The species having tetrahedral shape is

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(D) The geometry of coordination complexes depends on the nature of the ligand and the central metal ion. $1$. $[NiCl_4]^{2-}$: $Ni^{2+}$ has a $d^8$

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$[NiCl_4]^{2-}$

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(D) The geometry of coordination complexes depends on the nature of the ligand and the central metal ion. $1$. $[NiCl_4]^{2-}$: $Ni^{2+}$ has a $d^8$ configuration. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry. $2$. $[PdCl_4]^{2-}$,$[Ni(CN)_4]^{2-}$,and $[Pd(CN)_4]^{2-}$: These complexes involve $Pd^{2+}$ or strong field ligands like $CN^-$,which cause pairing of electrons. They undergo $dsp^2$ hybridization,resulting in a square planar geometry.

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

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Hybridizations of the central atom in $[Fe(H_2O)_6]^{3+}$ and $[Co(NH_3)_6]^{3+}$,respectively,are

The hybridisation involved in $[CoF_6]^{3-}$ is :

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 is the coordination number of the central metal ion if it forms an octahedral complex?

In a square planar complex of $Co^{2+}$ ion,the unpaired electron is present in which orbital?

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