Which of the following metal carbonyl complex | vedclass

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(C) $1$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state $(3d^5 4s^1)$. $CO$ is a strong field ligand,causing pairing. It is diamagnetic and octahedral $

Vedclass · Accepted Answer

$[Co(CO)_4]^- - 	ext{Diamagnetic} - 	ext{Tetrahedral}, sp^3$

Vedclass · Answer

(C) $1$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state $(3d^5 4s^1)$. $CO$ is a strong field ligand,causing pairing. It is diamagnetic and octahedral $(d^2sp^3)$.$2$. $[Fe(CO)_5]$: $Fe$ is in $0$ oxidation state $(3d^6 4s^2)$. All electrons pair up. It is diamagnetic and trigonal bipyramidal $(dsp^3)$.$3$. $[Co(CO)_4]^-$: $Co$ is in $-1$ oxidation state $(3d^8 4s^2)$. With $CO$ ligands,it becomes $3d^{10}$,which is diamagnetic and tetrahedral $(sp^3)$.$4$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. It is diamagnetic and tetrahedral $(sp^3)$.Therefore,option $C$ is correctly matched.

Which of the following metal carbonyl complexes is correctly matched with its magnetic property and geometry?

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