What type of geometry and magnetic nature is | vedclass

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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.

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

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

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