1. Forming Ionic Compounds (Metal + Non-Metal)

2. Ionic compounds Electrons are transferred from a metal to a non-metal forming two oppositely charged ions. Both have the electronic configuration of a noble gas (stable outer shell once they form an ion) The ionic bond is held together by electrostatic forces (opposite charges) They ultimately form an ionic lattice.

3. Making an ionic compound Na + F The name of this compound is sodium fluoride Na F

4. So what you get is… Sodium became Na + and transferred its electron to chlorine who became Cl -. Their opposite charges are what hold this molecule together

5. But many of these will form a 3D large lattice structure which is very strong.

6. Naming ionic compounds Always name your metal FIRST Next, write the name of the non-metal and change the ending to –ide. For example: –Nitrogen  –Oxygen  –Fluorine  –Chlorine  –Sulphur  –Phosphorus  –Bromine  nitride oxide fluoride chloride sulphide phosphide bromide

7. Lewis Structures Lewis Dot diagrams are far more simpler & less confusing than Bohr-Rutherford diagrams. Lewis Dot diagrams uses only dots to indicate valence electrons only Rules: 1.there should be as many dots around the atom as there are valence electrons 2.The dots are placed around the atom in sequence. 3.LESS than four valence electrons: place each one around the atom on their own 4.MORE than four valence electrons: go back and double up on the other dots

8. Lewis Dot Diagrams Less than four Ex Na Mg Al C More than four N O F Ne

9. Using Lewis Dot Diagrams to determine the chemical formula Be + Cl BeCl Cl

10. Using the criss-cross method 1.Write each element 2.Include the charge that each element makes 3.Criss cross the numbers only 4.Name the compound Na + Br Na Br 1+1- NaBr Sodium bromide

11. Using Transition Metals Transition metals are also used in the same way. They take on the valence charge as indicated on the periodic table. Use the criss-cross method only. Ex. Sn 2+ F 1-  SnF 2 Tin (II) fluoride Sn 2+ O 2-  Sn 2 O 2  SnOTin (II) oxide Sn 4+ F 1-  SnF 4 Tin (IV) fluoride