See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Hybridization Bond angle 2 H O 3 sp Bent (V – shape) 104.5o 2 H S 3 sp Bent (V – shape) 92o 3 ClF 3 sp d (Bent T-shape) Equatorial – axial = 87.5o Axial – axial = 175o 7 IF 3 3 sp d pentagonal bipyramidal Equatorial – equatorial = 72o Equatorial – axial = 90o Axial – axial = 180o 2 XeF 3 sp d Linear 180o AITS-PT-III-PCM(Sol.)-JEE(Main)/2021 Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 14 6 SF 3 2 sp d Octahedral 90o and 180o   5 TeF 3 2 sp d Square pyramidal < 79o 4 XeF 3 2 sp d Square planar 90o and 180o 3 O 2 sp (Bent V-shape) 117o 2 SO 2 sp (Bent V-shape) o 119 30 3 SO 2 sp Trigonal planar 120o SECTION – C