Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The ion that has $s{p}^3{d}^2$ hybridisation for the central atom, is

Imagine you are an architect, but instead of buildings, you are designing molecules. To predict the shape and structure of a molecule, you need to know its foundation: the hybridisation of its central atom. In this problem, we are on a quest to find the ion that boasts an $s{p}^3{d}^2$ hybridisation.

The Steric Number Formula

To find the hybridisation, we don't need to draw complex orbital diagrams every time. We have a powerful shortcut called the Steric Number ($H$) formula:

Let's break down this elegant equation: - $V$ represents the number of valence electrons on the central atom. - $M$ is the number of monovalent atoms (like Hydrogen or Halogens) directly attached to the central atom. - $C$ is the cationic charge (subtract it, because a positive charge means electrons were lost). - $A$ is the anionic charge (add it, because a negative charge means extra electrons were gained).

The resulting value, $H$, tells us the number of hybridised orbitals. An $H$ of 2 means $sp$, 3 means $s{p}^2$, 4 means $s{p}^3$, 5 means $s{p}^{3}d$, and 6 means $s{p}^3{d}^2$.

Let's put our candidates through the formula test.

1. $[IC{l}_2{]}^{-}$ and $[Br{F}_2{]}^{-}$ Both Iodine and Bromine are halogens, sitting proudly in Group 17, which means they have $V=7$ valence electrons. They are bonded to 2 monovalent halogens ($M=2$), and carry a $-1$ charge ($A=1$). An $H$ of 5 corresponds to $s{p}^{3}d$ hybridisation. Close, but not what we are looking for.

2. $[IC{l}_4{]}^{-}$ Here, the central Iodine ($V=7$) is bonded to 4 monovalent Chlorine atoms ($M=4$), with a $-1$ charge ($A=1$). Bingo! An $H$ of 6 perfectly matches $s{p}^3{d}^2$ hybridisation.

3. $[I{F}_6{]}^{-}$ Just to be absolutely certain, let's check the last option. Iodine ($V=7$) is bonded to 6 Fluorine atoms ($M=6$), with a $-1$ charge ($A=1$). An $H$ of 7 corresponds to $s{p}^3{d}^3$ hybridisation.

Now that we know $[IC{l}_4{]}^{-}$ is our winner, let's visualize it. With a steric number of 6, the foundational electron geometry is octahedral. However, Iodine is only bonded to 4 Chlorine atoms. This means the remaining 2 electron domains must be lone pairs.

To minimize the intense repulsion between these lone pairs, they position themselves as far apart as possible—at the axial positions (top and bottom). The 4 Chlorine atoms occupy the equatorial plane, resulting in a beautifully symmetric Square Planar molecular shape.

By mastering the steric number formula, you can instantly decode the hidden architecture of almost any molecule!