Geometrical shapes of the complexes formed by | vedclass

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Question

(B) $Ni^{2+}$ has a $3d^8$ electronic configuration.$1$. For $[NiCl_4]^{2-}$,$Cl^{-}$ is a weak field ligand,so it does not cause pairing of electrons

Vedclass · Accepted Answer

tetrahedral,square planar and octahedral

Vedclass · Answer

(B) $Ni^{2+}$ has a $3d^8$ electronic configuration.$1$. For $[NiCl_4]^{2-}$,$Cl^{-}$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry.$2$. For $[Ni(CN)_4]^{2-}$,$CN^{-}$ is a strong field ligand,which causes pairing of $3d$ electrons. The hybridization is $dsp^2$,resulting in a square planar geometry.$3$. For $[Ni(H_2O)_6]^{2+}$,$H_2O$ is a ligand that forms an octahedral complex with $Ni^{2+}$ involving $sp^3d^2$ hybridization.Thus,the shapes are tetrahedral,square planar,and octahedral,respectively. Therefore,option $B$ is correct.

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

Geometrical shapes of the complexes formed by the reaction of $Ni^{2+}$ with $Cl^{-}$,$CN^{-}$,and $H_2O$,respectively,are

Similar Questions

In which of the following complex ions is the central metal atom's $d$-orbital devoid of electrons?

Which of the following is a square planar complex ion?

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

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

Which of the following statements is not correct?

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