JEE Main 2026 — Chemistry Inorganic Chemistry

For Statement I:

The electronic configurations of the given transition metal ions are:

$\text{Ti}^{4+} : 3d^0$Ti4+:3d0 (colourless due to absence of unpaired electrons)

$\text{V}^{2+} : 3d^3$V2+:3d3 (coloured)

$\text{Mn}^{2+} : 3d^5$Mn2+:3d5 (coloured)

$\text{Fe}^{3+} : 3d^5$Fe3+:3d5 (coloured)

$\text{Cr}^{2+} : 3d^4$Cr2+:3d4 (coloured)

The pairs in which both ions are coloured are $[\text{V}^{2+}, \text{Mn}^{2+}]$[V2+,Mn2+], $[\text{Mn}^{2+}, \text{Fe}^{3+}]$[Mn2+,Fe3+], and $[\text{V}^{2+}, \text{Cr}^{2+}]$[V2+,Cr2+]. The pair $[\text{Ti}^{4+}, \text{V}^{2+}]$[Ti4+,V2+] contains $\text{Ti}^{4+}$Ti4+ which is colourless. Thus, there are exactly $3$3 pairs where both ions are coloured. Statement I is correct.

For Statement II:

The electronic configurations of the given f-block ions are:

$\text{La}^{3+} : [\text{Xe}] 4f^0$La3+:[Xe]4f0 (diamagnetic)

$\text{Yb}^{2+} : [\text{Xe}] 4f^{14}$Yb2+:[Xe]4f14 (diamagnetic)

$\text{Lu}^{3+} : [\text{Xe}] 4f^{14}$Lu3+:[Xe]4f14 (diamagnetic)

$\text{Ce}^{4+} : [\text{Xe}] 4f^0$Ce4+:[Xe]4f0 (diamagnetic)

$\text{Ac}^{3+} : [\text{Rn}] 5f^0$Ac3+:[Rn]5f0 (diamagnetic)

$\text{Lr}^{3+} : [\text{Rn}] 5f^{14}$Lr3+:[Rn]5f14 (diamagnetic)

All the given ions have either an $f^0$f0 or $f^{14}$f14 configuration, meaning they have no unpaired electrons and are diamagnetic. Therefore, in all $3$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