Question

Which of the following will be octahedral?
A) \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]
B) \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]
C) \[{\rm{PC}}{{\rm{l}}_{\rm{5}}}\]
D) \[{\rm{B}}{{\rm{O}}_{\rm{3}}}^{3 - }\]

Answer 1

Hint: An octahedral molecule has \[s{p^3}{d^2}\] hybridization. An octahedral molecule is to be identified by counting the groups surrounding the central atom. Also, we have to count the lone pairs. If the total count of the group is 6, then the molecule is \[s{p^3}{d^2}\]hybridized.

Complete step by step solution:Let’s discuss all the options one by one.
Option A is \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]. The Sulphur atom is bonded to six fluorine atoms. And there is no lone pair on the atom of Sulphur. Therefore, \[{\rm{S}}{{\rm{F}}_{\rm{6}}}\]is \[s{p^3}{d^2}\]hybridized and the molecular shape is octahedral.
Option B is\[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]. We can calculate the hybridization of B atom by the formula of \[\dfrac{{V + X - C + A}}{2}\] , where, V denotes the valence electrons, X is for the monovalent atoms, C is for the cationic charge and A is for the anionic charge.
For \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \],
\[H = \dfrac{{3 + 4 + 1}}{2} = 4\], so, \[{\rm{B}}{{\rm{F}}_{\rm{4}}}^ - \]is \[s{p^3}\] hybridized and the molecular shape is tetrahedral.
Option C is \[{\rm{PC}}{{\rm{l}}_{\rm{5}}}\]. Here, the molecule is \[s{p^3}d\] hybridized and the molecular shape is trigonal pyramidal.
Option D is\[{\rm{B}}{{\rm{O}}_{\rm{3}}}^{3 - }\]. We have to use the above formula to find the hybridization of Boron atoms. As there are no monovalent atoms present. Therefore,
\[H = \dfrac{{3 + 3}}{2} = 3\]
As H=3, the hybridization is \[s{p^2}\] and the molecular shape is trigonal planar.

Hence, the option A is right.

Note: The theory used above to determine the molecular shape is VSEPR theory. This theory guides to predict the molecular shape considering the lone pairs and bond pairs around the central atom. This theory also gives the molecular geometry.