Proof of the VSEPR shapes.












1












$begingroup$



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.










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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


















1












$begingroup$



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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$



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.










share|cite|improve this question











$endgroup$





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






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share|cite|improve this question













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share|cite|improve this question








edited Dec 15 '18 at 6:33







harshit54

















asked Dec 15 '18 at 6:24









harshit54harshit54

346113




346113












  • $begingroup$
    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










  • $begingroup$
    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












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