Match List I with List II: List I (Complexes) List II...

$Ni$ is in zero oxidation state in $[ Ni ( CO{)}_{4} ]$. So, the electronic configuration of $Ni$ is $3{d}^{8}4{s}^{2}$. As $CO$ is a strong ligand, it pushes all the electrons in the 3d orbital, therefore the hybridisation of $[ Ni ( CO{)}_{4} ]$ is sp³ and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons.

${\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2 +}$ ion, oxidation state of $Cu$ is +2 and its valence shell electronic configuration is $3{d}^{9}$. So, there would be a rearrangement of electrons in ${Cu}^{2 +}$ because of the ${NH}_{3}$ 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 $N$ 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 ${dsp}^{2}$.

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

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

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