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- 1. Find the central atom. Determine which orbitals are available for hybridization.
- 2. Count the valence electrons and add the electrons to account for the overall charge and the electrons
donated by bonded atoms (one for each single bond, two for each double bond, etc).
- 3. Divide the result by 2 to get the total number of electron pairs. For unusual compounds in which have an odd
number of total electrons, count the unpaired electron as an electron pair. Determine the number of lone pairs by
- 4. Minimize the electron pair-electron pair repulsion according to
This will mean that the linear geometry is favored for 5 pairs and 3 lone pairs, while the T-shaped is favored for 5
pairs and 2 lone pairs.
- 5. The arrangement is given by the total number of electron pairs, and the molecular geometry class by the number
of lone pairs.
| prs. |
arrangement |
lone pr. |
molecular geometry |
hybridization |
| 2 |
linear |
0 |
linear |
sp |
| 3 |
trigonal |
0 |
trigonal planar  | |
| |
planar |
1 |
bent |
|
| 4 |
tetrahedral |
0 |
tetrahedral |
 ,
 ,
 ,
 |
| |
|
1 |
trigonal pyramidal |
|
| |
|
2 |
bent |
|
| |
|
3 |
linear |
|
| 5 |
trigonal |
0 |
trigonal bipyramidal |
 |
| |
bipyramidal |
1 |
distorted pentahedral |
|
| |
|
2 |
T-shaped |
|
| |
|
3 |
linear |
|
| 6 |
octahedral |
0 |
octahedral |
 |
| |
|
1 |
square base pyramidal |
 |
| |
|
2 |
square planar |
 |
| 7 |
pent. bipyramidal |
0 |
pentagonal bipyramidal |
|
| species |
central |
bond |
total |
pairs |
lone |
geometry |
I | 7 |
2+1 |
10 |
5 |
3 |
linear |
| ICl3 |
7 |
3 |
10 |
5 |
2 |
T-shaped |
| IF5 |
7 |
5 |
12 |
6 |
2 |
square planar |
| PCl3F2 |
5 |
5 |
10 |
5 |
0 |
trigonal bipyramidal |
| PF3 |
5 |
3 |
8 |
4 |
1 |
trigonal pyramidal |
| SF2 |
6 |
2 |
8 |
4 |
2 |
bent |
| SO3 |
6 |
3 |
9 |
5 |
2 |
T-shaped |
| TeH2 |
6 |
2 |
8 |
4 |
2 |
bent |
© 1996-2007 Eric W. Weisstein
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