What is the bond order of the Be_{2} molecule | vedclass

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(C) The electronic configuration of the $Be_{2}$ molecule ($8$ electrons) is: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2}$.Here

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(C) The electronic configuration of the $Be_{2}$ molecule ($8$ electrons) is: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2}$.Here,the number of bonding electrons $(N_{b})$ is $4$ and the number of antibonding electrons $(N_{a})$ is $4$.The bond order is calculated as: $	ext{Bond Order} = \frac{N_{b} - N_{a}}{2} = \frac{4 - 4}{2} = 0$.Since the bond order is $0$,the $Be_{2}$ molecule does not exist.

What is the bond order of the $Be_{2}$ molecule?

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