as level chem

Im doing this question and its asking does CO or CH4 have a higher boiling point and Im unsure of. the type of intermolecular forces each has

Reply 1

Hellllpppp

20

CO has dipole-dipole forces as oxygen is much more electronegative than carbon and CH4 only has van der Waals. I hope this helps

Original post by Hellllpppp

CO has dipole-dipole forces as oxygen is much more electronegative than carbon and CH4 only has van der Waals. I hope this helps

Both of these interactions are types of Van der Waals force.

However, it depends on the definitions of your specification/syllabus

Methane has only London dispersion forces, while carbon monoxide has both London dispersion forces and permanent dipole-dipole interactions.

Reply 3

Hellllpppp

20

Original post by charco

Both of these interactions are types of Van der Waals force.

However, it depends on the definitions of your specification/syllabus

Methane has only London dispersion forces, while carbon monoxide has both London dispersion forces and permanent dipole-dipole interactions.

Oh yeah I missed the words permanent and I meant the strongest force is permanent dipole-dipole interactions. Sorry :smile:

For aqa spec you can put London but all the mark schemes have van der waals.

Reply 4

smellychicken

6

Original post by Kimik

Im doing this question and its asking does CO or CH4 have a higher boiling point and Im unsure of. the type of intermolecular forces each has

yh also to clarify (since i didnt know this when i was first learning this)- if youre asking urself why doesnt ch4 has (permanent) dipole-dipole forces like carbon monoxide? the difference in electronegativity between c and h is too negligible for the (permanent) dipole-dipole forces to be present. In (covalent) molecules you can always assume there are Van der Waals forces however there are some conditions for the other types of intermolecular forces (as mentioned in addition to Van der Waals forces, there are permanent dipole-dipole forces and then not relevant to this post but there is also hydrogen bonding). one further thing to clarify- i keep referring to PERMANENT dipole-dipole forces this is to not to confuse them w Van der Waals forces- these work by INDUCING a temporary dipole-dipole forces due to array/distribution of molecules. feel free to read more here: https://www.chemguide.co.uk/atoms/bonding/vdw.html

hope that helped!

Original post by smellychicken

yh also to clarify (since i didnt know this when i was first learning this)- if youre asking urself why doesnt ch4 has (permanent) dipole-dipole forces like carbon monoxide? the difference in electronegativity between c and h is too negligible for the (permanent) dipole-dipole forces to be present. In (covalent) molecules you can always assume there are Van der Waals forces however there are some conditions for the other types of intermolecular forces (as mentioned in addition to Van der Waals forces, there are permanent dipole-dipole forces and then not relevant to this post but there is also hydrogen bonding). one further thing to clarify- i keep referring to PERMANENT dipole-dipole forces this is to not to confuse them w Van der Waals forces- these work by INDUCING a temporary dipole-dipole forces due to array/distribution of molecules. feel free to read more here: https://www.chemguide.co.uk/atoms/bonding/vdw.html

hope that helped!

The lack of polarity of CH₄ has nothing to do with a negligible difference in electronegativity between carbon and hydrogen, and everything to do with the symmetry of the molecule causing the individual dipole vectors to resolve to zero.

The individual C-H bonds are polarised (not much, but polarised all the same), but the molecule overall is non-polar.

Reply 6

smellychicken

6

im pretty sure it is infact to do with electronegativity- this is why hydrocarbons also dont exhibit permanent dipole-dipole forces- although ig this is only MOST of the time.

here are some references:
https://en.wikipedia.org/wiki/Carbon%E2%80%93hydrogen_bond#:~:text=Using%20Pauling's%20scale%E2%80%94C%20(2.55,regarded%20as%20being%20non%2Dpolar.
https://www.quora.com/Why-are-C-H-bonds-considered-nonpolar
https://www.quora.com/Is-CH4-non-polar-Why-or-why-not

however, i do understand what youre saying due to shapes of molecules, the nature of CH4 with its 4 bonding pairs and therefore tetrahedral shape, even IF the bond was considered polar the dipoles would cancel each other out and therefore form a nonpolar molecule.

Reply 7

Kimik

OP

5

Original post by smellychicken

yh also to clarify (since i didnt know this when i was first learning this)- if youre asking urself why doesnt ch4 has (permanent) dipole-dipole forces like carbon monoxide? the difference in electronegativity between c and h is too negligible for the (permanent) dipole-dipole forces to be present. In (covalent) molecules you can always assume there are Van der Waals forces however there are some conditions for the other types of intermolecular forces (as mentioned in addition to Van der Waals forces, there are permanent dipole-dipole forces and then not relevant to this post but there is also hydrogen bonding). one further thing to clarify- i keep referring to PERMANENT dipole-dipole forces this is to not to confuse them w Van der Waals forces- these work by INDUCING a temporary dipole-dipole forces due to array/distribution of molecules. feel free to read more here: https://www.chemguide.co.uk/atoms/bonding/vdw.html

hope that helped!

Thanks that makes sense now :smile:

I understand the intermolecular forces they have but surely this means carbon monoxide has the higher boiling point because I thought Van Der Waals forces are the weakest intermolecular forces but I looked it up and it said carbon monoxide has a boiling point of -191 and methane has a boiling point of -161 which has slightly confused me

Original post by Kimik

Thanks that makes sense now :smile:

I understand the intermolecular forces they have but surely this means carbon monoxide has the higher boiling point because I thought Van Der Waals forces are the weakest intermolecular forces but I looked it up and it said carbon monoxide has a boiling point of -191 and methane has a boiling point of -161 which has slightly confused me

Sometimes chemistry throws up empirical evidence that contradicts the models that we have. When this happens, the model must be modified to account for the new observation/measurement.

The two molecules CH4 and CO both have London dispersion forces between particles, that are a function of relative mass, available surface area and packing. The second factor attenuates the first, more important factor. The packing is also a function of the shape of the molecule.

In this case the relative masses CH4 = 16, CO = 28, suggest that LDF are stronger in CO, so it seems that the model is failing. Also, CO should be CO polar, which would also contribute to intermolecular forces.

Bonding in carbon monoxide is not a simple triple bond, when the molecular orbital diagram is constructed, which could account for lack of polarity, but with the LDF issue this would not explain the overall picture.

Maybe the electrons in molecular pi orbitals repel from molecule to molecule - I don't know, but it seems that the three factors that would normally explain the boiling point cannot do so in this case.

EDIT: A little bit of research suggests that we are looking at a surface area issue. The length of the CO molecule is about 113pm whereas the height of the equilateral pyramid that is the methane molecule is about 400pm. This would give methane a much larger surface area on which the LDF can act.

(edited 3 years ago)

as level chem

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