COMEDK 2025 Afternoon Solved Paper

Concept:The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. The value depends on the oxidation state, geometry, and ligand field strength.Explanation:For each complex, first find the oxidation state and $d$-electron count.Complex 1: $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ Oxidation state of Fe: each CN⁻ is $-1$, total ligand charge $-6$, complex charge $-4$ ⇒ Fe = $+2$.Electronic configuration: $[\mathrm{Ar}]3d^6$.CN⁻ is a strong-field ligand, causing low-spin configuration in octahedral geometry.Low-spin $d^6$ gives all electrons paired: $t_{2g}^6$, $e_g^0$ ⇒ $n=0$.Magnetic moment: $\mu = \sqrt{0 \times (0+2)} = 0$ BM.Complex 2: $[\mathrm{MnCl}_4]^{2-}$Oxidation state of Mn: each Cl⁻ is $-1$, total ligand charge $-4$, complex charge $-2$ ⇒ Mn = $+2$.Electronic configuration: $[\mathrm{Ar}]3d^5$.Cl⁻ is a weak-field ligand, so high-spin in tetrahedral geometry (weak field always high-spin).High-spin $d^5$ has five unpaired electrons: $e^2 t_2^3$ ⇒ $n=5$.Magnetic moment: $\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM.Complex 3: $[\mathrm{CoCl}_4]^{2-}$Oxidation state of Co: each Cl⁻ is $-1$, total $-4$, complex charge $-2$ ⇒ Co = $+2$.Electronic configuration: $[\mathrm{Ar}]3d^7$.Cl⁻ is weak-field, tetrahedral geometry always weak-field, thus high-spin.High-spin $d^7$ in tetrahedral: $e^4 t_2^3$ ⇒ three unpaired electrons ⇒ $n=3$.Magnetic moment: $\mu = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ BM.Now compare the magnetic moments: $5.92 > 3.87 > 0$ BM.Thus the decreasing order is: $[\mathrm{MnCl}_4]^{2-} > [\mathrm{CoCl}_4]^{2-} > [\mathrm{Fe}(\mathrm{CN})_6]^{4-}$.Answer:The correct order is given by Option B: $\left[\mathrm{MnCl}_4\right]^{2-} > \left[\mathrm{CoCl}_4\right]^{2-} > \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$.