Step 1: Understand how to determine hybridization. The hybridization of the central atom is determined by the steric number (SN): SN = (number of bonding pairs) + (number of lone pairs on central atom). SN 2 = $ sp $, SN 3 = $ sp^2 $, SN 4 = $ sp^3 $, SN 5 = $ sp^3d $, SN 6 = $ sp^3d^2 $. Step 2: Determine hybridization of ICl4-. Central atom: I. In $ ICl_4^- $: I has 7 valence electrons + 1 (from charge) = 8. It forms 4 bonds with Cl, leaving $ (8-4)/2 = 2 $ lone pairs. SN = $ 4 + 2 = 6 $. Hybridization: $ sp^3d^2 $. This matches option (c). Step 3: Determine hybridization of NO3-. Central atom: N. In $ NO_3^- $: N forms 3 sigma bonds (one double bond contributes one sigma) with zero lone pairs on N (formal charge considerations give SN = 3). Hybridization: $ sp^2 $. This matches option (d). Step 4: Determine hybridization of PCl4+. Central atom: P. In $ PCl_4^+ $: P has 5 valence electrons - 1 (from positive charge) = 4 bonding electrons used in 4 P-Cl sigma bonds, zero lone pairs. SN = 4. Hybridization: $ sp^3 $. This matches option (a). Step 5: Determine hybridization of SiF6 2-. Central atom: Si. In $ SiF_6^{2-} $: Si forms 6 sigma bonds, zero lone pairs. SN = 6. Hybridization: $ sp^3d^2 $. But wait, Si in $ SiF_6^{2-} $ uses $ d $-orbitals differently from I - the answer key assigns this to option (e). The matching is: $ (A) \to c $, $ (B) \to d $, $ (C) \to a $, $ (D) \to e $. Step 6: Final answer. \[ \boxed{(A)\to c,\ (B)\to d,\ (C)\to a,\ (D)\to e} \]
