Which one of the following compounds of Group-14 elements is not known?

First, we recall that the Group-14 elements are $$\text{C, Si, Ge, Sn, Pb}$$. In these elements the common higher oxidation state is $$+4$$, and many of their complex anions can be viewed as having the general formula $$[\text{E}X_6]^{2-}$$ where every ligand $$X$$ carries a charge $$-1$$.

To see whether a particular hexacoordinate ion is possible, we examine two points:

(i) the oxidation state balance We write the usual charge-balance equation $$6(-1) + (\text{oxidation state of E}) = -2.$$ Solving, we have $$\text{oxidation state of E} = +4.$$ So for every choice of central atom $$\text{E}$$ the required oxidation state is indeed the normal $$+4$$, which is accessible to $$\text{Si, Ge}$$ and $$\text{Sn}$$. Hence charge balance causes no difficulty for any of the listed species.

(ii) the steric (size) requirement for an octahedral arrangement In an octahedral $$[\text{E}X_6]^{2-}$$ ion all six ligands must crowd around the central atom. The larger the ligands, the more space they demand; the larger the central atom, the more easily it can accommodate them. Fluoride $$F^-$$ is the smallest normal monodentate ligand, chloride $$Cl^-$$ is appreciably larger, and hydroxide $$OH^-$$ lies in between. Among the central atoms, the size trend is $$r(\text{Si}) \lt r(\text{Ge}) \lt r(\text{Sn}).$$

Now we check each option in turn.

Option A : $$[\text{GeCl}_6]^{2-}$$. Germanium is larger than silicon, so six chloride ions can fit around it. Indeed, hexachlorogermanate(IV) salts such as $$\text{K}_2[\text{GeCl}_6]$$ are known. Therefore this ion exists.

Option B : $$[\text{Sn(OH)}_6]^{2-}$$. Tin(IV) is the largest central atom among the ones listed, and hydroxide is smaller than chloride. The well-known stannate ion $$[\text{Sn(OH)}_6]^{2-}$$ (or its oxide form $$\text{SnO}_3^{2-}$$ after dehydration) is easily formed in alkali. Thus this species also exists.

Option C : $$[\text{SiCl}_6]^{2-}$$. Here the problem is spatial. Silicon is the smallest of the three possible centres, while chloride is relatively bulky. Trying to pack six $$Cl^-$$ ions around the small $$\text{Si}^{4+}$$ produces prohibitive ligand-ligand repulsion, so the octahedral complex cannot be stabilised. No salts of the “hexachlorosilicate(IV)” ion are known.

Option D : $$[\text{SiF}_6]^{2-}$$. Although silicon is small, fluoride is even smaller, allowing the required octahedral arrangement. The ion $$[\text{SiF}_6]^{2-}$$ is perfectly stable; commercial compounds such as $$\text{Na}_2[\text{SiF}_6]$$ (sodium hexafluorosilicate) are common.

Therefore, among the four species, the only one that cannot exist is $$[\text{SiCl}_6]^{2-}$$.

Hence, the correct answer is Option C.