Given below are two statements: Statement I: The number of pairs...

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