The complex  ${\left[\mathrm{Cr}(\mathrm{CN}{)}_6\right]}^{4-}$ involves
 

Cr in [Cr(CN)₆]⁴⁻ has an oxidation state of +2 , which makes it a d⁴ metal. Refer to a periodic table to see that. Since it's a d⁴ metal, it means it has four d electrons to distribute in its eg and t₂g orbitals

${\mathrm{Cr}}^{2+}\left(3{\mathrm{d}}^4\right)$ ion in the given complex has ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridization