Animated Solution for Chemistry - p-Block Elements: The reaction in which the hybridisation of the underlined atom is affected is

The concept of hybridization is like a molecular dance—atoms rearrange their electron orbitals to accommodate new bonding partners. But what happens when a chemical reaction forces an atom to change its dance routine entirely?

In this problem, we are on a mission to find the reaction where the central atom undergoes a change in its hybridization state. Let's break down each option and observe the molecular transformations.

Xenon's Transformation

Let's start with our first candidate: the reaction between Xenon tetrafluoride (${\text{XeF}}_4$) and Antimony pentafluoride (${\text{SbF}}_5$).

Xenon, a noble gas, typically minds its own business with 8 valence electrons. In ${\text{XeF}}_4$, it forms four single bonds with fluorine atoms. This leaves $8-4=4$ non-bonding electrons, which pair up to form 2 lone pairs.

To find the hybridization, we calculate the Steric Number (SN), which is the sum of bond pairs and lone pairs:

A steric number of 6 corresponds to an $s{p}^3{d}^2$ hybridization, giving the molecule a square planar shape.

Now, what happens when ${\text{SbF}}_5$ enters the scene? Antimony pentafluoride is a notorious Lewis acid—a strong fluoride ion acceptor. It aggressively snatches a fluoride ion (${\text{F}}^{-}$) from ${\text{XeF}}_4$.

This theft transforms ${\text{XeF}}_4$ into the $[{\text{XeF}}_3{]}^{+}$ cation. Let's recalculate the steric number for this new cation. Xenon now has a positive charge, meaning it has 7 valence electrons. It forms 3 bonds with the remaining fluorine atoms, leaving $7-3=4$ non-bonding electrons, or 2 lone pairs.

A steric number of 5 dictates an $s{p}^{3}d$ hybridization! The geometry shifts to a T-shape. We have found our answer: the hybridization of Xenon changes from $s{p}^3{d}^2$ to $s{p}^{3}d$.

Even though we caught the culprit, a good chemist always double-checks their work. Let's quickly review the other reactions.

Reaction B: Sulfuric acid (${\text{H}}_2{\text{SO}}_4$) reacts with $\text{NaCl}$ to form sodium sulfate (${\text{Na}}_2{\text{SO}}_4$). In both the reactant and the product, the central sulfur atom is bonded to four oxygen atoms with zero lone pairs. The steric number remains 4, so the hybridization stays firmly at $s{p}^3$.

Reaction C: Hypophosphorous acid (${\text{H}}_3{\text{PO}}_2$) undergoes disproportionation upon heating to yield phosphoric acid (${\text{H}}_3{\text{PO}}_4$) and phosphine (${\text{PH}}_3$). - In ${\text{H}}_3{\text{PO}}_2$ and ${\text{H}}_3{\text{PO}}_4$, phosphorus forms 4 sigma bonds (SN = 4). - In ${\text{PH}}_3$, phosphorus forms 3 sigma bonds and has 1 lone pair (SN = 4). Across all these molecules, the steric number is 4, meaning the phosphorus atom remains $s{p}^3$ hybridized.

Reaction D: Ammonia (${\text{NH}}_3$) accepts a proton to become the ammonium ion (${\text{NH}}_4^{+}$). Ammonia has 3 bond pairs and 1 lone pair (SN = 4). Ammonium has 4 bond pairs and 0 lone pairs (SN = 4). Once again, the nitrogen atom maintains its $s{p}^3$ hybridization.

By systematically calculating the steric number before and after each reaction, we can confidently conclude that only the reaction between ${\text{XeF}}_4$ and ${\text{SbF}}_5$ results in a change in hybridization. The noble gas Xenon is forced to adapt, proving that chemistry is truly a dynamic and ever-changing dance!