Question:

Which of the following does not exist?

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Silicon commonly forms stable fluoro-complexes (e.g., \([ \text{SiF}_6 ]^{2-}\)), but not stable hexachloro anions. - Larger Group 14 elements (Ge, Sn) can accommodate more substantial ligands in higher coordination states.
Updated On: Mar 10, 2025
  • \([ \text{GeCl}_6 ]^{2-}\)
  • \([ \text{SiF}_6 ]^{2-}\)
  • \([ \text{SiCl}_6 ]^{2-}\)
  • \([ \text{Sn(OH)}_6 ]^{2-}\)
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The Correct Option is C

Solution and Explanation

Step 1: Existence of Similar Complexes

  • \([ \text{SiF}_6 ]^{2-}\) (hexafluorosilicate) is well-known as a stable species.
  • \([ \text{GeCl}_6 ]^{2-}\) is also known, due to the relatively larger size of Ge allowing stable complex formation with six chlorines.
  • \([ \text{Sn(OH)}_6 ]^{2-}\) (stannate(II)) also exists, with tin in a high coordination environment.

Step 2: Why \([ \text{SiCl}_6 ]^{2-}\) Does Not Exist

  • Silicon does not form a stable \(\text{Si}^{2+}\) complex with six chloride ligands.
  • The large charge density and possible multiple-bond character with chlorine prevent a viable \([ \text{SiCl}_6 ]^{2-}\) ion.

Hence, \([ \text{SiCl}_6 ]^{2-}\) does not exist.

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