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(A) 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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$H_2^+$ and $He_2^-$

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(A) 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.SpeciesElectronic ConfigurationBond Order$H_2^+$$\sigma 1s^1$$(1-0)/2 = 0.5$$He_2^-$$\sigma 1s^2, \sigma^* 1s^1$$(2-1)/2 = 0.5$$He_2^{2-}$$\sigma 1s^2, \sigma^* 1s^2$$(2-2)/2 = 0$Thus,$H_2^+$ and $He_2^-$ have the same bond order of $0.5$.

Which of the following species have the same bond order: $H_2^+$,$He_2^-$,and $He_2^{2-}$?

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