The total number of sigma bonds present in P4O10 are :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 16
Every single bond is one σ bond; a double bond is one σ bond and one π bond. So count the bonds in the structure and take one σ bond from each of them.
In \( \ce{P4O10} \), the four phosphorus atoms sit at the corners of a tetrahedron:
Six oxygen atoms bridge the six edges of the tetrahedron (P–O–P). Each bridging oxygen forms two P–O single bonds: 6 × 2 = 12 σ bonds.
Each phosphorus also carries one terminal oxygen through a P=O double bond: 4 P=O bonds give 4 σ bonds (and 4 π bonds).
\( \displaystyle \text{Number of }\sigma\text{ bonds}=12+4=16 \)
So \( \ce{P4O10} \) has 16 σ bonds (and 4 π 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.