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

Step 1: Understand what is asked.
Each option shows a Group 14 halide gaining two more ligands to form a six coordinate complex. We must find the one that cannot happen.
Step 2: Look at silicon with fluorine.
Silicon is small but fluorine is also very small, so six fluorines fit around Si to give the stable $[SiF_6]^{2-}$.
Step 3: Look at silicon with chlorine.
Chlorine is much bigger than fluorine. Six large chlorines around the small Si crowd badly, so $[SiCl_6]^{2-}$ cannot form. This is the wrong reaction.
Step 4: Check germanium.
Ge is bigger than Si, so it has more room. $[GeCl_6]^{2-}$ forms fine.
Step 5: Check tin.
Sn is bigger still, so $[Sn(OH)_6]^{2-}$ forms without trouble.
Step 6: Pick the impossible one.
Only the small silicon cannot hold six bulky chlorides. \[ \boxed{SiCl_4+2Cl^-\rightarrow[SiCl_6]^{2-}\ \text{is not correct}} \]