Chemistry - Coordination Chemistry Question with Solution | TestHub

$\mathrm{Ni}$ is in zero oxidation state in $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ so the electronic configuration of $\mathrm{Ni}$ is $3{\mathrm{d}}^{8}4{\mathrm{s}}^2$. As $\mathrm{CO}$ is a strong ligand, it pushes all the electrons in the $3\mathrm{d}$ orbital, therefore the hybridisation of $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ is ${\mathrm{sp}}^3$ and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons.

In $[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4{]}^{2-}$, there is ${\mathrm{Ni}}^{2+}$ ion for which the electronic configuration in the valence shell is $3{\mathrm{d}}^{8}4{\mathrm{s}}^0$.

In presence of strong field ${\mathrm{CN}}^{-}$ ions, all the electrons are paired up.

The empty, $3\mathrm{d}, 3\mathrm{s}$ and two $4\mathrm{p}$ Orbitals undergo ${\mathrm{dsp}}^2$ hybridization to make bonds with ${\mathrm{CN}}^{-}$ ligands in square planar geometry.

The atomic number of $\mathrm{Co}$ is $27$ and its valence shell electronic configuration is $3{\mathrm{d}}^{7}4{\mathrm{s}}^2$.

$\mathrm{Co}$ is in $+3$ oxidation state in the complex ${\left[{\mathrm{CoF}}_6\right]}^{3-}$.

${\left[{\mathrm{CoF}}_6\right]}^{3-}$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridized and it is octahedral in shape.

Another ${\mathrm{Co}}^{3+}$ complex, $[\mathrm{Co}{\left(\mathrm{CN}\right)}_6{]}^{3-}$, is diamagnetic and has no unpaired electrons. The hybrid orbitals used to form this complex are ${\mathrm{d}}^2{\mathrm{sp}}^3$.

$\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ Hybridisation ${\mathrm{sp}}^3$

${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ Hybridisation ${\mathrm{dsp}}^2$

${\left[\mathrm{Co}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$ Hybridisation ${\mathrm{d}}^2{\mathrm{Sp}}^3$

${\left[\mathrm{Co}{\left(\mathrm{F}\right)}_6\right]}^{3-}$ Hybridisation ${\mathrm{sp}}^3{\mathrm{d}}^2$