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(B) The molecular orbital configuration of $O_2$ 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,

Vedclass · Accepted Answer

$A, D$

Vedclass · Answer

(B) The molecular orbital configuration of $O_2$ 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 of $O_2 = \frac{10 - 6}{2} = 2$.When an electron is added to $O_2$ to form $O_2^-$,it enters the antibonding $\pi^*$ orbital. The new configuration has $10$ bonding and $7$ antibonding electrons. Bond order = $\frac{10 - 7}{2} = 1.5$. Thus,bond order decreases (Statement $A$ is correct).When an electron is removed from $O_2$ to form $O_2^+$,it is removed from the antibonding $\pi^*$ orbital. The new configuration has $10$ bonding and $5$ antibonding electrons. Bond order = $\frac{10 - 5}{2} = 2.5$. Since bond order increases,bond length decreases (Statement $D$ is correct).Therefore,the correct set of statements is $A$ and $D$.

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