Match List I with List II: List I (Complexes) List II (Hybridisation) (A) Ni ( CO ) 4 I sp 3 (B) Cu NH 3 4 2…

$\mathrm{Ni}$ is in zero oxidation state in $\left[\mathrm{Ni}\right(\mathrm{CO}{)}_4]$. 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 3d orbital, therefore the hybridisation of $\left[Ni\right(CO{)}_4]$ is sp³ and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons.

${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$ ion, oxidation state of $\mathrm{Cu}$ is +2 and its valence shell electronic configuration is $3{\mathrm{d}}^9$.  So, there would be a rearrangement of electrons in ${\mathrm{Cu}}^{2+}$ because of the ${\mathrm{NH}}_3$ strong field ligand . And the last electron in the d-orbital would be out waiting for the N's electrons to fill up first. So, the 4 electron pairs from $\mathrm{N}$ would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the e^- from 3d would take place. So, you can say the hybridisation here would be ${\mathrm{dsp}}^2$.

${\left[\mathrm{Fe}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}$ ion, oxidation state of $\mathrm{Fe}$ is +2 and its valence shell electronic configuration is $3{\mathrm{d}}^6$. ${\mathrm{NH}}_3$ is strong field ligand with complex. So, no pairing will be there. So, you can say the hybridisation here would be ${\mathrm{d}}^2{\mathrm{sp}}^3$.

${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ ion, oxidation state of $\mathrm{Fe}$ is +2 and its valence shell electronic configuration is $3{\mathrm{d}}^6$. There are 4 unpaired electrons in 3d orbital.  So, you can say the hybridisation here would be ${\mathrm{sp}}^3{\mathrm{d}}^2$.

So, correct order of matching is given in the below table: