JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Coordination Compounds

For an octahedral aqua complex the crystal field stabilisation energy is $\text{CFSE}=(-0.4n_{t_{2g}}+0.6n_{e_g}){\Delta }_0$, with water treated as weak field so the ions are high spin. $[\mathrm{Ti}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ is $d^2$ ($t_{2g}^2$), giving $-0.8{\Delta }_0$. $[\mathrm{V}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ is $d^3$ ($t_{2g}^3$), giving $-1.2{\Delta }_0$. $[\mathrm{Mn}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$ is $d^4$ high spin ($t_{2g}^3{e}_g^1$), giving $-0.6{\Delta }_0$. $[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$ is $d^5$ high spin ($t_{2g}^3{e}_g^2$), giving $0$. Hence Ti(II) pairs with $-0.8$, V(II) with $-1.2$, Mn(III) with $-0.6$ and Fe(III) with $0$.