# Van Der Wal's force doesn't add up

• satisverborum20
In summary, as you go down a group in the periodic table, the boiling points of elements increase due to the increase in the number of electrons and the resulting stronger Van der Waals forces. This is because with more electrons, there is a greater chance of electron imbalances, leading to stronger London dispersion forces. Additionally, the increase in atomic radius also leads to greater polarizability and fluctuations in dipole moments, resulting in stronger Van der Waals interactions. Note that the atomic size and nuclear radius are not directly correlated.
satisverborum20
I was given the explanation that as you go down a group in the periodic table, the boiling points of the elements increase as the radius of the atom increase. There are more electrons and therefore the greater distance they can move in which the temporary dipoles can cover and so the Van Der wal's force is bigger.

And here comes the question. I thought that Van der wal's force is a temporary imbalance of electrons in an atom in which one end is slightly negative than the other. Therefore if you have larger atomic radius then wouldn't there be smaller van der wal's force as they electrons are more likely to spread out over a bigger area creating lots of little dipoles which are insignificant to the overall charge of the atom or cancel each other out instead of creating a big temporary dipoles by being concentrated on one side.

Can anyone help me out or is there something I'm missing?

More electrons means that there will be more electron imbalances. Therefore, the London dispersion forces will be stronger.

But if the nuclear radius increase wouldn't this counteract the van der wal's forces because with bigger radius there would be less chance of all the electrons causing an imbalance as they would be more likely to "spread out".

Bigger atom => greater polarizability => greater dipole moment fluctuations => greater van der waals interactions.

For a rule of thumb relationship between atomic size and polarizability, see Griffiths' Introduction to Electrodynamics.

PS: Note that the atomic size is not correlated to the nuclear radius in any way.

## What is Van Der Wal's force?

Van Der Wal's force, also known as the London dispersion force, is a weak intermolecular force that results from temporary dipoles formed between atoms and molecules due to fluctuations in their electron distributions.

## Why doesn't Van Der Wal's force add up?

Van Der Wal's force is a cumulative effect of multiple temporary dipoles, and the strength of this force is dependent on the number of atoms/molecules present. However, at very close distances, the repulsive forces between atoms/molecules start to outweigh the attractive Van Der Wal's force, resulting in a net repulsive force.

## How is Van Der Wal's force calculated?

Van Der Wal's force is calculated using the formula F = -(C/r^6), where F is the force, C is a constant, and r is the distance between the atoms/molecules. This formula is based on the theory of quantum mechanics.

## What are the implications of Van Der Wal's force not adding up?

The fact that Van Der Wal's force does not add up at very close distances has significant implications in fields such as nanotechnology and surface science. It can affect the stability of materials and the behavior of nanoparticles and molecules at the nanoscale.

## Can Van Der Wal's force be manipulated?

Yes, Van Der Wal's force can be manipulated by changing the distance between atoms/molecules or by altering the electron distributions through the use of polarizable materials or external electric fields. Researchers are also exploring the potential of using Van Der Wal's force for various applications, such as controlling molecular self-assembly and creating new materials.

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