The hybridization of the central atom in the anion [ICl₄]⁻ is:
(A) sp³
(B) sp³d
(C) sp³d²
(D) dsp²
Correct: C
ICl₄⁻ has iodine as the central atom. Iodine in ICl₄⁻ has 7 valence electrons; adding one for the negative charge gives 8 electrons. Four I–Cl bonds use 4 electrons, leaving 4 electrons as two lone pairs. The steric number is 6 (4 bonding pairs + 2 lone pairs), which corresponds to octahedral electron geometry. The molecular geometry is square planar. The hybridization is sp³d². Note: Some textbooks describe the bonding in hypervalent iodine compounds using d-orbital participation (sp³d²), while modern explanations invoke 3-center-4-electron bonds or s-p mixing. For JEE purposes, sp³d² is the expected answer. Answer: C.