The shapes of XeF4, XeF5^- and SnCl2 are : — Chemical Bonding Chemistry Question

💡 Solution & Explanation

Step 1: XeF4 has 4 bond pairs and 2 lone pairs on Xenon, which gives it a square planar shape. Step 2: XeF5^- has Xenon (8 valence e^- + 1 charge = 9) forming 5 bonds, leaving 4 non-bonding electrons (2 lone pairs). Steric number = 7, giving a pentagonal planar shape. SnCl2 has Tin (4 valence e^-) forming 2 bonds, leaving 1 lone pair, resulting in a bent (angular) shape. Step 3: Therefore, the shapes are square planar, pentagonal planar, and angular respectively, which corresponds to option (c).