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(B) $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^{6}$. $NH_{3}$ is a strong field ligand $(SFL)$, causing pairing. Hybridization is $d^{2}sp^{3}$, which

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$A$, $B$ & $D$ only

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(B) $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^{6}$. $NH_{3}$ is a strong field ligand $(SFL)$, causing pairing. Hybridization is $d^{2}sp^{3}$, which is an inner orbital complex. (Correct)$(B)$ $[MnCl_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. $Cl^{-}$ is a weak field ligand $(WFL)$. Hybridization is $sp^{3}d^{2}$, which is an outer orbital complex. (Correct)$(C)$ $[CoF_{6}]^{3-}$: $Co^{3+}$ is $3d^{6}$. $F^{-}$ is a $WFL$. Hybridization is $sp^{3}d^{2}$, which is an outer orbital complex. (Incorrect as stated $d^{2}sp^{3}$)$(D)$ $[FeF_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $F^{-}$ is a $WFL$. Hybridization is $sp^{3}d^{2}$, which is an outer orbital complex. (Correct)$(E)$ $[Ni(CN)_{4}]^{2-}$: $Ni^{2+}$ is $3d^{8}$. $CN^{-}$ is a $SFL$. Hybridization is $dsp^{2}$, which is an inner orbital complex. (Incorrect as stated $sp^{3}$)Therefore, the correct statements are $A$, $B$, and $D$.

List-$I$ (Complex/Species)	List-$II$ (Shape and magnetic moment)
$A$. $[Ni(CO)_4]$	$I$. Tetrahedral,$2.8 \ BM$
$B$. $[Ni(CN)_4]^{2-}$	$II$. Square planar,$0 \ BM$
$C$. $[NiCl_4]^{2-}$	$III$. Tetrahedral,$0 \ BM$
$D$. $[MnBr_4]^{2-}$	$IV$. Tetrahedral,$5.9 \ BM$

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