Electron Groups and Platonic Solids
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In my Chemistry class, lately I have been learning about Lewis Structures for molecules, and how the arrangements of groups of electrons on each molecule repel each other to form the molecule into a specific shape. For example, consider the following diagram below of a familiar molecule:
The dots represent electrons, and the dashes represent bonds between atoms, each of which contains two electrons.
One might think that this molecule should assume a straight, linear shape with one $H$ atom on each side, but because each bond (the short lines) consists of electrons, the two electron pairs above $O$ repel them, causing the molecule to bend into this shape. Now consider this smelly example:
Each bond consists of electrons with equal charges, so they will all repel each other with forces that vary equally based on the distance between each respective bond. Thus, they will tend towards a shape that maximizes the distance between each respective bond in the molecule, and this shape turns out to be a tetrahedron (with $C$ at the center and $H$ at the vertices). Another example is the following:
Each of the bonds and $Cl$ atoms, containing electrons, will repel each other, and try to assume a position of maximum pairwise distance. As with the previous example, they will assume a uniform "lattice-like" distribution about the sphere, which, for $6$ atoms, turns into an octahedron.
Now I begin leave the realm of chemistry and enter the realm of spherical geometry and pure mathematics. When there are $4$ outer atoms, they will distribute themselves uniformly about the sphere with radius equal to the bond length centered at the central atom, forming the platonic solid with $4$ vertices, or the tetrahedron. When there are $6$ it forms the platonic solid with $6$ vertices, or the octahedron. Though I'm not sure this can happen in real life, it doesn't take much imagination to guess that if we had $8$ outer atoms, it would assume the shape of a cube, and if we had $12$ outer atoms, it would assume the shape of an icosahedron (and for $20$ outer atoms, a dodecahedron, of course).
QUESTION: What if there were $7$ outer molecules? Or $9$? Or some other number that is not equal to the number of vertices of any platonic solid?
Formulated mathematically (and borrowing some equations from physics), this question is:
Suppose that $n$ points are distributed around a sphere such that for any of these points $X$, the vectors pointing from each of the other $n-1$ point towards $X$, each with magnitude equal to a constant times the reciprocal of the squared distance between $X$ and the respective point, have a sum equal to a normal vector of the sphere. How can these points be distributed for an arbitrary value of $nne 4,6,8,12,20$?
optimization vectors spherical-geometry chemistry platonic-solids
asked Nov 16 at 22:40
Frpzzd
19.8k638101
add a comment |
up vote
1
down vote
favorite
In my Chemistry class, lately I have been learning about Lewis Structures for molecules, and how the arrangements of groups of electrons on each molecule repel each other to form the molecule into a specific shape. For example, consider the following diagram below of a familiar molecule:
The dots represent electrons, and the dashes represent bonds between atoms, each of which contains two electrons.
One might think that this molecule should assume a straight, linear shape with one $H$ atom on each side, but because each bond (the short lines) consists of electrons, the two electron pairs above $O$ repel them, causing the molecule to bend into this shape. Now consider this smelly example:
Each bond consists of electrons with equal charges, so they will all repel each other with forces that vary equally based on the distance between each respective bond. Thus, they will tend towards a shape that maximizes the distance between each respective bond in the molecule, and this shape turns out to be a tetrahedron (with $C$ at the center and $H$ at the vertices). Another example is the following:
Each of the bonds and $Cl$ atoms, containing electrons, will repel each other, and try to assume a position of maximum pairwise distance. As with the previous example, they will assume a uniform "lattice-like" distribution about the sphere, which, for $6$ atoms, turns into an octahedron.
Now I begin leave the realm of chemistry and enter the realm of spherical geometry and pure mathematics. When there are $4$ outer atoms, they will distribute themselves uniformly about the sphere with radius equal to the bond length centered at the central atom, forming the platonic solid with $4$ vertices, or the tetrahedron. When there are $6$ it forms the platonic solid with $6$ vertices, or the octahedron. Though I'm not sure this can happen in real life, it doesn't take much imagination to guess that if we had $8$ outer atoms, it would assume the shape of a cube, and if we had $12$ outer atoms, it would assume the shape of an icosahedron (and for $20$ outer atoms, a dodecahedron, of course).
QUESTION: What if there were $7$ outer molecules? Or $9$? Or some other number that is not equal to the number of vertices of any platonic solid?
Formulated mathematically (and borrowing some equations from physics), this question is:
Suppose that $n$ points are distributed around a sphere such that for any of these points $X$, the vectors pointing from each of the other $n-1$ point towards $X$, each with magnitude equal to a constant times the reciprocal of the squared distance between $X$ and the respective point, have a sum equal to a normal vector of the sphere. How can these points be distributed for an arbitrary value of $nne 4,6,8,12,20$?
optimization vectors spherical-geometry chemistry platonic-solids
asked Nov 16 at 22:40
Frpzzd
19.8k638101
socratic.org/questions/…
– Andrei
Nov 16 at 22:50
1
Do I understand correctly that we have points $x_1,dots,x_n in S^2$ and $c > 0$ such that for each $y in S^2 setminus { x_1,dots,x_n }$ with the property $lVert y - x_i rVert = c lVert y - x_i rVert^{-2}$ (i.e. $lVert y - x_i rVert^3 =c$) for $i =1,dots,n$, we have $s(y) = sum_{i=1}^n (y - x_i) ne 0$? Note that any $s ne 0$ is a normal vector of $S^2$.
– Paul Frost
Nov 16 at 23:04
@PaulFrost No, $yin{x_1,x_2,...,x_n}$, not its complement. I will clarify.
– Frpzzd
Nov 16 at 23:05
1
For $n < 4$, things are simple (just put them equally spaced on a circle). You might want to read this: I suspect that the answer to your question is either "they can't" or "exactly the solutions to the Thompson problem", with the difference being a matter of subtleties of wording. If that's so, then for $n = 5$, the answer is apparently a triangular dipyramid.
– user3482749
Nov 16 at 23:18
Although 7 is hard to think about, 8 is not as obvious as a cube. Take a look a the picture here
– Narlin
Nov 17 at 4:37
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In my Chemistry class, lately I have been learning about Lewis Structures for molecules, and how the arrangements of groups of electrons on each molecule repel each other to form the molecule into a specific shape. For example, consider the following diagram below of a familiar molecule:
The dots represent electrons, and the dashes represent bonds between atoms, each of which contains two electrons.
One might think that this molecule should assume a straight, linear shape with one $H$ atom on each side, but because each bond (the short lines) consists of electrons, the two electron pairs above $O$ repel them, causing the molecule to bend into this shape. Now consider this smelly example:
Each bond consists of electrons with equal charges, so they will all repel each other with forces that vary equally based on the distance between each respective bond. Thus, they will tend towards a shape that maximizes the distance between each respective bond in the molecule, and this shape turns out to be a tetrahedron (with $C$ at the center and $H$ at the vertices). Another example is the following:
Each of the bonds and $Cl$ atoms, containing electrons, will repel each other, and try to assume a position of maximum pairwise distance. As with the previous example, they will assume a uniform "lattice-like" distribution about the sphere, which, for $6$ atoms, turns into an octahedron.
Now I begin leave the realm of chemistry and enter the realm of spherical geometry and pure mathematics. When there are $4$ outer atoms, they will distribute themselves uniformly about the sphere with radius equal to the bond length centered at the central atom, forming the platonic solid with $4$ vertices, or the tetrahedron. When there are $6$ it forms the platonic solid with $6$ vertices, or the octahedron. Though I'm not sure this can happen in real life, it doesn't take much imagination to guess that if we had $8$ outer atoms, it would assume the shape of a cube, and if we had $12$ outer atoms, it would assume the shape of an icosahedron (and for $20$ outer atoms, a dodecahedron, of course).
QUESTION: What if there were $7$ outer molecules? Or $9$? Or some other number that is not equal to the number of vertices of any platonic solid?
Formulated mathematically (and borrowing some equations from physics), this question is:
Suppose that $n$ points are distributed around a sphere such that for any of these points $X$, the vectors pointing from each of the other $n-1$ point towards $X$, each with magnitude equal to a constant times the reciprocal of the squared distance between $X$ and the respective point, have a sum equal to a normal vector of the sphere. How can these points be distributed for an arbitrary value of $nne 4,6,8,12,20$?
optimization vectors spherical-geometry chemistry platonic-solids
asked Nov 16 at 22:40
Frpzzd
19.8k638101
In my Chemistry class, lately I have been learning about Lewis Structures for molecules, and how the arrangements of groups of electrons on each molecule repel each other to form the molecule into a specific shape. For example, consider the following diagram below of a familiar molecule:
The dots represent electrons, and the dashes represent bonds between atoms, each of which contains two electrons.
One might think that this molecule should assume a straight, linear shape with one $H$ atom on each side, but because each bond (the short lines) consists of electrons, the two electron pairs above $O$ repel them, causing the molecule to bend into this shape. Now consider this smelly example:
Each bond consists of electrons with equal charges, so they will all repel each other with forces that vary equally based on the distance between each respective bond. Thus, they will tend towards a shape that maximizes the distance between each respective bond in the molecule, and this shape turns out to be a tetrahedron (with $C$ at the center and $H$ at the vertices). Another example is the following:
Each of the bonds and $Cl$ atoms, containing electrons, will repel each other, and try to assume a position of maximum pairwise distance. As with the previous example, they will assume a uniform "lattice-like" distribution about the sphere, which, for $6$ atoms, turns into an octahedron.
Now I begin leave the realm of chemistry and enter the realm of spherical geometry and pure mathematics. When there are $4$ outer atoms, they will distribute themselves uniformly about the sphere with radius equal to the bond length centered at the central atom, forming the platonic solid with $4$ vertices, or the tetrahedron. When there are $6$ it forms the platonic solid with $6$ vertices, or the octahedron. Though I'm not sure this can happen in real life, it doesn't take much imagination to guess that if we had $8$ outer atoms, it would assume the shape of a cube, and if we had $12$ outer atoms, it would assume the shape of an icosahedron (and for $20$ outer atoms, a dodecahedron, of course).
QUESTION: What if there were $7$ outer molecules? Or $9$? Or some other number that is not equal to the number of vertices of any platonic solid?
Formulated mathematically (and borrowing some equations from physics), this question is:
Suppose that $n$ points are distributed around a sphere such that for any of these points $X$, the vectors pointing from each of the other $n-1$ point towards $X$, each with magnitude equal to a constant times the reciprocal of the squared distance between $X$ and the respective point, have a sum equal to a normal vector of the sphere. How can these points be distributed for an arbitrary value of $nne 4,6,8,12,20$?
optimization vectors spherical-geometry chemistry platonic-solids
optimization vectors spherical-geometry chemistry platonic-solids
asked Nov 16 at 22:40
Frpzzd
19.8k638101
asked Nov 16 at 22:40
Frpzzd
19.8k638101
edited Nov 16 at 23:06
asked Nov 16 at 22:40
Frpzzd
19.8k638101
asked Nov 16 at 22:40
Frpzzd
19.8k638101
asked Nov 16 at 22:40
Frpzzd
19.8k638101
19.8k638101
socratic.org/questions/…
– Andrei
Nov 16 at 22:50
1
Do I understand correctly that we have points $x_1,dots,x_n in S^2$ and $c > 0$ such that for each $y in S^2 setminus { x_1,dots,x_n }$ with the property $lVert y - x_i rVert = c lVert y - x_i rVert^{-2}$ (i.e. $lVert y - x_i rVert^3 =c$) for $i =1,dots,n$, we have $s(y) = sum_{i=1}^n (y - x_i) ne 0$? Note that any $s ne 0$ is a normal vector of $S^2$.
– Paul Frost
Nov 16 at 23:04
@PaulFrost No, $yin{x_1,x_2,...,x_n}$, not its complement. I will clarify.
– Frpzzd
Nov 16 at 23:05
1
For $n < 4$, things are simple (just put them equally spaced on a circle). You might want to read this: I suspect that the answer to your question is either "they can't" or "exactly the solutions to the Thompson problem", with the difference being a matter of subtleties of wording. If that's so, then for $n = 5$, the answer is apparently a triangular dipyramid.
– user3482749
Nov 16 at 23:18
Although 7 is hard to think about, 8 is not as obvious as a cube. Take a look a the picture here
– Narlin
Nov 17 at 4:37
add a comment |
socratic.org/questions/…
– Andrei
Nov 16 at 22:50
1
Do I understand correctly that we have points $x_1,dots,x_n in S^2$ and $c > 0$ such that for each $y in S^2 setminus { x_1,dots,x_n }$ with the property $lVert y - x_i rVert = c lVert y - x_i rVert^{-2}$ (i.e. $lVert y - x_i rVert^3 =c$) for $i =1,dots,n$, we have $s(y) = sum_{i=1}^n (y - x_i) ne 0$? Note that any $s ne 0$ is a normal vector of $S^2$.
– Paul Frost
Nov 16 at 23:04
@PaulFrost No, $yin{x_1,x_2,...,x_n}$, not its complement. I will clarify.
– Frpzzd
Nov 16 at 23:05
1
For $n < 4$, things are simple (just put them equally spaced on a circle). You might want to read this: I suspect that the answer to your question is either "they can't" or "exactly the solutions to the Thompson problem", with the difference being a matter of subtleties of wording. If that's so, then for $n = 5$, the answer is apparently a triangular dipyramid.
– user3482749
Nov 16 at 23:18
Although 7 is hard to think about, 8 is not as obvious as a cube. Take a look a the picture here
– Narlin
Nov 17 at 4:37
socratic.org/questions/…
– Andrei
Nov 16 at 22:50
socratic.org/questions/…
– Andrei
Nov 16 at 22:50
1
1
Do I understand correctly that we have points $x_1,dots,x_n in S^2$ and $c > 0$ such that for each $y in S^2 setminus { x_1,dots,x_n }$ with the property $lVert y - x_i rVert = c lVert y - x_i rVert^{-2}$ (i.e. $lVert y - x_i rVert^3 =c$) for $i =1,dots,n$, we have $s(y) = sum_{i=1}^n (y - x_i) ne 0$? Note that any $s ne 0$ is a normal vector of $S^2$.
– Paul Frost
Nov 16 at 23:04
Do I understand correctly that we have points $x_1,dots,x_n in S^2$ and $c > 0$ such that for each $y in S^2 setminus { x_1,dots,x_n }$ with the property $lVert y - x_i rVert = c lVert y - x_i rVert^{-2}$ (i.e. $lVert y - x_i rVert^3 =c$) for $i =1,dots,n$, we have $s(y) = sum_{i=1}^n (y - x_i) ne 0$? Note that any $s ne 0$ is a normal vector of $S^2$.
– Paul Frost
Nov 16 at 23:04
@PaulFrost No, $yin{x_1,x_2,...,x_n}$, not its complement. I will clarify.
– Frpzzd
Nov 16 at 23:05
@PaulFrost No, $yin{x_1,x_2,...,x_n}$, not its complement. I will clarify.
– Frpzzd
Nov 16 at 23:05
1
1
For $n < 4$, things are simple (just put them equally spaced on a circle). You might want to read this: I suspect that the answer to your question is either "they can't" or "exactly the solutions to the Thompson problem", with the difference being a matter of subtleties of wording. If that's so, then for $n = 5$, the answer is apparently a triangular dipyramid.
– user3482749
Nov 16 at 23:18
For $n < 4$, things are simple (just put them equally spaced on a circle). You might want to read this: I suspect that the answer to your question is either "they can't" or "exactly the solutions to the Thompson problem", with the difference being a matter of subtleties of wording. If that's so, then for $n = 5$, the answer is apparently a triangular dipyramid.
– user3482749
Nov 16 at 23:18
Although 7 is hard to think about, 8 is not as obvious as a cube. Take a look a the picture here
– Narlin
Nov 17 at 4:37
Although 7 is hard to think about, 8 is not as obvious as a cube. Take a look a the picture here
– Narlin
Nov 17 at 4:37
add a comment |
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socratic.org/questions/…
– Andrei
Nov 16 at 22:50
1
Do I understand correctly that we have points $x_1,dots,x_n in S^2$ and $c > 0$ such that for each $y in S^2 setminus { x_1,dots,x_n }$ with the property $lVert y - x_i rVert = c lVert y - x_i rVert^{-2}$ (i.e. $lVert y - x_i rVert^3 =c$) for $i =1,dots,n$, we have $s(y) = sum_{i=1}^n (y - x_i) ne 0$? Note that any $s ne 0$ is a normal vector of $S^2$.
– Paul Frost
Nov 16 at 23:04
@PaulFrost No, $yin{x_1,x_2,...,x_n}$, not its complement. I will clarify.
– Frpzzd
Nov 16 at 23:05
1
For $n < 4$, things are simple (just put them equally spaced on a circle). You might want to read this: I suspect that the answer to your question is either "they can't" or "exactly the solutions to the Thompson problem", with the difference being a matter of subtleties of wording. If that's so, then for $n = 5$, the answer is apparently a triangular dipyramid.
– user3482749
Nov 16 at 23:18
Although 7 is hard to think about, 8 is not as obvious as a cube. Take a look a the picture here
– Narlin
Nov 17 at 4:37