Which one of the following has the regular tetrahedral structure
A) \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]
B) \[{\rm{S}}{{\rm{F}}_{\rm{4}}}\]
C) \[{\rm{Xe}}{{\rm{F}}_{\rm{4}}}\]
D) \[{\left[ {{\rm{Ni(CN}}{{\rm{)}}_{\rm{4}}}} \right]^{2 - }}\]

Hint: Here, first we will calculate hybridization of the central atom of all the compounds. If the value of hybridization of a molecule is two, then the molecule is sp hybridized and the shape of the molecule is linear.

Formula Used:\[H = \dfrac{{V + X - C + A}}{2}\]
Here, H stands for hybridization, V is count of valence electrons, X is the count of monovalent groups bonded to the central atom, C stands for cationic charge and A stands for charge of anion.

Complete step by step solution:Let’s find out hybridization of each compound.
For \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \], Count of valence electrons=3, count of monovalent atoms=4, anionic charge=1
 \[H = \dfrac{{3 + 4 + 1}}{2} = 4\]
So, the value of H is 4, this indicates that hybridization B atom is \[s{p^3}\] . So, this molecule has a tetrahedral shape.
Now,
For \[{\rm{S}}{{\rm{F}}_{\rm{4}}}\], Count of valence electrons=6, count of monovalent atoms=4
Therefore, \[H = \dfrac{{6 + 4}}{2} = 5\]
The value of H is 5, this indicates that hybridization of the S atom is not\[s{p^3}d\] . So, this molecule has see saw shape.
For\[{\rm{Xe}}{{\rm{F}}_{\rm{4}}}\], Count of valence electrons=8, count of monovalent atoms=4
Therefore, \[H = \dfrac{{8 + 4}}{2} = 6\]
The value of H is 6, this indicates that hybridization of the Xe atom is \[s{p^3}{d^2}\] . So, this molecule has a square planar shape.
For \[{\left[ {{\rm{Ni(CN}}{{\rm{)}}_{\rm{4}}}} \right]^{2 - }}\], using the valence bond theory, the hybridization is \[ds{p^2}\] hybridized. So, its structure is square planar.

Therefore, option A is right.

Note: It is to be noted that, if the hybridization is \[s{p^3}\], the electron geometry of the molecule is tetrahedral. And if the molecule has\[s{p^2}\] hybridization, then the molecule's electron geometry is trigonal planar.