JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Coordination Compounds

In an octahedral field the pairing turns on the metal $d$-count and the ligand strength. $[\mathrm{Mn}(\mathrm{CN}{)}_6{]}^{3-}$ is $\mathrm{M}{\mathrm{n}}^{3+}$, $d^4$; even with strong-field $\mathrm{C}{\mathrm{N}}^{-}$ (low spin $t_{2g}^4$) one unpaired electron remains, so it is paramagnetic. $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$ is $\mathrm{C}{\mathrm{o}}^{3+}$, $d^6$; strong-field $\mathrm{N}{\mathrm{H}}_3$ gives $t_{2g}^6{e}_g^0$ with all electrons paired, so it is diamagnetic. $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$ is $\mathrm{F}{\mathrm{e}}^{2+}$, $d^6$; strong-field $\mathrm{C}{\mathrm{N}}^{-}$ gives $t_{2g}^6$, again fully paired and diamagnetic. $[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_3{\mathrm{F}}_3]$ is $\mathrm{C}{\mathrm{o}}^{3+}$, $d^6$, but the weak-field ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{F}}^{-}$ give high spin $t_{2g}^4{e}_g^2$ with four unpaired electrons, so it is paramagnetic. The diamagnetic complexes are $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$ and $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$.