Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: In which of the following pairs, the two species are not isostructural?

When we talk about molecules being isostructural, we are simply asking: Do they look the same in 3D space? Do they share the exact same geometric shape? To answer this, we rely on the elegant VSEPR (Valence Shell Electron Pair Repulsion) Theory.

The core idea of VSEPR is that electron pairs around a central atom repel each other and will arrange themselves as far apart as possible to minimize this repulsion. To find the shape, we first need to calculate the Steric Number ($Z$), which tells us the total number of hybridized orbitals (bond pairs + lone pairs) around the central atom.

The magic formula is: Where: - $V$ = Number of valence electrons on the central atom. - $M$ = Number of monovalent atoms (like H, F, Cl) attached to it. (Note: Divalent atoms like Oxygen are ignored here!). - $C$ = Cationic charge (subtract it, because electrons are lost). - $A$ = Anionic charge (add it, because electrons are gained).

Let's put this formula to the test and hunt down the pair that breaks the rule!

Option (a): ${\text{CO}}_3^{2-}$ and ${\text{NO}}_3^{-}$ For the carbonate ion (${\text{CO}}_3^{2-}$), Carbon is in Group 14, so $V=4$. Oxygen is divalent, so $M=0$. The charge is $-2$, so $A=2$. A steric number of 3 means $s{p}^2$ hybridization. With 3 bond pairs and 0 lone pairs, the shape is perfectly Trigonal Planar. For the nitrate ion (${\text{NO}}_3^{-}$), Nitrogen has $V=5$. The charge is $-1$, so $A=1$. Again, $Z=3$, meaning it is also Trigonal Planar. They are isostructural.

Option (b): ${\text{PCl}}_4^{+}$ and ${\text{SiCl}}_4$ For ${\text{PCl}}_4^{+}$, Phosphorus has $V=5$. Chlorine is monovalent, so $M=4$. The charge is $+1$, so $C=1$. A steric number of 4 means $s{p}^3$ hybridization. With 4 bond pairs and 0 lone pairs, the shape is Tetrahedral. For ${\text{SiCl}}_4$, Silicon has $V=4$ and $M=4$. Also Tetrahedral. They are isostructural.

Option (c)

Option (c): ${\text{PF}}_5$ and ${\text{BrF}}_5$ At first glance, both have 5 fluorine atoms. You might think they are identical. Let's look closer. For ${\text{PF}}_5$, Phosphorus has $V=5$ and $M=5$. A steric number of 5 means $s{p}^{3}d$ hybridization. Since all 5 are bond pairs, there are 0 lone pairs. The geometry is Trigonal Bipyramidal.

Now, for ${\text{BrF}}_5$. Bromine is a halogen, so it has $V=7$. It is bonded to 5 fluorines, so $M=5$. A steric number of 6 means $s{p}^3{d}^2$ hybridization! But wait, it only has 5 fluorine atoms attached. This means out of the 6 hybrid orbitals, 5 are bond pairs and 1 is a lone pair.

In an octahedral electron geometry, one lone pair distorts the shape into a Square Pyramidal geometry.

Clearly, a Trigonal Bipyramidal shape is vastly different from a Square Pyramidal shape. Therefore, ${\text{PF}}_5$ and ${\text{BrF}}_5$ are NOT isostructural.

Option (d): ${\text{AlF}}_6^{3-}$ and ${\text{SF}}_6$ For ${\text{AlF}}_6^{3-}$, Aluminum has $V=3$, $M=6$, and $A=3$. For ${\text{SF}}_6$, Sulfur has $V=6$ and $M=6$. Both have a steric number of 6 with 0 lone pairs, making them both perfectly Octahedral. They are isostructural.

Our analysis confirms that Option (c) is the correct answer!