## electron pair repulsion

Higher-dimensional geometry (previously "Polyshapes").

### electron pair repulsion

We're learning about this in chemistry right now. What happens in 3D goes like this:

2 pairs align themselves along a line
3 pairs point at the corners of an equilateral triangle
4 pairs point at the corners of a regular tetrahedron
5 pairs have the equilateral triangle plus the line, perpendicular to each other
6 pairs have three lines, each perpendicular to each other

Now what I'd like to know is if you can do more than 6 pairs in 3D, and what happens in 4D, after 5 pairs (2-4 are the same, 5 is obviously going to point at the corners of a regular pentachoron).

Keiji

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pairs?

when you say what the pairs are doing, do you mean that a pair acts as a single entity? or is there a specifit configuration for each pair in each groups configuration?
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batmanmg
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Pretty much yeah the pair acts as one entity, because they are between atoms in covalent bonds. Not sure what would happen in a double or triple bond though.

Keiji

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in doubled they would probably end up so the line between the pairs is perpendicular to a ray from the center of the configuration.

clueless for the tripled pairs though.

(first time i've used "ray" for geometrical purposes since learning the word when learning angles in elementary school )
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batmanmg
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I was never very good at chemistry. Pairs of what, exactly?

PWrong
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Pairs of electrons. Hence the title, "ELECTRON PAIR repulsion".

Keiji

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But it's not just electrons, otherwise they would just repel each other.

So where do these come in?
they are between atoms in covalent bonds.

PWrong
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Okay. Here's an example: methane.

There are four pairs of electrons around the C, and they want to be as far away from each other as possible, because electrons repel. But the protons in hydrogen want to still be close to their electrons. So what happens is you get the carbon atom in the middle, with the four hydrogen atoms at the corners of an imaginary tetrahedron (centered at the carbon atom).

Keiji

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Oh ok, we're talking about carbon chains. I was pretty good at those.

With 5 electron pairs, would you be talking about one carbon and 5 hydrogens? I thought you couldn't have that :?.I don't think there is any symmetric way to arrange five points in 3D.

PWrong
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### Re: electron pair repulsion

iNVERTED wrote:We're learning about this in chemistry right now. What happens in 3D goes like this:

2 pairs align themselves along a line
3 pairs point at the corners of an equilateral triangle
4 pairs point at the corners of a regular tetrahedron
5 pairs have the equilateral triangle plus the line, perpendicular to each other
6 pairs have three lines, each perpendicular to each other

thats the octahedron right?

i guess the only symetry thats needed is rotational?
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batmanmg
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No it isn't the octahedron. 6 is the octahedron. 5 is a triangular bipyramid.

Keiji

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### Re: electron pair repulsion

batmanmg wrote:thats the octahedron right?

The six pairs is an octahedron. The five pairs is a triangular dipyramid (or bipyramid, depending on your prefixial preference).

i guess the only symetry thats needed is rotational?

I don't know. I had suspected it would be the configuration of minimum potential of electrons on a sphere. But, alas, the bipyramid threw that one off.
pat
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### Re: electron pair repulsion

pat wrote:
batmanmg wrote:thats the octahedron right?

The six pairs is an octahedron. The five pairs is a triangular dipyramid (or bipyramid, depending on your prefixial preference).

I used bipyramid for clarity, because "dipyramid" is used in rotope naming: http://tetraspace.alkaline.org/wiki/ind ... le=Rotopes

Keiji

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This is the problem of placing circles on a sphere. So you can for example, ask, how far can you place points on a sphere so that the distances are maximal.

For 2d, it is simply the polygons.

For 3d, one can use the 2d cases, or

4. tetrahedron or square
5. triangular tegum
6. octahedron, triangular prism
7. pentagonal tegum.
8. cube.
9.

For four dimensions one can set any number of points up in this way, by starting off with a p×p bi-prism, and then (for some a), mark off 1,a 2,2a, 3,3a &c. This will give a uniformly distributed set of points. One also has

Code: Select all
`   5.   pentachoron     p=5, a=2   6.   bi-triangular tegum   p=6, a=3   7    p=7, a=2   8    16-cell    p-4, a=1,   done mod 2   9,   p=3, a=1  bi-triangular prism 10,   p=10, a=3   = pentachoron + dual 11    p=11, a=3 13    p=13, a=5  (very interesting figure) ... 24    24choron.`
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wendy
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wendy wrote:For 3d, one can use the 2d cases, or

4. tetrahedron or square
5. triangular tegum
6. octahedron, triangular prism
7. pentagonal tegum.
8. cube.
9.

So what comes after 8? Pentagon plus two lines?

And yes... tegum is a much better word than bipyramid

Keiji