The knowledge of intermolecular forces can be used to explain for the different boiling point (or melting point) trend of compounds which exist as a simple discrete molecule. These covalent compounds break intermolecular forces when they undergo a change in physical state (e.g. Solid to liquid or liquid to gas) and in this process these compound's covalent bond (between atoms) remain intact.
In an earlier post, I briefly described to you the different intermolecular forces. In this write up, I will aim to show how the knowledge of intermolecular forces can be applied to account for the different boiling point trend (or melting) in a series of simple discrete molecules.
Do note that melting point can also be affected by how the molecules are packed. Smaller molecules can be packed more densely and hence when melting them, more intermolecular forces are broken and thus giving a larger melting point despite its intermolecular forces is weaker.
(A) Comparing id-id against pd-pd against Hydrogen bonding
We will compare the different types of intermolecular forces when (1) The Mr of the molecules is relatively similar and (2) The shape of the molecule is relatively the same.
In (1) this implies that the number of electrons in the series of compounds is the similar and hence id-id (which is dependent on number of electrons) will be similar. In (2) it will ensure that the electron cloud size is similar (hence id-id) is similar.
When this happens, Hydrogen bonding is the strongest while id-id is the weakest.
(B) Comparing the Mr of the molecule
When the a pair of molecules respective Mr is very different. This implies that the number of electrons the present in the pair is also very different. Hence, the size of the electron cloud will be different too.
Hence, this result in the larger molecule to be polarised easily and hence its induce dipole moment is larger and hence its id-id interaction is stronger and therefore having a higher boiling point.
(C) Comparing the shape of the molecule (e.g. Branched or unbranched)
Unbranched molecules such as CH3CH2CH2CH3 have an elongated electron cloud which is easier to be polarised. Hence, its induced dipole moment will be large and hence, id-id interaction is stronger. While unbranched molecules such as (CH3)4C have a spherical electron cloud which is less easily polarised and hence smaller induced dipole and thus weaker id-id.
(D) If Mr, shape of molecules and IMF are similar/identical
For both molecules satisfy this subtitle and both have hydrogen bonding. Then there are two main factors. (1) Extensive-ness of hydrogen bonding. For example H2O has more extensive hydrogen bonding than HF and NH3 because it has the best (most equal) lone pair to hydrogen ratio. (2) Using the definition of Hydrogen bonding as illustrated by this diagram, HF has a stronger hydrogen bonding than NH3 because F is more electronegative than N and hence making the d+ on H in HF to be larger.
When both molecules satisfy this subtitle and both have pd-pd interactions. There are also two main factors. (1) Electronegativity of the atom. More electronegative atom results in a larger dipole moment. (2) More electronegative atoms which create a dipole moment in the same direction. This will result in the molecule to have a larger overall net dipole moment.
Needless to say, when both have only id-id interaction, the boiling point should be the same.
As this is a length entry. A mind map, which shows you how to sequence your thoughts to construct the concise and comprehensive argument is found here.
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Article written by Kwok YL 2009.
In an earlier post, I briefly described to you the different intermolecular forces. In this write up, I will aim to show how the knowledge of intermolecular forces can be applied to account for the different boiling point trend (or melting) in a series of simple discrete molecules.
Do note that melting point can also be affected by how the molecules are packed. Smaller molecules can be packed more densely and hence when melting them, more intermolecular forces are broken and thus giving a larger melting point despite its intermolecular forces is weaker.
(A) Comparing id-id against pd-pd against Hydrogen bonding
We will compare the different types of intermolecular forces when (1) The Mr of the molecules is relatively similar and (2) The shape of the molecule is relatively the same.
In (1) this implies that the number of electrons in the series of compounds is the similar and hence id-id (which is dependent on number of electrons) will be similar. In (2) it will ensure that the electron cloud size is similar (hence id-id) is similar.
When this happens, Hydrogen bonding is the strongest while id-id is the weakest.
(B) Comparing the Mr of the molecule
When the a pair of molecules respective Mr is very different. This implies that the number of electrons the present in the pair is also very different. Hence, the size of the electron cloud will be different too.
Hence, this result in the larger molecule to be polarised easily and hence its induce dipole moment is larger and hence its id-id interaction is stronger and therefore having a higher boiling point.
(C) Comparing the shape of the molecule (e.g. Branched or unbranched)
Unbranched molecules such as CH3CH2CH2CH3 have an elongated electron cloud which is easier to be polarised. Hence, its induced dipole moment will be large and hence, id-id interaction is stronger. While unbranched molecules such as (CH3)4C have a spherical electron cloud which is less easily polarised and hence smaller induced dipole and thus weaker id-id.
(D) If Mr, shape of molecules and IMF are similar/identical
For both molecules satisfy this subtitle and both have hydrogen bonding. Then there are two main factors. (1) Extensive-ness of hydrogen bonding. For example H2O has more extensive hydrogen bonding than HF and NH3 because it has the best (most equal) lone pair to hydrogen ratio. (2) Using the definition of Hydrogen bonding as illustrated by this diagram, HF has a stronger hydrogen bonding than NH3 because F is more electronegative than N and hence making the d+ on H in HF to be larger.
When both molecules satisfy this subtitle and both have pd-pd interactions. There are also two main factors. (1) Electronegativity of the atom. More electronegative atom results in a larger dipole moment. (2) More electronegative atoms which create a dipole moment in the same direction. This will result in the molecule to have a larger overall net dipole moment.
Needless to say, when both have only id-id interaction, the boiling point should be the same.
As this is a length entry. A mind map, which shows you how to sequence your thoughts to construct the concise and comprehensive argument is found here.
-- -- -- -- --
Article written by Kwok YL 2009.
Disclaimer and remarks:
- If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.
- This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.
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