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Molecular orbitals (MOs) are formed when atomic orbitals combine as atoms approach each other. Think of it as combining the energy levels of electrons in an atom to form energy levels that span multiple atoms, which are the molecules.
When two atomic orbitals merge, you get two molecular orbitals: a lower energy 'bonding' molecular orbital and a higher energy 'antibonding' molecular orbital. The bonding MO tends to strengthen the bond by attracting electrons, while the antibonding MO works against forming a stable bond.

• Bonding MO: Denoted as \( \sigma \) or \( \pi \), lower energy and favorable for bond formation.
• Antibonding MO: Denoted with an asterisk, such as \( \sigma^* \) or \( \pi^* \), higher energy.

Electrons fill these molecular orbitals starting from the lowest energy levels, similar to filling atomic orbitals. Understanding this setup is key to predicting molecular stability and bond order.

The \( \mathrm{He}_2 \) molecule is an example where simply knowing about molecular orbitals is not enough for stability. Helium, being a noble gas, has a full set of electrons in its first shell, therefore does not easily form stable bonds.
In the \( \mathrm{He}_2 \) molecule, each helium atom contributes two electrons; hence, there are four electrons to place in molecular orbitals.

• First two electrons go into the low-energy bonding MO.
• Remaining two electrons fill the high-energy antibonding MO.

Calculating its bond order comes out to be zero:\[\text{Bond order} = \frac{(2) - (2)}{2} = 0\]The zero bond order indicates that \( \mathrm{He}_2 \) does not have a stable bond under normal circumstances.

The \( \mathrm{He}_2^+ \) ion has some interesting properties differing from \( \mathrm{He}_2 \) due to its missing electron. This ion forms by ionizing \( \mathrm{He}_2 \), meaning one electron is removed, leaving three electrons to work with.
As we apply molecular orbital theory here:

• Two electrons fill the \( \sigma_{1s} \) bonding MO.
• Only one electron goes into the \( \sigma_{1s}^* \) antibonding MO.

Now, let's calculate the bond order:\[\text{Bond order} = \frac{(2) - (1)}{2} = \frac{1}{2}\]This \( \frac{1}{2} \) bond order suggests a weak but existent bond. The removal of an electron from the antibonding orbital decreases electron-electron repulsion and helps in slightly stabilizing the bond in \( \mathrm{He}_2^+ \).

Understanding bond stability is all about predicting how likely a molecule is to stay intact. In terms of molecular orbital theory, bond order is a useful metric.
Bond order helps predict bond stability and can be calculated using: \[\text{Bond order} = \frac{\text{(number of electrons in bonding MOs)} - \text{(number of electrons in antibonding MOs)}}{2}\]Bond order greater than zero indicates a stable or somewhat stable bond. Here's what different bond orders imply:

• Zero bond order: No stable bond, as seen in \( \mathrm{He}_2 \).
• Fractional/positive bond order: Suggests some level of stability, as in \( \mathrm{He}_2^+ \), which has a bond order of \( \frac{1}{2} \).
• Greater bond order: The higher it is, the more stable the bond, typically found in stronger chemical bonds.

In conclusion, the bond stability of a molecule is tied to its bond order, and this concept is vital for predicting the behavior of molecules in various chemical reactions.