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Chemical Bonding and Molecular Structure question

2023 · 24 Jan · Shift 2 · Q3
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  5. /2023 · 24 Jan · Shift 2 · Q3

Chemical Bonding and Molecular Structure question

2023 · 24 Jan · Shift 2 · Q3

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : N2;N2+;O2;O2+\mathrm{{N_2};N_2^ + ;{O_2};O_2^ + }N2​;N2+​;O2​;O2+​ ?
  1. A
    0, 1, 0, 1
  2. B
    2, 1, 0, 1
  3. C
    0, 1, 2, 1
  4. D
    2, 1, 2, 1
View written solutionFree

Correct answer: C

  1. Write the MO ordering

For second-period homonuclear diatomic molecules:

  • For B2,C2,N2\mathrm{B_2, C_2, N_2}B2​,C2​,N2​:
    σ(2s)<σ∗(2s)<π(2px)=π(2py)<σ(2pz)\sigma(2s)<\sigma^*(2s)<\pi(2p_x)=\pi(2p_y)<\sigma(2p_z)σ(2s)<σ∗(2s)<π(2px​)=π(2py​)<σ(2pz​)
  • For O2,F2\mathrm{O_2, F_2}O2​,F2​:
    σ(2s)<σ∗(2s)<σ(2pz)<π(2px)=π(2py)<π∗(2px)=π∗(2py)\sigma(2s)<\sigma^*(2s)<\sigma(2p_z)<\pi(2p_x)=\pi(2p_y)<\pi^*(2p_x)=\pi^*(2p_y)σ(2s)<σ∗(2s)<σ(2pz​)<π(2px​)=π(2py​)<π∗(2px​)=π∗(2py​)

We now find the highest occupied molecular orbital (HOMO) in each species and count its unpaired electron(s).


  1. N2\mathrm{N_2}N2​

Each N has 7 electrons, so N2\mathrm{N_2}N2​ has 141414 electrons total.

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

The HOMO is σ(2pz)\sigma(2p_z)σ(2pz​), which contains 2 paired electrons.

So, number of unpaired electrons in HOMO of N2\mathrm{N_2}N2​ is: 000


  1. N2+\mathrm{N_2^+}N2+​

Removing one electron from N2\mathrm{N_2}N2​ removes it from the HOMO, i.e. from σ(2pz)\sigma(2p_z)σ(2pz​).

So configuration becomes: σ(2pz)1\sigma(2p_z)^1σ(2pz​)1 for the HOMO.

Hence the HOMO has one unpaired electron.

So, number of unpaired electrons in HOMO of N2+\mathrm{N_2^+}N2+​ is: 111


  1. O2\mathrm{O_2}O2​

Each O has 8 electrons, so O2\mathrm{O_2}O2​ has 161616 electrons total.

For oxygen, MO order is: σ(2s)<σ∗(2s)<σ(2pz)<π(2px)=π(2py)<π∗(2px)=π∗(2py)\sigma(2s)<\sigma^*(2s)<\sigma(2p_z)<\pi(2p_x)=\pi(2p_y)<\pi^*(2p_x)=\pi^*(2p_y)σ(2s)<σ∗(2s)<σ(2pz​)<π(2px​)=π(2py​)<π∗(2px​)=π∗(2py​)

Configuration: σ(1s)2 σ∗(1s)2 σ(2s)2 σ∗(2s)2 σ(2pz)2 π(2px)2 π(2py)2 π∗(2px)1 π∗(2py)1\sigma(1s)^2\,\sigma^*(1s)^2\,\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(1s)2σ∗(1s)2σ(2s)2σ∗(2s)2σ(2pz​)2π(2px​)2π(2py​)2π∗(2px​)1π∗(2py​)1

The HOMO consists of the degenerate orbitals π∗(2px)\pi^*(2p_x)π∗(2px​) and π∗(2py)\pi^*(2p_y)π∗(2py​), each with one electron.

Thus, in the highest occupied molecular orbital level, there are: 222 unpaired electrons.


  1. O2+\mathrm{O_2^+}O2+​

Removing one electron from O2\mathrm{O_2}O2​ removes it from the HOMO π∗\pi^*π∗ level.

So configuration becomes: π∗(2p)3\pi^*(2p)^3π∗(2p)3 spread over two degenerate π∗\pi^*π∗ orbitals. One orbital has a pair and the other has one single electron.

Thus, number of unpaired electrons in HOMO of O2+\mathrm{O_2^+}O2+​ is: 111


  1. Final sequence

For N2; N2+; O2; O2+\mathrm{N_2;\ N_2^+;\ O_2;\ O_2^+}N2​; N2+​; O2​; O2+​

the number of unpaired electrons in the HOMO are: 0, 1, 2, 10,\ 1,\ 2,\ 10, 1, 2, 1

This matches Option C.


  1. Comparison with stored answer

Stored correct answer: C
Derived answer: C

Hence, the derived answer agrees with the stored correct answer.

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