Crystal field stabilization energy for high spin d^4 octahedral...

Key Idea:- In case of high spin complex, $\Delta_0$ is small. Thus, the energy required to pair up the fourth electron with the electrons of lower energy d-orbitals would be higher than that required to place the electrons in the higher d-orbital. Thus, pairing does not occur. For high spin $\mathrm{d}^4$ octahedral complex,
$\therefore$ Crystal field stabilisation energy$$\begin{aligned}& =(-3 \times0.4+1 \times0.6) \Delta_{{}^{\circ}} \\ & =(-1.2+0.6) \Delta_0 \\ & =-0.6 \Delta_0 \end{aligned}$$