Lone Pair Placement  [ENDORSED]

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Fiona Grant 1I
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Joined: Fri Apr 06, 2018 11:03 am

Lone Pair Placement

Postby Fiona Grant 1I » Fri Jun 01, 2018 3:14 pm

I remember that we learned about how for some molecular shapes, it doesn't matter where you choose to place the lone pair, whereas for others, it does. For example, I believe that in molecules with a tetrahedral shape, the placement of the lone pair doesn't matter, but in molecules with trigonal bipyramidal shape, it does matter. I'm a little confused about this concept. When the placement of the lone pairs does matter, how exactly do you determine where the lone pairs have to be placed?

Natalie Noble 1G
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Re: Lone Pair Placement  [ENDORSED]

Postby Natalie Noble 1G » Fri Jun 01, 2018 3:14 pm

So the rule is to make the lone pair in the site that will have the least repulsion, since having the least repulsion is the most stable arrangement. For tetrahedral, it does not matter which area around the central atom is the lone pair, because all of the would cause equal repulsion. In trigonal bipyramidal, there is a trigonal planar shape with one atom sticking up and one below. Since they are not equally spaced (not the same bond angles), the site that is given the lone pair could cause more or less repulsion effect. If you put the lone pair on one of the sites in the trigonal planar looking shape, they will cause the least repulsion. So visualize the shape and determine which bonding site would cause the least damage (repulsion).

AnthonyDis1A
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Joined: Fri Apr 06, 2018 11:05 am

Re: Lone Pair Placement

Postby AnthonyDis1A » Sun Jun 03, 2018 1:03 am

As mentioned in the previous post, some molecules like the trigonal bipyramidal shape have multiple intersecting planes of bonded atoms. In the horizontal plane of the trigonal bipyramidal shape, the bond angles are larger, and so I think the lone pair would probably cause less repulsion if placed between those atoms, since they have more space to occupy and less interference from other bonds in the molecule.


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