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Q.Show by drawing its molecular orbital diagram why O₂ is paramagnetic. OR

(i) Draw the canonicals of CO₃²⁻.
(ii) Why is boiling point of H₂O greater than that of H₂S? [2 + 1]
West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Subjective· 3mImportance★★★★★
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Figure — Molecular-orbital energy-level diagram of O2 (16 electrons). Left and right columns are the 2s/2p at
Figure — Molecular-orbital energy-level diagram of O2 (16 electrons). Left and right columns are the 2s/2p at

Molecular orbital theory correctly predicts O2's paramagnetism — something Lewis/VSEPR structures cannot explain — because filling the degenerate antibonding π∗\pi^* orbitals leaves two electrons unpaired.

Building the molecular orbital diagram of O2O_2 (16 electrons total) by filling MOs in increasing energy order:

σ1s2 σ1s∗2 σ2s2 σ2s∗2 σ2pz2 π2px2=π2py2 π2px∗1=π2py∗1\sigma_{1s}^2\ \sigma^{*2}_{1s}\ \sigma_{2s}^2\ \sigma^{*2}_{2s}\ \sigma_{2p_z}^2\ \pi_{2p_x}^2 = \pi_{2p_y}^2\ \pi^{*1}_{2p_x} = \pi^{*1}_{2p_y}

The key feature is the last two electrons: they must go into the doubly-degenerate antibonding π2px∗\pi^*_{2p_x} and π2py∗\pi^*_{2p_y} orbitals. By Hund's rule, these two orbitals of equal energy are each singly occupied first, with parallel spins, before any pairing occurs — leaving two unpaired electrons.

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