Multiple choice

Match the Column I molecules/molecular ions with Column II bond orders and Column III magnetic properties:

   
Molecular ion
Bond order
Magnetic character
(i) H2+ a. 2.5 x. Paramagnetic
(ii) C2 b. 3.0 y. Diamagnetic
(iii) CN- c. 0.5
(iv) O2- d. 1.0
(v) NO e. 2.0
(vi) F2 f. 1.5

  1. i - a - y, ii - d - x, iii - b - x, iv - c - y, v - f - x, vi - e - y

  2. i - e - y, ii - c - x, iii - f - x, iv - b - y, v - a - x, vi - d - y

  3. i - c - y, ii - e - x, iii - b - x, iv - f - y, v - a - y, vi - d - x

  4. i - a - x, ii - d - y, iii - b - y, iv - c - x, v - f - y, vi - e - x

  5. i - c - x, ii - e - y, iii - b - y, iv - f - x, v - a - x, vi - d - y
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

This option is correct because all the molecular ions are correctly matched with their bond order and magnetic properties. According to molecular orbital energy level diagram: for example of energy levels of molecular orbitals of O2- : σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2, π2p x 2=π2py2, π*2px2=π*2py1 Bond order = 1/2 (number of bonding electrons-number of antibonding electrons) = 1/2 (10-7) = 1.5 And due to the presence unpaired electron in π*2py1 molecular orbital, this molecule is paramagnetic.

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