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

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

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? H2 HF HCl HBr H2O NH3 CH4 H H H F H Cl H Br H H C H H N H H O H H H

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

Group 7, Family 17 Halogen ns2np5 F Cl Br I

What patterns do you notice about the # of bonds ( - ) and number of dots ( ∙ ) for F, Cl, Br and I? F2 Cl2 Br2 I2 HF HCl HBr HI H H H H OF2 OCl2 OBr2 O O O

Group 6 or Family 16 Elements

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

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

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

Group 4, Family 14: C, Si,

Group 4, Family 14: C, Si

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

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-.

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-.

Link to Notes

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

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

Electron Count CS2 example

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

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.

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

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

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.

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- - 10 bonding e- = 30 dots to be placed; fill outside atoms first

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- + 1 e- - 8 bonding e- = 28 dots to be placed; fill outside atoms first

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- - 12 bonding e- = 36 dots to be placed

Link to 9-8 Key 9-8 Blank

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