Which of the following is correctly matched?

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(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. It is tetrahedral $(s

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$Ni(CO)_4$ and $[Co(CO)_4]^{-}$ $-$ both are tetrahedral and diamagnetic

Vedclass · Answer

(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. It is tetrahedral $(sp^3)$ and diamagnetic.$2$. In $[Co(CO)_4]^{-}$,$Co$ is in $-1$ oxidation state $(3d^8 4s^2)$. It is isoelectronic with $Ni(CO)_4$,hence it is also tetrahedral $(sp^3)$ and diamagnetic.$3$. $[Fe(CN)_6]^{3-}$ is paramagnetic due to one unpaired electron.$4$. $[Ni(CN)_4]^{2-}$ is square planar $(dsp^2)$.$5$. $[Co(H_2O)_6]^{3+}$ is $sp^3d^2$ (outer orbital complex) due to weak field ligand $H_2O$.

List-$I$	List-$II$
$A$. $[MnBr_4]^{2-}$	$I$. $d^2sp^3$ and diamagnetic
$B$. $[FeF_6]^{3-}$	$II$. $sp^3d^2$ and paramagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$III$. $sp^3$ and diamagnetic
$D$. $[Ni(CO)_4]$	$IV$. $sp^3$ and paramagnetic

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