Construct an MO diagram for the He+2 ion.
a.Determine if the He+2 ion is stable or unstable?
b.What is its bond order?
The correct answer and explanation is :
To construct the Molecular Orbital (MO) diagram for the ( \text{He}_2^+ ) ion, follow these steps:
MO Diagram Construction:
- Atomic Orbitals: Each helium atom has an electron configuration of ( 1s^2 ). In ( \text{He}_2^+ ), we consider two helium atoms, but since one electron is removed, we have three electrons in total.
- Molecular Orbitals Formation:
- The ( 1s ) orbitals of the two helium atoms combine to form:
- A bonding molecular orbital (( \sigma_{1s} ))
- An antibonding molecular orbital (( \sigma_{1s}^* ))
- Electron Filling: The three electrons are placed in these orbitals following the Aufbau principle, Pauli-exclusion principle, and Hund’s rule.
- Two electrons fill the bonding ( \sigma_{1s} ) orbital.
- The third electron goes into the antibonding ( \sigma_{1s}^* ) orbital.
Bond Order Calculation:
The bond order is given by the formula:
[
\text{Bond Order} = \frac{(\text{Number of bonding electrons}) – (\text{Number of antibonding electrons})}{2}
]
From the MO diagram:
- Bonding electrons: 2 (in ( \sigma_{1s} ))
- Antibonding electrons: 1 (in ( \sigma_{1s}^* ))
[
\text{Bond Order} = \frac{2 – 1}{2} = 0.5
]
Stability of ( \text{He}_2^+ ):
A molecule is considered stable if it has a positive bond order. Since ( \text{He}_2^+ ) has a bond order of 0.5, it suggests a weakly bound species, meaning it is more stable than ( \text{He}_2 ) (which has a bond order of 0) but still relatively weak compared to molecules with higher bond orders.
This suggests that ( \text{He}_2^+ ) can exist, but it is not very stable. The single bond formed is weak and can dissociate easily.
Conclusion:
- Bond Order: 0.5
- Stability: Weakly stable but more stable than ( \text{He}_2 ).
- Physical Implication: The bond is weak, meaning ( \text{He}_2^+ ) exists only under certain conditions (such as in plasma or ionized gases).