NEETChemistryChemical Bonding and Molecular Structure
Match List-I with List-II: List-I (Species) List-II (Highest Occupied Molecular Orbital) (A) B ₂ (I) _ 2p_z (B) N ₂ (II) ^ * _ 2p (C) O ₂ (III) ^ * _ 2p_z (D) Ne ₂ (IV) _ 2p Choose the correct answer from the options given below:
Options
- A(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
- B(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
- C(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
- D(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
Correct answer
B. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Step-by-step solution
The total number of electrons and the Highest Occupied Molecular Orbital (HOMO) for each species are as follows: B ₂ (10 electrons): The molecular orbital configuration is _ 1s ^2 ^ * _ 1s ^2 _ 2s ^2 ^ * _ 2s ^2 _ 2p_x ^1 = _ 2p_y ^1 . The HOMO is _ 2p . N ₂ (14 electrons): The molecular orbital configuration is _ 1s ^2 ^ * _ 1s ^2 _ 2s ^2 ^ * _ 2s ^2 _ 2p_x ^2 = _ 2p_y ^2 _ 2p_z ^2 . The HOMO is _ 2p_z . O ₂ (16 electrons): The molecular orbital configuration is _ 1s ^2 ^ * _ 1s ^2 _ 2s ^2 ^ * _ 2s ^2 _ 2p_z ^2 _