Question

SiF$_6^{2-}$ exists but CF$_6^{2-}$ does not. Offer reason.

• A. Due to difference in atomic radius (Si = 1.17 \AA, C = 0.77 \AA)
• B. Si is more electronegative than carbon (Pauling scale)
• C. C lacks d-orbital but Si possesses vacant 3d-orbital
• D. C cannot extend its coordination number beyond 4 but Si does so

Hint:

The availability of vacant d-orbitals in silicon allows it to form complexes with coordination numbers greater than 4, unlike carbon, which cannot do so due to the absence of d-orbitals.

Answer 1

The Correct Option is C

Step 1: Consideration of electronic configuration.
The ability of an element to form complexes with coordination numbers greater than 4 depends on the availability of vacant d-orbitals. Silicon (Si) has vacant 3d orbitals that can participate in bonding, allowing it to form \(\text{SiF}_6^{2-}\) with a coordination number of 6.
Step 2: Carbon’s limitation.
Carbon (C), however, does not have vacant d-orbitals. This limits its ability to extend its coordination number beyond 4. As a result, \(\text{CF}_6^{2-}\) is not stable and does not exist.
Step 3: Comparing atomic radii.
Although the atomic radius of Si (1.17 \(\AA\)) is larger than C (0.77 \(\AA\)), the key reason for the stability of \(\text{SiF}_6^{2-}\) is the availability of the 3d orbital in Si, not just the difference in atomic radii.
Conclusion:
The correct explanation is that carbon lacks the 3d orbitals, whereas silicon possesses vacant 3d orbitals that enable it to form complexes like \(\text{SiF}_6^{2-}\).