Hydrolysis of chlorides occurs when a chloride compound reacts with water, typically forming hydrochloric acid and the corresponding hydroxide or oxide. This happens when the central atom has vacant d-orbitals or can expand its octet, allowing nucleophilic attack by water molecules.

Let's analyze each option:

SnCl₄: Tin(IV) chloride. Tin has vacant d-orbitals and can expand its octet. It hydrolyzes vigorously: $SnCl_{4} + 2 H_{2} O \to SnO_{2} + 4 HCl$

PbCl₄: Lead(IV) chloride. Lead also has the ability to expand its octet. It is unstable and hydrolyzes easily: $PbCl_{4} + 2 H_{2} O \to PbO_{2} + 4 HCl$

CCl₄: Carbon tetrachloride. Carbon, being a second-period element, has no low-lying d-orbitals available to accept a lone pair from a water molecule in a nucleophilic attack. Its octet is complete and cannot be expanded. Therefore, it does not undergo hydrolysis under normal conditions. This is a key exception.

SiCl₄: Silicon tetrachloride. Silicon has vacant 3d orbitals, which it can use to expand its octet. It hydrolyzes rapidly: $SiCl_{4} + 2 H_{2} O \to SiO_{2} + 4 HCl$

Final Answer: CCl₄ (Carbon tetrachloride) is the chloride that cannot get hydrolyzed.

Related Topic: Hydrolysis of Covalent Chlorides

The tendency of a covalent chloride to hydrolyze depends primarily on the availability of vacant orbitals on the central atom. If the central atom can accept a lone pair from a water molecule (acting as a Lewis acid), hydrolysis occurs. Elements like Si, Ge, Sn, and P have accessible d-orbitals and their chlorides hydrolyze. Carbon, in the second period, has no available d-orbitals, making CCl₄ resistant to hydrolysis.

Key Formula/Concept

The general reaction for the hydrolysis of a tetrachloride is: $MCl_{4} + 2 H_{2} O \to MO_{2} + 4 HCl$ (where M is an element like Si, Sn, etc.)

This reaction is not possible for carbon due to the absence of d-orbitals and the high stability of the C-Cl bond.