Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct statement about ${\text{ICl}}_5$ and ${\text{ICl}}_4^{-}$ is

To determine the molecular geometry of any covalent molecule or ion, we rely on the Valence Shell Electron Pair Repulsion (VSEPR) theory. The first step is always to calculate the steric number, or hybridization index ($H$), for the central atom.

The formula is elegantly simple:

Here, $V$ represents the number of valence electrons on the central atom, $M$ is the number of monovalent atoms directly attached to it, $C$ is the cationic (positive) charge, and $A$ is the anionic (negative) charge. Once we have $H$, we can determine the base electron geometry and then subtract the number of bond pairs to find the number of lone pairs. These lone pairs act like invisible balloons, taking up space and pushing the bonded atoms into their final molecular shape.

The Square Pyramidal Geometry

Let's apply our formula to Iodine pentachloride (${\text{ICl}}_5$). The central atom is Iodine, a halogen, which means it has $7$ valence electrons ($V=7$). It is bonded to $5$ monovalent chlorine atoms ($M=5$). Since the molecule is neutral, both $C$ and $A$ are zero.

Plugging these into our equation:

A steric number of $6$ corresponds to an $s{p}^3{d}^2$ hybridization, which gives us an octahedral base electron geometry. However, we only have $5$ surrounding chlorine atoms. This means there is exactly $1$ lone pair ($6-5=1$).

In an octahedral setup, all six positions are initially equivalent. Placing one lone pair at any position pushes the remaining five bonds away slightly. The resulting molecular shape is Square Pyramidal, with four chlorines forming a square base and one chlorine at the apex.

The Square Planar Geometry

Now, let's examine the tetrachloroiodate ion (${\text{ICl}}_4^{-}$). Again, the central Iodine has $7$ valence electrons ($V=7$). This time, it is bonded to $4$ monovalent chlorine atoms ($M=4$). Crucially, the ion carries a $-1$ charge, meaning it has gained an extra electron ($A=1$).

Let's calculate the steric number:

Fascinatingly, the steric number is still $6$, meaning the hybridization remains $s{p}^3{d}^2$ with an octahedral base geometry. But here is the catch: we only have $4$ bond pairs. This leaves us with $2$ lone pairs ($6-4=2$).

To minimize the intense repulsion between these two bulky lone pairs, they position themselves as far apart as possible—at opposite axial positions (an angle of $180^{\circ }$). This forces the four chlorine atoms to occupy the equatorial positions, resulting in a perfectly flat, Square Planar shape.

Isolobal but not Isostructural

By comparing our results, we see that ${\text{ICl}}_5$ is square pyramidal, while ${\text{ICl}}_4^{-}$ is square planar.

Because both species share the same $s{p}^3{d}^2$ hybridization, they are termed isolobal. However, because their final physical shapes are entirely different due to the varying number of lone pairs, they are not isostructural. Understanding this subtle distinction is key to mastering chemical bonding!