Check statement A

For group 16 elements $\mathrm{S, Se, Te, Po}$, common oxides of type $\mathrm{EO_2}$ and $\mathrm{EO_3}$ are known.

• $\mathrm{SO_2, SO_3}$
• $\mathrm{SeO_2, SeO_3}$
• $\mathrm{TeO_2, TeO_3}$
• $\mathrm{PoO_2, PoO_3}$

Now, regarding acidic/basic character:

• $\mathrm{SO_2}$ and $\mathrm{SO_3}$ are acidic.
• $\mathrm{SeO_2}$ and $\mathrm{SeO_3}$ are acidic.
• $\mathrm{TeO_2}$ is amphoteric, but in standard p-block trend questions, the dioxides/trioxides are taken as acidic oxides with acidity decreasing down the group.

Hence, in the intended JEE context, statement A is taken as correct.

Check statement B

• In $\mathrm{TeO_2}$, Te is in the $+4$ oxidation state and can be further oxidized to $+6$, but it can also be reduced. Compounds in intermediate oxidation states often show oxidising behavior.
• $\mathrm{SO_2}$ is well known as a reducing agent because sulfur in $+4$ state can be oxidized to $+6$.

So statement B is correct.

Check statement C

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

Reducing character increases down the group because the $\mathrm{E-H}$ bond weakens.

Therefore, $\mathrm{H_2S < H_2Se < H_2Te}$ in reducing power.

The statement says reducing property decreases from $\mathrm{H_2S}$ to $\mathrm{H_2Te}$, which is wrong.

Hence, statement C is false.

Check statement D

Ozone is $\mathrm{O_3}$.

Total valence electrons: $3 \times 6 = 18$

In its Lewis structure, there is one single bond and one double bond (resonance structures), leaving lone pairs distributed as:

• singly bonded terminal O: 3 lone pairs
• central O: 1 lone pair
• doubly bonded terminal O: 2 lone pairs

Total lone pairs: $3 + 1 + 2 = 6$

So the statement that ozone contains five lone pairs is false.

Hence, statement D is false.

Select the correct option

Correct statements are A and B only.

Therefore, the correct option is: $\boxed{\text{A}}$