Class 11 NEET Chemistry Some P block Elements Questions

Concept:The question asks for the incorrect statement, so each statement about Group 13, 14 and 16 elements must be tested against known periodic trends and bonding behaviour.Explanation:Statement A (option A): Carbon has the ability to form $p\pi - p\pi$ multiple bonds with itself.This is true, as seen in $\mathrm{C}=\mathrm{C}$ (ethene) and $\mathrm{C}\equiv\mathrm{C}$ (ethyne), because carbon is small enough for effective sideways overlap of $p$-orbitals.Statement B (option B): $\mathrm{ECl}_3$ is a monomer for $\mathrm{E}=\mathrm{B}$ and a dimer for $\mathrm{E}=\mathrm{Al}$.This is true: $\mathrm{BCl}_3$ exists as a monomer because boron is too small to accommodate four large $\mathrm{Cl}$ atoms around it, whereas $\mathrm{AlCl}_3$ dimerises to $\mathrm{Al}_2\mathrm{Cl}_6$ through bridging chlorine atoms to complete the octet of aluminium.Statement C (option C): The order of catenation in Group 14 is $\mathrm{C} \gg \mathrm{Si} > \mathrm{Ge} \approx \mathrm{Sn}$.This is true, since $\mathrm{E}-\mathrm{E}$ bond strength decreases sharply down the group, giving carbon by far the greatest tendency for catenation.Statement D (option D): Oxygen exhibits only $-2$ oxidation state.This is false.Oxygen shows an oxidation state of $0$ in $\mathrm{O}_2$, $-2$ in oxides such as $\mathrm{H}_2\mathrm{O}$, $-1$ in peroxides such as $\mathrm{H}_2\mathrm{O}_2$, $-\tfrac{1}{2}$ in superoxides such as $\mathrm{KO}_2$, and even $+2$ in $\mathrm{OF}_2$.Hence oxygen does not exhibit only the $-2$ oxidation state.Answer:D (option D) — the statement that oxygen exhibits only $-2$ oxidation state is the incorrect one.