Exams › JEE Advanced › Chemistry
Correct answer: II < III < IV < I < V
Central atom hybridizations: CO3²-: 3 regions of electron density -> sp2 (33%). XeF4: 6 regions (4 F + 2 lone pairs) -> sp3d2 (1/6 * 100 = 16.7%). I3⁻: 5 regions (2 bonds + 3 lone pairs) -> sp3d (20%). NCl3: 4 regions (3 bonds + 1 lone pair) -> sp3 (25%). BeCl2: 2 regions -> sp (50%). Increasing order of s-character: 16.7% < 20% < 25% < 33% < 50% => XeF4 < I3⁻ < NCl3 < CO3²- < BeCl2 => II < III < IV < I < V.