Valence bond theory pictures electrons staying on their own atoms, but it cannot explain why O2 is paramagnetic. Molecular Orbital Theory (MOT), developed by Hund and Mulliken, treats the molecule as a whole.
Forming molecular orbitals. When atomic orbitals of comparable energy and proper symmetry combine, they form the same number of molecular orbitals (MOs) that belong to the entire molecule. Two atomic orbitals give:
- a low-energy bonding MO (constructive overlap, electron density between the nuclei), e.g. sigma, written as the sum of the atomic orbitals;
- a high-energy antibonding MO (destructive overlap, a node between the nuclei), e.g. sigma* (sigma star), written as the difference.
Electrons fill MOs by the Aufbau principle, Pauli exclusion and Hund's rule, just as for atoms.
Bond order measures bond strength and stability:
$\text{Bond order} = \frac{N_b - N_a}{2}$,
where $N_b$ and $N_a$ are the numbers of electrons in bonding and antibonding MOs. A positive bond order means a stable molecule; a zero bond order (as in He2) means no molecule forms.
Magnetic behaviour. If all MO electrons are paired, the species is diamagnetic; if there are unpaired electrons, it is paramagnetic. The triumph of MOT is O2: its two highest electrons occupy two degenerate pi* orbitals singly (Hund's rule), so O2 has two unpaired electrons and is paramagnetic - exactly what experiment shows.
Worked bond orders. H2: 2 bonding, 0 antibonding, bond order = (2-0)/2 = 1. N2: 10 bonding, 4 antibonding, bond order = (10-4)/2 = 3 (matches the triple bond, diamagnetic). O2: 10 bonding, 6 antibonding, bond order = (10-6)/2 = 2 (double bond, paramagnetic).
Hydrogen bonding. When hydrogen is bonded to a small, highly electronegative atom (F, O or N), the bond is strongly polar and the partly positive H is attracted to a lone pair on an electronegative atom of a neighbouring molecule. This electrostatic attraction is a hydrogen bond (about 10-40 kJ/mol) - far weaker than a covalent bond but the strongest intermolecular force.
- Intermolecular H-bonding (between molecules) raises boiling point, explaining why H2O, HF and NH3 boil far higher than expected, why ice is less dense than water (open cage structure), and why ethanol is miscible with water.
- Intramolecular H-bonding (within one molecule), as in ortho-nitrophenol, lowers boiling point compared with the para isomer.
Calculate the bond order of the hydrogen molecule H2.
Solution- Step 1: H2 has 2 electrons; both occupy the bonding sigma1s orbital.
- Step 2: Bonding electrons Nb = 2, antibonding electrons Na = 0.
- Step 3: Bond order = (Nb - Na)/2 = (2 - 0)/2.
- Step 4: Bond order = 1.
Answer: Bond order of H2 is 1 (a single bond, diamagnetic).
Show, using MO theory, why the helium molecule He2 does not exist.
Solution- Step 1: He2 would have 4 electrons.
- Step 2: Two fill the bonding sigma1s and two fill the antibonding sigma*1s.
- Step 3: Nb = 2 and Na = 2.
- Step 4: Bond order = (2 - 2)/2 = 0, so no net bond forms.
Answer: Bond order is 0, hence He2 does not exist as a stable molecule.
Calculate the bond order of N2 and state its magnetic behaviour.
Solution- Step 1: N2 has 14 electrons; 10 occupy bonding MOs and 4 occupy antibonding MOs.
- Step 2: Bond order = (Nb - Na)/2 = (10 - 4)/2.
- Step 3: Bond order = 3, consistent with a triple bond.
- Step 4: All electrons are paired, so N2 is diamagnetic.
Answer: Bond order of N2 is 3 and it is diamagnetic.
Explain why O2 is paramagnetic and calculate its bond order.
Solution- Step 1: O2 has 16 electrons; 10 are in bonding and 6 in antibonding MOs.
- Step 2: Bond order = (10 - 6)/2 = 2 (a double bond).
- Step 3: The last two electrons enter two degenerate pi* orbitals singly (Hund's rule).
- Step 4: Two unpaired electrons make O2 paramagnetic.
Answer: O2 has bond order 2 and is paramagnetic due to two unpaired electrons in pi* orbitals.
Account for the fact that water has a much higher boiling point (100 degrees C) than H2S (about -60 degrees C).
Solution- Step 1: Both are bent hydrides of group 16, so size suggests H2S should boil higher.
- Step 2: Oxygen is small and highly electronegative, so water forms strong O-H...O hydrogen bonds.
- Step 3: Sulphur is larger and less electronegative, so H2S has no significant hydrogen bonding.
- Step 4: Breaking the extensive H-bond network in water needs much more energy, raising its boiling point.
Answer: Extensive intermolecular hydrogen bonding in water (absent in H2S) gives it the much higher boiling point.
ortho-Nitrophenol is more volatile (lower boiling point) than para-nitrophenol. Explain in terms of hydrogen bonding.
Solution- Step 1: In ortho-nitrophenol the -OH and -NO2 groups are adjacent.
- Step 2: This allows intramolecular hydrogen bonding within the same molecule (a chelate ring).
- Step 3: In para-nitrophenol the groups are far apart, so only intermolecular H-bonding (between molecules) occurs.
- Step 4: Intermolecular bonding links molecules together, raising the boiling point of the para isomer; the ortho isomer stays more volatile.
Answer: Intramolecular H-bonding in the ortho isomer leaves it more volatile; intermolecular H-bonding in the para isomer raises its boiling point.