The nature of oxide $$(TeO_2)$$ and hydride $$(TeH_2)$$ formed by Te, respectively are :

Tellurium belongs to Group 16. Its common oxide is $$TeO_2$$ (tellurium in the $$+4$$ oxidation state) and its hydride is $$TeH_2$$.

Case 1: Nature of $$TeO_2$$
• In the +4 state a chalcogen atom can either be oxidised to the +6 state or be reduced to 0/-2. • For the lighter members (S, Se) the +4 species are generally better reducing agents because they are more readily oxidised to +6. • For heavier Te, the +4 compound is more easily reduced to the elemental state; the half-reaction is $$TeO_2 + 4H^+ + 2e^- \;\rightarrow\; Te + 2H_2O \qquad E^{\circ} \approx +0.55\;V$$ A positive reduction potential means $$Te^{IV}$$ tends to accept electrons, i.e. $$TeO_2$$ behaves as an oxidising agent.

Hence $$TeO_2$$ is predominantly oxidising (and amphoteric in acid-base behaviour).

Case 2: Nature of $$TeH_2$$
• Acidity of Group 16 hydrides increases down the group because the E-H bond weakens: $$H_2O \lt H_2S \lt H_2Se \lt H_2Te$$ • Therefore $$TeH_2$$ is the most acidic hydride of the group (it can liberate $$H_2S$$ from sulfides, etc.).

Combining the two results:
  • $$TeO_2$$ - oxidising
  • $$TeH_2$$ - acidic

Option A (Oxidising and acidic) is correct.

Final answer : Option A