COMEDK 2025 Evening Solved Paper

Concept:The magnetic moment of a complex is determined by the number of unpaired electrons ($n$), given by $\mu = \sqrt{n(n+2)}$.Explanation:Find the oxidation state and $d$-electron count for each iron ion.For $\left[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}$: Fe is $+2$, $d^6$; $\mathrm{H}_2\mathrm{O}$ is a weak‑field ligand → high‑spin, $t_{2g}^4 e_g^2$, $n=4$.For $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: Fe is $+2$, $d^6$; $\mathrm{CN}^-$ is a strong‑field ligand → low‑spin, $t_{2g}^6 e_g^0$, $n=0$.For $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$: Fe is $+3$, $d^5$; strong‑field → low‑spin, $t_{2g}^5 e_g^0$, $n=1$.For $\left[\mathrm{FeF}_6\right]^{3-}$: Fe is $+3$, $d^5$; $\mathrm{F}^-$ is a weak‑field ligand → high‑spin, $t_{2g}^3 e_g^2$, $n=5$.Increasing order of $n$ is $0 < 1 < 4 < 5$, matching the order $\mathrm{B} < \mathrm{C} < \mathrm{A} < \mathrm{D}$.Since magnetic moment increases with $n$, the same order holds.Answer:Option B: $\mathrm{B} < \mathrm{C} < \mathrm{A} < \mathrm{D}$