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(C) $1$. The central metal ion is $Ni^{2+}$. The atomic number of $Ni$ is $28$,so its electronic configuration is $[Ar] 3d^8 4s^2$.$2$. In $[NiCl_4]^{

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Tetrahedral,Paramagnetic

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(C) $1$. The central metal ion is $Ni^{2+}$. The atomic number of $Ni$ is $28$,so its electronic configuration is $[Ar] 3d^8 4s^2$.$2$. In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$,so the configuration of $Ni^{2+}$ is $[Ar] 3d^8$.$3$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.$4$. The $Ni^{2+}$ ion undergoes $sp^3$ hybridization using one $4s$ and three $4p$ orbitals.$5$. This results in a tetrahedral geometry.$6$. Since there are two unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

List $-I$	List $-II$
$A$. $[PtCl_4]^{2-}$	$I$. $sp^3d$
$B$. $BrF_5$	$II$. $d^2sp^3$
$C$. $PCl_5$	$III$. $dsp^2$
$D$. $[Co(NH_3)_6]^{3+}$	$IV$. $sp^3d^2$

What type of geometry and magnetic nature is present in the complex $[NiCl_4]^{2-}$?

Similar Questions

Explain on the basis of valence bond theory that $[Ni(CN)_4]^{2-}$ ion with square planar structure is diamagnetic and the $[NiCl_4]^{2-}$ ion with tetrahedral geometry is paramagnetic.

Cobalt chloride when dissolved in water forms a pink-colored complex $X$ which has an octahedral geometry. This solution,upon treatment with concentrated $HCl$,forms a deep blue complex $Y$ which has a $Z$ geometry. $X$,$Y$,and $Z$,respectively,are:

Match List $-I$ with List $-II$:

List $-I$ List $-II$

$A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$

$B$. $BrF_5$ $II$. $d^2sp^3$

$C$. $PCl_5$ $III$. $dsp^2$

$D$. $[Co(NH_3)_6]^{3+}$ $IV$. $sp^3d^2$

Which of the following complexes has $dsp^2$ hybridisation?

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

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