Nitrogen forms \(N_2\), phosphorus forms \(P_2\) but phosphorus is readily converted to \(P_4\) because:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(p\pi -p\pi \) bonding in phosphorus is weak
Nitrogen has a unique ability to form pπ–pπ multiple bonds with itself because of its small size and high electronegativity, so it exists as N≡N (\( \ce{N2} \)). The heavier elements of group 15 do not form pπ–pπ bonds, because their atomic orbitals are so large and diffuse that they cannot overlap effectively. A P≡P molecule (\( \ce{P2} \)) is therefore much less stable than single bonds, and phosphorus forms P4: a tetrahedron of four P atoms held by six P–P single bonds (white phosphorus).
Why the other options are incorrect
Option (1) Triple bond is present in phosphorus atoms — Incorrect. \( \ce{P4} \) has only P–P single bonds; a triple bond is what phosphorus avoids.
Option (3) \(p\pi -p\pi \) bonding in phosphorus is strong — Incorrect. If it were strong, phosphorus would stay as \( \ce{P2} \) like \( \ce{N2} \).
Option (4) Multiple bonds are easily formed in phosphorus — Incorrect. The reverse; poor overlap of the large 3p orbitals makes multiple bonds weak, so phosphorus prefers single bonds.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Anomalous Properties of Nitrogen – “… Nitrogen has unique ability to form pπ-pπ multiple bonds with itself and with other elements having small size and high electronegativity (e.g., C, O). Heavier elements of this group do not form pπ-pπ bonds as their atomic orbitals are so large and diffuse that they cannot have effective overlapping.”