JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Coordination Compounds

With four different ligands the square-planar $[\mathrm{Pd}(\mathrm{F})(\mathrm{Cl})(\mathrm{Br})(\mathrm{I}){]}^{2-}$ has three geometrical isomers, so $n=3$ and the second complex is $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$. Here iron is $\mathrm{F}{\mathrm{e}}^{3+}$ with a $3d^5$ configuration, and the strong-field $\mathrm{C}{\mathrm{N}}^{-}$ forces a low-spin arrangement $t_{2g}^5{e}_g^0$ with just one unpaired electron. The spin-only moment is $\sqrt{1(1+2)}=1.73$ BM, and the crystal-field stabilisation energy is $5\times (-0.4{\Delta }_0)=-2.0{\Delta }_0$.