Influence of the neighbouring molecules in a crystal at the example of XeF4
$begingroup$
For reasons of symmetry I (without deeper chemical knowledge) supposed, that $ce{XeF4}$ has a shape like $ce{CH4}$. But according to Wikipedia its
crystalline structure was determined by both NMR spectroscopy and X-ray crystallography in 1963. The structure is square planar, as has been confirmed by neutron diffraction studies.
In a crystal the neighbor atoms have some influence on another and sometimes the chemical formula shows the numerical ratio between the elements but not draw the full picture of all interactions between the neighbor elements. Does the influence of the neighboring molecules in a $ce{XeF4}$ crystal makes the square planar shap? Is it imaginable that an isolated $ce{XeF4}$ molecule has a tetrahedral shape?
Additional question. What is the crystalline structure of $ce{XeF4}$?
crystal-structure solid-state-chemistry vsepr-theory noble-gases
edited Jan 27 at 14:07
andselisk
17.4k656117
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
$endgroup$
add a comment |
$begingroup$
For reasons of symmetry I (without deeper chemical knowledge) supposed, that $ce{XeF4}$ has a shape like $ce{CH4}$. But according to Wikipedia its
crystalline structure was determined by both NMR spectroscopy and X-ray crystallography in 1963. The structure is square planar, as has been confirmed by neutron diffraction studies.
In a crystal the neighbor atoms have some influence on another and sometimes the chemical formula shows the numerical ratio between the elements but not draw the full picture of all interactions between the neighbor elements. Does the influence of the neighboring molecules in a $ce{XeF4}$ crystal makes the square planar shap? Is it imaginable that an isolated $ce{XeF4}$ molecule has a tetrahedral shape?
Additional question. What is the crystalline structure of $ce{XeF4}$?
crystal-structure solid-state-chemistry vsepr-theory noble-gases
edited Jan 27 at 14:07
andselisk
17.4k656117
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
$endgroup$
$begingroup$
It seems to me that you imagine that some "square planar XeF4" will stick together in the same plane, but it is obviously not the case because fluorine atoms repel each others. In fact, if you take two "square-planar XeF4" on top of each other, one will be shifted so one of its fluorine atom is right above its neighbor Xe atom. And because of this arrangement, each Xe atom forms an octahedron with 6 surrounding fluorine atoms.
$endgroup$
– SteffX
Jan 27 at 14:20
$begingroup$
@SteffX Why not await two fluorine neigbors above and two below?
$endgroup$
– HolgerFiedler
Jan 27 at 14:38
$begingroup$
@SteffX Wow, andselisks answer indeed shows 2 F above and 2 F below :-)
$endgroup$
– HolgerFiedler
Jan 27 at 14:52
add a comment |
$begingroup$
For reasons of symmetry I (without deeper chemical knowledge) supposed, that $ce{XeF4}$ has a shape like $ce{CH4}$. But according to Wikipedia its
crystalline structure was determined by both NMR spectroscopy and X-ray crystallography in 1963. The structure is square planar, as has been confirmed by neutron diffraction studies.
In a crystal the neighbor atoms have some influence on another and sometimes the chemical formula shows the numerical ratio between the elements but not draw the full picture of all interactions between the neighbor elements. Does the influence of the neighboring molecules in a $ce{XeF4}$ crystal makes the square planar shap? Is it imaginable that an isolated $ce{XeF4}$ molecule has a tetrahedral shape?
Additional question. What is the crystalline structure of $ce{XeF4}$?
crystal-structure solid-state-chemistry vsepr-theory noble-gases
edited Jan 27 at 14:07
andselisk
17.4k656117
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
$endgroup$
For reasons of symmetry I (without deeper chemical knowledge) supposed, that $ce{XeF4}$ has a shape like $ce{CH4}$. But according to Wikipedia its
crystalline structure was determined by both NMR spectroscopy and X-ray crystallography in 1963. The structure is square planar, as has been confirmed by neutron diffraction studies.
In a crystal the neighbor atoms have some influence on another and sometimes the chemical formula shows the numerical ratio between the elements but not draw the full picture of all interactions between the neighbor elements. Does the influence of the neighboring molecules in a $ce{XeF4}$ crystal makes the square planar shap? Is it imaginable that an isolated $ce{XeF4}$ molecule has a tetrahedral shape?
Additional question. What is the crystalline structure of $ce{XeF4}$?
crystal-structure solid-state-chemistry vsepr-theory noble-gases
crystal-structure solid-state-chemistry vsepr-theory noble-gases
edited Jan 27 at 14:07
andselisk
17.4k656117
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
edited Jan 27 at 14:07
andselisk
17.4k656117
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
edited Jan 27 at 14:07
andselisk
17.4k656117
edited Jan 27 at 14:07
andselisk
17.4k656117
edited Jan 27 at 14:07
andselisk
17.4k656117
17.4k656117
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
asked Jan 27 at 14:00
HolgerFiedlerHolgerFiedler
206128
206128
$begingroup$
It seems to me that you imagine that some "square planar XeF4" will stick together in the same plane, but it is obviously not the case because fluorine atoms repel each others. In fact, if you take two "square-planar XeF4" on top of each other, one will be shifted so one of its fluorine atom is right above its neighbor Xe atom. And because of this arrangement, each Xe atom forms an octahedron with 6 surrounding fluorine atoms.
$endgroup$
– SteffX
Jan 27 at 14:20
$begingroup$
@SteffX Why not await two fluorine neigbors above and two below?
$endgroup$
– HolgerFiedler
Jan 27 at 14:38
$begingroup$
@SteffX Wow, andselisks answer indeed shows 2 F above and 2 F below :-)
$endgroup$
– HolgerFiedler
Jan 27 at 14:52
add a comment |
$begingroup$
It seems to me that you imagine that some "square planar XeF4" will stick together in the same plane, but it is obviously not the case because fluorine atoms repel each others. In fact, if you take two "square-planar XeF4" on top of each other, one will be shifted so one of its fluorine atom is right above its neighbor Xe atom. And because of this arrangement, each Xe atom forms an octahedron with 6 surrounding fluorine atoms.
$endgroup$
– SteffX
Jan 27 at 14:20
$begingroup$
@SteffX Why not await two fluorine neigbors above and two below?
$endgroup$
– HolgerFiedler
Jan 27 at 14:38
$begingroup$
@SteffX Wow, andselisks answer indeed shows 2 F above and 2 F below :-)
$endgroup$
– HolgerFiedler
Jan 27 at 14:52
$begingroup$
It seems to me that you imagine that some "square planar XeF4" will stick together in the same plane, but it is obviously not the case because fluorine atoms repel each others. In fact, if you take two "square-planar XeF4" on top of each other, one will be shifted so one of its fluorine atom is right above its neighbor Xe atom. And because of this arrangement, each Xe atom forms an octahedron with 6 surrounding fluorine atoms.
$endgroup$
– SteffX
Jan 27 at 14:20
$begingroup$
It seems to me that you imagine that some "square planar XeF4" will stick together in the same plane, but it is obviously not the case because fluorine atoms repel each others. In fact, if you take two "square-planar XeF4" on top of each other, one will be shifted so one of its fluorine atom is right above its neighbor Xe atom. And because of this arrangement, each Xe atom forms an octahedron with 6 surrounding fluorine atoms.
$endgroup$
– SteffX
Jan 27 at 14:20
$begingroup$
@SteffX Why not await two fluorine neigbors above and two below?
$endgroup$
– HolgerFiedler
Jan 27 at 14:38
$begingroup$
@SteffX Why not await two fluorine neigbors above and two below?
$endgroup$
– HolgerFiedler
Jan 27 at 14:38
$begingroup$
@SteffX Wow, andselisks answer indeed shows 2 F above and 2 F below :-)
$endgroup$
– HolgerFiedler
Jan 27 at 14:52
$begingroup$
@SteffX Wow, andselisks answer indeed shows 2 F above and 2 F below :-)
$endgroup$
– HolgerFiedler
Jan 27 at 14:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $ce{F-Xe-F}$ slightly deviates from $90°$.
Figure 1. Packing of $ce{XeF4}$ molecules. Color code: $color{#90E050}{Largebullet}~ce{F}$;$color{#429EB0}{Largebullet}~ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11%$ (ICSD #27467)
The molecule has symmetry $D_mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $ce{F1-Xe-F2}$ is $pu{86 ± 30°}$ and the two independent distances $ce{Xe}$ to $ce{F}$ do not differ significantly and have an average value of $pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $ce{F...F}$ contact is $pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $ce{XeF4}$. Since all of the intermolecular $ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $pu{2.7 Å}$ it is not surprising that a second phase of $ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.
Fig. 1. The structure of the $ce{XeF4}$ molecule in the solid.
Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6%$ and stated that
The structure consists of a molecular packing of square planar molecules of $ce{XeF4}$.
With a comment on [1]:
Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.
References
- Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
- Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
$endgroup$
1
$begingroup$
This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.
$endgroup$
– Andrew
Jan 27 at 14:52
$begingroup$
@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.
$endgroup$
– HolgerFiedler
Jan 27 at 14:59
$begingroup$
I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)
$endgroup$
– Andrew
Jan 27 at 15:13
1
$begingroup$
For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).
$endgroup$
– Andrew
Jan 27 at 15:21
add a comment |
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1 Answer
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$begingroup$
There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $ce{F-Xe-F}$ slightly deviates from $90°$.
Figure 1. Packing of $ce{XeF4}$ molecules. Color code: $color{#90E050}{Largebullet}~ce{F}$;$color{#429EB0}{Largebullet}~ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11%$ (ICSD #27467)
The molecule has symmetry $D_mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $ce{F1-Xe-F2}$ is $pu{86 ± 30°}$ and the two independent distances $ce{Xe}$ to $ce{F}$ do not differ significantly and have an average value of $pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $ce{F...F}$ contact is $pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $ce{XeF4}$. Since all of the intermolecular $ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $pu{2.7 Å}$ it is not surprising that a second phase of $ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.
Fig. 1. The structure of the $ce{XeF4}$ molecule in the solid.
Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6%$ and stated that
The structure consists of a molecular packing of square planar molecules of $ce{XeF4}$.
With a comment on [1]:
Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.
References
- Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
- Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
$endgroup$
1
$begingroup$
This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.
$endgroup$
– Andrew
Jan 27 at 14:52
$begingroup$
@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.
$endgroup$
– HolgerFiedler
Jan 27 at 14:59
$begingroup$
I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)
$endgroup$
– Andrew
Jan 27 at 15:13
1
$begingroup$
For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).
$endgroup$
– Andrew
Jan 27 at 15:21
add a comment |
$begingroup$
There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $ce{F-Xe-F}$ slightly deviates from $90°$.
Figure 1. Packing of $ce{XeF4}$ molecules. Color code: $color{#90E050}{Largebullet}~ce{F}$;$color{#429EB0}{Largebullet}~ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11%$ (ICSD #27467)
The molecule has symmetry $D_mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $ce{F1-Xe-F2}$ is $pu{86 ± 30°}$ and the two independent distances $ce{Xe}$ to $ce{F}$ do not differ significantly and have an average value of $pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $ce{F...F}$ contact is $pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $ce{XeF4}$. Since all of the intermolecular $ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $pu{2.7 Å}$ it is not surprising that a second phase of $ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.
Fig. 1. The structure of the $ce{XeF4}$ molecule in the solid.
Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6%$ and stated that
The structure consists of a molecular packing of square planar molecules of $ce{XeF4}$.
With a comment on [1]:
Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.
References
- Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
- Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
$endgroup$
1
$begingroup$
This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.
$endgroup$
– Andrew
Jan 27 at 14:52
$begingroup$
@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.
$endgroup$
– HolgerFiedler
Jan 27 at 14:59
$begingroup$
I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)
$endgroup$
– Andrew
Jan 27 at 15:13
1
$begingroup$
For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).
$endgroup$
– Andrew
Jan 27 at 15:21
add a comment |
$begingroup$
There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $ce{F-Xe-F}$ slightly deviates from $90°$.
Figure 1. Packing of $ce{XeF4}$ molecules. Color code: $color{#90E050}{Largebullet}~ce{F}$;$color{#429EB0}{Largebullet}~ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11%$ (ICSD #27467)
The molecule has symmetry $D_mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $ce{F1-Xe-F2}$ is $pu{86 ± 30°}$ and the two independent distances $ce{Xe}$ to $ce{F}$ do not differ significantly and have an average value of $pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $ce{F...F}$ contact is $pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $ce{XeF4}$. Since all of the intermolecular $ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $pu{2.7 Å}$ it is not surprising that a second phase of $ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.
Fig. 1. The structure of the $ce{XeF4}$ molecule in the solid.
Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6%$ and stated that
The structure consists of a molecular packing of square planar molecules of $ce{XeF4}$.
With a comment on [1]:
Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.
References
- Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
- Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
$endgroup$
There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $ce{F-Xe-F}$ slightly deviates from $90°$.
Figure 1. Packing of $ce{XeF4}$ molecules. Color code: $color{#90E050}{Largebullet}~ce{F}$;$color{#429EB0}{Largebullet}~ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11%$ (ICSD #27467)
The molecule has symmetry $D_mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $ce{F1-Xe-F2}$ is $pu{86 ± 30°}$ and the two independent distances $ce{Xe}$ to $ce{F}$ do not differ significantly and have an average value of $pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $ce{F...F}$ contact is $pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $ce{XeF4}$. Since all of the intermolecular $ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $pu{2.7 Å}$ it is not surprising that a second phase of $ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.
Fig. 1. The structure of the $ce{XeF4}$ molecule in the solid.
Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6%$ and stated that
The structure consists of a molecular packing of square planar molecules of $ce{XeF4}$.
With a comment on [1]:
Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.
References
- Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
- Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
edited Jan 27 at 14:49
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
answered Jan 27 at 14:44
andseliskandselisk
17.4k656117
17.4k656117
1
$begingroup$
This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.
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– Andrew
Jan 27 at 14:52
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@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.
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– HolgerFiedler
Jan 27 at 14:59
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I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)
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– Andrew
Jan 27 at 15:13
1
$begingroup$
For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).
$endgroup$
– Andrew
Jan 27 at 15:21
add a comment |
1
$begingroup$
This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.
$endgroup$
– Andrew
Jan 27 at 14:52
$begingroup$
@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.
$endgroup$
– HolgerFiedler
Jan 27 at 14:59
$begingroup$
I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)
$endgroup$
– Andrew
Jan 27 at 15:13
1
$begingroup$
For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).
$endgroup$
– Andrew
Jan 27 at 15:21
1
1
$begingroup$
This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.
$endgroup$
– Andrew
Jan 27 at 14:52
$begingroup$
This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.
$endgroup$
– Andrew
Jan 27 at 14:52
$begingroup$
@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.
$endgroup$
– HolgerFiedler
Jan 27 at 14:59
$begingroup$
@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.
$endgroup$
– HolgerFiedler
Jan 27 at 14:59
$begingroup$
I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)
$endgroup$
– Andrew
Jan 27 at 15:13
$begingroup$
I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)
$endgroup$
– Andrew
Jan 27 at 15:13
1
1
$begingroup$
For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).
$endgroup$
– Andrew
Jan 27 at 15:21
$begingroup$
For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).
$endgroup$
– Andrew
Jan 27 at 15:21
add a comment |
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$begingroup$
It seems to me that you imagine that some "square planar XeF4" will stick together in the same plane, but it is obviously not the case because fluorine atoms repel each others. In fact, if you take two "square-planar XeF4" on top of each other, one will be shifted so one of its fluorine atom is right above its neighbor Xe atom. And because of this arrangement, each Xe atom forms an octahedron with 6 surrounding fluorine atoms.
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– SteffX
Jan 27 at 14:20
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@SteffX Why not await two fluorine neigbors above and two below?
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– HolgerFiedler
Jan 27 at 14:38
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@SteffX Wow, andselisks answer indeed shows 2 F above and 2 F below :-)
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– HolgerFiedler
Jan 27 at 14:52