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Lesson 7: VSEPR

# VSEPR for 6 electron clouds

In this video, we apply VSEPR theory to molecules and ions with six groups or “clouds” of electrons around the central atom. To minimize repulsions, six electron clouds will always adopt a octahedral electron geometry. Depending on how many of the clouds are lone pairs, the molecular geometry will be octahedral (no lone pairs), square pyramidal (one lone pair), or square planer (two lone pairs). Created by Jay.

## Want to join the conversation?

• At , why doesn't the lone pair repel the fluorine atoms to create a bond angle less than 90 degrees?
• It does, the prediction made in the video was wrong. In BrF₅, the four F atoms that are predicted to be in the same plane with each other actually have bond angles of 89.5° with each other and have a bond angle of 84.8° with the remaining atom of F (the F that is linear with the lone pair and the Br).

So, you are correct. Unfortunately, some resources on VESPR don't take into account the fact that lone pairs take up more space than bond and "push" the bonds out of ideal geometries.
• Can I place one lone pair in the axial side and one at an equitorial side for XeF4 ?
• No. The 3D structure is unstable in that conformation. The lone pairs are more negatively charged than the bonds, so the lone pairs will try to be as far apart as possible. This would place them either both axial, or opposite equatorial. They cannot be both axial and equatorial.
• the lone pair must be on equitorial . but in BrF5 molecule it is on axial line?i
• Octahedral or square bipyramidal has a 90* bond angle between the axial and equitorial atoms both. Regarless of how you spin the molecule it looks the same from all sides. Just pick off an atom to get Square Pyramidal. (Axial and Equatorial don't really axis for Octahedral - play with a model kit)

Trigonal bipyramidal DOES have axial (90*) and equatorial (120*) bond angles. If you remove an equatorial, you still have 2 atoms forming tight 90* bonds to both the axials (Electron repulsion = yikes). If you remove an axial, you have the three equatorials with no 90* bonds formed to the axial thats removed (less electron repulsion). Try it on a model kit
• How come the lone pairs aren't assigned an equatorial position in this example? In the previous videos it was mentioned that lone pairs should be assigned an equatorial position? :/
• It may be because all the Fluorines are the same distance from each other here so it doesn't matter where the lone pairs are placed in relation to the Fluorines. When they have their free choice of where to go, they choose positions that are furtherest away from each other as they repel each other.
• With 6 electron clouds, can't there be cases with 3 lone pairs and 4 lone pairs as well? If so, what are the shapes of those?
• Can't there be 3 THREE lone pairs and 3 bond pairs ??
What will be the structure then??
Example -> XeF3(-1)
• A molecule can be formed using 3 lone pairs and 3 bonds but the central atom cannot have 3 lone pairs and 3 bonds.

The reason being is because then the electron the central atom would have would be ((1/2) * bonds = 3) + (1 for each lone pair electron (2 * 3 = 6)) = 9.

So in short having 3 pairs of shared electrons and 3 pairs of lone electrons would mean it would have 9 valence electrons which is invalid.

Hope this helps,
- Convenient Colleague
• At how do we figure out that the electron cloud geometry was a regular octahedron? Is it worked out experimentally or is it something that we could deduce just by the number of electron density regions?
• It was determined experimentally. This was made possible due to development of advanced technology such as spectroscopy.
• In drawing the structures, it is always mentioned "[...] is following the octet rule". How do I know which atom follows the octet rule? Don't they all?
(1 vote)
• Only the atoms from C to Si must follow the octet rule.
Atoms after Si often follow the octet rule, but they may "expand their octets" and disobey the octet rule.