Identify the incorrect statement from the following:

Year	2026	Total
Questions	2	2

Year

2026

Total

Questions

Solution & Explanation

Core Logic

(1) Phosphorus and Arsenic possess vacant d-orbitals. They can act as $π$ $\pi$-acceptor ligands, forming $dπ-dπ$ $d\pi-d\pi$ backbonds with transition metals. (Correct) (2) Nitrogen (Period 2) does NOT possess vacant d-orbitals in its valence shell. Oxygen also does not have d-orbitals. Therefore, Nitrogen cannot form a $dπ-pπ$ $d\pi-p\pi$ bond with oxygen. (Incorrect) (3) Nitrogen forms strong $pπ-pπ$ $p\pi-p\pi$ multiple bonds with itself (e.g., in $N \equiv N$ $N \equiv N$). (Correct) (4) Phosphorus, arsenic, and antimony exhibit catenation (e.g., $P₄$ $P_4$), although to a lesser extent than carbon. (Correct)

Step 1: Final Conclusion

Statement (2) is incorrect because nitrogen lacks d-orbitals.

Pattern Recognition

Period 2 elements (Li, Be, B, C, N, O, F) NEVER form $dπ-pπ$ $d\pi-p\pi$ bonds using their own d-orbitals because they do not have accessible 2d orbitals.

Chapter Mix

Class 12 Chemistry: The p-Block Elements Class 11 Chemistry: Chemical Bonding and Molecular Structure

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Q62 neet_2026_03_may_morning Anomalous Properties of Elements

Identify the incorrect statement from the following:

A. Carbon has the ability to form $pπ - pπ$ $p\pi - p\pi$ multiple bond with itself.
B. ECl3 (E = B and Al) is a monomer when E = B and a dimer when E = Al.
C. Oxygen exhibits only -2 oxidation state.
D. The order of catenation property of Group 14 elements is C >> Si > Ge \approx Sn.

Solution

Core Logic

(1) Carbon can form $pπ - pπ$ $p\pi - p\pi$ multiple bonds with itself. It is observed in $C=C$ $C=C$ and $C \equiv C$ $C \equiv C$. (Statement is correct) (2) $ECl₃$ $ECl_3$ (E = B and Al): $BCl₃$ $BCl_3$ does not form a dimer due to small size of Boron. $AlCl₃$ $AlCl_3$ forms a dimer $Al₂Cl₆$ $Al_2Cl_6$ to complete its octet. (Statement is correct) (3) Oxygen exhibits an oxidation state of 0 in $O₂$ $O_2$, -1 in peroxides (e.g., $H₂O₂$ $H_2O_2$), -1/2 in superoxides (e.g., $KO₂$ $KO_2$), and +2 in $OF₂$ $OF_2$. Therefore, stating it exhibits only -2 is incorrect. (4) Catenation property of group 14 is highest for Carbon due to strong C-C bonds: $C gg Si > Ge \approx Sn$ $C \gg Si > Ge \approx Sn$. (Statement is correct)

Step 1: Final Conclusion

Statement (3) is incorrect because Oxygen exhibits multiple oxidation states depending on the compound.

Pattern Recognition

Absolute words like 'only' in oxidation state statements for p-block elements are often false. Oxygen in peroxides, superoxides, and fluorides deviates from the standard -2.

Chapter Mix

Class 11 Chemistry: The p-Block Elements Class 11 Chemistry: Redox Reactions

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More The p-Block Elements Previous-Year Questions

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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Identify the incorrect statement from the following:

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More The p-Block Elements Previous-Year Questions

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

More The p-Block Elements Questions — neet_2026_03_may_morning

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