Which of the following is correctly matched?

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(A) $1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridi

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$[Fe(CN)_6]^{4-}$ and $[Fe(CN)_6]^{3-}$ - Both are octahedral with $d^2sp^3$ hybridisation

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(A) $1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridisation (inner orbital complex).$2$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. $CN^-$ causes pairing,resulting in $d^2sp^3$ hybridisation.$3$. Both are octahedral and $d^2sp^3$ hybridised. Thus,option $A$ is correct.$4$. $[Ni(CO)_4]$ is tetrahedral ($sp^3$,diamagnetic),but $[Ni(CN)_4]^{2-}$ is square planar ($dsp^2$,diamagnetic). Option $B$ is incorrect.$5$. $[Ni(CO)_4]$ and $[Co(CO)_4]^{-}$ are both isoelectronic and tetrahedral ($sp^3$,diamagnetic). Option $C$ is also technically correct in terms of geometry and hybridisation,but $A$ is the standard textbook example for $d^2sp^3$ comparison.$6$. $[Co(H_2O)_6]^{3+}$ is $d^2sp^3$ (low spin),while $[Cr(H_2O)_6]^{3+}$ is $d^2sp^3$ (inner orbital). However,$A$ is the most definitive match.

List - $I$ (Complex)	List – $II$ (Geometry)
$(a)$ $[Co(NH_3)_6]^{3+}$	$(i)$ Trigonal bipyramidal
$(b)$ $[NiCl_4]^{2-}$	(ii) Octahedral
$(c)$ $[Ni(CN)_4]^{2-}$	(iii) Tetrahedral
$(d)$ $[Fe(CO)_5]$	(iv) Square planar

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