Among [Ni(CO)_4],[Ni(CN)_4]^{2-},and [NiCl_4] | vedclass

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(D) $1$. 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 $3d$,res

Vedclass · Accepted Answer

$sp^3, dsp^2, sp^3$ (Atomic number of $Ni = 28$)

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(D) $1$. 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 $3d$,resulting in $sp^3$ hybridization.$2$. For $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of $3d$ electrons,resulting in $dsp^2$ hybridization.$3$. For $[NiCl_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,no pairing occurs,resulting in $sp^3$ hybridization.Therefore,the hybridization states are $sp^3, dsp^2, sp^3$ respectively. The correct option is $D$.

$A$. $[CoF_6]^{3-}$	$M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$	$N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$	$P$. $d^2sp^3$,diamagnetic

Among $[Ni(CO)_4]$,$[Ni(CN)_4]^{2-}$,and $[NiCl_4]^{2-}$ species,the hybridization states at the $Ni$ atom are,respectively:

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Match the correct option.
$A$. $[CoF_6]^{3-}$ $M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$ $N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$ $O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$ $P$. $d^2sp^3$,diamagnetic

The geometry of $Ni(CO)_4$ and $Ni(PPh_3)_2Cl_2$ are

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