LEWIS DOT STRUCTURES1.1. Arrange the symbols such that the least electronegative element is in the center and the other elements are surrounding the central atom.

O C OO C O

2.2. Give each of the elements their appropriate number of valence electrons (dots). Remember the number of valence electrons for a representative element is the same as the group number.

. . . .. . . .: O : C : O :: O : C : O :

. . . .. . . .3. Keep track of the total numbr of valence electrons for the compound by adding the valence electrons from each atom. If the compound is an ion then add electrons (dots) for each negative charge or subtract electrons (dots) for each positive charge.

4 for C and 6 for O (twice) = 16 electrons4 for C and 6 for O (twice) = 16 electrons

LEWIS DOT STRUCTURES4. 4. Now move the dots around so that you have 8 dots (the octet rule) around each element (do not forget the exceptions) while at the same time keeping the dots in pairs. Electrons, at this point, exist as pairs (the buddy system).

5.5. EXCEPTIONS TO THE OCTET RULE: Group I, II, and III need only 2, 4, and 6 electrons, respectively, around that atom.

6.6. If there are too few pairs to give each atom eight electrons, change the single bonds between two atoms to either double or triple bonds by moving the unbonded pairs of electrons next to a bonding pair.

. . . .. . . .: O : : C : : O :: O : : C : : O :

LEWIS DOT STRUCTURES7.7. Once the octet rule has been satisfied for each atom in the molecule then you may replace each pair of dots between two atoms with a dash.

. . . .. . . .: O = C = O :: O = C = O :

8.8. Now check your structure by a) count the total number of electrons to make sure you did not lose or gain electrons during the process.

b) Use FORMAL CHARGE (FC) calculations as a guideline to the correct structure. A zero formal charge is usually a good indication of a stable structure.

FC (X) = # of valence electrons - (1/2 bonding electrons + nonbonding electrons)FC (X) = # of valence electrons - (1/2 bonding electrons + nonbonding electrons)

For our example: FC(C) = 4 - (1/2 8 + 0) = 0

FC(O) = 6 - (1/2 4 + 4) = 0

The Basics: The Basics: Drawing Lewis StructuresDrawing Lewis Structures

Rules

a) A hydrogen atom always forms one bond. Hydrogen is always a terminal atom in a Lewis diagram – an atom that is bonded to only one other atom.

b) A carbon atom normally forms four bonds

c) When several carbon atoms appear in the same molecule, the are often bonded to each other. (No cyclic compounds in Chem 60/68)

Step 1: Calculate the total number of valence electrons in the molecule or ion

Step 2: Determine the central atom(s) of the molecule or ion – usually it’s the least electronegative atom.

Step 3: Draw a tentative diagram for the molecule or ion.

Draw the Lewis Structure for the following molecules.

1. H2O

Oxygen has 6 valence electrons & Hydrogen has 1 valence electron for a total of 8 electrons.

. . . .

H : O : HH : O : H . .. .

2. COOxygen has 6 valence electrons & Carbon has 4 valence electrons for a total of 10 electrons.

: C : : : O :: C : : : O :

Draw the Lewis Structure for the following molecules.

3. BH3 Boron has 3 valence electrons & Hydrogen has 1 valence electron for a total of 6 electrons

H : B : HH : B : H . .. . HH

4. NH3 Nitrogen has 5 valence electrons & Hydrogen has 1 valence electron for a total of 8 electrons

. . . .

H : N : HH : N : H . .. . HH

p. 362

p. 374

p. 362

p. 375

p. 363

Cations

p. 366

Anions

p. 366

Resonance

LEWIS DOT STRUCTURESPredict the most stable structure: Predict the most stable structure: ONCONC-- or or OCNOCN-- oror NOC NOC--

. . .. . . .. .. ... . .. . . .. .. .. :O::N::C: :O::N::C: oror :O::C::N: :O::C::N: oror :N::O::C: :N::O::C:1) Total electrons is:6 e- for O + 5 e- for N + 4 e- for C + 1 e- for negative charge = 16 e- total. All structures fulfill the octet rule.

2) FC (X) = # of valence electrons - (1/2 bonding electrons + nonbonding electrons)FC (X) = # of valence electrons - (1/2 bonding electrons + nonbonding electrons)

structure#1: structure #3:

FC(C) = 4 - (1/2 4 + 4) = -2 FC(C) = 4 - (1/2 4 + 4) = -2

FC(O) = 6 - (1/2 4 + 4) = 0 FC(O) = 6 - (1/2 8 + 0) = +2

FC(N) = 5 -(1/2 8 + 0) = +1 FC(N) = 5 -(1/2 4 + 4) = -1

structure#2:

FC(C) = 4 - (1/2 8 + 0) = 0

FC(O) = 6 - (1/2 4 + 4) = 0

FC(N) = 5 -(1/2 4 + 4) = -1

structure #2 has the combination with the lowest structure #2 has the combination with the lowest formal charge. It also has the negative formal formal charge. It also has the negative formal charge on one of the more electronegative charge on one of the more electronegative atomsatoms. Calculate the formal charge for the most

stable structure: . . . . :O:C:::N::O:C:::N: . .. .

(-1, 0, 0)(-1, 0, 0)

Practice Problem #13Practice Problem #13

ClO2- SiH4

AsH3

.. .. .... .. ..: O : Cl : O :: O : Cl : O : .. .. .... .. ..

HH ¨̈

H : Si : H H : Si : H ¨̈ HH

....H : As : HH : As : H ¨̈ HH

Group Study Problem #13Group Study Problem #131. Draw the Lewis structure for the following

a) H2S b) PH3 c) CH2O

d) NO2- e) H2CO3 f) CBr4

g) CH2FCl h) C2H2 I) O3

2. Calculate the formal charge for “c”, “d”, “f”, and “I”.