SbF5 reacts with XeF4 and XeF6 to form ionic compounds [XeF3^+][SbF6^-] and [XeF5^+][SbF6^-] then mo — Chemical Bonding Chemistry Question

💡 Solution & Explanation

Step 1: Determine the geometry of the cationic species [XeF3+]. Xe has 8 valence electrons, minus 1 for the positive charge = 7. Three form Xe-F bonds, leaving 2 lone pairs. Steric number = 3 bond pairs + 2 lone pairs = 5 (sp3d hybridization, T-shaped geometry). Step 2: Determine the geometry of [XeF5+]. Xe has 8 valence electrons, minus 1 = 7. Five form Xe-F bonds, leaving 1 lone pair. Steric number = 5 bond pairs + 1 lone pair = 6 (sp3d2 hybridization, square pyramidal geometry). Step 3: Consequently, the molecular shapes of [XeF3+] and [XeF5+] are bent T-shape and square pyramidal respectively, matching option (b).