Question

Identify, from the fơllowing, thẻ diamagnetic, tetrahedral complex

• A. ${\left[Ni(Cl{)}_4\right]}^{2-}$
• B. ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{3-}$
• C. ${\left[Ni(CN{)}_4\right]}^{2-}$
• D. $\left[Ni(CO{)}_4\right]$

Answer 1

Answer: (D)

$Niin[Ni(Cl){]}^{2-}$ exist as $N{i}^{2+}$ ion.



$\because Cl$ is a weak field ligand (high spin). It will not go for pairing.

Hence, configuration of ${\left[Ni(Cl{)}_4\right]}^{2-}$ is



Tetrahedral with two unpaired electrons (i.e., paramagnetic) with $s{p}^3$ hybridisation.

(2) In ${\left[CO{\left(C_2{O}_4\right)}_3\right]}^{3-},{\left(C_2{O}_4\right)}_3^{2-}$ is a bidentate ligand thus, give octahedral structure.

(3) In ${\left[Ni(CN{)}_4\right]}^{2-}$, $Ni$ exist as $N{i}^{2+}$ ion.



$\because C{N}^{-}$ is a strong field ligand (low spin), the unpaired electrons of 3rd orbital go for pairing to give ${\left[Ni(CN{)}_4\right]}^{2-}$ as follows:



Hence, the structure of ${\left[Ni(CN{)}_4\right]}^{2-}$ is square planar with $ds{p}^2$ hybridisation having diamagnetic nature.

In $\left[Ni(CO{)}_4\right]$, Ni has zero oxidation state.

i.e.

When CO approaches to Ni-metal, electron of Ni, go for rearrangement to give tetrahedral structure with no unpaired electrons (diamagnetic nature) and $s{p}^3$ hybridisation.