Question:
Which one of the following compounds of Group-14 elements is not known ?
Which one of the following compounds of Group-14 elements is not known ?
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Silicon's inability to form \([SiCl_6]^{2-}\) is a textbook example of steric hindrance preventing the formation of a high coordination number complex.
Updated On: Feb 11, 2026
- $[SiF_6]^{2-}$
- $[SiCl_6]^{2-}$
- $[GeCl_6]^{2-}$
- $[Sn(OH)_6]^{2-}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The existence of a complex depends on the size of the central atom and the size of the ligands (steric hindrance).
Step 2: Detailed Explanation:
1. Silicon is a relatively small atom in Group 14.
2. Fluorine is a small ligand, so six \(F^-\) ions can fit around the Silicon atom in \([SiF_6]^{2-}\).
3. Chlorine is a large ligand. Six large Chloride ions cannot be accommodated around the small Silicon atom due to significant steric repulsion.
4. As we move down to \(Ge\) and \(Sn\), the atomic size increases, allowing them to accommodate larger ligands like \(Cl^-\) or \(OH^-\).
Therefore, \([SiCl_6]^{2-}\) is not known.
Step 3: Final Answer:
The correct option is (B).
The existence of a complex depends on the size of the central atom and the size of the ligands (steric hindrance).
Step 2: Detailed Explanation:
1. Silicon is a relatively small atom in Group 14.
2. Fluorine is a small ligand, so six \(F^-\) ions can fit around the Silicon atom in \([SiF_6]^{2-}\).
3. Chlorine is a large ligand. Six large Chloride ions cannot be accommodated around the small Silicon atom due to significant steric repulsion.
4. As we move down to \(Ge\) and \(Sn\), the atomic size increases, allowing them to accommodate larger ligands like \(Cl^-\) or \(OH^-\).
Therefore, \([SiCl_6]^{2-}\) is not known.
Step 3: Final Answer:
The correct option is (B).
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