Both [Ni(CO)_4] and [Ni(CN)_4]^{2-} are diama | vedclass

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(C) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pa

Vedclass · Accepted Answer

$sp^3, dsp^2$

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(C) In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,leading to $sp^3$ hybridization and tetrahedral geometry.In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The electronic configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,which forces the $3d$ electrons to pair up,resulting in $dsp^2$ hybridization and square planar geometry.

$A$. $[CoF_6]^{3-}$	$M$. $sp^3d^2$,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

Column-$I$ Complex	Column-$II$ Structure / Geometry / Property
$A$. $[Ni(CN)_4]^{2-}$	$I$. Tetrahedral / Paramagnetic
$B$. $[Ni(CO)_4]$	$II$. Tetrahedral / Diamagnetic
$C$. $[NiCl_4]^{2-}$	$III$. Square planar / Diamagnetic

Both $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are diamagnetic. The hybridization of nickel in these complexes,respectively,are

Similar Questions

Identify the hybridization of the central metal atom in the $[Fe(CO)_5]$ complex.

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

Define the following term: Coordination polyhedron.

Match the following:

Column-$I$ Complex Column-$II$ Structure / Geometry / Property

$A$. $[Ni(CN)_4]^{2-}$ $I$. Tetrahedral / Paramagnetic

$B$. $[Ni(CO)_4]$ $II$. Tetrahedral / Diamagnetic

$C$. $[NiCl_4]^{2-}$ $III$. Square planar / Diamagnetic

The correct match is:

What type of hybridization is exhibited by $[CoF_6]^{3-}$?

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