Question

Which of the following reaction is not correct?

• A. \(SiF_{4}+2F^{-}\rightarrow[SiF_{6}]^{2-} \)
• B. \(GeCl_{4}+2Cl^{-}\rightarrow[GeCl_{6}]^{2-} \)
• C. \(Sn(OH)_{4}+2OH^{-}\rightarrow[Sn(OH)_{6}]^{2-} \)
• D. \(SiCl_{4}+2Cl^{-}\rightarrow[SiCl_{6}]^{2-} \)

Hint:

Always consider the atomic radius of the central element when assessing coordination stability; larger ligands require larger central atoms to minimize inter-ligand repulsion.

Answer 1

The Correct Option is D

Concept: The stability of complex ions for Group 14 elements is governed by the ability of the central atom to expand its coordination sphere, which depends on its atomic radius and the steric bulk of the surrounding ligands.

Step 1: Evaluate the coordination of Silicon.
Silicon has a relatively small atomic radius. While it can accommodate six small fluorine ligands to form the stable \([SiF_6]^{2-}\) complex (due to the high electronegativity and small size of fluorine), it cannot accommodate six larger chloride ligands.

Step 2: Analyze steric repulsion.
The chlorine atom is much larger than the fluorine atom. If six chlorine atoms were to bond with silicon, the steric repulsion between the chloride ligands would be too great for the small silicon atom to handle, rendering the complex \([SiCl_6]^{2-}\) unstable.

Step 3: Compare with other elements.
Elements like Germanium (Ge) and Tin (Sn) have larger atomic radii than Silicon. Their larger size allows them to accommodate six large ligands without as much steric hindrance, making complex ions like \([GeCl_6]^{2-}\) and \([Sn(OH)_6]^{2-}\) chemically feasible.

Step 4: Conclusion.
Therefore, the formation of the \([SiCl_6]^{2-}\) complex is the reaction that is not correct due to excessive steric hindrance.