Answer: B. Each covalent bond uses 2 bonding electrons; a double bond uses 4; a triple bond uses 6.
Ammonia, \(\text{NH}_3\): three N–H single bonds = \(3 \times 2 = 6\) electrons.
Carbon dioxide, \(\text{CO}_2\): two C=O double bonds = \(2 \times 4 = 8\) electrons.
Nitrogen, \(\text{N}_2\): one N≡N triple bond = 6 electrons.
Oxygen, \(\text{O}_2\): one O=O double bond = 4 electrons.
Row B (6, 8, 6, 4) is the only match. Lone pairs are not bonding electrons and are not counted here.
Every O Level Pure Chemistry TYS MCQ solution video, plus full Sec 3 / Sec 4 topic recordings.
Find out more