Arrange the following in the decreasing order of their van der Waals constant $^{'} a^{'}$ values:
$H_{2} O, C O_{2}, A r$

H_{2} O > C O_{2} > A r

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A r > H_{2} O > C O_{2}

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A r = H_{2} O = C O_{2}

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H_{2} O > A r > C O_{2}

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Solution

The correct option is A $H_{2} O > C O_{2} > A r$
Van der Waals constant 'a' is a measure of intermolecular forces of attractions. Stronger is the force of attraction, greater will be ‘ $a$ ’ and easily a gas can be liquified. In $H_{2} O$ , hydrogen bonding exists, hence $H_{2} O$ possesses strongest intermolecular forces of attraction. In $C O_{2}$ , due to resonance, it is easily polarizable and hence stronger intermolecular forces of attraction. $A r$ has only van der Waals forces of attraction. So, the forces of attraction for $H_{2} O$ are higher, followed by $C O_{2}$ and then $A r$ .
Hence, the values of $a$ are in the order: $H_{2} O > C O_{2} > A r$ .

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