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Chemical Bonding and Molecular Structure question

2000 · Q110
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Chemical Bonding and Molecular Structure question

2000 · Q110

NEETChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Which species does not exhibit paramagnetism?
  1. A
    N2+
  2. B
    O2−-−
  3. C
    CO
  4. D
    NO
View written solutionFree

Correct answer: C

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.



(A) N2+ has 13 electrons.

Moleculer orbital configuration of N2+

= σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz1{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}σ1s2​σ1s2∗​σ2s2​σ2s2∗​π2px2​​=π2py2​​σ2pz1​​

Here 1 unpaired electrons present, so it is paramagnetic.

(B) Molecular orbital configuration of O2−O_2^ - O2−​ (17 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py1∗{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * σ1s2​σ1s2∗​σ2s2​σ2s2∗​σ2pz2​​π2px2​​=π2py2​​π2px2​∗​=π2py1​∗​

Here 1 unpaired electrons present, so it is paramagnetic.

(A) CO has 14 electrons.

Moleculer orbital configuration of CO

= σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz2{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}σ1s2​σ1s2∗​σ2s2​σ2s2∗​π2px2​​=π2py2​​σ2pz2​​

Here 0 unpaired electrons present, so it is diamagnetic.

(D) Molecular orbital configuration of NO (15 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * σ1s2​σ1s2∗​σ2s2​σ2s2∗​σ2pz2​​π2px2​​=π2py2​​π2px1​∗​=π2pyo​∗​

Here 1 unpaired electrons present, so it is paramagnetic.

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