1. #1 Count the total number of valence electrons in the molecule C: 4 electrons x 1 atom = 4 valence electrons H: 1 electron x 4 atoms = 4 valence electrons Total = 8 valence electrons

2. #2 Find the number of “octet” electrons for the molecule. C: 8 octet electrons x 1 atom = 8 octet electrons H: 2 octet electrons x 4 atoms = 8 octet electrons Total = 16 octet electrons Rules: Hydrogen always has 2 octet electrons Beryllium always has 4 octet electrons Boron wants 6 for neutral molecules (8 if it’s an anion)

3. #3 Subtract the number of valence electrons from the number of octet electrons to find the number of bonding electrons. 16 – 8 = 8 bonding electrons

4. #4 Divide the number of bonding electrons by 2 to find the number of bonds in the molecule. 8 / 2 = 4 bonds in CH 4 Why 2? Because there are two electrons in every covalent bond!

5. #5 Draw an arrangement of atoms that has the number of bonds you found in step 4. Of course, there are more rules… Hydrogen and Halogens = bond once! Oxygen’s family and Beryllium bond twice in neutral molecules Nitrogen’s family and Boron bond 3 times in neutral molecules Carbon’s family = bonds 4 times!

6. More hints… If you bond everything together and you have bonds left over, look for double/triple bonds. The atom nearest to the left side of the periodic table is probably in the middle of the molecule.

7. H 2 O CCl 4 NH 3 CO 2 PCL 3

8. N 2 CH 4 C 2 H 6 CH 3 OH BH 3

9. PBr 3 N 2 H 2 C 2 H 2 SiH 4 C 2 H 4

10. HF HCl CCl 2 F 2