The chloride that CANNOT get hydrolysed is:

We begin by recalling what is meant by “hydrolysis of a chloride”. Hydrolysis is a reaction with water in which the chloride $$MCl_n$$ is converted into an oxide, hydroxide or oxy-chloride of the element $$M$$ with simultaneous liberation of $$HCl$$.

For covalent chlorides, the key requirement for hydrolysis is the presence of vacant low-energy orbitals on the central atom that can accept lone-pair electrons donated by water’s oxygen. These vacant orbitals are usually the $$d$$-orbitals that first become available from the third period onward.

Let us examine each option one by one while stating explicitly whether such vacant $$d$$-orbitals exist.

Option A is $$PbCl_4$$. Lead lies in the sixth period and possesses accessible $$6d$$ orbitals. Water can donate an electron pair to these orbitals, so $$PbCl_4$$ readily undergoes hydrolysis.

Option C is $$SnCl_4$$. Tin is a fifth-period element with vacant $$5d$$ orbitals. Hence $$SnCl_4$$ is easily hydrolysed:

$$SnCl_4 + 2H_2O \; \longrightarrow \; SnO_2 + 4HCl$$

Option D is $$SiCl_4$$. Silicon belongs to the third period, and beginning with the third period the $$3d$$ subshell is available (though higher in energy than $$3s$$ and $$3p$$). These vacant $$3d$$ orbitals can still accept electron density from water, therefore $$SiCl_4$$ is also hydrolysed:

$$SiCl_4 + 2H_2O \; \longrightarrow \; SiO_2 + 4HCl$$

Option B is $$CCl_4$$. Carbon is a second-period element; its valence shell consists only of the $$2s$$ and $$2p$$ orbitals. There are no vacant $$d$$-orbitals in the second period. Because carbon cannot expand its octet to accommodate the lone-pair donation from water, the required coordinate bond formation cannot take place, and hydrolysis does not proceed. Consequently, $$CCl_4$$ remains completely unreactive toward water under ordinary conditions.

Thus, among the four chlorides listed, the only one that cannot get hydrolysed is $$CCl_4$$.

Hence, the correct answer is Option 2.