NEETChemistryChemical Bonding and Molecular Structure
Match the chemical species in List-I with their corresponding bond order and properties in List-II. List-I List-II (A) Be ₂ (I) B.O. = 0.5, exists (B) He ₂^+ (II) B.O. = 2, paramagnetic (C) C ₂ (III) B.O. = 0, does not exist (D) O ₂ (IV) B.O. = 2, diamagnetic Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
- B(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
- C(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
- D(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Correct answer
D. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Step-by-step solution
According to Molecular Orbital Theory, Bond Order (B.O.) is calculated as 1 2 (N_b - N_a) . (A) Be ₂ (8 electrons): Configuration is (1s)^2 ^ * (1s)^2 (2s)^2 ^ * (2s)^2 . B.O. = 1 2 (4 - 4) = 0 . Since B.O. is zero, the molecule does not exist. (B) He ₂^+ (3 electrons): Configuration is (1s)^2 ^ * (1s)^1 . B.O. = 1 2 (2 - 1) = 0.5 . The species exists. (C) C ₂ (12 electrons): Configuration is (1s)^2 ^ * (1s)^2 (2s)^2 ^ * (2s)^2 (2p_x)^2 = (2p_y)^2 . B.O. = 1 2 (8 - 4) = 2 . All electrons are paired, so it is diamag