Give the electron configuration,bond order,magnetic property,and energy diagram for the beryllium $(Be_{2})$ molecule and discuss its existence.
(N/A) $Be$ $(Z=4)$. The total number of electrons in $Be_{2}$ is $8$.
Electron configuration in $MO$ for $Be_{2}$:
$(\sigma 1s)^{2}(\sigma^{*} 1s)^{2}(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}$ $OR$ $KK(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}$.
Magnetic Property: All electrons are paired,so it is diamagnetic.
Bond order $(BO) = \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (4 - 4) = 0$.
Since the bond order is zero,$Be_{2}$ is unstable and does not exist.
The energy diagram for the $Be_{2}$ molecule is shown below.
Electron configuration in $MO$ for $Be_{2}$:
$(\sigma 1s)^{2}(\sigma^{*} 1s)^{2}(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}$ $OR$ $KK(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}$.
Magnetic Property: All electrons are paired,so it is diamagnetic.
Bond order $(BO) = \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (4 - 4) = 0$.
Since the bond order is zero,$Be_{2}$ is unstable and does not exist.
The energy diagram for the $Be_{2}$ molecule is shown below.
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