Question

Compare the bond order of \(\mathrm{He}_{2}\) and \(\mathrm{He}_{2}^{+}\).

Short answer

The bond order of He₂ is 0, and the bond order of He₂⁺ is \(\frac{1}{2}\). Thus, He₂⁺ has a higher bond order and a more stable bond compared to the non-existing bond in He₂.

Step-by-step solution

Determine the atomic orbitals of Helium

For He₂ and He₂⁺, each helium atom has 2 electrons in its 1s orbital. So, we have a total of 4 electrons for He₂ and 3 electrons for He₂⁺.

Construct molecular orbitals for He₂

When two atomic orbitals combine, they form two molecular orbitals: one is lower in energy (bonding MO) and one is higher in energy (antibonding MO). In the case of He₂, both the orbitals of helium atoms combine, creating a bonding MO (σ₁s) and an antibonding MO (σ₁s*).

Fill the molecular orbitals with electrons for He₂

According to Hund's rule and the Aufbau principle, we fill the lowest energy MOs first. We have 4 electrons in total for He₂: 1. First, place 2 electrons in the bonding MO (σ₁s). 2. Second, place the remaining 2 electrons in the antibonding MO (σ₁s*).

Calculate bond order for He₂

Bond order is calculated using the formula: \[ \textrm{Bond order} = \frac{\textrm{(Number of electrons in bonding MOs) - (Number of electrons in antibonding MOs)}}{2} \] For He₂: \[ \textrm{Bond order} = \frac{(2) - (2)}{2} = 0 \]

Construct molecular orbitals for He₂⁺

The molecular orbitals for He₂⁺ are the same as those for He₂: a bonding MO (σ₁s) and an antibonding MO (σ₁s*).

Fill the molecular orbitals with electrons for He₂⁺

This time, we have only 3 electrons in total for He₂⁺: 1. First, place 2 electrons in the bonding MO (σ₁s). 2. Second, place the remaining 1 electron in the antibonding MO (σ₁s*).

Calculate bond order for He₂⁺

Using the same formula as earlier, we can calculate the bond order for He₂⁺: \[ \textrm{Bond order} = \frac{(2) - (1)}{2} = \frac{1}{2} \] Now that we have calculated the bond orders for both He₂ and He₂⁺, we can compare them.

Compare bond orders for He₂ and He₂⁺

We found that the bond order of He₂ is 0, while the bond order of He₂⁺ is 1/2. Since 1/2 > 0, He₂⁺ has a higher bond order than He₂. This means He₂⁺ has a more stable bond compared to the non-existing bond in He₂.

Key concepts

Molecular Orbitals

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.

He₂ Molecule

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.

He₂⁺ Ion

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 \( \mathrm{He}_2^+ \).

Bond Stability

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.