Question

Which of the following two are isostructural?
A. $Xe{{F}_{2}},I{{F}_{2}}^{-}$
B.$N{{H}_{3}},B{{F}_{3}}$
C.$C{{O}_{3}}^{2-}$
D.$PC{{l}_{3}},IC{{l}_{5}}$

Answer 1

Hint: Isostructural compounds are those compounds which have similar structure. Find the shape of the compounds to check whether the compounds are isostructural or not with the help of VSEPR theory. Isoelectronic compounds are also isostructural.

Complete answer:
As you have learnt about the isostructural compounds in your chemistry lessons that this type of compounds have similar structure.
Let us take one of the cases from the given options,
(A). $Xe{{F}_{2}},I{{F}_{2}}^{-}$
-$Xe{{F}_{2}}$
$Xe{{F}_{2}}$ this is a covalent compound with 3 atoms, one of Xe atoms and two fluorine atoms. $Xe$ have eight electrons in their outermost orbital and fluorine have 7 electrons
So, from here we can say that this compound will have two pairs of bond atoms and three pairs of lone pair electrons. And the hybridization of this compound will be $s{{p}^{3}}d$ with geometry of trigonal bipyramidal but due to the presence of lone pair it gets distorted from its original geometry.
So, the structure of $Xe{{F}_{2}}$ will be,
 

The bond angle between them is 180◦ and gives $Xe{{F}_{2}}$a linear geometry.
In the case of $I{{F}_{2}}^{-}$ it is a covalent compound with 3 atoms, one of I atoms and two fluorine atoms. I have seven electrons in their outermost orbital and fluorine also have 7 electrons.
So, from here we can say that this compound will have two pairs of bond atoms and three pairs of lone pair electrons. And the hybridization of this compound will be $s{{p}^{3}}d$ with geometry of trigonal bipyramidal but due to the presence of lone pair it gets distorted from its original geometry.
So, the structure of $I{{F}_{2}}^{-}$ is,

One extra lone pair comes due to the presence of negative charge on the compound.
The bond angle between them is also 180◦ and has a linear geometry.
So $Xe{{F}_{2}},I{{F}_{2}}^{-}$ are isostructural with each other.
In option B), $N{{H}_{3}}$ has trigonal bipyramidal structure and $B{{F}_{3}}$ has trigonal planar structure as shown below,

In option C), since there is only one compound that is carbonate ion, it has trigonal planar similar to that of $B{{F}_{3}}$ molecule.
In option D), $PC{{l}_{3}}$ has trigonal pyramidal shape which includes a lone pair of electrons and $IC{{l}_{5}}$ has square pyramidal shape as shown below,

Thus the correct option will be (A).

Note:
The shapes of the molecules should be similar to be an isostructural compound. It is not necessary that hybridization of the compound should also be similar to be an isostructural compound because one hybridization can have more than one shape, hybridization can be different.