Question

Elements in the periodic table are classified into groups as they show similarities in properties with the elements of the respective groups. Identify the element which shows anomalous behavior so that it can be placed in two different groups of the periodic table.

• A. Hydrogen (H)
• B. Gold (Au)
• C. Helium (He)
• D. Carbon (C)

Answer 1

The correct option is A Hydrogen (H)

The element hydrogen (H) can form hydrogen halides meaning, it can form $H^{+}$ ions in HF, HCl, HBr, etc. It resembles group - I elements, Li, Na, K, etc. as they form $L{i}^{+},N{a}^{+},K^{+}$, etc. in metal halides (LiCl, NaCl, KBr, etc.). So, it can be placed in the group - I along with Li, Na, K. Hydrogen also resembles halogens as it can form Hydrides i.e., it can form $H^{-}$ ions in NaH, LiH, etc. Halogens can form such halide ions $(F^{-},C{l}^{-},B{r}^{-},etc.)$ in NaF, NaCl, KBr, etc. So, hydrogen can also be placed in ${17}^{th}$ group along with F, Cl, Br, I, etc. as it resembles halogens also. This behavior of hydrogen makes it difficult to place it in a single group.