COMEDK 2025 Morning Solved Paper

Concept:Higher oxidation states of transition metals increase covalent character and acidic nature of their oxides.Explanation:Ionic compounds of $\mathrm{Sc}^{3+}$ (electron configuration $[\mathrm{Ar}]\,3d^0$) have no $d$ electrons, so colour cannot arise from $d$–$d$ transitions.$\mathrm{Cu}^{+}$ (electron configuration $[\mathrm{Ar}]\,3d^{10}$) has a fully filled $d$ subshell, so $d$–$d$ transitions are not possible.Thus option A is false.For paramagnetic nature, the number of unpaired electrons in high‑spin configurations:$\mathrm{V}^{2+}\,(3d^3)$ has 3 unpaired electrons.$\mathrm{Cr}^{2+}\,(3d^4)$ and $\mathrm{Fe}^{2+}\,(3d^6)$ both have 4 unpaired electrons.$\mathrm{Mn}^{2+}\,(3d^5)$ has 5 unpaired electrons.The correct order is $\mathrm{V}^{2+} < \mathrm{Cr}^{2+} = \mathrm{Fe}^{2+} < \mathrm{Mn}^{2+}$, so option B is false.As the oxidation number of a transition metal rises, the cation becomes highly polarising, making the compound more covalent (ionic character decreases).Higher‑oxidation‑state oxides tend to be acidic (e.g., $\mathrm{Mn}_2\mathrm{O}_7$ is acidic), while lower‑oxidation‑state oxides are basic.Thus option C is correct.Cobalt ($Z=27$) in the $+3$ oxidation state: $\mathrm{Co}^{3+}$ has the configuration $[\mathrm{Ar}]\,3d^6$, not $3d^5$. So option D is false.Answer:Option C is the correct statement.