Question
Explain molecular orbital theory (MOT). How do bonding and antibonding orbitals form? Calculate the bond order of O2, N2, and O2- and predict their magnetic properties.
(JEE Main, JEE Advanced, CBSE 11 — MOT is one of the most asked topics in chemical bonding)
Solution — Step by Step
When two atomic orbitals combine, they form two molecular orbitals:
- Bonding MO (sigma, pi): Constructive interference — electron density increases between nuclei, stabilises the molecule. Lower energy than parent AOs.
- Antibonding MO (sigma, pi):** Destructive interference — electron density decreases between nuclei, a node appears. Higher energy than parent AOs.
The number of MOs formed always equals the number of AOs combined.
For molecules with (B2, C2, N2): the orbitals are lower in energy than :
\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}For molecules with (O2, F2): the is lower than :
\sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}where = electrons in bonding MOs, = electrons in antibonding MOs.
N2 (14 electrons): , → Bond order = (triple bond, diamagnetic)
O2 (16 electrons): , → Bond order = (double bond, paramagnetic — 2 unpaired electrons in )
O2- (17 electrons): , → Bond order = (paramagnetic — 1 unpaired electron)
graph TD
A["Atomic Orbitals"] --> B["Constructive overlap"]
A --> C["Destructive overlap"]
B --> D["Bonding MO - Lower energy"]
C --> E["Antibonding MO - Higher energy"]
F["Bond Order = Nb - Na / 2"] --> G{"Bond Order > 0?"}
G -->|"Yes"| H["Molecule is stable"]
G -->|"No"| I["Molecule does not exist"]
J["Unpaired e- in MO?"] -->|"Yes"| K["Paramagnetic"]
J -->|"No"| L["Diamagnetic"]Why This Works
MOT succeeds where Valence Bond Theory fails — it correctly predicts the paramagnetism of O2. VBT predicts O2 should be diamagnetic (all electrons paired), but experiments show O2 is attracted to a magnet. MOT explains this: the two highest-energy electrons in O2 occupy two degenerate orbitals with parallel spins (Hund’s rule), giving two unpaired electrons.
Bond order predicts stability and bond strength: higher bond order means shorter bond, stronger bond, and higher bond dissociation energy. N2 has the highest bond order (3) among diatomic molecules, which is why it is exceptionally stable and inert.
Alternative Method
For JEE, memorise these bond orders and magnetic properties:
| Species | Bond Order | Magnetic |
|---|---|---|
| N2 | 3 | Diamagnetic |
| O2 | 2 | Paramagnetic |
| O2+ | 2.5 | Paramagnetic |
| O2- | 1.5 | Paramagnetic |
| O2 2- | 1 | Diamagnetic |
| F2 | 1 | Diamagnetic |
The O2 family is the most tested. Each electron added to O2 goes into antibonding orbitals, reducing the bond order by 0.5.
Common Mistake
The energy ordering of MOs changes at oxygen. Students often use the same order for all molecules. For N2 and below (B2, C2), the orbitals are lower than . For O2 and above, is lower. Using the wrong order gives wrong electron configurations and wrong magnetic predictions.
Also, bond order can be a fraction (like 1.5 for O2-). Students sometimes round to the nearest integer — do not do this. A bond order of 1.5 is meaningful and correct.