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NEET CHEMISTRYMedium

Which orbital does the additional electron occupy when O2O_{2} is converted to the O2O_{2}^{-} ion?

A

π\pi orbital

B

σ\sigma^{*} orbital

C

σ\sigma orbital

D

π\pi^{*} orbital

Step-by-Step Solution

The ground-state electronic configuration of the oxygen molecule (O2O_{2}), which contains 16 electrons, is (σ1s)2(σ1s)2(σ2s)2(σ2s)2(σ2pz)2(π2px2=π2py2)(π2px1=π2py1)(\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}^{1}) . In this configuration, the highest occupied molecular orbitals (HOMO) are the degenerate π\pi^{*} antibonding orbitals, each containing one unpaired electron . When O2O_{2} is converted to the superoxide ion (O2O_{2}^{-}), it gains one additional electron. According to the Aufbau principle, this incoming electron will occupy the lowest available energy state, which is one of the partially filled π\pi^{*} antibonding orbitals .

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