Question

Among the following species, identify the isostructural pairs.
$N{F_3}$, $NO_3^ - $, $B{F_3}$, ${H_3}{O^ + }$,$H{N_3}$
A.$\left[ {N{F_3},NO_3^ - } \right]$ and $\left[ {B{F_3},{H_3}{O^ + }} \right]$
B.$\left[ {N{F_3},H{N_3}} \right]$ and $\left[ {NO_3^ - ,B{F_3}} \right]$
C.$\left[ {N{F_3},{H_3}{O^ + }} \right]$ and $\left[ {NO_3^ - ,B{F_3}} \right]$
D.$\left[ {N{F_3},{H_3}{O^ + }} \right]$ and $\left[ {H{N_3},B{F_3}} \right]$

Answer 1

Hint: To find the isostructural species among the given compounds, we must find the hybridisation of the central atom of each species and the arrangement of the bond pairs and lone pairs on the central atom. To do that, you must recall the VSEPR (Valence shell electron pair repulsion) theory. It suggests that all valence shell electron pairs surrounding the central atom arrange themselves in such a manner so as to be as far away from each other as possible.

Complete step by step solution:
The species which have similar chemical structures are known as isostructural chemical compounds.
Considering $N{F_3}$ : We know that nitrogen has five valence electrons. In the given compound nitrogen is forming three bonds with fluorine atoms. Thus, nitrogen has three bond pairs and one lone pair. Its hybridisation is $s{p^3}$ and the shape is pyramidal.
Considering $NO_3^ - $ : Nitrogen has five valence electrons. The compound carries one negative charge and nitrogen is making four bonds with oxygen atoms, out of which one is a double bond. The hybridisation is $s{p^2}$ and shape is triangular planar
Considering $B{F_3}$ : Boron has three valence electrons and forms a single bond with each fluorine atom. The hybridisation is $s{p^2}$ and the shape is triangular planar
Considering ${H_3}{O^ + }$ : oxygen has six valence electrons. It forms two covalent bonds with two hydrogen atoms and a coordinate bond with the third hydrogen. In total, it has three bond pairs and a lone pair. The hybridisation is $s{p^3}$ and the shape is pyramidal.
$N{F_3}$ and ${H_3}{O^ + }$ are isostructural and $NO_3^ - $ and $B{F_3}$ are isostructural.

Thus, the correct answer is option (C).

Note: During bond formation, the atomic orbitals of an atom are mixed in such a manner as to produce equivalent orbitals. This mixing of orbitals is known as hybridisation. The arrangement of these hybrid orbitals according to the VSEPR theory gives us the shape of the molecule.