dot structures!. one way of doing dot structures is to follow a few simple rules:
TRANSCRIPT
DOTSTRUCTURES!
One way of doing dot structuresis to follow a few simple rules:
1. Count the total number ofvalence electrons from each atom
2. Arrange the molecule: place theouter atoms around the central atom.
3. Place two electrons between each atom;add the rest of the electrons so that
each atom has an octet.
4. If one or more of the atoms cannot getan octet with the available electrons,
then rearrange the electrons into double ortriple bonds as needed for the octets.
EXAMPLE:
SO3sulfur trioxide
step 1: How many total valence electronsdoes SO3 have?
6 + 6 + 6 + 6 = 24
step 1: How many total valence electronsdoes SO3 have?
24
24
So, distribute 24 valence electrons around SO3
so that each atom has an octet!
24electrons
step 2:arrange the outer atomsaround thecentral atom
S
24electrons
step 2:arrange the outer atomsaround thecentral atom
S O
24electrons
step 2:arrange the outer atomsaround thecentral atom
S OO
24electrons
step 2:arrange the outer atomsaround thecentral atom
O S OO
24electrons
step 2:arrange the outer atomsaround thecentral atom
O OO
step 3:put electronsin the bonds
24electrons
S
O OO
1
23electrons
left
S
step 3:put electronsin the bonds
O OO
1
2
22electrons
left
S
step 3:put electronsin the bonds
O OO
1
2
3
21electrons
left
S
step 3:put electronsin the bonds
O OO
1
2
34
20electrons
left
S
step 3:put electronsin the bonds
O OO
1
2
34
5
19electrons
left
S
step 3:put electronsin the bonds
O OO
1
2
34
5
6
18electrons
left
S
step 3:put electronsin the bonds
O OO
1
2
34
5
6
18electrons
left
S
O OO
then,distribute therest to formthe octets
1
2
34
5
6
18electrons
left
S
O OO
1
2
34
5
6
87
16electrons
left
S
O OO
1
2
34
5
6
87
9
10
14electrons
left
S
O OO
1
2
34
5
6
87
9
10
1112
12electrons
left
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
10electrons
left
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
8electrons
left
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
6electrons
left
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
4electrons
left
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
2electrons
left
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
step 4: check tosee that eachatom has an octet
O OO
oxygen doesnot have anoctet
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
O OO
sulfur can sharethese with oxygenand still countthem in its octet
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19 20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
1920
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
2021
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
oxygen nowhas an octet
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
sulfur still hasits octet
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
S
O OO
1
2
34
5
6
87
9
10
1112
14 13
15
16
17
18
19
20
21
22
23 24
S
O OOS
O OOS
(don’t forget resonance!)
O OOS
(don’t forget shapes!)
O O
OS
O
SO
O
SO3O
SO
O
!SO3O
SO
O