Chalcogen group elements are:

The periodic table is arranged in vertical columns known as groups. Group $$16$$ is traditionally called the chalcogen or oxygen family. By definition, every element lying in group $$16$$ is called a chalcogen.

Listing the members of group $$16$$ in order of increasing atomic number, we have

$$\text{O}\;(Z = 8),\;\text{S}\;(Z = 16),\;\text{Se}\;(Z = 34),\;\text{Te}\;(Z = 52),\;\text{Po}\;(Z = 84),\;\text{and}\;\text{Lv}\;(Z = 116).$$

Now we compare each option with this authentic list:

Option A: $$\text{Se},\;\text{Te},\;\text{Po}$$ All three—selenium, tellurium and polonium—are present in the list above, so every element in this option is indeed a chalcogen.

Option B: $$\text{O},\;\text{Ti},\;\text{Po}$$ While $$\text{O}$$ and $$\text{Po}$$ are chalcogens, $$\text{Ti}$$ (titanium) belongs to group $$4$$, not group $$16$$, so this set is not exclusively chalcogenic.

Option C: $$\text{Se},\;\text{Tb},\;\text{Pu}$$ Here, $$\text{Se}$$ is a chalcogen, but $$\text{Tb}$$ (terbium) is a lanthanide and $$\text{Pu}$$ (plutonium) is an actinide. Hence this option fails the requirement.

Option D: $$\text{S},\;\text{Te},\;\text{Pm}$$ Although $$\text{S}$$ and $$\text{Te}$$ are chalcogens, $$\text{Pm}$$ (promethium) is again a lanthanide, not a group $$16$$ element. Therefore the set is invalid.

Only Option A contains elements exclusively from group $$16$$ without exception.

Hence, the correct answer is Option A.