Identify the element and its group trend

Tellurium, $\mathrm{Te}$, belongs to Group 16 (chalcogens).

We need the nature of:

• oxide: $\mathrm{TeO_2}$
• hydride: $\mathrm{TeH_2}$

Nature of $\mathrm{TeO_2}$

In $\mathrm{TeO_2}$, oxygen is $-2$, so Te is in the $+4$ oxidation state.

For heavier p-block elements, lower oxidation states often show both oxidising and reducing behavior, but here the standard textbook trend for Group 16 dioxides is:

• $\mathrm{SO_2}$ is mainly reducing
• down the group, the dioxides become less reducing and more oxidising

Thus $\mathrm{TeO_2}$ is taken as oxidising in character.

Nature of $\mathrm{TeH_2}$

Hydrides of Group 16 are: $\mathrm{H_2O,\ H_2S,\ H_2Se,\ H_2Te}$

Their acidic character increases down the group because the $\mathrm{E-H}$ bond weakens down the group, making release of $\mathrm{H^+}$ easier.

Therefore: $\mathrm{H_2O < H_2S < H_2Se < H_2Te}$ in acidic strength.

So $\mathrm{TeH_2}$ (i.e. $\mathrm{H_2Te}$) is acidic.

Match with the options

• Oxide $\mathrm{TeO_2}$: oxidising
• Hydride $\mathrm{TeH_2}$: acidic

Hence the correct option is: $\boxed{\text{C: Oxidising and acidic}}$

Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.