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From my book, it says that 'without van der Waals (London Dispersion) forces, it would be impossible to liquefy noble gases.'

Why is this the case?

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  • $\begingroup$ a noble gas can be liquefied due to instantaneous dipole induce dipole interaction and here london force provide dipole which can be considered as induce dipole $\endgroup$
    – karan
    Nov 16, 2019 at 19:33

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I'd say that because noble gases are non-polar atoms, the only force that applies to them is the London force.

If you want a liquid phase, you need to have forces between the atoms / molecules that constitute this liquid (forces too strong would lead to solid state).

So because London force is the only one that applies on noble gases molecules, they are necessary to liquify to gases.

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    $\begingroup$ I think this is half the answer. The Van Der Waahl's forces can't be the only forces because the noble gases will also form solids. So there must be something else. In solids the atoms are close enough to actually have have some weak chemical bonding (overlapping of atomic orbitals). In a liquid the atoms are moving around too much for such weak molecular bonding to occur. So in a liquid the Van Der Waahl's forces provide the necessary interatomic attractive force. Ar melts at 83.81 K and boils at 87.302 K. $\endgroup$
    – MaxW
    Nov 7, 2015 at 23:30

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