What is the bond order of the O_2 molecule? | vedclass

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Question

(C) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the $O_2$ molecule ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2,

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the $O_2$ molecule ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2}$ Where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons. $N_b = 10$ $N_a = 6$ $	ext{Bond Order} = \frac{10 - 6}{2} = \frac{4}{2} = 2$. Thus,the bond order of $O_2$ is $2$.

What is the bond order of the $O_2$ molecule?

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