JEE Main · 2021 · Shift-ImediumPB11-052

Which one of the following compounds of Group-14 elements is not known?

p-Block Elements (Class 11) · Class 11 · JEE Main Previous Year Question

Question

Which one of the following compounds of Group-14 elements is not known?

Options
  1. a

    [GeCl6]2−[\mathrm{GeCl_6}]^{2-}

  2. b

    [Sn(OH)6]2−[\mathrm{Sn(OH)_6}]^{2-}

  3. c

    [SiCl6]2−[\mathrm{SiCl_6}]^{2-}

    ✓
  4. d

    [SiF6]2−[\mathrm{SiF_6}]^{2-}

Correct Answerc

[SiCl6]2−[\mathrm{SiCl_6}]^{2-}

Detailed Solution

Strategy: Identify which Group 14 hexahalide or hexahydroxide complex is NOT known.

Step 1: Ability to Form Hexacoordinate Complexes Group 14 elements from Si downward can form 6-coordinate species by utilising d orbitals:

  • [\ceSiF6]2−[\ce{SiF6}]^{2-}: Well-known ✅ (F is small enough for 6-coordinate Si)
  • [\ceGeCl6]2−[\ce{GeCl6}]^{2-}: Known ✅
  • [\ceSn(OH)6]2−[\ce{Sn(OH)6}]^{2-}: Known (stannate ion) ✅

Step 2: Why [\ceSiCl6]2−[\ce{SiCl6}]^{2-} Is Not Known Chlorine is much larger than fluorine. Six \ceCl\ce{Cl} atoms cannot fit around the relatively small silicon atom — the steric repulsion is too great. Therefore, [\ceSiCl6]2−[\ce{SiCl6}]^{2-} is not known despite [\ceSiF6]2−[\ce{SiF6}]^{2-} being stable (F is smaller). ❌

Answer: (C) [\ceSiCl6]2−\boxed{\text{Answer: (C) }[\ce{SiCl6}]^{2-}}

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