JEE Main 2026 — d and f Block Elements Question with Solution

For Statement I:
The electronic configurations of the given transition metal ions are:
${\text{Ti}}^{4+}:3d^0$ (colourless due to absence of unpaired electrons)
${\text{V}}^{2+}:3d^3$ (coloured)
${\text{Mn}}^{2+}:3d^5$ (coloured)
${\text{Fe}}^{3+}:3d^5$ (coloured)
${\text{Cr}}^{2+}:3d^4$ (coloured)

The pairs in which both ions are coloured are $[{\text{V}}^{2+},{\text{Mn}}^{2+}]$, $[{\text{Mn}}^{2+},{\text{Fe}}^{3+}]$, and $[{\text{V}}^{2+},{\text{Cr}}^{2+}]$. The pair $[{\text{Ti}}^{4+},{\text{V}}^{2+}]$ contains ${\text{Ti}}^{4+}$ which is colourless. Thus, there are exactly $3$ pairs where both ions are coloured. Statement I is correct.

For Statement II:
The electronic configurations of the given f-block ions are:
${\text{La}}^{3+}:[\text{Xe}]4f^0$ (diamagnetic)
${\text{Yb}}^{2+}:[\text{Xe}]4f^{14}$ (diamagnetic)
${\text{Lu}}^{3+}:[\text{Xe}]4f^{14}$ (diamagnetic)
${\text{Ce}}^{4+}:[\text{Xe}]4f^0$ (diamagnetic)
${\text{Ac}}^{3+}:[\text{Rn}]5f^0$ (diamagnetic)
${\text{Lr}}^{3+}:[\text{Rn}]5f^{14}$ (diamagnetic)

All the given ions have either an $f^0$ or $f^{14}$ configuration, meaning they have no unpaired electrons and are diamagnetic. Therefore, in all $3$ pairs, both ions are diamagnetic. Statement II is correct.

Both Statement I and Statement II are correct.

Answer: Both Statement I and Statement II are correct