For a ${\mathrm{d}}^4$ metal ion in an octahedral field, the correct electronic configuration is:

When ${∆}_{\circ }<\mathrm{P},$ the electrons will remain unpaired in t_2g and e_g orbitals because energy gap between the orbitals is small. This lead to the formation of a high spin complex. Hence, electronic arrangement becomes ${\mathrm{t}}_{2\mathrm{g}}^3{\mathrm{e}}_{\mathrm{g}}^1.$