Arrange the following gases in increasing order of van der Waals constant 'a'
(A) Ar
(B) CH$$_4$$
(C) H$$_2$$O
(D) C$$_6$$H$$_6$$
Choose the correct option from the following.

The van der Waals constant $$a$$ measures the magnitude of intermolecular attractive forces.

Stronger intermolecular forces $$\Rightarrow$$ larger value of $$a$$.

Given gases:

$$Ar,\ CH_4,\ H_2O,\ C_6H_6$$

Analysis:

For argon: $$Ar$$

Being a noble gas, it has very weak intermolecular forces.

Hence it has the smallest value of $$a$$.

For methane: $$CH_4$$

Non-polar molecule with weak van der Waals forces, but stronger than Ar.

For water: $$H_2O$$

Strong hydrogen bonding is present, so intermolecular attraction is much larger.

For benzene: $$C_6H_6$$

Large molecular size and highly polarizable electron cloud produce very strong intermolecular attractions.

Hence benzene has the largest value of $$a$$.

Therefore the increasing order of van der Waals constant $$a$$ is:

$${Ar < CH_4 < H_2O < C_6H_6}$$

Hence, Option D is correct