91Ó°ÊÓ

: For option a, we have the following species: \(\mathrm{H}_{2}^{+}, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2-}\). Let's analyze each one. 1. \(\mathrm{H}_{2}^{+}\): There is one valence electron in total. Fill the molecular orbitals according to the energy levels: - One electron goes into the \(\sigma_{1s}\) bonding orbital. There are more bonding electrons than anti-bonding electrons (1 > 0), so \(\mathrm{H}_{2}^{+}\) is stable. 2. \(\mathrm{H}_{2}\): There are two valence electrons in total. Fill the molecular orbitals according to the energy levels: - Both electrons go into the \(\sigma_{1s}\) bonding orbital. There are more bonding electrons than anti-bonding electrons (2 > 0), so \(\mathrm{H}_{2}\) is stable. 3. \(\mathrm{H}_{2}^{-}\): There are three valence electrons in total. Fill the molecular orbitals according to the energy levels: - Two electrons go into the \(\sigma_{1s}\) bonding orbital. - One electron goes into the \(\sigma_{1s}^{*}\) anti-bonding orbital. There are more bonding electrons than anti-bonding electrons (2 > 1), so \(\mathrm{H}_{2}^{-}\) is stable. 4. \(\mathrm{H}_{2}^{2-}\): There are four valence electrons in total. Fill the molecular orbitals according to the energy levels: - Two electrons go into the \(\sigma_{1s}\) bonding orbital. - Two electrons go into the \(\sigma_{1s}^{*}\) anti-bonding orbital. There are an equal number of bonding and anti-bonding electrons (2 = 2), so \(\mathrm{H}_{2}^{2-}\) is not stable. For option a, the stable diatomic species are \(\mathrm{H}_{2}^{+}, \mathrm{H}_{2},\) and \(\mathrm{H}_{2}^{-}\).

: For option b, we have the following species: \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\). Let's analyze each one. 1. \(\mathrm{He}_{2}^{2+}\): There are two valence electrons in total. Fill the molecular orbitals according to the energy levels: - Both electrons go into the \(\sigma_{1s}\) bonding orbital. There are more bonding electrons than anti-bonding electrons (2 > 0), so \(\mathrm{He}_{2}^{2+}\) is stable. 2. \(\mathrm{He}_{2}^{+}\): There are three valence electrons in total. Fill the molecular orbitals according to the energy levels: - Two electrons go into the \(\sigma_{1s}\) bonding orbital. - One electron goes into the \(\sigma_{1s}^{*}\) anti-bonding orbital. There are more bonding electrons than anti-bonding electrons (2 > 1), so \(\mathrm{He}_{2}^{+}\) is stable. 3. \(\mathrm{He}_{2}\): There are four valence electrons in total. Fill the molecular orbitals according to the energy levels: - Two electrons go into the \(\sigma_{1s}\) bonding orbital. - Two electrons go into the \(\sigma_{1s}^{*}\) anti-bonding orbital. There are an equal number of bonding and anti-bonding electrons (2 = 2), so \(\mathrm{He}_{2}\) is not stable. For option b, the stable diatomic species are \(\mathrm{He}_{2}^{2+}\) and \(\mathrm{He}_{2}^{+}\). In conclusion, the stable diatomic species predicted by the molecular orbital model in the given options are: \(\mathrm{H}_{2}^{+}, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{He}_{2}^{2+}\), and \(\mathrm{He}_{2}^{+}\).