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According to my knowledge, polarization occurs when in a compound there is a difference of polarity its constituents.

But in the case of noble gases, there is a single component. What exactly is polarization of noble gases and how does it occur?

NotEvans.
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Esha Mukhopadhyay
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  • You've (or your textbook has) generalised the problem to the point where an answer makes no physical sense any more. Van-der-Waals forces have different components depending on the observed system (i.e. London forces, dipole-dipole interactions, etc.). Read and understand, don't try to memorise rules. – Karl May 07 '17 at 12:55
  • i too know van der waal forces differ from point of observed system. but the issue is in case of noble gases i do not how dipole dipole interactions occur? – Esha Mukhopadhyay May 07 '17 at 13:01
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    The answer is London dispersion: Induced dipoles. They are one central part of the van-der-Waals concept, which you seem to have missed. – Karl May 07 '17 at 13:03
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    You probably mean polarisability rather than polarisation. This is a quantum effect and due to the fluctuating electron density any atom has. A transient dipole is formed as the electron density is, instantaneously, not symmetrical. Over time this transient dipole averages to zero but while it exists this produces an electric field that can induce a dipole in a nearby atom. It works out that the time average of the interaction between the two induced dipoles is not zero which generates an attractive (dispersion) force between them. London's eqn. describes the interaction. – porphyrin May 07 '17 at 15:06
  • ok so if we consider this method then any atom can be polarized i guess? but m sure the time period for which the noble gases remain a dipole would b vry small then. so is that enlarged by bringing a continous source of electric field? – Esha Mukhopadhyay May 07 '17 at 17:33
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    Yes, all atoms can show this effect. The polarisation scales roughly as electron radius cubed, so approx as volume. Thus it is far larger in heavier atoms, iodine vs helium for example. The time period is very small but that does not matter as the important point is that on average the interaction is not zero. – porphyrin May 07 '17 at 17:55

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Although this applies to all noble gases, consider helium. There are two electrons moving around the nucleus. The movement of those electrons is symmetrical and during a defined amount of time, the atom is not polarised. However, the momentary positions of those electrons can polarise the atom. This can happen when both electrons are relatively close, on the same one side of the atom. In this case a momentary dipole is formed, since there is separation of two charges, nucleus being positive and the pair of electrons being the negative pole.

Αντώνιος Κελεσίδης
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  • but according to the atomic model electrons are present as electron clouds. their location cant be predicted according to heisenberg's uncertainty principle. – Esha Mukhopadhyay May 07 '17 at 12:50
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    To Esha: You can think of the electrons as being present as clouds, but the cloud's density changes over space and time; at some times, the two electron's clouds may both have high densities in the same region of space, which is analogous to saying the two single electron particles are close together. Whether you think about them as waves or particles, fluctuations in particle location/cloud density occur – iammax Jul 06 '17 at 16:06