 LEM › A molecule is composed of atoms that are bound together by sharing pairs of e- using the atomic orbitals of the bound atoms. › Lone pairs › Bonding pairs

1. Lewis structure 2. VSEPR model (geometry of molecule) 3. Type of atomic orbitals 1. Share e- 2. Hold lone pairs

 Duet rule › Share 2 e- › Hydrogen and helium  Octet rule › Share 8 e-

Determine the total number of valence electrons to be combined. C: 1 x 4ve- +)Cl: 4 x 7ve- Total ve-

Arrange the atoms to form a dot structure for the molecule. If carbon is present, it is the central atom. Otherwise, the least-electronegative atom is central (except hydrogen, which is never central).

Surround it with the other atoms in a symmetrical fashion. Connect the outside atoms to the central atom with two dots.

Add unshared pairs of electrons so that each other nonmetal is surrounded by 8 electrons and each hydrogen atom shares a pair of electrons. Ve- available:

Count the electrons in the structure. Step #5: Ve- available:

Find the difference of ve-needed from step 3 and ve- available from step 5. ve- needed (step3): -) ve- available (step5): Difference:

If the difference is 0, all single bonds. If the difference is 2, one double. If 4, one triple or 2 double. ve- needed (step3): -) ve- available (step5): Difference: If the difference is 2 or 4….. Double or triple bond

Ex: SO 2 (ws7) S: 6ve- O: 6ve- ve needed = 18 Ve- available = 20 O-S-O Difference: 2 1 double bond O-S=O ve- needed (step3): -) ve- available (step5): Difference:

 Draw polyatomic ions  Ex. NH 4 + ve- needed (step3): -) ve- available (step5): Difference:

 Draw polyatomic ions  Ex. SO 4 2- ve- needed (step3): -) ve- available (step5): Difference:

1. HF 2. N 2 3. NO + (cation  minus 1) 4. SO 2 5. CO 3 2- (anion  add 2)

 Boron › Not have a complete octet › E- deficient  2 nd row in PT › Never break octet rule  Period 3 and beyond › Exceed the octet rule by using d orbitals

 S exceed the octet rule  SF 6  F need to follow octet   S need 12 e- (use d orbital)

 P: 1x5=5ve-  Cl: 5x7=35ve-  Total = 40ve-  Arrange atoms  P needs 5 bonding pairs  Count all e-  40ve-

 I: 3x7=21ve-+1=22ve- needed  Arrange atoms  Assume all follow octet rule  Count all e-  20ve-  Add 2more ve- around central atom

a. ClF 3 b. XeO 3 c. RnCl 2 d. BeCl 2 e. ICl 4 -

a. ClF 3 b. XeO 3 c. RnCl 2 d. BeCl 2 e. ICl 4 -

 When more than one valid Lewis structure can be written for a particular molecule  Connect w/double-headed arrows  Does not mean “flips”  Usually 1 or 2 double bonds in a structure

 Difference b/w the # of ve- on the free atom and the # of ve- assigned to the atom in the molecule

 Formal Charge = (# of ve- on free atom) – (# of ve- assigned to the atom in the molecule)  # of ve- assigned to a given atom = (# of lone pair e-)-1/2 (# of shared electrons)

 # of ve- assigned to a given atom = (# of lone pair e-) + 1/2 (# of shared electrons) › For OFor S › = 6 + 1/2(2)=7= 0 + ½(8) = 4  Formal Charge = (# of ve- on free atom) – (# of ve- assigned to the atom in the molecule) › = 6 – 7 = -1For S = = 2

 # of ve- assigned to a given atom = (# of lone pair e-) + 1/2 (# of shared electrons) › For O (single)For S › 6 + 1/2(2)=7= 0 + ½(12) = 6 › For O (double) › 4 + ½(4)=6  Formal Charge = (# of ve- on free atom) – (# of ve- assigned to the atom in the molecule) › = 6 – 7 = -1(single)For S = = 0 › = 6 – 6 = 0 (double)

 Which is better? › Atoms in molecules try to achieve formal charges as close to zero as possible › Any negative formal charges are expected to reside on the most electronegative atoms.  A: O = -1, S = 2  B: O = -1, 0, S = 0  B is preferred (lower formal charges) A B