According to Molecular Orbital (MO) Theory, two atoms mix their orbitals to form one that is spread out over both atoms. The mixing of two atomic orbitals depends on the signs of their wave functions. Addition of two orbitals can lead to bonding MO and anti-bonding MO. On this page, the bonding in a H2 molecule is explained using MO Theory.
Two H atoms in their ground state configuration come together and form a single bond. The bond formation stabilizes both atoms and, therefore, is lower in energy than the atomic orbitals. This is also observed in Valence Bond Theory, which implies that each H atom in H2 shares its electron with one another, so that both can achieve the stable configuration of He.
On top of that, MO Theory allows one to compute the amount of energy released from a bond formation and a distance between two bonded atoms as well as predict the magnetic property of a molecule (or a substance). For H2, the bond strength is -432 kJ/mol, and the bond length is 74 angstrom (or 74 pm). H2 is a diamagnetic molecule because the electrons paired up; therefore, it is not attracted by a magnetic field.
Figure1: MO diagram for H2
The filling of electrons in MO is consistent with the aufbau principle, Hund's rule, and Pauli exclusion principle.
Bonding and Anti-bonding molecular orbitals in H2
Each H atom has a 1s atomic orbital. When two H atoms come to a proper proximity, their 1s orbitals interact and produce two molecular orbitals: a bonding MO and an anti-bonding MO.
If the electrons are in phase, they have a constructive interference. This results in a bonding sigma MO (σ1s). This MO has an increased probability of finding electrons in the bonding region.
Figure 2: Schematic representation of the bonding molecular orbital σ(1s)
If the electrons are out of phase, they have a destructive interference. This results in an anti-bonding sigma MO (σ*1s). This MO has a decreased probability of finding electrons in the bonding region. (Valence Bond Theory does not explain this phenomenon.)
Figure 3: Schematic represenation of antibonding molecular orbital σ*(1s)
Note that there is a nodal plane in the anti-bonding MO.
Bond order = 1/2 (#e- in bonding MO - #e- in antibonding MO)
For H2, bond order = 1/2 (2-0) = 1, which means H2has only one bond. The antibonding orbital is empty. Thus, H2 is a stable molecule.
Again, in the MO, there is no unpaired electron, so H2 is diamagnetic.
- Chang, Raymond. Physical Chemistry for the Biosciences. Sausalito, CA: University Science Books, 2005. 458 - 460.
- Housecroft, Catherine E. and Alan G. Sharpe. Inorganic Chemistry. 3rd ed. England: Pearson - Prentice Hall, 2008. 33 - 36.
- What does the MO of H2+ look like? What is its bond order? What is its magnetic property? Explain.
- What does the MO of H2- look like? What is its bond order? What is its magnetic property? Explain.
- Which one is the most stable: H2, H2+, or H2-? Why?
- When a hydrogen atom accepts an electron, it becomes a hydride H-. Theoretically would it be possible to form a molecule from two hydrides, that is to form H22-? Why?
Bond order = 1/2 (1-0) = 1/2
Paramagnetic because it has one unpaired e- in the σ(1s) orbital.
Bond order = 1/2 (2-1) = 1/2
Paramagnetic because it has one unpaired e- in the σ*(1s) orbital.
3- H2 is the most stable because it has the highest bond order (1), in comparison with the bond orders (1/2) of H2+ and H2-.
4- Theoretically it would not be possible to form a molecule from two hydrides because the anti-bonding and bonding orbitals would cancel each other out. So, the bond order is zero. Because the antibonding ortibal is filled, it destabilizes the structure, making the "molecule" H22- very non-stable.
Bond order = 1/2 (2-2) = 0 ---> no bond formation. Thus, this molecule doesn't exist.