2. Sodium Chlorine

3. Sodium Chlorine PNPN PNPN Electron configuration

4. Sodium Chlorine P P Electrons Protons

5. Sodium Chlorine P P Electrons Protons Charge Electrons Protons Charge

6. Sodium Chlorine P 11 P 17 Electrons 11 Protons 11 Charge Electrons 17 Protons 17 Charge I wish my outside shell was empty I wish my outside shell was full Atom 0 0 Ion Chloride Ion 10 18 +1

7. So one sodium atom can make one chlorine atom happy So one magnesium atom can make two chlorine atoms happy Gives 2 e- Accepts 1 e-

8. Magnesium Chlorine 2+

10. Group 7 Group 1 reacting with Group 7 Group 1

11. Group 6 Group 1 reacting with Group 6 Group 1

12. Group 5 Group 1 reacting with Group 5 Group 1

13. Group 7 Group 2 reacting with Group 7 Group 2

14. Group 6 Group 2 reacting with Group 6 Group 2

15. Group 5 Group 2 reacting with group 5 Group 2 Group 5 Group 2

16. Group 4 Can Give four or accept four

17. How many Chlorines could an Aluminium make happy? How many Oxygens could an Aluminium make happy? How many Chlorines could a Beryllium make happy? How many Lithiums could an Oxygen help out? What is the formula of Sodium fluoride? What is the formula of magnesium fluoride? What is the formula of Aluminium fluoride? Try these questions What is the formula of Aluminium oxide?

18. Sodium Chlorine P 11 P 17 Electrons 11 Protons 11 Charge Electrons 17 Protons 17 Charge Atom 0 0 Ion Chloride Ion 10 18 +1

19. Sodium P 11 Electrons 11 Protons 11 Charge Atom 0 Ion 10 +1

20. In each of the following cases Place the missing electron(s) Na  Na + Mg  Mg +2 S  S -2 F 2  2F -1 O 2  2O -2

23. Na  Na + Loses an electron but gains a positive charge

24. Cl + e -  Cl -

25. 2 Na + FeCl 2  2 NaCl + Fe

26. Before learning how to write balanced net ionic equations, we must first learn about electrolytes. Electrolytes If an aqueous solution of a compound conducts electricity, it is called an electrolyte. The ability to conduct electricity results from the dissociation of the compound into ions in solution. There are two types of electrolytes:

27. 1.Strong Electrolytes - in aqueous solution, these compounds dissociate 100% and exist as ions in solutions. They include strong acids (HCl, HBr, HI, HNO 3, H 2 SO 4 and HClO 4 ), strong bases (all the Group IA and IIA hydroxides) and all soluble salts. On the next slide is a general solubility table and in your notes is a more extensive table of common polyatomic ions

29. 1.Weak Electrolytes and/or nonelectrolytes - in aqueous solution, these compounds yield few to no ions in solution and should be represented by the molecular formula. They include: water, weak acids (any acid that isn't strong), weak bases (any base that isn't strong) and nonelectrolytes. “ Once you have learned the types of electrolytes, writing a balanced equation is best explained by example. ”

30. Example #1: Write the balanced net ionic equation for the reaction of aqueous sodium hydroxide and aqueous hydrochloric acid Step #1: Write the balanced GENERAL EQUATION - In order to write this equation, you must decide what the products are. This example problem is an acid-base reaction. The products will be a salt (NaCl) and water. After you have written the reaction, it must be balanced.

31. Step #2: Write the TOTAL IONIC EQUATION - Here, each reactant and product is studied to determine whether it dissociates in solution. If it is a strong electrolyte, it is written as ions. If it isn't a strong electrolyte it is written as a molecule. Because NaOH, HCl and NaCl are strong electrolytes they are written as ions. Water is a nonelectrolyte and should be written as a molecule.

32. Step #3: Write the NET IONIC EQUATION - Each species that does not undergo a change is called a "spectator ion". These species are removed from the equation leaving the balanced net ionic equation In this example, Na+ and Cl- are spectator ions. They do not undergo change in the reaction. Therefore, they are removed.

33. Step 1 Write a balanced net ionic equation for the reaction of solid sodium hydroxide with nitric acid. NaOH (s) + HNO 3(aq)  NaNO 3(aq) +H 2 O (l) Step 2 Full Ionic equation NaOH + H + + NO 3 -  Na + + NO 3 - + H 2 O It is important to point out that solids are not considered separate ions Step 3 Cancel out spectator ions NaOH + H +  Na + + H 2 O

34. Step 1 Write a balanced net ionic equation for the reaction of Sodium with Water. 2 Na (s) + 2H 2 O (l)  2 NaOH (aq) +H 2(g) Step 2 Full Ionic equation 2Na (s) + 2H 2 O  2Na + + 2OH - + H 2(g) It is important to point out that solids are not considered separate ions Step 3 Cancel out spectator ions 2Na (s) + 2H 2 O  2Na + + 2OH - + H 2(g)

35. We Do Step 1 Write a balanced net ionic equation for the reaction of Aqueous Silver nitrate (AgNO 3 ) with Aqueous Sodium Chloride. Step 2 Full Ionic equation Step 3 Cancel out spectator ions