Which of the following species has the highes | vedclass

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(B) The bond energy is directly proportional to the bond order. The bond order can be calculated using the formula: $	ext{Bond Order} = \frac{N_b - N

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$O_{2}^{+}$

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(B) The bond energy is directly proportional to the bond order. The bond order can be 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.$(A)$ $O_{2}^{2-}$: Total electrons = $18$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^2$. Bond order = $\frac{10-8}{2} = 1$.$(B)$ $O_{2}^{+}$: Total electrons = $15$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1$. Bond order = $\frac{10-5}{2} = 2.5$.$(C)$ $O_{2}^{-}$: Total electrons = $17$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^1$. Bond order = $\frac{10-7}{2} = 1.5$.$(D)$ $O_{2}$: Total electrons = $16$. Configuration: $KK(\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^1, (\pi^* 2p_y)^1$. Bond order = $\frac{10-6}{2} = 2$.Comparing the bond orders: $O_{2}^{+} (2.5) > O_{2} (2) > O_{2}^{-} (1.5) > O_{2}^{2-} (1)$.Since $O_{2}^{+}$ has the highest bond order,it has the highest bond energy.

Which of the following species has the highest bond energy?

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