Give the electron configuration,bond order,magnetic property,and $MO$ energy diagram for the Helium $(He_{2})$ molecule.
(N/A) $He$ $(Z=2)$,so the total number of electrons in $He_{2} = 4$.
Electron configuration in $MO$ for $He_{2}$: $(\sigma_{1s})^{2}(\sigma_{1s}^{*})^{2}$.
All electrons are paired in $He_{2}$,so it is diamagnetic.
Bond order $= \frac{1}{2}(N_{b} - N_{a}) = \frac{1}{2}(2 - 2) = 0$.
Since the bond order in $He_{2}$ is zero,it is unstable and does not exist.
The $MO$ energy diagram of $He_{2}$ is shown below.
Electron configuration in $MO$ for $He_{2}$: $(\sigma_{1s})^{2}(\sigma_{1s}^{*})^{2}$.
All electrons are paired in $He_{2}$,so it is diamagnetic.
Bond order $= \frac{1}{2}(N_{b} - N_{a}) = \frac{1}{2}(2 - 2) = 0$.
Since the bond order in $He_{2}$ is zero,it is unstable and does not exist.
The $MO$ energy diagram of $He_{2}$ is shown below.
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