The metal Ni (Z = 28) combines with a unident | vedclass

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(C) In the complex $[NiX_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. Since the complex is

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$2$; Tetrahedral

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(C) In the complex $[NiX_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. Since the complex is paramagnetic,the ligand $X^-$ is a weak field ligand that does not cause pairing of electrons. Thus,the $3d^8$ configuration remains as $t_{2g}^6 e_g^2$,resulting in $2$ unpaired electrons. Due to the $sp^3$ hybridization,the geometry of the complex is tetrahedral.

The metal $Ni$ $(Z = 28)$ combines with a unidentate ligand $X^-$ to form a paramagnetic complex $[NiX_4]^{2-}$. What are its geometry and the number of unpaired electrons in $Ni$?

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