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Given,${{He}}_{{2}}^{{{ + 2}}}$ ion is more stable than ${{H}}{{{e}}_{{2}}}$ molecule. Explain.

Last updated date: 19th Jul 2024
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Hint: Helium is a colorless, odorless, tasteless, non-toxic, inert monatomic gas that belongs to the noble gas section of the periodic table. It has the lowest boiling point of all the elements. In the observable universe, helium is the second lightest and most abundant element.

Complete answer:
There are only two valence electrons in the ${{He}}_{{2}}^{{{ + 2}}}$ ion (two from each He atom minus two for the $ + 2$ charge). The molecular orbital diagram can now be filled in.
The two electrons occupy the bonding $\left( {{{\sigma 1s}}} \right)$ orbital, which is the lowest energy molecular orbital, resulting in an $\left( {{{\sigma 1s}}} \right)$ $2$ electron configuration. So, the bond order is
${{ = }}\,\dfrac{{{1}}}{{{2}}}\left[ {{{{N}}_{{b}}}\,{{ - }}\,{{{N}}_{{a}}}} \right]\,\,{{ = }}\,\dfrac{{\left( {{{2}}\,{{ - }}\,{{0}}} \right)}}{{{2}}}\,{{ = }}\,\dfrac{{{2}}}{{{2}}}\,{{ = }}\,{{1}}$
As a result, ${{He}}_{{2}}^{{{ + 2}}}$ is expected to contain a single ${{He}}\,\,{{ - }}\,{{He}}$ bond. As a result, it should be a stable species.
Let us take a look at the ${{H}}{{{e}}_{{2}}}$ molecules, which are made up of two ${{He}}$ atoms with the ${{1}}{{{s}}^{{2}}}$ electronic configuration.
Both the ${{\sigma 1s}}$ bonding and $\left( {{{\sigma *1s}}} \right)$ antibonding orbitals must contain two electrons with a total of four valence electrons. This results in an electronic configuration of \[{\left( {{{\sigma 1s}}} \right)^{{2}}}{\left( {{{\sigma 1s}}} \right)^{{2}}}\].
So, the bond order is
Bond order ${{ = }}\,\dfrac{{{1}}}{{{2}}}\left[ {{{{N}}_{{b}}}\,{{ - }}\,{{{N}}_{{a}}}} \right]\,\,{{ = }}\,\dfrac{{\left( {{{2}}\,{{ - }}\,2} \right)}}{{{2}}}\,{{ = }}\,0$
The ${{H}}{{{e}}_{{2}}}$ molecule has no net covalent bond, indicating that it is not a stable species.
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