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JEE MainChemistryChemical Bonding and Molecular Structure

Match List I with List II regarding the changes observed when one electron is removed from the neutral diatomic species to form a monocation. List I (Species) List II (Observation upon ionization) (A) NO (I) Bond length increases, species remains paramagnetic (B) O ₂ (II) Bond length decreases, species becomes diamagnetic (C) N ₂ (III) Bond length increases, species becomes paramagnetic (D) B ₂ (IV) Bond length decre

Options

  1. A(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
  2. B(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  3. C(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  4. D(A)-(III), (B)-(I), (C)-(II), (D)-(IV)

Correct answer

A. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

Step-by-step solution

According to Molecular Orbital Theory, bond length is inversely proportional to bond order. (A) NO ( 15 e ^- ) has a bond order of 2.5 and is paramagnetic. Removing an electron from the ^ * antibonding orbital forms NO ^+ ( 14 e ^- ), which has a bond order of 3.0 and is diamagnetic. Bond order increases, so bond length decreases. (A) matches (II). (B) O ₂ ( 16 e ^- ) has a bond order of 2.0 and is paramagnetic. Removing an electron from the ^ * antibonding orbital forms O ₂^+ ( 15 e ^- ), which has a bond order of

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