Which of the following pairs of species have | vedclass

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(B) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$

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$O_2^+, N_2^-$

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(B) The bond order is calculated using the formula: $	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.For $O_2^-$: Total electrons = $8 + 8 + 1 = 17$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $\frac{10 - 7}{2} = 1.5$.For $N_2^+$: Total electrons = $7 + 7 - 1 = 13$. Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $\frac{9 - 4}{2} = 2.5$.For $O_2^+$: Total electrons = $15$. Bond order = $\frac{10 - 5}{2} = 2.5$.For $N_2^-$: Total electrons = $15$. Bond order = $\frac{10 - 5}{2} = 2.5$.Comparing the options,$O_2^+$ and $N_2^-$ both have a bond order of $2.5$.

Which of the following pairs of species have the same bond order?

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