London Forces/Halogens Homework: Boiling Point Trend & Prediction

[Physics345]

Homework Statement

Consider the following boiling point data for these halogen molecules.

F2 18 electrons BP -188

Cl2 34 electrons BP -34

Br2 70 electrons BP 59

I2 106 electrons BP 184a) Explain the trend in the boiling points of the halogens by describing the intermolecular forces present.

b) Plot a graph of boiling point against the number of electrons. What force is directly related to the number of electrons?

c) Use your graph to predict the boiling point of astatine

Homework Equations

none

The Attempt at a Solution

Hope to hear back from you guys soon thanks!

 

[Ygggdrasil]

Your answer somewhere should describe they physical principle for why electrons with a greater number of electrons experience greater London dispersion forces.

You should show your work for how you estimated the boiling point of astatine from the number of electrons. Currently, it looks like you just Googled the boiling point.

 

[Physics345]

Your answer somewhere should describe they physical principle for why electrons with a greater number of electrons experience greater London dispersion forces.

You should show your work for how you estimated the boiling point of astatine from the number of electrons. Currently, it looks like you just Googled the boiling point.

I actually did google it. It didn't make sense to estimate the BP since they didn't do it in their examples but considering the trend of the graph an estimated number would be slightly below double the BP of Iodine, making me estimate 350 degrees. I'll get back to you with a physical property in a moment.

 

[Physics345]

London Forces, are caused by negative electrons attracting in one molecule, as well as attracting the positive electrons from another molecule. When there are more electrons/protons pulling each other, the London forces also become stronger. Therefore London forces rising/decreasing is proportional to the even amount of electrons/protons.

But where should I place it under the graph? or in part a ?

 

[Ygggdrasil]

London Forces, are caused by negative electrons attracting in one molecule, as well as attracting the positive electrons from another molecule.

Is there such a thing as positive electrons?

A key concept to consider in your answer is how the polarizability of a molecule enables dispersion forces (induced dipole-induced dipole interactions).

Question (b) asks which force varies with the number of electrons, so that would be a good opportunity to describe why the strength of dispersion forces increases with the number of electrons a molecule has.

 

[Physics345]

Is there such a thing as positive electrons?

A key concept to consider in your answer is how the polarizability of a molecule enables dispersion forces (induced dipole-induced dipole interactions).

Question (b) asks which force varies with the number of electrons, so that would be a good opportunity to describe why the strength of dispersion forces increases with the number of electrons a molecule has.

Oops, I wrote that up quickly, I meant to say protons good catch!
Also what do you mean by dispersion forces? I thought only London forces apply to this scenario.

 

[Ygggdrasil]

Oops, I wrote that up quickly, I meant to say protons good catch!
Also what do you mean by dispersion forces? I thought only London forces apply to this scenario.

They are different names for the same phenomena: https://en.wikipedia.org/wiki/London_dispersion_force

 

[Physics345]

Awesome! that really helped clarify things. To be honest I only just started chem two days ago. You have been a great help with my grasping of the subject and I greatly appreciate you taking the time and giving me advice on all my threads. Now onto my new and improved answer!

London forces are caused by negative electrons attracting in one molecule, as well as attracting the positive protons from another molecule. When there are more electrons/protons pulling each other, the London forces also become stronger. Therefore London forces rising/decreasing are proportional to the even amount of electrons/protons. Also it is important to note, the strength of the London forces is also increased for larger molecules but in a smaller manner because in turn the polarizability is also increased since larger molecules have more dispersed electron clouds.

 

[Ygggdrasil]

The answer is still not correct. The net effect of having more electrons and more protons cancels out; yes, more electrons in the first atom experience a greater pull from the greater number of protons in the second atom, but this is canceled out by the greater number of electrons in the second atom. The answer really has to do with how tightly the electrons are held by each atom and the extent to which the electron clouds can deformed by an external electric field. I would suggest re-reading your textbook's section on London dispersion forces.

 

[Physics345]

Okay I just finished re reading my text. Honestly all it says relating to physical properties of London forces, is relayed in the form of a graph that has all the forces. The London forces section of the graph says 0.5-to-2 kJ/mol. Are we referring to the equal distribution of electrons causing a temporary dipole in the molecule? or the fact that the electrons attract the protons?
Also on a side note I thought that part a explains that the electrons/protons cancel each others pulls.

 

[Physics345]

Okay after doing research on google I came to a much better understanding my textbook, is very vague and cheap. It's a horrible source of information the way the information is listed is complete garbage. The connections between concepts are way to spread out. Anyways here's my new answer =)

New answer
Inter-molecular forces and boiling points are related to the polarization and dipole-to-dipole of the molecules in question, and the polarization of the molecules is directly related to the electrons of the molecules and the space/radius the electrons are spread. Therefore, the boiling points are attributed to the strength of the bond between the atoms, and the net dipole-to-dipole moment which occurs between the atoms when the electrons attract each other from positive end of polar molecule/atom to the negative end of another polar molecule/atom.

 

[mjc123]

But dihalogen molecules are not polar.

 

[Physics345]

But dihalogen molecules are not polar.

Well technically it is partially polar/non polar, but for a general chem course this type of answer will suffice. The CH bond is not really polar, and the other three bonds are polar, therefore we can say it is relatively polar that would still be correct.

 

[Ygggdrasil]

New answer
Inter-molecular forces and boiling points are related to the polarization and dipole-to-dipole of the molecules in question, and the polarization of the molecules is directly related to the electrons of the molecules and the space/radius the electrons are spread. Therefore, the boiling points are attributed to the strength of the bond between the atoms, and the net dipole-to-dipole moment which occurs between the atoms when the electrons attract each other from positive end of polar molecule/atom to the negative end of another polar molecule/atom.

This answer seems more correct, but it is not explained very clearly or logically.

The boiling point of a solution is related to the strength of the inter-molecular forces between the molecules that compose the solution. In the case of the dihalogens, the main intermolecular force between the molecules are London dispersion forces. The strength of London disperson forces are related to the polarizability of the molecule. The polarizability of the molecule is related to the size of the molecule (which is related to the number of electrons). The more electrons the molecule has, the more weakly held the outermost electrons are, allowing them to be more easily influenced by an external electric field.

the net dipole-to-dipole moment which occurs between the atoms when the electrons attract each other from positive end of polar molecule/atom to the negative end of another polar molecule/atom.

From this part, it still does not seem like you understand the London dispersion force. It's unclear what you mean when you say "dipole-to-dipole moment" as this is not a term chemists typically use. Furthermore, as the other commenter mentioned, the dihalogens are not polar, so the last part of the explanation is also incorrect. In most cases, the dipole-dipole interactions between two polar compounds are stronger than the London dispersion forces they experience, so if the dihalogens were polar, the effects of the London dispersion forces would be minor compared to the other intermolecular forces that the molecules experience.

For more info in dispersion forces, see:
https://ch301.cm.utexas.edu/section2.php?target=imfs/forces/forces-attraction-all.php
https://chem.libretexts.org/Core/Ph...c_Interactions/London_Dispersion_Interactions

 

[Physics345]

For more info in dispersion forces, see:
https://ch301.cm.utexas.edu/section2.php?target=imfs/forces/forces-attraction-all.php
https://chem.libretexts.org/Core/Ph...c_Interactions/London_Dispersion_Interactions

I'll be sure to read them in a little bit, I'm currently running a few errands as soon as I get home I'll get on it. Hopefully I'll get a better understanding then. I appreciate you taking your time writing all that out.

 

[Physics345]

Okay I figured out the issue, I did some relatively intensive reading on inter molecular forces allowing me to see what's wrong here.