Which of the following is the correct order for increasing bond angle ? — Chemical Bonding Chemistry Question

💡 Solution & Explanation

Step 1: Analyze the periodic trends of bond angles for each option. Step 2: For group 15 hydrides, the angle decreases down the group (NH3 > PH3 > AsH3 > SbH3), making (a) incorrect. For (b), the correct order is OF2 (~103°) < H2O (104.5°) < Cl2O (~111°), so (b) is incorrect. For (d), all boron trihalides are perfectly symmetric trigonal planar molecules with identical bond angles of exactly 120°, making (d) incorrect. Step 3: For isoelectronic cationic hydrides of group 16 (H3O^+, H3S^+, H3Se^+, H3Te^+), as the electronegativity of the central atom increases (Te < Se < S < O), the bond angle increases due to increased repulsion of bonding pairs close to the central atom. Therefore, the correct increasing order of bond angle is H3Te^+ < H3Se^+ < H3S^+ < H3O^+, which matches option (c).