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(C) $1$. For $[Co(CO)_4]^-$,the oxidation state of $Co$ is $-1$. The electronic configuration is $3d^8 4s^2$. Since $CO$ is a strong field ligand,it c

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$[Co(CO)_4]^-$ and $[Ni(CO)_4]$

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(C) $1$. For $[Co(CO)_4]^-$,the oxidation state of $Co$ is $-1$. The electronic configuration is $3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons. The geometry is $sp^3$ (tetrahedral) and it is diamagnetic due to the pairing of all electrons.$2$. For $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration is $3d^8 4s^2$. $CO$ causes pairing of electrons,resulting in $sp^3$ hybridization (tetrahedral) and it is diamagnetic.$3$. Other options like $[NiCl_4]^{2-}$ are tetrahedral but paramagnetic,and $[Ni(CN)_4]^{2-}$ is square planar and diamagnetic.

Which of the following pairs of complex compounds are both tetrahedral and diamagnetic?

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$[Mn(CO)_4 NO]$ is diamagnetic because

$STATEMENT-1$: $[Fe(H_2O)_5NO]SO_4$ is paramagnetic.
$STATEMENT-2$: The $Fe$ in $[Fe(H_2O)_5NO]SO_4$ has three unpaired electrons.

The geometry and magnetic property of $[NiCl_4]^{2-}$,respectively are

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