Which statement is correct?

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(B) $Ni^{2+}$ ion has a $3d^8$ electronic configuration.In $[NiCl_4]^{2-}$, $Cl^-$ is a weak field ligand. It does not cause the pairing of electrons

Vedclass · Accepted Answer

$[NiCl_4]^{2-}$ is an outer orbital complex whereas $[Ni(CN)_4]^{2-}$ is an inner orbital complex.

Vedclass · Answer

(B) $Ni^{2+}$ ion has a $3d^8$ electronic configuration.In $[NiCl_4]^{2-}$, $Cl^-$ is a weak field ligand. It does not cause the pairing of electrons in the $3d$ orbitals. Therefore, the complex undergoes $sp^3$ hybridization, utilizing the $4s$ and $4p$ orbitals, making it an outer orbital complex.In $[Ni(CN)_4]^{2-}$, $CN^-$ is a strong field ligand. It causes the pairing of electrons in the $3d$ orbitals, leaving one $3d$ orbital vacant. Therefore, the complex undergoes $dsp^2$ hybridization, utilizing the $3d$, $4s$, and $4p$ orbitals, making it an inner orbital complex.

List-$I$ (Complex/ion)	List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$	$I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$	$II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$	$III$. Square planar
$D$. $[Fe(CO)_5]$	$IV$. Tetrahedral

Similar Questions

$[Co(NH_3)_6]^{+2}$ is paramagnetic in nature due to the presence of unpaired $e^-$ in

Match List-$I$ with List-$II$:
List-$I$ (Complex/ion) List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$ $I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$ $II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$ $III$. Square planar
$D$. $[Fe(CO)_5]$ $IV$. Tetrahedral

What type of hybridization is present in $[Co(NH_3)_6]^{3+}$ complex?

$A$ square planar complex is formed by the hybridization of which atomic orbitals?

Which complex compound possesses $sp^3d^2$ hybridisation?

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