MF + XeF4 'A' (M+ = Alkali metal cation)<br>The state of hybridisation of the central atom in 'A' an β p Block Elements Chemistry Question
Question
MF + XeF4 $\rightarrow$ 'A' (M+ = Alkali metal cation)<br>The state of hybridisation of the central atom in 'A' and shape of the species are:
π‘ Solution & Explanation
Step 1: The reaction of alkali metal fluorides (MF) with xenon tetrafluoride (XeF4) produces an ionic adduct M+[XeF5-], where the central atom of the anion is Xenon. Step 2: In the [XeF5]- anion, Xenon has 8 valence electrons plus 1 negative charge, making 9 electrons. It forms 5 single bonds with Fluorine, leaving 4 non-bonding electrons (2 lone pairs). The steric number is 5 (bond pairs) + 2 (lone pairs) = 7, which corresponds to sp3d3 hybridization. Step 3: According to VSEPR theory, a species with 5 bond pairs and 2 lone pairs in pentagonal bipyramidal electronic geometry adopts a pentagonal planar molecular shape. Thus, the correct option is (c).