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Chemical Bonding and Molecular Structure question

2021 · 26 Aug · Shift 2 · Q2
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Chemical Bonding and Molecular Structure question

2021 · 26 Aug · Shift 2 · Q2

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
The bond order and magnetic behaviour of O2−O_2^ -O2−​ ion are respectively :
  1. A
    1.5 and paramagnetic
  2. B
    1.5 and diamagnetic
  3. C
    2 and diamagnetic
  4. D
    1 and paramagnetic
View written solutionFree

Correct answer: A

  1. Write the molecular orbital configuration of O2O_2O2​

For O2O_2O2​ and related species, the valence MO filling is:

σ(2s)2 σ∗(2s)2 σ(2pz)2 π(2px)2=π(2py)2 π∗(2px)1 π∗(2py)1\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2=\pi(2p_y)^2\,\pi^*(2p_x)^1\,\pi^*(2p_y)^1σ(2s)2σ∗(2s)2σ(2pz​)2π(2px​)2=π(2py​)2π∗(2px​)1π∗(2py​)1

So, in O2O_2O2​, there are two unpaired electrons in the degenerate π∗\pi^*π∗ orbitals.

  1. Form O2−O_2^-O2−​ by adding one electron

The ion O2−O_2^-O2−​ has one extra electron compared to O2O_2O2​. That extra electron goes into one of the antibonding π∗\pi^*π∗ orbitals.

Hence for O2−O_2^-O2−​, the antibonding π∗\pi^*π∗ electrons become 333 in total.

  1. Calculate bond order

Bond order is:

Bond order=Nb−Na2\text{Bond order} = \frac{N_b - N_a}{2}Bond order=2Nb​−Na​​

where:

  • NbN_bNb​ = number of bonding electrons
  • NaN_aNa​ = number of antibonding electrons

For O2−O_2^-O2−​ (considering valence MOs):

  • Bonding electrons: 888
  • Antibonding electrons: 555

Therefore,

Bond order=8−52=32=1.5\text{Bond order} = \frac{8-5}{2} = \frac{3}{2} = 1.5Bond order=28−5​=23​=1.5
  1. Determine magnetic behaviour

In O2−O_2^-O2−​, the two π∗\pi^*π∗ orbitals contain 333 electrons total. This means one orbital has a pair and the other has one unpaired electron.

So, O2−O_2^-O2−​ has one unpaired electron, hence it is paramagnetic.

  1. Check options
  • A: 1.51.51.5 and paramagnetic ✅
  • B: 1.51.51.5 and diamagnetic ❌
  • C: 222 and diamagnetic ❌
  • D: 111 and paramagnetic ❌

Therefore, the correct answer is A.

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