Proof of the VSEPR shapes.
$begingroup$
The VSEPR(Valence Shell Electron Pair Repulsion) predicts that the atoms arrange themselves in such a way that the electron pair repulsion is minimized. Thus the atoms arrange themselves in space in such a manner so as to be as far apart as possible.
What I was wondering was that how do one prove that the shapes are correctly predicted?
For example if looking at the molecule CH4, how do i prove that for maximum distance between the H atoms, they have to arrange themselves around the C atom in a tetrahedral shape.
Restatement:
If I have a ball fixed in space and have four other balls that I have to arrange around it, provided that the 4 balls are at the same finite distance from the fixed ball, how do i prove that the four balls must form a tetrahedral for having the maximum distance between them?
This question was originally posted on Chemistry SE, where a user told me post this question here.
geometry 3d chemistry
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
$endgroup$
|
show 2 more comments
$begingroup$
The VSEPR(Valence Shell Electron Pair Repulsion) predicts that the atoms arrange themselves in such a way that the electron pair repulsion is minimized. Thus the atoms arrange themselves in space in such a manner so as to be as far apart as possible.
What I was wondering was that how do one prove that the shapes are correctly predicted?
For example if looking at the molecule CH4, how do i prove that for maximum distance between the H atoms, they have to arrange themselves around the C atom in a tetrahedral shape.
Restatement:
If I have a ball fixed in space and have four other balls that I have to arrange around it, provided that the 4 balls are at the same finite distance from the fixed ball, how do i prove that the four balls must form a tetrahedral for having the maximum distance between them?
This question was originally posted on Chemistry SE, where a user told me post this question here.
geometry 3d chemistry
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
$endgroup$
$begingroup$
I believe you were told wrong. If anyone here is able to answer your chemistry question, it will be pure luck.
$endgroup$
– Ben W
Dec 15 '18 at 6:28
$begingroup$
You can ask in chemistry.stackexchange.com instead.
$endgroup$
– tarit goswami
Dec 15 '18 at 6:32
$begingroup$
@taritgoswami I already did and also provided the link.
$endgroup$
– harshit54
Dec 15 '18 at 6:33
1
$begingroup$
Given four points (not coplanar) they will form a tetrahedron (not necessarily regular). But since the four H- atoms have the same attributes, then you will necessarily have a regular tetrahedron. In some sense the answer is obvious, but if you want to show this from the level of the many-body Schrodinger equation then that is a whole other story.
$endgroup$
– Jacky Chong
Dec 15 '18 at 6:39
$begingroup$
@JackyChong But why a tetrahedron and not any other shape like a square? Does the answer involve some very complex math?
$endgroup$
– harshit54
Dec 15 '18 at 7:15
|
show 2 more comments
$begingroup$
The VSEPR(Valence Shell Electron Pair Repulsion) predicts that the atoms arrange themselves in such a way that the electron pair repulsion is minimized. Thus the atoms arrange themselves in space in such a manner so as to be as far apart as possible.
What I was wondering was that how do one prove that the shapes are correctly predicted?
For example if looking at the molecule CH4, how do i prove that for maximum distance between the H atoms, they have to arrange themselves around the C atom in a tetrahedral shape.
Restatement:
If I have a ball fixed in space and have four other balls that I have to arrange around it, provided that the 4 balls are at the same finite distance from the fixed ball, how do i prove that the four balls must form a tetrahedral for having the maximum distance between them?
This question was originally posted on Chemistry SE, where a user told me post this question here.
geometry 3d chemistry
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
$endgroup$
The VSEPR(Valence Shell Electron Pair Repulsion) predicts that the atoms arrange themselves in such a way that the electron pair repulsion is minimized. Thus the atoms arrange themselves in space in such a manner so as to be as far apart as possible.
What I was wondering was that how do one prove that the shapes are correctly predicted?
For example if looking at the molecule CH4, how do i prove that for maximum distance between the H atoms, they have to arrange themselves around the C atom in a tetrahedral shape.
Restatement:
If I have a ball fixed in space and have four other balls that I have to arrange around it, provided that the 4 balls are at the same finite distance from the fixed ball, how do i prove that the four balls must form a tetrahedral for having the maximum distance between them?
This question was originally posted on Chemistry SE, where a user told me post this question here.
geometry 3d chemistry
geometry 3d chemistry
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
edited Dec 15 '18 at 6:33
harshit54
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
asked Dec 15 '18 at 6:24
harshit54harshit54
348113
348113
$begingroup$
I believe you were told wrong. If anyone here is able to answer your chemistry question, it will be pure luck.
$endgroup$
– Ben W
Dec 15 '18 at 6:28
$begingroup$
You can ask in chemistry.stackexchange.com instead.
$endgroup$
– tarit goswami
Dec 15 '18 at 6:32
$begingroup$
@taritgoswami I already did and also provided the link.
$endgroup$
– harshit54
Dec 15 '18 at 6:33
1
$begingroup$
Given four points (not coplanar) they will form a tetrahedron (not necessarily regular). But since the four H- atoms have the same attributes, then you will necessarily have a regular tetrahedron. In some sense the answer is obvious, but if you want to show this from the level of the many-body Schrodinger equation then that is a whole other story.
$endgroup$
– Jacky Chong
Dec 15 '18 at 6:39
$begingroup$
@JackyChong But why a tetrahedron and not any other shape like a square? Does the answer involve some very complex math?
$endgroup$
– harshit54
Dec 15 '18 at 7:15
|
show 2 more comments
$begingroup$
I believe you were told wrong. If anyone here is able to answer your chemistry question, it will be pure luck.
$endgroup$
– Ben W
Dec 15 '18 at 6:28
$begingroup$
You can ask in chemistry.stackexchange.com instead.
$endgroup$
– tarit goswami
Dec 15 '18 at 6:32
$begingroup$
@taritgoswami I already did and also provided the link.
$endgroup$
– harshit54
Dec 15 '18 at 6:33
1
$begingroup$
Given four points (not coplanar) they will form a tetrahedron (not necessarily regular). But since the four H- atoms have the same attributes, then you will necessarily have a regular tetrahedron. In some sense the answer is obvious, but if you want to show this from the level of the many-body Schrodinger equation then that is a whole other story.
$endgroup$
– Jacky Chong
Dec 15 '18 at 6:39
$begingroup$
@JackyChong But why a tetrahedron and not any other shape like a square? Does the answer involve some very complex math?
$endgroup$
– harshit54
Dec 15 '18 at 7:15
$begingroup$
I believe you were told wrong. If anyone here is able to answer your chemistry question, it will be pure luck.
$endgroup$
– Ben W
Dec 15 '18 at 6:28
$begingroup$
I believe you were told wrong. If anyone here is able to answer your chemistry question, it will be pure luck.
$endgroup$
– Ben W
Dec 15 '18 at 6:28
$begingroup$
You can ask in chemistry.stackexchange.com instead.
$endgroup$
– tarit goswami
Dec 15 '18 at 6:32
$begingroup$
You can ask in chemistry.stackexchange.com instead.
$endgroup$
– tarit goswami
Dec 15 '18 at 6:32
$begingroup$
@taritgoswami I already did and also provided the link.
$endgroup$
– harshit54
Dec 15 '18 at 6:33
$begingroup$
@taritgoswami I already did and also provided the link.
$endgroup$
– harshit54
Dec 15 '18 at 6:33
1
1
$begingroup$
Given four points (not coplanar) they will form a tetrahedron (not necessarily regular). But since the four H- atoms have the same attributes, then you will necessarily have a regular tetrahedron. In some sense the answer is obvious, but if you want to show this from the level of the many-body Schrodinger equation then that is a whole other story.
$endgroup$
– Jacky Chong
Dec 15 '18 at 6:39
$begingroup$
Given four points (not coplanar) they will form a tetrahedron (not necessarily regular). But since the four H- atoms have the same attributes, then you will necessarily have a regular tetrahedron. In some sense the answer is obvious, but if you want to show this from the level of the many-body Schrodinger equation then that is a whole other story.
$endgroup$
– Jacky Chong
Dec 15 '18 at 6:39
$begingroup$
@JackyChong But why a tetrahedron and not any other shape like a square? Does the answer involve some very complex math?
$endgroup$
– harshit54
Dec 15 '18 at 7:15
$begingroup$
@JackyChong But why a tetrahedron and not any other shape like a square? Does the answer involve some very complex math?
$endgroup$
– harshit54
Dec 15 '18 at 7:15
|
show 2 more comments
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$begingroup$
I believe you were told wrong. If anyone here is able to answer your chemistry question, it will be pure luck.
$endgroup$
– Ben W
Dec 15 '18 at 6:28
$begingroup$
You can ask in chemistry.stackexchange.com instead.
$endgroup$
– tarit goswami
Dec 15 '18 at 6:32
$begingroup$
@taritgoswami I already did and also provided the link.
$endgroup$
– harshit54
Dec 15 '18 at 6:33
1
$begingroup$
Given four points (not coplanar) they will form a tetrahedron (not necessarily regular). But since the four H- atoms have the same attributes, then you will necessarily have a regular tetrahedron. In some sense the answer is obvious, but if you want to show this from the level of the many-body Schrodinger equation then that is a whole other story.
$endgroup$
– Jacky Chong
Dec 15 '18 at 6:39
$begingroup$
@JackyChong But why a tetrahedron and not any other shape like a square? Does the answer involve some very complex math?
$endgroup$
– harshit54
Dec 15 '18 at 7:15