12 neighbors - Cu,Ag,Au,Al,Pb Hexagonal Closest Packing (HCP) -12 neighbors, but diff shape - Mg,Zn,Cd Quiz tomorrow: 1.Balance equation: 2.Mass-mass: 10 grams of ____ reacts with excess ____. How much ____ produced? 3.Density: Givens: grams and cm3 4.Temp conversion: ˚C = ___K CH7 HW#2 8-12

  20. Ch7 HW#2 8 – 12 8) Why do metals tend to form cations? 9) Electron configurations: a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10 b) Gold (I) Au+1Au: [Xe] 6s24f145d9 6s14f145d10 c) Cadmium Cd+2 Cd: [Kr] 5s24d10 d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10 10) Ductile - Malleable - 11) Electron configurations for +3 charged: a) Chromium Cr: [Ar] 4s23d4 b) Manganese Mn[Ar] 4s23d5 c) Iron Fe: [Ar] 4s23d6

  21. Ch7 HW#2 8 – 12 8) Why do metals tend to form cations? Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons 9) Electron configurations: a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10 b) Gold (I) Au+1Au: [Xe] 6s24f145d9 6s14f145d10 c) Cadmium Cd+2 Cd: [Kr] 5s24d10 d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10 10) Ductile - Malleable - 11) Electron configurations for +3 charged: a) Chromium Cr: [Ar] 4s23d4 b) Manganese Mn[Ar] 4s23d5 c) Iron Fe: [Ar] 4s23d6

  22. Ch7 HW#2 8 – 12 8) Why do metals tend to form cations? Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons 9) Electron configurations: a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10 Cu+1: [Ar] 3d10 b) Gold (I) Au+1Au: [Xe] 6s24f145d9 6s14f145d10 Au+1: [Xe] 4f145d10 c) Cadmium Cd+2 Cd: [Kr] 5s24d10 d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10 10) Ductile - Malleable - 11) Electron configurations for +3 charged: a) Chromium Cr: [Ar] 4s23d4 b) Manganese Mn[Ar] 4s23d5 c) Iron Fe: [Ar] 4s23d6

  23. Ch7 HW#2 8 – 12 8) Why do metals tend to form cations? Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons 9) Electron configurations: a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10 Cu+1: [Ar] 3d10 b) Gold (I) Au+1Au: [Xe] 6s24f145d9 6s14f145d10 Au+1: [Xe] 4f145d10 c) Cadmium Cd+2 Cd: [Kr] 5s24d10 Cd+2[Kr] 4d10 d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10 Hg+2:[Xe] 4f145d10 10) Ductile - Malleable - 11) Electron configurations for +3 charged: a) Chromium Cr: [Ar] 4s23d4 b) Manganese Mn[Ar] 4s23d5 c) Iron Fe: [Ar] 4s23d6

  24. Ch7 HW#2 8 – 12 8) Why do metals tend to form cations? Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons 9) Electron configurations: a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10 Cu+1: [Ar] 3d10 b) Gold (I) Au+1Au: [Xe] 6s24f145d9 6s14f145d10 Au+1: [Xe] 4f145d10 c) Cadmium Cd+2 Cd: [Kr] 5s24d10 Cd+2[Kr] 4d10 d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10 Hg+2:[Xe] 4f145d10 10) Ductile - a metal can be drawn into a wire Malleable - a metal can be pounded into a sheet (or shape) 11) Electron configurations for +3 charged: a) Chromium Cr: [Ar] 4s23d4 b) Manganese Mn[Ar] 4s23d5 c) Iron Fe: [Ar] 4s23d6

  25. Ch7 HW#2 8 – 12 8) Why do metals tend to form cations? Easier to lose 1 – 3 electrons rather than gain 5 – 7 electrons 9) Electron configurations: a) Copper (I) Cu+1 Cu: [Ar] 4s2Cd9 [Ar] 4s13d10 Cu+1: [Ar] 3d10 b) Gold (I) Au+1Au: [Xe] 6s24f145d9 6s14f145d10 Au+1: [Xe] 4f145d10 c) Cadmium Cd+2 Cd: [Kr] 5s24d10 Cd+2[Kr] 4d10 d) Mercury (II) Hg+2 Hg: [Xe] 6s24f145d10 Hg+2:[Xe] 4f145d10 10) Ductile - a metal can be drawn into a wire Malleable - a metal can be pounded into a sheet (or shape) 11) Electron configurations for +3 charged: a) Chromium Cr: [Ar] 4s23d4Cr+3: [Ar] 3d3 b) Manganese Mn[Ar] 4s23d5Mn+3: [Ar] 3d4 c) Iron Fe: [Ar] 4s23d6Fe+3: [Ar] 3d5

  26. 12) Use dots to combine: a) Cu(I) & Cl: Cu Cl b) Cu(II) & Cl: Cu Cl Cl c) Fe (II) & O Fe O d) Fe (III) & O O Fe O Fe O

  27. 12) Use dots to combine: a) Cu(I) & Cl: Cu ClCuCl b) Cu(II) & Cl: Cu Cl CuCl2 Cl c) Fe (II) & O Fe O d) Fe (III) & O O Fe O Fe O

  28. 12) Use dots to combine: a) Cu(I) & Cl: Cu ClCuCl b) Cu(II) & Cl: Cu Cl CuCl2 Cl c) Fe (II) & O Fe O FeO d) Fe (III) & O Fe2O3 O Fe O Fe O

  29. Ch7 Rev WS 1. Use the periodic table to find the number of valence electrons in an atom. a. sodium b. carbon c. phosphorus Draw the electron dot formulas of these representative elements: K AI O CI 2. Describe the formation of a cation from an atom of a metallic element, using the octet rule and the importance of noble-gas electron configurations. Describe the formation of the sodium ion using an electron dot structure.

  30. 3. Describe the formation of an atom of a non metallic element. Describe the formation of the chloride ion using an electron dot structure. 4. List the characteristics of an ionic bond.

  31. 5. Explain the electrical conductivity of melted and of aqueous solutions of ionic compounds, using the characteristics of ionic compounds. 6.Explain the physical properties of metals, using theory of metallic bonding.

  32. B. Questions 22. Write electron dot structures for the following atoms: a. silicon b. rubidium c. barium d. tin e. iodine f. arsenic Si RbBaSn I As 23. complete the following table. outer electron outer electron formula of ion type of ion config of atom config of ion Se K Ca Br N

  33. 24. Use electron dot formulas to determine chemical formulas of the ionic compound formed when the fallowing elements combine. a. strontium and fluorine b. magnesium and chlorine c. sodium and oxygen 25. Same for aluminum and nitrogen and then name it.

  34. Formulas of Ionic Compounds Elements exchange electrons in ionic bonds . Cations form from metals by the loss of valence electrons. Anions form from nonmetals by the gain of electrons. Ionic bonds form as the result of oppositely charged ions attracting one another. An ionic compound always contains at least one positive ion (cation) and one negative ion (anion.) These ions must combine in such a way as to produce a neutral compound. The formula unit of an ionic compound represents the smallest sample of an ionic compound that has the composition of that compound. This formula unit will reflect the balance of charges of the compound’s ions. The use of electron dot formulas is a helpful tool in predicting formulas of ionic compounds. This worksheet will help to show you how to write formulas of various ionic compounds. 1. How many valence electrons does the element iodine have? What is the formula for iodine’s most stable ion?

  35. Example B Remember that metals lose valence electrons, and nonmetals gain electrons in order to achieve electron configurations resembling those of noble gases. Sufficient numbers of atoms of each element must be included in the formula so that the number of electrons lost by one element is equal to the number of electrons gained by the other element. Sulfur needs two electrons to fill its octet of electrons. Sodium has only one electron to lose. 2. Determine the formula of the ionic compound formed when barium and phosphorus combine.

  36. 3. How many valence electrons does the element gallium have? 4. Write the formula for the ion formed when nitrogen gains electrons to attain a noble gas configuration. 5. What is the formula for the compound formed when astatine and strontium combine?

  37. Lab7.1 – Models - due in 2 days - Ch7 Rev WS due at beginning of period

  38. Ch8.1 – Covalent Bonds Single Covalent Bonds – one pair of electrons shared between 2 atoms. Ex1) H + H HHH-H one shared pair Structural formula Each dash represents one shared pair of electrons. . . : Ex2) F2 Ex3) H2O

  39. Ch8.1 – Covalent Bonds Single Covalent Bonds – one pair of electrons shared between 2 atoms. Ex1) H + H HHH-H one shared pair Structural formula Each dash represents one shared pair of electrons. . . : .. .. .. .. . . : : : : Ex2) F2 F + F F–F Unshared pairs of electrons (nonbonding pairs) Take up more space then the shared pairs. .. .. .. .. Ex3) H2O

  40. Double Covalent Bonds – 2 shared pairs of electrons. Triple Covalent Bonds – 3 shared pairs of electron. Ex4) O2 Ex5) N2

  41. Covalent Compounds Ex6) NH3 Ex7)CH4 Ex8) CO2 Ch8 HW # 1

  42. Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2 ClCl Br Br I I 2) How many unshared pairs are in each halogen molecule 3) Why necessary to form double & triple bond sometimes? 4) How many electrons does each atom contribute in: Double Bond? Triple Bond?

  43. Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2 ClCl Br Br I I Cl–Cl Br – Br I – I 2) How many unshared pairs are in each halogen molecule 3) Why necessary to form double & triple bond sometimes? 4) How many electrons does each atom contribute in: Double Bond? Triple Bond?

  44. Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2 ClCl Br Br I I Cl–Cl Br – Br I – I 2) How many unshared pairs are in each halogen molecule 3 pairs per atom, 6 pairs per molecule 3) Why necessary to form double & triple bond sometimes? 4) How many electrons does each atom contribute in: Double Bond? Triple Bond?

  45. Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2 ClCl Br Br I I Cl–Cl Br – Br I – I 2) How many unshared pairs are in each halogen molecule 3 pairs per atom, 6 pairs per molecule 3) Why necessary to form double & triple bond sometimes? For stability, to satisfy the octet rule 4) How many electrons does each atom contribute in: Double Bond? Triple Bond?

  46. Ch8 HW#1 1 – 5 1) Dot structures for diatomic molecules a) Chlorine, Cl2 b) Bromine, Br2 c) Iodine, I2 ClCl Br Br I I Cl–Cl Br – Br I – I 2) How many unshared pairs are in each halogen molecule 3 pairs per atom, 6 pairs per molecule 3) Why necessary to form double & triple bond sometimes? For stability, to satisfy the octet rule 4) How many electrons does each atom contribute in: Double Bond? 2 electrons per atom Triple Bond? 3 electrons per atom

  47. 5) Dot structures a) H2S H S H H b) PH3 P H H c) ClFCl F

  48. 5) Dot structures a) H2S H S H S H-S-H H H H b) PH3 P H H c) ClFCl F

  49. 5) Dot structures a) H2S H S H S H-S-H H H H HH b) PH3 P H P H P-H H HH c) ClFCl F

  50. 5) Dot structures a) H2S H S H S H-S-H H H H HH b) PH3 P H P H P-H H HH c) ClFCl F Cl F Cl-F

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