Question

alculate the bond order of O2 -​

Answer 1

Answer:

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Explanation:

answer is 2

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Answer 2

Molecular Orbital configuration of O₂⁻

$\\ O_{2}^{ - } : (\sigma 1s ^{2} ) < ( \sigma ^{*} {1s}^{2} ) < (\sigma {2s}^{2} ) < (\sigma ^{*} {2s}^{2} ) < (\pi {2p _{x}}^{2} =\pi {2p _{y} }^{2} ) < (\sigma {2p _{z}}^{2} ) < (\pi {}^{* } {2p _{x}}^{2} =\pi {}^{*} {2p _{y} }^{1} )$

Thus,

$B.O = \frac{10 - 7}{2} = \frac{3}{2}$

Therefore, Bond order is 1.5