COMEDK 2025 Morning Solved Paper

Concept:Determine hybridization and magnetic moment from electron configuration and ligand field strength.Explanation:$\mathrm{Ni}(\mathrm{CO})_4$: $\mathrm{Ni}^0$ has $3d^{10}$ configuration. $\mathrm{CO}$ is a strong‑field ligand, forcing all electrons to pair. Geometry is tetrahedral ($\mathrm{sp}^3$). Magnetic moment $\mu = 0$ BM $\rightarrow$ matches R.$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$: $\mathrm{Ni}^{2+}$ is $d^8$; $\mathrm{CN}^-$ is strong‑field, giving square‑planar geometry ($\mathrm{dsp}^2$). All electrons paired, $\mu = 0$ BM $\rightarrow$ matches S.$\left[\mathrm{NiCl}_4\right]^{2-}$: $\mathrm{Ni}^{2+}$ is $d^8$; $\mathrm{Cl}^-$ is weak‑field, giving tetrahedral ($\mathrm{sp}^3$). Two unpaired electrons: $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$ BM $\rightarrow$ matches Q.$\left[\mathrm{MnBr}_4\right]^{2-}$: $\mathrm{Mn}^{2+}$ is $d^5$; $\mathrm{Br}^-$ is weak‑field, giving tetrahedral ($\mathrm{sp}^3$). Five unpaired electrons: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM $\rightarrow$ matches P.Answer:A = R, B = S, C = Q, D = P (Option A).