The splitting of some hydrated ions [M(H₂O)₆]²⁺ is, in cm⁻¹: V²⁺ 12400, Ni²⁺ 8500 and Mn²⁺ 7800…

Step 1. Vanadium. $\text{V}^{2+}$ is $t_{2g}^3$, so $\text{CFSE} = -1.2 \times 12400 = -14880\ \text{cm}^{-1}$. In kJ/mol that is $-14880 \times 0.01196 = -178.0\ \text{kJ/mol}$.

Step 2. Nickel. $\text{Ni}^{2+}$ is $t_{2g}^6 e_g^2$, so $\text{CFSE} = -1.2 \times 8500 = -10200\ \text{cm}^{-1} = -122.0\ \text{kJ/mol}$.

Step 3. Manganese. $\text{Mn}^{2+}$ is $d^5$. In a weak field it is $t_{2g}^3 e_g^2$, so $\text{CFSE} = (-1.2 + 1.2)\,\Delta_o = 0$. One ion has zero CFSE.