The bond order and magnetic behaviour of O_{2 | vedclass

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Question

(A) The electronic configuration of $O_{2}^{-}$ ($17$ electrons) is: $(\sigma_{1s})^{2}(\sigma_{1s}^{*})^{2}(\sigma_{2s})^{2}(\sigma_{2s}^{*})^{2}(\si

Vedclass · Accepted Answer

$1.5$ and paramagnetic

Vedclass · Answer

(A) The electronic configuration of $O_{2}^{-}$ ($17$ electrons) is: $(\sigma_{1s})^{2}(\sigma_{1s}^{*})^{2}(\sigma_{2s})^{2}(\sigma_{2s}^{*})^{2}(\sigma_{2p_{z}})^{2}(\pi_{2p_{x}}^{2} = \pi_{2p_{y}}^{2})(\pi_{2p_{x}}^{*2} = \pi_{2p_{y}}^{*1})$.$	ext{Bond order} = \frac{N_b - N_a}{2} = \frac{10 - 7}{2} = 1.5$.Since there is one unpaired electron in the $\pi^{*}$ orbital,the ion is paramagnetic.

The bond order and magnetic behaviour of $O_{2}^{-}$ ion are,respectively:

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