What is the maximum distance over which van der Waals forces act?

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The VDW radius places a lower limit on the distance between atoms mutually attracted to one another. I am looking for the upper limit: what is the name of the concept or theory that describes the maximum distance over which the various attractive forces (Keesom, Debye, London...) become operable / take effect?



This question is similar to the following Q, which didn't get fully answered:



Range of distance for Van der Waals force










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    I think editing chemistry.stackexchange.com/questions/107870/… would be preferable to asking new question.
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    – Mithoron
    Jan 12 at 22:57















7












$begingroup$


The VDW radius places a lower limit on the distance between atoms mutually attracted to one another. I am looking for the upper limit: what is the name of the concept or theory that describes the maximum distance over which the various attractive forces (Keesom, Debye, London...) become operable / take effect?



This question is similar to the following Q, which didn't get fully answered:



Range of distance for Van der Waals force










share|improve this question









$endgroup$











  • $begingroup$
    I think editing chemistry.stackexchange.com/questions/107870/… would be preferable to asking new question.
    $endgroup$
    – Mithoron
    Jan 12 at 22:57













7












7








7


2



$begingroup$


The VDW radius places a lower limit on the distance between atoms mutually attracted to one another. I am looking for the upper limit: what is the name of the concept or theory that describes the maximum distance over which the various attractive forces (Keesom, Debye, London...) become operable / take effect?



This question is similar to the following Q, which didn't get fully answered:



Range of distance for Van der Waals force










share|improve this question









$endgroup$




The VDW radius places a lower limit on the distance between atoms mutually attracted to one another. I am looking for the upper limit: what is the name of the concept or theory that describes the maximum distance over which the various attractive forces (Keesom, Debye, London...) become operable / take effect?



This question is similar to the following Q, which didn't get fully answered:



Range of distance for Van der Waals force







intermolecular-forces






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asked Jan 12 at 19:07









charlie K3charlie K3

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523











  • $begingroup$
    I think editing chemistry.stackexchange.com/questions/107870/… would be preferable to asking new question.
    $endgroup$
    – Mithoron
    Jan 12 at 22:57
















  • $begingroup$
    I think editing chemistry.stackexchange.com/questions/107870/… would be preferable to asking new question.
    $endgroup$
    – Mithoron
    Jan 12 at 22:57















$begingroup$
I think editing chemistry.stackexchange.com/questions/107870/… would be preferable to asking new question.
$endgroup$
– Mithoron
Jan 12 at 22:57




$begingroup$
I think editing chemistry.stackexchange.com/questions/107870/… would be preferable to asking new question.
$endgroup$
– Mithoron
Jan 12 at 22:57










3 Answers
3






active

oldest

votes


















9












$begingroup$

There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.



The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.



The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.



Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.






share|improve this answer











$endgroup$












  • $begingroup$
    this clarifies things quite a bit, thanks. Further: in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces. Beyond this distance, they don't "see" each other and thus remain separated? I understand the exact distance value varies.
    $endgroup$
    – charlie K3
    Jan 12 at 19:40










  • $begingroup$
    VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated
    $endgroup$
    – Ian Bush
    Jan 13 at 7:29


















5












$begingroup$

As jheindel has previously pointed the interaction has no limit. What you are probably looking for is the distance at which the interaction energy becomes smaller than thermal energy at a given temperature. When the interaction is smaller then thermal motions govern what happens not the intermolecular interaction. Thermal energy at $300$ K is $k_BT = 1.38cdot10^-23cdot 300 = 4.1cdot 10^-19$ J.



The interactions all have the general form $E=V/(epsilon r^n)$ where $V$ is scaling term: for example for charge - charge interaction $n=1$ and $V=q_1q_2e^2/(4piepsilon_0)$ where $q$ are the charge signs, $e$ the charge on the electron and $epsilon _0$ the permittivity of free space $8.854cdot10^-12$. Finally $epsilon$ is the solvent dielectric constant, cyclohexane = 2, water = 78.



The interaction energy between two charges, say sodium and chlorine ions, in vacuum separated by 0.3 nm is approximately $190k_BT$, but in water this only $2.4k_BT$ so the distance at which the intermolecular interaction is important depends very much on the conditions. The reason that $epsilon$ has such an effect is that the high dielectric solvent attenuates the electric field around an ion or dipole.



There are other types of interaction ion-dipole, dipole-dipole, induced dipole-induced dipole and so on that have energies that depend, where appropriate, on the magnitude of the dipole, polarisability, the relative dipole orientation and various powers of separation $1/r^n$. For example for dipole-dipole interaction with fixed dipole directions the relative dipole angle is important and $n=3$ but for freely rotating dipoles the dipole angle is averaged out resulting in $n=6$ (Keesom energy). London dispersion force (induced dipole-induced dipole) depends on each molecule's polarisability and has $n=6$,in comparison charge-dipole interaction has $n=4$.



If you look these up to calculate particular interactions note that many authors quote values in vacuum and so ignore the solvent dielectric constant. You can include this by changing $epsilon_0to epsilon_0epsilon$.






share|improve this answer









$endgroup$




















    0












    $begingroup$

    Indeed there are various decriptions of VdW forces. For some, this class includes dipole-dipole, dipole-instantaneous dipole, and instantaneous dipole-instantaneous dipole. Of course, these forces are not all of the same intensity.
    But for all, whatever the exponent of r, there is no limit, and this is why one speaks of energy instead, as the one needed for putting the two chemical species at an infinite distance.
    By the way, I would suggest that when one is giving an answer, references would be given as well.
    And here, I suggest:
    Physical Chemistry, McQuarrie
    The chemical bond, Linus Pauling ;-)






    share|improve this answer









    $endgroup$












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      3 Answers
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      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

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      active

      oldest

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      9












      $begingroup$

      There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.



      The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.



      The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.



      Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.






      share|improve this answer











      $endgroup$












      • $begingroup$
        this clarifies things quite a bit, thanks. Further: in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces. Beyond this distance, they don't "see" each other and thus remain separated? I understand the exact distance value varies.
        $endgroup$
        – charlie K3
        Jan 12 at 19:40










      • $begingroup$
        VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated
        $endgroup$
        – Ian Bush
        Jan 13 at 7:29















      9












      $begingroup$

      There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.



      The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.



      The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.



      Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.






      share|improve this answer











      $endgroup$












      • $begingroup$
        this clarifies things quite a bit, thanks. Further: in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces. Beyond this distance, they don't "see" each other and thus remain separated? I understand the exact distance value varies.
        $endgroup$
        – charlie K3
        Jan 12 at 19:40










      • $begingroup$
        VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated
        $endgroup$
        – Ian Bush
        Jan 13 at 7:29













      9












      9








      9





      $begingroup$

      There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.



      The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.



      The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.



      Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.






      share|improve this answer











      $endgroup$



      There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.



      The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.



      The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.



      Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Jan 12 at 23:13

























      answered Jan 12 at 19:30









      jheindeljheindel

      7,8342451




      7,8342451











      • $begingroup$
        this clarifies things quite a bit, thanks. Further: in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces. Beyond this distance, they don't "see" each other and thus remain separated? I understand the exact distance value varies.
        $endgroup$
        – charlie K3
        Jan 12 at 19:40










      • $begingroup$
        VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated
        $endgroup$
        – Ian Bush
        Jan 13 at 7:29
















      • $begingroup$
        this clarifies things quite a bit, thanks. Further: in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces. Beyond this distance, they don't "see" each other and thus remain separated? I understand the exact distance value varies.
        $endgroup$
        – charlie K3
        Jan 12 at 19:40










      • $begingroup$
        VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated
        $endgroup$
        – Ian Bush
        Jan 13 at 7:29















      $begingroup$
      this clarifies things quite a bit, thanks. Further: in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces. Beyond this distance, they don't "see" each other and thus remain separated? I understand the exact distance value varies.
      $endgroup$
      – charlie K3
      Jan 12 at 19:40




      $begingroup$
      this clarifies things quite a bit, thanks. Further: in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces. Beyond this distance, they don't "see" each other and thus remain separated? I understand the exact distance value varies.
      $endgroup$
      – charlie K3
      Jan 12 at 19:40












      $begingroup$
      VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated
      $endgroup$
      – Ian Bush
      Jan 13 at 7:29




      $begingroup$
      VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated
      $endgroup$
      – Ian Bush
      Jan 13 at 7:29











      5












      $begingroup$

      As jheindel has previously pointed the interaction has no limit. What you are probably looking for is the distance at which the interaction energy becomes smaller than thermal energy at a given temperature. When the interaction is smaller then thermal motions govern what happens not the intermolecular interaction. Thermal energy at $300$ K is $k_BT = 1.38cdot10^-23cdot 300 = 4.1cdot 10^-19$ J.



      The interactions all have the general form $E=V/(epsilon r^n)$ where $V$ is scaling term: for example for charge - charge interaction $n=1$ and $V=q_1q_2e^2/(4piepsilon_0)$ where $q$ are the charge signs, $e$ the charge on the electron and $epsilon _0$ the permittivity of free space $8.854cdot10^-12$. Finally $epsilon$ is the solvent dielectric constant, cyclohexane = 2, water = 78.



      The interaction energy between two charges, say sodium and chlorine ions, in vacuum separated by 0.3 nm is approximately $190k_BT$, but in water this only $2.4k_BT$ so the distance at which the intermolecular interaction is important depends very much on the conditions. The reason that $epsilon$ has such an effect is that the high dielectric solvent attenuates the electric field around an ion or dipole.



      There are other types of interaction ion-dipole, dipole-dipole, induced dipole-induced dipole and so on that have energies that depend, where appropriate, on the magnitude of the dipole, polarisability, the relative dipole orientation and various powers of separation $1/r^n$. For example for dipole-dipole interaction with fixed dipole directions the relative dipole angle is important and $n=3$ but for freely rotating dipoles the dipole angle is averaged out resulting in $n=6$ (Keesom energy). London dispersion force (induced dipole-induced dipole) depends on each molecule's polarisability and has $n=6$,in comparison charge-dipole interaction has $n=4$.



      If you look these up to calculate particular interactions note that many authors quote values in vacuum and so ignore the solvent dielectric constant. You can include this by changing $epsilon_0to epsilon_0epsilon$.






      share|improve this answer









      $endgroup$

















        5












        $begingroup$

        As jheindel has previously pointed the interaction has no limit. What you are probably looking for is the distance at which the interaction energy becomes smaller than thermal energy at a given temperature. When the interaction is smaller then thermal motions govern what happens not the intermolecular interaction. Thermal energy at $300$ K is $k_BT = 1.38cdot10^-23cdot 300 = 4.1cdot 10^-19$ J.



        The interactions all have the general form $E=V/(epsilon r^n)$ where $V$ is scaling term: for example for charge - charge interaction $n=1$ and $V=q_1q_2e^2/(4piepsilon_0)$ where $q$ are the charge signs, $e$ the charge on the electron and $epsilon _0$ the permittivity of free space $8.854cdot10^-12$. Finally $epsilon$ is the solvent dielectric constant, cyclohexane = 2, water = 78.



        The interaction energy between two charges, say sodium and chlorine ions, in vacuum separated by 0.3 nm is approximately $190k_BT$, but in water this only $2.4k_BT$ so the distance at which the intermolecular interaction is important depends very much on the conditions. The reason that $epsilon$ has such an effect is that the high dielectric solvent attenuates the electric field around an ion or dipole.



        There are other types of interaction ion-dipole, dipole-dipole, induced dipole-induced dipole and so on that have energies that depend, where appropriate, on the magnitude of the dipole, polarisability, the relative dipole orientation and various powers of separation $1/r^n$. For example for dipole-dipole interaction with fixed dipole directions the relative dipole angle is important and $n=3$ but for freely rotating dipoles the dipole angle is averaged out resulting in $n=6$ (Keesom energy). London dispersion force (induced dipole-induced dipole) depends on each molecule's polarisability and has $n=6$,in comparison charge-dipole interaction has $n=4$.



        If you look these up to calculate particular interactions note that many authors quote values in vacuum and so ignore the solvent dielectric constant. You can include this by changing $epsilon_0to epsilon_0epsilon$.






        share|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          As jheindel has previously pointed the interaction has no limit. What you are probably looking for is the distance at which the interaction energy becomes smaller than thermal energy at a given temperature. When the interaction is smaller then thermal motions govern what happens not the intermolecular interaction. Thermal energy at $300$ K is $k_BT = 1.38cdot10^-23cdot 300 = 4.1cdot 10^-19$ J.



          The interactions all have the general form $E=V/(epsilon r^n)$ where $V$ is scaling term: for example for charge - charge interaction $n=1$ and $V=q_1q_2e^2/(4piepsilon_0)$ where $q$ are the charge signs, $e$ the charge on the electron and $epsilon _0$ the permittivity of free space $8.854cdot10^-12$. Finally $epsilon$ is the solvent dielectric constant, cyclohexane = 2, water = 78.



          The interaction energy between two charges, say sodium and chlorine ions, in vacuum separated by 0.3 nm is approximately $190k_BT$, but in water this only $2.4k_BT$ so the distance at which the intermolecular interaction is important depends very much on the conditions. The reason that $epsilon$ has such an effect is that the high dielectric solvent attenuates the electric field around an ion or dipole.



          There are other types of interaction ion-dipole, dipole-dipole, induced dipole-induced dipole and so on that have energies that depend, where appropriate, on the magnitude of the dipole, polarisability, the relative dipole orientation and various powers of separation $1/r^n$. For example for dipole-dipole interaction with fixed dipole directions the relative dipole angle is important and $n=3$ but for freely rotating dipoles the dipole angle is averaged out resulting in $n=6$ (Keesom energy). London dispersion force (induced dipole-induced dipole) depends on each molecule's polarisability and has $n=6$,in comparison charge-dipole interaction has $n=4$.



          If you look these up to calculate particular interactions note that many authors quote values in vacuum and so ignore the solvent dielectric constant. You can include this by changing $epsilon_0to epsilon_0epsilon$.






          share|improve this answer









          $endgroup$



          As jheindel has previously pointed the interaction has no limit. What you are probably looking for is the distance at which the interaction energy becomes smaller than thermal energy at a given temperature. When the interaction is smaller then thermal motions govern what happens not the intermolecular interaction. Thermal energy at $300$ K is $k_BT = 1.38cdot10^-23cdot 300 = 4.1cdot 10^-19$ J.



          The interactions all have the general form $E=V/(epsilon r^n)$ where $V$ is scaling term: for example for charge - charge interaction $n=1$ and $V=q_1q_2e^2/(4piepsilon_0)$ where $q$ are the charge signs, $e$ the charge on the electron and $epsilon _0$ the permittivity of free space $8.854cdot10^-12$. Finally $epsilon$ is the solvent dielectric constant, cyclohexane = 2, water = 78.



          The interaction energy between two charges, say sodium and chlorine ions, in vacuum separated by 0.3 nm is approximately $190k_BT$, but in water this only $2.4k_BT$ so the distance at which the intermolecular interaction is important depends very much on the conditions. The reason that $epsilon$ has such an effect is that the high dielectric solvent attenuates the electric field around an ion or dipole.



          There are other types of interaction ion-dipole, dipole-dipole, induced dipole-induced dipole and so on that have energies that depend, where appropriate, on the magnitude of the dipole, polarisability, the relative dipole orientation and various powers of separation $1/r^n$. For example for dipole-dipole interaction with fixed dipole directions the relative dipole angle is important and $n=3$ but for freely rotating dipoles the dipole angle is averaged out resulting in $n=6$ (Keesom energy). London dispersion force (induced dipole-induced dipole) depends on each molecule's polarisability and has $n=6$,in comparison charge-dipole interaction has $n=4$.



          If you look these up to calculate particular interactions note that many authors quote values in vacuum and so ignore the solvent dielectric constant. You can include this by changing $epsilon_0to epsilon_0epsilon$.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 13 at 10:57









          porphyrinporphyrin

          17.4k2954




          17.4k2954





















              0












              $begingroup$

              Indeed there are various decriptions of VdW forces. For some, this class includes dipole-dipole, dipole-instantaneous dipole, and instantaneous dipole-instantaneous dipole. Of course, these forces are not all of the same intensity.
              But for all, whatever the exponent of r, there is no limit, and this is why one speaks of energy instead, as the one needed for putting the two chemical species at an infinite distance.
              By the way, I would suggest that when one is giving an answer, references would be given as well.
              And here, I suggest:
              Physical Chemistry, McQuarrie
              The chemical bond, Linus Pauling ;-)






              share|improve this answer









              $endgroup$

















                0












                $begingroup$

                Indeed there are various decriptions of VdW forces. For some, this class includes dipole-dipole, dipole-instantaneous dipole, and instantaneous dipole-instantaneous dipole. Of course, these forces are not all of the same intensity.
                But for all, whatever the exponent of r, there is no limit, and this is why one speaks of energy instead, as the one needed for putting the two chemical species at an infinite distance.
                By the way, I would suggest that when one is giving an answer, references would be given as well.
                And here, I suggest:
                Physical Chemistry, McQuarrie
                The chemical bond, Linus Pauling ;-)






                share|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Indeed there are various decriptions of VdW forces. For some, this class includes dipole-dipole, dipole-instantaneous dipole, and instantaneous dipole-instantaneous dipole. Of course, these forces are not all of the same intensity.
                  But for all, whatever the exponent of r, there is no limit, and this is why one speaks of energy instead, as the one needed for putting the two chemical species at an infinite distance.
                  By the way, I would suggest that when one is giving an answer, references would be given as well.
                  And here, I suggest:
                  Physical Chemistry, McQuarrie
                  The chemical bond, Linus Pauling ;-)






                  share|improve this answer









                  $endgroup$



                  Indeed there are various decriptions of VdW forces. For some, this class includes dipole-dipole, dipole-instantaneous dipole, and instantaneous dipole-instantaneous dipole. Of course, these forces are not all of the same intensity.
                  But for all, whatever the exponent of r, there is no limit, and this is why one speaks of energy instead, as the one needed for putting the two chemical species at an infinite distance.
                  By the way, I would suggest that when one is giving an answer, references would be given as well.
                  And here, I suggest:
                  Physical Chemistry, McQuarrie
                  The chemical bond, Linus Pauling ;-)







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Jan 13 at 10:23









                  user50565user50565

                  762




                  762



























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