JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Coordination Compounds

The unpaired count of each central ion is fixed by its $d$-configuration and the field strength of the ligand. In $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$ iron is $\mathrm{F}{\mathrm{e}}^{3+}$ ($d^5$) and the strong-field cyanide forces pairing to $t_{2g}^5$, leaving one unpaired electron. In $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$ the same $d^5$ ion meets weak-field fluoride and stays high spin, $t_{2g}^3{e}_g^2$, with five unpaired electrons. In $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$ cobalt is $\mathrm{C}{\mathrm{o}}^{3+}$ ($d^6$) and, high spin with fluoride, is $t_{2g}^4{e}_g^2$ with four unpaired electrons. In $[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}$ manganese is $\mathrm{M}{\mathrm{n}}^{3+}$ ($d^4$) and the strong cyanide field gives $t_{2g}^4$ with two unpaired electrons. The unpaired electrons are therefore $1,5,4$ and $2$ respectively.