1. Why do atoms form chemical bonds?
Lowest Energy Arrangement = Completely Filled Valence Electron Level
Atoms gain, lose or SHARE valence electrons in order to obtain a completely filled valence electron level.
First energy level: Filled = 1s2
All other energy levels = ns2np6
DUET RULE ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
OCTET RULE

2. # of Bonds formed = # of electrons needed to fill outer energy level

3. H H H F H Cl H Br H H C H H N H H O H H H
What patterns do you notice about #’s of bonds ( ) and nonbonding electrons (dots) and total valence e- on H?
H HF HCl HBr
H2O NH CH4 H H H F H Cl H Br H H C H H N H H O H H H

5. Duet Rule
Hydrogen always makes 1 covalent bond with no nonbonding electrons (dots) in order to fill the first energy level. H H H H H H H F H F

6. Group 7, Family 17 Halogen
ns2np5 F Cl Br I

7. What patterns do you notice about the # of bonds ( - ) and number of dots ( ∙ ) for F, Cl, Br and I?
F Cl Br I2
HF HCl HBr HI
H H H H
OF OCl OBr2 O O O

8. Group 6 or Family 16 Elements

9. Group 6 Elements: O, S, Se H O H H S H H Se H H C O H H H O H C H S Se

10. Group 5, Family 15: N, P, As,

11. Group 5: N, P, As H N H H H As P H H H H H N N H N N H

12. Group 4, Family 14: C, Si,

13. Group 4, Family 14: C, Si

14. For each pattern, identify the element that could represent X
Choices: A) H B) F C) C D) N E) O
E) O
1) X
A) H
2) X
D) N
3) X
4) X
C) C
B) F
5) X

15. In hydroxide -1 ion, why does O make 1 bond instead of the usual 2?
Ans: -1 charge means O has gained 1 valence e- and acts like an atom with 7 valence e- instead of an atom with 6 valence e-.

16. In ammonium + ion, why does N make 4 bonds here instead of the usual 3?
Ans: + charge means N has lost 1 valence e- and acts like an atom with 4 valence e- instead of an atom with 5 valence e-.

17. Link to Notes

18. Electron Count Method- for hard problems and octet exceptions
Step #1: Determine Central Atom
Formula Method: Single atom in formula =
# of Bonds Method:
Step #2: Connect all atoms with single bonds coming off central atom: C
C =4 bonds, H =1, Cl =1 H Cl C Cl H

19. Electron Count Method- for hard problems and octet exceptions
Count the total # of valence electrons for the molecule by adding the # from each atom
Determine dots available by subtracting off 2 e- for each bond from total H C Cl Cl C:
1 atom x 4 valence e- = 4 e- H H:
2 atoms x 1 valence e- = 2 e-
Cl:
2 atoms x 7valence e- = 14 e-
4 bonds x 2 e - / bond = 8 bond e -
Total valence e - = 20 e-
- 8 bonding e- = 12 dots to be placed

20. Electron Count CS2 example

21. Exceptions to the Octet Rule
ELECTRON DEFICIENT (FEWER THAN 8 VALENCE e-)
EXPANDED OCTETS (MORE THEN 8 VALENCE e-)

22. ELECTRON DEFICIENT (central atom = less than 8)
MEMORIZE: BeX2 and BX3 Where X = H, F, Cl, Br, I
Example #1: BeCl2
NOT IONIC (although metal + nonmetal)
NO DOUBLE BONDS (although Be does not have octet; halogens never double bond)
Be is stable with only 4 valence electrons; B is stable with only 6 valence electrons.

23. Electron Count Method (for octet-exception rule problems or very hard structures)
Formula Method = Be (single atom in formula)
1) Determine Central Atom :
2) Connect each outside atom with single bonds to central atom
3) Count the total # of valence electrons for the molecule by adding the # from each atom
4) Determine dots available by subtracting off 2 e- for each bond from total
NOTE: Be is stable with only 4 valence e- ; (less than normal 8); no double bond with Cl Be Cl Cl
Be:
1 atom x 2 valence e- = 2 e-
Cl:
2 atoms x 7 valence e- = 14 e-
2 bonds x 2 e - / bond = 4 bond e -
Total valence e - = 16 e-
- 4 bonding e- = 12 dots to be placed; fill outside atoms first

24. Electron Deficient (less than 8) Example #2: BF3
Central Atom = B (single atom in formula) F
NOTE: B is stable with only 6 valence e- ; (less than normal 8); no double bond with F F B F B:
1 atom x 3 valence e- = 3 e- F:
3 atoms x 7 valence e- = 21 e-
3 bonds x 2 e - / bond = 6 bond e -
Total valence e - = 24 e-
- 6 bonding e- = 18 dots to be placed; fill outside atoms first

25. Expanded Octets: (Central Atom has more than 8)
Central atoms from periods 3,4,5,6,7 (3rd row down) can expand octet by utilizing empty d orbitals.
Don’t have to memorize, problem will force you to expand octet.

26. Expanded Octet (more than 8) Example #1: PCl5
Central Atom = P (single atom in formula) Cl
NOTE: P is stable with 10 valence e- ; (more than normal 8) Cl Cl P Cl Cl P:
1 atom x 5 valence e- = 5 e-
Cl:
5 atoms x 7 valence e- = 35 e-
5 bonds x 2 e - / bond = 10 bond e -
Total valence e - = 40 e-
bonding e- = 30 dots to be placed; fill outside atoms first

27. Expanded Octet (more than 8) Example #2: IF4 -1
Central Atom = I (single atom in formula)
NOTE: I is stable with 12 valence e- ; (more than normal 8) F F I F F I:
1 atom x 7 valence e- = 7 e-
Add 1 e- for negative charge F:
4 atoms x 7 valence e- = 28 e-
4 bonds x 2 e - / bond = 8 bond e -
Total valence e - = 35 e e-
- 8 bonding e- = 28 dots to be placed; fill outside atoms first

28. Expanded Octet (more than 8) Example #2: SF6
Central Atom = S (single atom in formula) F
NOTE: S is stable with 12 valence e- ; (more than normal 8) F F S F F F S:
1 atom x 6 valence e- = 6 e- F:
6 atoms x 7 valence e- = 42 e-
6 bonds x 2 e - / bond = 12 bond e -
Total valence e - = 48 e-
bonding e- = 36 dots to be placed

29. Link to 9-8 Key
9-8 Blank

30. Draw the Lewis Dot Structure for CH2Cl2
3) Arrange dots so that each atom has filled outer energy level (Duet Rule for H, Octet Rule for Cl) H Cl C Cl H