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Lewis Structures.

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Presentation on theme: "Lewis Structures."— Presentation transcript:

1 Lewis Structures

2 5 rules for Lewis Structures
Find the total # of valence electrons for the molecule Find the center atom (the element with the least # of atoms) Draw bonds. Subtract 2 electrons from the total # of valence electrons for each bond drawn.

3 Distribute electrons around each atom to give a total of 8 electrons except H, Al, B, & Be
If there are not enough electrons to give 8 around each atom, create double & triple bonds.

4 Example 1: CF4 1. Total valence electrons = C + F
2. Center atom (least # of atoms) = C write C in the middle draw bonds (lines) to connect four F atoms F F C F 3. Subtract 2 times (# of bonds)- each bond uses 2 electrons 32 – 2(4) = 24 electrons 24 electrons left to distribute around the F atoms one line means 2 electrons

5 4. Distribute left over 24 electrons to F
F — C — F C needs no electrons because it has 4 bonds (octet) Now each F has 8 electrons around it (octet) This is the Lewis structure of CF4 It shows how many bonds and where e- are located Lone pairs of e- not involved in bonding

6 Example 2: NH (1) = 8 electrons H can have 2 e- (H is one of the exceptions) 8 – 2(3 bonds) = 2 electrons left for N H N H H

7 Example 3: H2S 2(1) + 6 = 8 electrons H can have 2 e- (H is one of the exceptions) 8 – 2(2 bonds) = 4 electrons left for S H S H

8 Ions With Ions we add or take away electrons
Example 4: NH lost 1e- 5 + 4(1) - 1 = 8 electrons 8-2(4 bonds) = 0 electrons to distribute Put ions in brackets and the charge on the outside

9 Double Bonds Example 5: CO2 - What’s wrong with Carbon?
Carbon does not have 8 electrons, so electrons will bond differently O – C – O Now carbon and oxygen have 8 electrons O – C – O “Double Bond”

10 Triple Bond Example 6: CO = 10 electrons 10 – 2(1 bond) = 8 electrons left for C or O If we try putting 4 electrons on C and 4 on O, then they will each only have 6 electrons. Try doubling or tripling the bonds if you do not have an octet. C – O

11 Incomplete Octet B Cl Cl
Example 7: BCl (7) = 24 electrons 24 – 2(3 bonds) = 18 electrons left for Cl Boron does not have 8e- around it. It is incomplete but it is okay; it is one of our exceptions. Cl B Cl

12 Go to the back of the Lewis ws for this next slide

13 Does your diagram look like the model A, B, or C?
-1 -1 -1

14 Are the other diagrams wrong
Are the other diagrams wrong? Discuss this with your neighbor why or why not? Do you think there are more molecules that have more than one valid structure like NO3-?

15 Do you know what molecule this is?
Ozone which is NOT an “O” or “ring” shape!!!


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