The question asks which of the given halides cannot be hydrolyzed. To determine this, we need to understand the concept of hydrolysis of halides and the properties of the compounds listed.
Hydrolysis is the chemical breakdown of a compound due to reaction with water. In the case of halides, hydrolysis involves the reaction of the halide with water, leading to the formation of an acid and a base.
Carbon tetrachloride ($CCl_4$) does not undergo hydrolysis due to the lack of available d-orbitals in carbon for back-bonding with oxygen from water. Carbon is in the +4 oxidation state, and in carbon tetrachloride, it is in a tetrahedral configuration, leaving no room for any increase in coordination number. Hence, $CCl_4$ cannot hydrolyze.
Silicon tetrachloride can be hydrolyzed. Silicon can expand its coordination number because it has vacant d-orbitals to accommodate more electrons. When $SiCl_4$ reacts with water, it forms silicic acid and hydrochloric acid:
\(SiCl_4 + 2H_2O \rightarrow Si(OH)_4 + 4HCl\)
Similar to silicon, germanium also has d-orbitals, allowing hydrolysis. Thus, $GeCl_4$ can undergo hydrolysis:
\(GeCl_4 + 2H_2O \rightarrow Ge(OH)_4 + 4HCl\)
Tin can utilize its d-orbitals for hydrolysis, similar to silicon and germanium. When $SnCl_4$ is hydrolyzed, tin hydroxide and hydrochloric acid are formed:
\(SnCl_4 + 2H_2O \rightarrow Sn(OH)_4 + 4HCl\)
After analyzing all the options, we conclude that \(CCl_4\) is the only compound that cannot be hydrolyzed due to the absence of d-orbitals and incapacity to expand its coordination number. The rest of the compounds have vacant d-orbitals necessary for hydrolysis.