Question

Which one of the following molecules has the octahedral structure?
(a)- $BeC{{l}_{2}}$
(b)- $S{{F}_{6}}$
(c)- $B{{F}_{3}}$
(d)- $P{{F}_{5}}$
(e)- $C{{F}_{4}}$

Answer 1

The shape of the compound is determined by the number of bonds formed by the central atom. Octahedral means there must be six bonds formed by the central atom and these must be single bonds.

Complete answer:
According to the number of elements joined to the central atom. There are many shapes of the compound like linear, trigonal planar, tetrahedral, square planar, pentagonal bipyramidal, octahedral, etc.
These are divided by the number of bonds, type of bonds, and the number of lone pairs in the molecule.
In $BeC{{l}_{2}}$, there are two bonds, and both the bonds are single bonds and there are no lone pairs in the compound. So, the structure of the compound is linear.

In $S{{F}_{6}}$, there are six bonds and all the bonds are single bonds and there are no lone pairs in the compound. So, the structure of the compound is octahedral.

In $B{{F}_{3}}$, there are three bonds and all the bonds are single bonds and there are no lone pairs in the compound. So, the structure of the compound is trigonal planar.

In $P{{F}_{5}}$, there are five bonds and all the bonds are single bonds and there are no lone pairs in the compound. So, the structure of the compound is trigonal bipyramidal.

In $C{{F}_{4}}$, there are five bonds and all the bonds are single bonds and there are no lone pairs in the compound. So, the structure of the compound is tetrahedral.

Therefore, the correct answer is an option (b).

Note:
In all the compounds, there is no lone pair so, there is no repulsion between the bonds. But due to the presence of lone pairs, there will be repulsion in the bonds and the shape of the compound will be changed.