$[Cr(NH_3)_6]^{3+}$ is paramagnetic while $[Ni(CN)_4]^{2-}$ is diamagnetic. Explain why?

(N/A) In $[Cr(NH_3)_6]^{3+}$,$Cr$ is in the $+3$ oxidation state,which corresponds to a $d^3$ configuration.
$NH_3$ is a ligand that forms an octahedral complex. The $Cr^{3+}$ ion undergoes $d^2sp^3$ hybridization.
Since there are $3$ unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state,which corresponds to a $d^8$ configuration.
$CN^-$ is a strong field ligand that causes the pairing of electrons in the $3d$ orbitals.
After pairing,the $Ni^{2+}$ ion undergoes $dsp^2$ hybridization to form a square planar complex.
As there are no unpaired electrons,the complex is diamagnetic.