Chemistry - Coordination Chemistry Question with Solution | TestHub

(a) Ni in ${\left[\mathrm{Ni}(\mathrm{Cl}{)}_4\right]}^{2-}$ exist as ${\mathrm{Ni}}^{2+}$ ion. Cl is a weak field ligand (high spin). It will not cause pairing of electrons. Hence, configuration of ${\left[\mathrm{Ni}(\mathrm{Cl}{)}_4\right]}^{2-}$ isTetrahedral with two unpaired electrons (i.e., paramagnetic) and has $s{p}^3$ hybridisation. (b) In ${\left[\mathrm{Co}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]}^{3-},{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3^{2-}$ is a bidentate ligand thus, give octahedral structure. (c) In ${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$, Ni exist as ${\mathrm{Ni}}^{2+}$ ion. ${\mathrm{CN}}^{-}$is a strong field ligand (low spin). Causes pairing of electrons of $3d$ orbital.Hence, the structure of is square planar and it is diamagnetic. (d) In $\left[\mathrm{Ni}(\mathrm{CO}{)}_4\right]$, Ni has zero oxidation state. i.e.,CO is a strong field ligand causes rearrangement and pairing of electrons of $3d$ and $4s$ orbital.Structure of $\mathrm{Ni}\left({\mathrm{CO}}_4\right)$ is tetrahedral with $s{p}^3$ hybridisation and it is diamagnetic. Hence, (4) is the correct answer.