Match List-I with List-II LIST-I Coordination Complex LIST-II...

${\left[Cr ( CN{)}_{6}\right]}^{3 -}$ ion, oxidation state of $Cr$ is +3 and its valence shell electronic configuration is $3{d}^{3}$. There are $3$ unpaired electrons in $3d$ orbital.

(A). ${\left[Cr ( CN{)}_{6}\right]}^{3 -}$

No. of unpaired electrons $= 3$

${\left[Fe{\left({H}_{2}O\right)}_{6}\right]}^{2 +}$ 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 ${sp}^{3}{d}^{2}$.

(B). ${\left[Fe{\left({H}_{2}O\right)}_{6}\right]}^{2 +}$

No. of unpaired electrons $= 4$

${\left[Co{\left({NH}_{3}\right)}_{6}\right]}^{3 +}$ in this oxidation state of central metal atom is $+ 3$ and it has no unpaired electrons.
(C). ${\left[Co{\left({NH}_{3}\right)}_{6}\right]}^{3 +}$

No. of unpaired electrons $= 0$

${\left[Ni{\left({NH}_{3}\right)}_{6}\right]}^{2 +}$ the oxidation state of $Ni$ is ${Ni}^{2 +}$ and it has two unpaired electrons.
(D) ${\left[Ni{\left({NH}_{3}\right)}_{6}\right]}^{2 +}$

No. of unpaired electrons $= 2$
So the correct option among the given is $A$.