Question

Arrange the following in the decreasing order of their van der Waals constant ${}^{\prime }{a}^{\prime }$ values:
$H_{2}O,C{O}_2,Ar$

• A. $H_{2}O>C{O}_2>Ar$
• B. $Ar>H_{2}O>C{O}_2$
• C. $Ar=H_{2}O=C{O}_2$
• D. $H_{2}O>Ar>C{O}_2$

Answer 1

The correct option is A $H_{2}O>C{O}_2>Ar$

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. $Ar$ 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 $Ar$.
Hence, the values of $a$ are in the order: $H_{2}O>C{O}_2>Ar$.