Question

Identify the complex ion which does not exist.

• A. \([\mathrm{SiF_6}]^{2-}\)
• B. \([\mathrm{GeCl_6}]^{2-}\)
• C. \([\mathrm{Sn(OH)_6}]^{2-}\)
• D. \([\mathrm{SiCl_6}]^{2-}\)

Hint:

Remember the important complexes: \[ \boxed{ [\mathrm{SiF_6}]^{2-}, \quad [\mathrm{GeCl_6}]^{2-}, \quad [\mathrm{Sn(OH)_6}]^{2-} } \] are stable, whereas \[ \boxed{ [\mathrm{SiCl_6}]^{2-} } \] does not exist.

Answer 1

The Correct Option is D

Step 1: Recall the complex formation of Group 14 elements. Silicon readily forms the fluoro complex \[ \boxed{[\mathrm{SiF_6}]^{2-}} \] because fluoride ion is small and highly electronegative. Similarly, \[ [\mathrm{GeCl_6}]^{2-} \] and \[ [\mathrm{Sn(OH)_6}]^{2-} \] are known stable complex ions.

Step 2: Identify the non-existent complex. Silicon cannot expand its coordination effectively with the larger chloride ions to form \[ [\mathrm{SiCl_6}]^{2-}. \] Hence, \[ \boxed{[\mathrm{SiCl_6}]^{2-}} \] does not exist. Therefore, the correct option is \(\boxed{(D)}\).