NEETChemistryChemical Bonding and Molecular Structure
Match List I with List II. List I (Species) List II (Properties) A. He ₂ I. Bond order = 0, does not exist B. C ₂ II. Bond order = 2, diamagnetic C. O ₂ III. Bond order = 2, paramagnetic D. N ₂ IV. Bond order = 3, diamagnetic Choose the correct answer from the options given below:
Options
- AA-I, B-III, C-II, D-IV
- BA-I, B-IV, C-III, D-II
- CA-I, B-II, C-III, D-IV
- DA-II, B-I, C-III, D-IV
Correct answer
C. A-I, B-II, C-III, D-IV
Step-by-step solution
According to Molecular Orbital Theory: A. He ₂ (4 electrons): Configuration is 1s² ^ * 1s² . Bond order = 2 - 2 2 = 0 . The molecule does not exist. B. C ₂ (12 electrons): Configuration is 1s² ^ * 1s² 2s² ^ * 2s² 2p_ x ² = 2p_ y ² . Bond order = 8 - 4 2 = 2 . All electrons are paired, so it is diamagnetic. C. O ₂ (16 electrons): Configuration ends with ^ * 2p_ x ¹ = ^ * 2p_ y ¹ . Bond order = 10 - 6 2 = 2 . It has two unpaired electrons, so it is paramagnetic. D. N ₂ (14 electrons): Configuration ends with 2p_ z ²