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(D) Let us analyze the changes in bond order $(BO)$ and number of unpaired electrons $(n)$: $1$. $N_2 (14e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2

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$O_2^{-} 	o O_2$

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(D) Let us analyze the changes in bond order $(BO)$ and number of unpaired electrons $(n)$: $1$. $N_2 (14e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $BO = 3, n = 0$. $N_2^{-2} (16e^-): \dots \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = 2, n = 2$. (Both decrease/increase differently) $2$. $B_2 (10e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1 = \pi 2p_y^1$. $BO = 1, n = 2$. $B_2^{-} (11e^-): \dots \pi 2p_x^2 = \pi 2p_y^1$. $BO = 1.5, n = 1$. $3$. $O_2^{-} (17e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $BO = 1.5, n = 1$. $O_2^{2-} (18e^-): \dots \pi^* 2p_x^2 = \pi^* 2p_y^2$. $BO = 1, n = 0$. $4$. $O_2^{-} (17e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $BO = 1.5, n = 1$. $O_2 (16e^-): \dots \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = 2, n = 2$. In the change $O_2^{-} 	o O_2$,the bond order increases from $1.5$ to $2$ and the number of unpaired electrons increases from $1$ to $2$.

In which change do both the bond order and the number of unpaired electrons increase?

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