Question

Which one of the following has the regular tetrahedral structure?

• A. $B{F}_4^{-}$
• B. $S{F}_4$
• C. $Xe{F}_4$
• D. $\left[Ni\right(CN{)}_4{]}^{2-}$
  

Answer 1

The correct option is A $B{F}_4^{-}$

Boron has $s{p}^3$ hybridisation with 4 bond pairs and no lone pairs. $B{F}_4^{-}$ has regular tetrahedral geometry. $S{F}_4$ has $s{p}^{3}d$ hybridization. $Xe{F}_4$ has $s{p}^3{d}^2$ hybridization. $\left[Ni\right(CN{)}_4{]}^{-2}$ is $ds{p}^2$ hybridized.