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Why does a tetrahedral molecule like methane have a dipole moment of zero?
Why does sulfate have this structure?Why do C-F bond has less dipole moment than C-Cl bond?What is lowest energy state, how does an atom gain it and what does it have to do with hybridization?Are the Phosphate and Sulfate ion diagrams with double bonds an invalid picture?Understanding the bond dissociation energy of poly-atomic moleculesRelationship between Lewis Theory and Redox theory?How to explain the structure of HNO3 with a Molecular Orbital diagram?What is the most polar covalent bond?Ethylene hybridizationForming the resonance structures for the nitrite anion
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$begingroup$
I was studying chemical bonding when I noticed something odd.
We say compounds like $ceCCl4$ and $ceCH4$ have a tetrahedral geometry (which is a 3D structure) but when we talk about their dipole moments, we say they have no dipole moment. We give the reason that as the H atoms are opposite each other (hence assuming it to be a 2D structure), they cancel out their bond moments.
But why? In the beginning we saw that they are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero.
But certainly I am wrong as these values have been calculated scientifically. So can someone please point out where i am going wrong in my understanding?
inorganic-chemistry bond
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
$endgroup$
add a comment |
$begingroup$
I was studying chemical bonding when I noticed something odd.
We say compounds like $ceCCl4$ and $ceCH4$ have a tetrahedral geometry (which is a 3D structure) but when we talk about their dipole moments, we say they have no dipole moment. We give the reason that as the H atoms are opposite each other (hence assuming it to be a 2D structure), they cancel out their bond moments.
But why? In the beginning we saw that they are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero.
But certainly I am wrong as these values have been calculated scientifically. So can someone please point out where i am going wrong in my understanding?
inorganic-chemistry bond
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
$endgroup$
$begingroup$
Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-)
$endgroup$
– Karl
2 hours ago
add a comment |
$begingroup$
I was studying chemical bonding when I noticed something odd.
We say compounds like $ceCCl4$ and $ceCH4$ have a tetrahedral geometry (which is a 3D structure) but when we talk about their dipole moments, we say they have no dipole moment. We give the reason that as the H atoms are opposite each other (hence assuming it to be a 2D structure), they cancel out their bond moments.
But why? In the beginning we saw that they are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero.
But certainly I am wrong as these values have been calculated scientifically. So can someone please point out where i am going wrong in my understanding?
inorganic-chemistry bond
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
$endgroup$
I was studying chemical bonding when I noticed something odd.
We say compounds like $ceCCl4$ and $ceCH4$ have a tetrahedral geometry (which is a 3D structure) but when we talk about their dipole moments, we say they have no dipole moment. We give the reason that as the H atoms are opposite each other (hence assuming it to be a 2D structure), they cancel out their bond moments.
But why? In the beginning we saw that they are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero.
But certainly I am wrong as these values have been calculated scientifically. So can someone please point out where i am going wrong in my understanding?
inorganic-chemistry bond
inorganic-chemistry bond
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
edited 7 hours ago
Karsten Theis
8,19110 silver badges52 bronze badges
8,19110 silver badges52 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
asked 9 hours ago
guccigucci
394 bronze badges
394 bronze badges
$begingroup$
Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-)
$endgroup$
– Karl
2 hours ago
add a comment |
$begingroup$
Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-)
$endgroup$
– Karl
2 hours ago
$begingroup$
Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-)
$endgroup$
– Karl
2 hours ago
$begingroup$
Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-)
$endgroup$
– Karl
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Karsten's answer is excellent, but here is a figure that shows the mathematics involved:
The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.
Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - theta$ ($approx 70.52877940 ^circ$) away from being directly "under" the central atom, so each contributes $cos(1 - theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.
But $cos(1 - theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
$endgroup$
1
$begingroup$
Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.
$endgroup$
– Ed V
8 hours ago
$begingroup$
(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)
$endgroup$
– andselisk♦
6 hours ago
2
$begingroup$
@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)
$endgroup$
– Karsten Theis
6 hours ago
|
show 2 more comments
$begingroup$
Orientation 1
Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.
For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.
If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.
If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.
Orientation 2
Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0.
The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.
Orientation 3
In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.
Orientation 4
Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Karsten's answer is excellent, but here is a figure that shows the mathematics involved:
The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.
Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - theta$ ($approx 70.52877940 ^circ$) away from being directly "under" the central atom, so each contributes $cos(1 - theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.
But $cos(1 - theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
$endgroup$
1
$begingroup$
Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.
$endgroup$
– Ed V
8 hours ago
$begingroup$
(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)
$endgroup$
– andselisk♦
6 hours ago
2
$begingroup$
@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)
$endgroup$
– Karsten Theis
6 hours ago
|
show 2 more comments
$begingroup$
Karsten's answer is excellent, but here is a figure that shows the mathematics involved:
The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.
Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - theta$ ($approx 70.52877940 ^circ$) away from being directly "under" the central atom, so each contributes $cos(1 - theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.
But $cos(1 - theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
$endgroup$
1
$begingroup$
Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.
$endgroup$
– Ed V
8 hours ago
$begingroup$
(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)
$endgroup$
– andselisk♦
6 hours ago
2
$begingroup$
@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)
$endgroup$
– Karsten Theis
6 hours ago
|
show 2 more comments
$begingroup$
Karsten's answer is excellent, but here is a figure that shows the mathematics involved:
The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.
Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - theta$ ($approx 70.52877940 ^circ$) away from being directly "under" the central atom, so each contributes $cos(1 - theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.
But $cos(1 - theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
$endgroup$
Karsten's answer is excellent, but here is a figure that shows the mathematics involved:
The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.
Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - theta$ ($approx 70.52877940 ^circ$) away from being directly "under" the central atom, so each contributes $cos(1 - theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.
But $cos(1 - theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
edited 7 hours ago
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
answered 8 hours ago
Ed VEd V
6171 silver badge11 bronze badges
6171 silver badge11 bronze badges
1
$begingroup$
Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.
$endgroup$
– Ed V
8 hours ago
$begingroup$
(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)
$endgroup$
– andselisk♦
6 hours ago
2
$begingroup$
@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)
$endgroup$
– Karsten Theis
6 hours ago
|
show 2 more comments
1
$begingroup$
Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).
$endgroup$
– Karsten Theis
8 hours ago
1
$begingroup$
Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.
$endgroup$
– Ed V
8 hours ago
$begingroup$
(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)
$endgroup$
– andselisk♦
6 hours ago
2
$begingroup$
@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)
$endgroup$
– Karsten Theis
6 hours ago
1
1
$begingroup$
Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!
$endgroup$
– Karsten Theis
8 hours ago
$begingroup$
Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!
$endgroup$
– Karsten Theis
8 hours ago
1
1
$begingroup$
If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).
$endgroup$
– Karsten Theis
8 hours ago
$begingroup$
If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).
$endgroup$
– Karsten Theis
8 hours ago
1
1
$begingroup$
Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.
$endgroup$
– Ed V
8 hours ago
$begingroup$
Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.
$endgroup$
– Ed V
8 hours ago
$begingroup$
(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)
$endgroup$
– andselisk♦
6 hours ago
$begingroup$
(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)
$endgroup$
– andselisk♦
6 hours ago
2
2
$begingroup$
@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)
$endgroup$
– Karsten Theis
6 hours ago
$begingroup$
@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)
$endgroup$
– Karsten Theis
6 hours ago
|
show 2 more comments
$begingroup$
Orientation 1
Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.
For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.
If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.
If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.
Orientation 2
Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0.
The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.
Orientation 3
In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.
Orientation 4
Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
$endgroup$
add a comment |
$begingroup$
Orientation 1
Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.
For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.
If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.
If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.
Orientation 2
Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0.
The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.
Orientation 3
In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.
Orientation 4
Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
$endgroup$
add a comment |
$begingroup$
Orientation 1
Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.
For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.
If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.
If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.
Orientation 2
Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0.
The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.
Orientation 3
In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.
Orientation 4
Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
$endgroup$
Orientation 1
Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.
For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.
If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.
If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.
Orientation 2
Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0.
The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.
Orientation 3
In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.
Orientation 4
Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
edited 8 hours ago
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
answered 9 hours ago
Karsten TheisKarsten Theis
8,19110 silver badges52 bronze badges
8,19110 silver badges52 bronze badges
add a comment |
add a comment |
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$begingroup$
Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-)
$endgroup$
– Karl
2 hours ago