Question

Would you expect \(\mathrm{He}_{2} \oplus\) to be more stable than \(\mathrm{He}_{2}\) ? Than \(\mathrm{He}\) ?

Short answer

In conclusion, \(\mathrm{He}_{2} \oplus\) is more stable than \(\mathrm{He}_{2}\) due to bond formation. However, it is difficult to conclusively say if it is more stable than \(\mathrm{He}\) without further details such as bond dissociation energy or other factors that contribute to their stabilities.

Step-by-step solution

Identify electronic configurations

To compare the stabilities of these species, we first need to determine their electronic configurations. Here are the electronic configurations of the ground state: - \(\mathrm{He}\): 1s\(^2\) - \(\mathrm{He}_{2}\): (1s\(_{A}$$)^2\) (1s\(_{B}$$)^2\) where A and B represent the two helium atoms. - \(\mathrm{He}_{2} \oplus\): (1s\(_{A}$$)^2\) (1s\(_{B}$$)^1\) (one electron is removed)

Analyze bond formation

Now, let's analyze the bond formation for \(\mathrm{He}_{2}\) and \(\mathrm{He}_{2} \oplus\): - \(\mathrm{He}_{2}\): There are four electrons, and since the 1s orbital can only accommodate two electrons, there are no available orbitals for bond formation, making \(\mathrm{He}_{2}\) unstable and less likely to exist. - \(\mathrm{He}_{2} \oplus\): There are three electrons. The two electrons from 1s\(_{A}\) and one electron from 1s\(_{B}\) can occupy molecular orbitals, promoting bond formation and making \(\mathrm{He}_{2} \oplus\) more stable than \(\mathrm{He}_{2}\).

Compare with \(\mathrm{He}\) stability

Finally, let's compare the stability of \(\mathrm{He}_{2} \oplus\) and \(\mathrm{He}\). Because \(\mathrm{He}\) is a noble gas and has a filled 1s orbital, it is relatively stable. However, it is necessary to consider the bond dissociation energy to compare their stability: - \(\mathrm{He}_{2} \oplus\): The bond dissociation energy (energy required to break the bond) is positive, meaning it is stable. - \(\mathrm{He}\): No bond exists, and the species is overall stable due to its full 1s orbital.

Conclusion

In conclusion, we would expect \(\mathrm{He}_{2} \oplus\) to be more stable than \(\mathrm{He}_{2}\) because it has a bond formation that makes it more stable. While \(\mathrm{He}_{2} \oplus\) is also stable, it is difficult to conclusively say if it is more stable than \(\mathrm{He}\) without knowing further details such as bond dissociation energy or other factors that contribute to their stabilities.

Key concepts

Electronic Configuration

In understanding molecular stability, examining the electronic configuration is vital. Electronic configuration refers to the arrangement of electrons in an atom or molecule's orbitals. For helium (\(\mathrm{He}\)), the configuration is \(1s^2\), meaning it has two electrons filling its 1s orbital.

When considering helium diatomics like \(\mathrm{He}_{2}\), each helium contributes an electron pair, resulting in \((1s_A)^2 (1s_B)^2\). This pairing fills the sigma bonding orbital and the sigma antibonding orbital, making a net zero bond order, leading to instability. However, \(\mathrm{He}_{2}^{\oplus}\), formed by removing one electron, has an electronic configuration of \((1s_A)^2 (1s_B)^1\).

This altered state redistributes electrons, allowing partial occupancy of the sigma bonding orbital without filling the antibonding orbital, creating a more favorable condition for stability than \(\mathrm{He}_{2}\).

Bond Formation

Bond formation is a central factor in determining molecular stability. When atoms come together, they can share electrons to form a bond, which lowers the system's energy and increases its stability. However, excess electrons can occupy antibonding orbitals, countering this stability.

For \(\mathrm{He}_{2}\), with its configuration \((1s_A)^2 (1s_B)^2\), no net bond forms because the presence of electrons in the sigma antibonding molecular orbitals neutralizes any bonding interaction from the filled sigma bonding orbitals. This configuration results in no bond order and is theoretically unstable.

In contrast, for \(\mathrm{He}_{2}^{\oplus}\), having the configuration \((1s_A)^2 (1s_B)^1\), there is a potential to form a weak bond. Here, the absence of a fourth electron in the antibonding orbital decreases anti-bonding effects, effectively contributing a positive bond order (albeit weak) and thus providing more stability than neutral \(\mathrm{He}_{2}\).

Bond Dissociation Energy

Bond dissociation energy (BDE) is the energy required to break a chemical bond in a molecule into its separate atoms. It is a crucial indicator of a molecule's stability; a high BDE suggests a stable bond, while a low BDE suggests an unstable bond.

For \(\mathrm{He}_{2}\), the bond dissociation energy isn’t a useful measure as it exhibits no significant bonding to break. This setup renders \(\mathrm{He}_{2}\) as essentially non-existent.

However, for \(\mathrm{He}_{2}^{\oplus}\), a bond does form due to the unpaired electron configuration, resulting in a small but positive BDE. This low yet positive bond dissociation energy means that \(\mathrm{He}_{2}^{\oplus}\) requires energy input to break its weakly forming bonds, highlighting its relatively stable character compared to the non-existing bond in \(\mathrm{He}_{2}\). When evaluating \(\mathrm{He}\), its stability derives not from bond formation but from the full s orbital in atomic helium, solidifying stability without depending on BDE.