Question

Identify the molecule in which the arrangement of election pairs around the central atom is octahedral and shape is not octahedral

• A. $S{F}_6$
• B. $Xe{F}_6$
• C. $Br{F}_5$
• D. $Xe{O}_2{F}_4$

Answer 1

Answer: (C)

The arrangement of electron pairs around the central atom is octahedral and shape is not octahedral can be identified with the help of presence of lone pair $(lp)$ of electrons and bond pair $(bp)$ of electrons associated with the central atom.

The structure of given species are:

(i) $S{F}_6$ Central atom sulphur has $6$ valence electrons,

which are bonded with $6F$ -atoms,

thus have octahedral shape



(ii) XeF ${}_6$ Central atom, $Xe$ have $8$ valence electrons,

out of which $6$ are bonded with $6F$-atoms and one $lp$ of electron is associated with $Xe$-atom.

Thus, has distorted octahedral shape.



$Br{F}_5$ Central atom $(Br)$ have $7$ valence electrons,

out of which $5$ are bonded with $5F$-atoms and one $lp$ of electron is associated with $Br$-atom having square pyramidal shape.



$Xe{O}_2{F}_4$ Central atom, $Xe$ have $8$ valence electrons,

out of which $4$ are bonded through $4$ -sigma $(\sigma )$ bonds with $4F$ -atoms and two oxygen atoms each having one sigma $(\sigma )$ bond and one pi $(\pi )$ bond is associated with $Xe$-atom.

Thus, has octahedral shape.