Match the complex ions in List-I with their corresponding hybridization in List-II.

List-I (Complex ion) (A) $[Ni(CN)_4]^{2-}$ (B) $[CoF_6]^{3-}$ (C) $[Co(C_2O_4)_3]^{3-}$ (D) $[NiCl_4]^{2-}$

List-II (Hybridisation) (I) $sp^3$ (II) $d^2sp^3$ (III) $sp^3d^2$ (IV) $dsp^2$

Choose the correct answer from the options given below:

According to Valence Bond Theory : (A) $[Ni(CN)_4]^{2-}$: The central metal ion is $Ni^{2+}$ with a $3d^8$ configuration. $CN^-$ is a strong field ligand, causing the pairing of unpaired $d$ electrons. This vacates one $3d$ orbital. The hybridization is $dsp^2$ forming a square planar complex . (B) $[CoF_6]^{3-}$: The central metal ion is $Co^{3+}$ with a $3d^6$ configuration. $F^-$ is a weak field ligand, so no pairing of $d$ electrons occurs. It uses outer $4d$ orbitals for hybridization, which is $sp^3d^2$ (outer orbital octahedral complex) . (C) $[Co(C_2O_4)_3]^{3-}$: The central metal ion is $Co^{3+}$ ($3d^6$). The oxalate ion ($C_2O_4^{2-}$) acts as a strong field ligand for $Co^{3+}$ causing pairing of electrons. It uses inner $3d$ orbitals for hybridization, which is $d^2sp^3$ (inner orbital octahedral complex) . (D) $[NiCl_4]^{2-}$: The central metal ion is $Ni^{2+}$ ($3d^8$). $Cl^-$ is a weak field ligand, so no pairing occurs. The hybridization involves one $4s$ and three $4p$ orbitals, resulting in $sp^3$ hybridization (tetrahedral complex) . Therefore, the correct matching is A-IV, B-III, C-II, D-I.