Which of the following complexes is tetrahedr | vedclass

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Question

(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. It forms $dsp^2$ hybridization,which is square planar

Vedclass · Accepted Answer

$[Ni(CO)_4]$

Vedclass · Answer

(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. It forms $dsp^2$ hybridization,which is square planar and diamagnetic.$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand. It forms $sp^3$ hybridization,which is tetrahedral and paramagnetic.$3$. $[Ni(CO)_4]$: $Ni$ is $d^8s^2$. $CO$ is a strong field ligand,forcing electrons to pair. It forms $sp^3$ hybridization,which is tetrahedral and diamagnetic.$4$. $[Fe(CO)_5]$: $Fe$ is $d^6s^2$. It forms $dsp^3$ hybridization,which is trigonal bipyramidal and diamagnetic.Therefore,the complex that is both tetrahedral and diamagnetic is $[Ni(CO)_4]$.

Which of the following complexes is tetrahedral as well as diamagnetic?

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