SbF5 reacts with XeF4 to form an adduct. The shapes of cation and anion in the adduct are respective — p Block Elements Chemistry Question

💡 Solution & Explanation

Step 1: In the reaction of XeF4 with the strong Lewis acid SbF5, XeF4 acts as a fluoride donor to yield the ionic adduct [XeF3]+[SbF6]-. Step 2: The cation [XeF3]+ has 3 bond pairs and 2 lone pairs on Xenon (steric number 5), which corresponds to sp3d hybridization and a T-shape. The anion [SbF6]- has 6 bond pairs on Antimony, representing regular octahedral geometry. Step 3: Consequently, the shapes of the cation and anion are T-shaped and octahedral respectively, matching option (b).