d and f Block Elements Part 3

Concept:Ions are colored in aqueous solution if they have partially filled $d$-orbitals, enabling $d$‑$d$ transitions.Ions with $d^0$ or $d^{10}$ configuration are colourless.A reducing agent donates electrons and gets oxidised; $\mathrm{Eu}^{2+}$ is a strong reducing agent because it readily oxidises to $\mathrm{Eu}^{3+}$.Explanation:Statement I:Check each pair for partially filled $d$-orbitals.$[\mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}]$:$\mathrm{Sc}^{3+}$ (Z=21) has $[\mathrm{Ar}]\,3d^0$ → colourless.$\mathrm{Ti}^{3+}$ (Z=22) has $[\mathrm{Ar}]\,3d^1$ → coloured.So both are not coloured.$[\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}]$:$\mathrm{Mn}^{2+}$ has $[\mathrm{Ar}]\,3d^5$ → coloured.$\mathrm{Cr}^{2+}$ has $[\mathrm{Ar}]\,3d^4$ → coloured.Both are coloured. ✅$[\mathrm{Cu}^{2+}, \mathrm{Zn}^{2+}]$:$\mathrm{Cu}^{2+}$ has $[\mathrm{Ar}]\,3d^9$ → coloured.$\mathrm{Zn}^{2+}$ has $[\mathrm{Ar}]\,3d^{10}$ → colourless.Both not coloured.$[\mathrm{Ni}^{2+}, \mathrm{Ti}^{4+}]$:$\mathrm{Ni}^{2+}$ has $[\mathrm{Ar}]\,3d^8$ → coloured.$\mathrm{Ti}^{4+}$ has $[\mathrm{Ar}]\,3d^0$ → colourless.Both not coloured.Only 1 pair has both coloured, not 3.Hence Statement I is false.Statement II:$\mathrm{Ce}^{4+}$ is a strong oxidising agent (gets reduced to $\mathrm{Ce}^{3+}$).$\mathrm{Th}^{4+}$ is in a high oxidation state and is not a strong reducing agent.$\mathrm{Gd}^{3+}$ is a very stable common oxidation state; not strongly reducing.$\mathrm{Eu}^{2+}$ readily oxidises to $\mathrm{Eu}^{3+}$, making it the strongest reducing agent among the given species.Thus $\mathrm{Th}^{4+}$ is not the strongest reducing agent.Statement II is false.Answer:Both Statement I and Statement II are false.Correct option: B.