### Core Logic Let's map the electronic configurations and count the unpaired d-orbital electrons for each option: * For pair (1): V^2+ implies [Ar] 3d^3 4s^0 implies 3 text unpaired electrons$V^{2+} \implies [Ar] 3d^3 4s^0 \implies 3 \text{ unpaired electrons}$ Co^2+ implies [Ar] 3d^7 4s^0 implies t_2g^5 e_g^2 implies 3 text unpaired electrons$Co^{2+} \implies [Ar] 3d^7 4s^0 \implies t_{2g}^5 e_g^2 \implies 3 \text{ unpaired electrons}$ Both ions contain exactly 3 unpaired electrons. * For other ions: Ti^2+ implies [Ar] 3d^2 implies 2 text unpaired e-, quad Fe^3+ implies [Ar] 3d^5 implies 5 text unpaired e-$Ti^{2+} \implies [Ar] 3d^2 \implies 2 \text{ unpaired e-}, \quad Fe^{3+} \implies [Ar] 3d^5 \implies 5 \text{ unpaired e-}$ Cr^2+ implies [Ar] 3d^4 implies 4 text unpaired e-, quad Ti^3+ implies [Ar] 3d^1 implies 1 text unpaired e-$Cr^{2+} \implies [Ar] 3d^4 \implies 4 \text{ unpaired e-}, \quad Ti^{3+} \implies [Ar] 3d^1 \implies 1 \text{ unpaired e-}$ Mn^2+ implies [Ar] 3d^5 implies 5 text unpaired e-$Mn^{2+} \implies [Ar] 3d^5 \implies 5 \text{ unpaired e-}$ ### Pattern Recognition D-orbital counts follow a predictable symmetry: a 3d^n$3d^n$ system contains the same number of unpaired electrons as a 3d^10-n$3d^{10-n}$ system under high-spin conditions. This explains why 3d^3$3d^3$ (V^2+$V^{2+}$) and 3d^7$3d^7$ (Co^2+$Co^{2+}$) match perfectly with 3 unpaired electrons each. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements