COMEDK 2025 Afternoon Solved Paper

Concept:The question asks to identify the two incorrect statements among four given about lanthanoids.Explanation:Statement A: $\mathrm{Ce}^{4+}$ has no unpaired electrons ($[\mathrm{Xe}]\,4f^0$), so it is diamagnetic. $\mathrm{Sm}^{3+}$ has five unpaired electrons ($[\mathrm{Xe}]\,4f^5$), so it is paramagnetic. Hence statement A is correct.Statement B: Due to lanthanide contraction, the atomic sizes of 5d transition metals (atomic numbers >71) are very close to those of the 4d elements directly above them in the same group. Hence statement B is correct.Statement C: Lanthanoids react with hot water to produce $\mathrm{H}_2$ gas, not $\mathrm{O}_2$, and their hydroxides $\mathrm{Ln}(\mathrm{OH})_3$ are sparingly soluble, not water soluble. The correct reaction is: $$2\mathrm{Ln} + 6\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Ln}(\mathrm{OH})_3 + 3\mathrm{H}_2$$ Thus statement C is incorrect.Statement D: The general electronic configuration of lanthanoids is not strictly $(n-2)f^{1-14}5d^0 6s^2$ for $n=6$. Many lanthanoids (e.g., La, Ce, Gd, Lu) have a $5d^1$ electron. For example: La: $[\mathrm{Xe}]\,5d^1 6s^2$, Ce: $[\mathrm{Xe}]\,4f^1 5d^1 6s^2$, Gd: $[\mathrm{Xe}]\,4f^7 5d^1 6s^2$. Hence statement D is incorrect.Therefore, the incorrect statements are C and D.Answer:Option A (C & D)