LIST- I contains metal species and LIST- II contains their...

$\left(I\right){\left[Cr{\left(CN\right)}_{6}\right]}^{4 -}$

${Cr}^{2} = \left[Ar\right]3{d}^{4}4{s}^{0}$

It is ${d}^{2}{sp}^{3}$ hybridised low spin complex as ${CN}^{-} is a strong field ligand.</p><p>\left(II\right){\left[{RuCl}_{6}\right]}^{2 -}$

Low spin complex as Ru belongs to 4d series, as elements belonging to 4d and 5d series forms low spin complexes even with weak field ligand.

${Ru}^{+ 4} = \left[Kr\right]4{d}^{4}5{s}^{0}$

${t}_{2g}$ set contains $4$ electrons.

\left(III\right){\left[Cr{\left({H}_{2}O\right)}_{6}\right]}^{2 +}$</p><p>${Cr}^{+ 2} = \left[Ar\right]3{d}^{4}4{s}^{0}$</p><p>Its configuration is${t}_{2g}^{3}{e}_{g}^{1}$</p><p>It has$4$unpaired${e}^{-}$as${H}_{2}O$is weak field ligand.</p><p>So, its Magnetic moment,$\mu = 4 . 9B . M$</p><p>\left(IV\right){\left[Fe\left({H}_{2}O\right)6\right]}^{2 +}$

${Fe}^{2 +} = \left[Ar\right]3{d}^{6}4{s}^{0}$

$= {t}_{2g}^{4}{e}_{g}^{2}$

It has $4$ unpaired ${e}^{-}$, its Magnetic moment, $\mu = 4 . 9B . M .</p></blockquote>$