JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Coordination Compounds

Crystal field stabilisation energy follows $(-0.4n_{t_{2g}}+0.6n_{e_g}){\Delta }_0$. In $[\mathrm{Cu}(\mathrm{NH}{}_3){}_6{]}^{2+}$, $\mathrm{Cu}{}^{2+}$ is $d^9$ with $t_{2g}^6{e}_g^3$, giving $-0.6{\Delta }_0$. In $[\mathrm{Ti}(\mathrm{H}{}_2\mathrm{O}){}_6{]}^{3+}$, $\mathrm{Ti}{}^{3+}$ is $d^1$ with $t_{2g}^1$, giving $-0.4{\Delta }_0$. In $[\mathrm{Fe}(\mathrm{CN}){}_6{]}^{3-}$, $\mathrm{Fe}{}^{3+}$ is $d^5$ and, with the strong-field cyanide, is low spin $t_{2g}^5$, giving $-2.0{\Delta }_0$. In $[\mathrm{NiF}{}_6{]}^{4-}$, $\mathrm{Ni}{}^{2+}$ is $d^8$ with $t_{2g}^6{e}_g^2$, giving $-1.2{\Delta }_0$. Therefore the hexaamminecopper ion pairs with $-0.6$, the hexaaquatitanium ion with $-0.4$, the hexacyanoferrate ion with $-2.0$, and the hexafluoronickelate ion with $-1.2$.