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.

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(N/A) In both complexes,$Ni$ is in the $+2$ oxidation state,which corresponds to a $d^8$ electronic configuration.
For $[Ni(CN)_4]^{2-}$:
$CN^-$ is a strong field ligand. It causes the pairing of the two unpaired $3d$ electrons. This results in $dsp^2$ hybridization,leading to a square planar geometry. Since all electrons are paired,the complex is diamagnetic.
For $[NiCl_4]^{2-}$:
$Cl^-$ is a weak field ligand. It does not cause the pairing of the two unpaired $3d$ electrons. Therefore,the complex undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry. Due to the presence of two unpaired electrons,the complex is paramagnetic.

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Match the List-$I$ with List-$II$:
List-$I$ (Complex/Species) List-$II$ (Shape and magnetic moment)
$A$. $[Ni(CO)_4]$ $I$. Tetrahedral,$2.8 \ BM$
$B$. $[Ni(CN)_4]^{2-}$ $II$. Square planar,$0 \ BM$
$C$. $[NiCl_4]^{2-}$ $III$. Tetrahedral,$0 \ BM$
$D$. $[MnBr_4]^{2-}$ $IV$. Tetrahedral,$5.9 \ BM$

Choose the correct answer from the options given below:

$[PdCl_2(PMe_3)_2]$ is a diamagnetic complex of $Pd(II)$. How many unpaired electrons are present in the analogous complex of $Ni(II)$?

Which set of $d-$ orbitals is used by the central metal during the formation of $MnO_4^-$?

Which of the following has a square planar geometry? $(Co = 27, Ni = 28, Fe = 26, Pt = 78)$

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