The number of unpaired electrons in [NiCl_{4} | vedclass

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Question

(B) $1$. For $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired elec

Vedclass · Accepted Answer

$2, 0, 1$

Vedclass · Answer

(B) $1$. For $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired electrons.$2$. For $Ni(CO)_{4}$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$ orbitals,resulting in $0$ unpaired electrons.$3$. For $[Cu(NH_{3})_{4}]^{2+}$: $Cu$ is in $+2$ oxidation state $(3d^9)$. It has $1$ unpaired electron.Thus,the number of unpaired electrons are $2, 0, 1$ respectively. The correct option is $B$.

The number of unpaired electrons in $[NiCl_{4}]^{2-}$,$Ni(CO)_{4}$ and $[Cu(NH_{3})_{4}]^{2+}$ respectively are

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