Final answer:
The correct answer is A. C (Carbon) and Sn (Tin), as both elements belong to Group 14 (IVA) of the periodic table, which means they share similar chemical properties.
Step-by-step explanation:
The pairs of elements belonging to the same group on the periodic table are those with similar chemical properties due to having the same number and distribution of valence electrons. Looking at the options provided, let's analyze each:
- A. C (Carbon) and Sn (Tin) both belong to Group 14 (IVA).
- B. K (Potassium) is an alkali metal in Group 1 (IA), while Be (Beryllium) is an alkaline earth metal in Group 2 (IIA).
- C. O (Oxygen) is a chalcogen in Group 16 (VIA), whereas F (Fluorine) is a halogen in Group 17 (VIIA).
- D. Mn (Manganese) is a transition metal in Group 7 (VIIB), while Mo (Molybdenum) is also a transition metal, but in Group 6 (VIB).
Therefore, the correct answer is A. C and Sn, since Carbon and Tin are both members of Group 14 in the periodic table.