Draw MOT diagram of O₂ and explain its paramagnetism

Question

Draw the Molecular Orbital Theory (MOT) energy level diagram for O₂. Use it to explain why O₂ is paramagnetic, and calculate its bond order.

Solution — Step by Step

Oxygen has atomic number 8. Electronic configuration: $1s^2 2s^2 2p^4$ .

Each O atom contributes 8 electrons, so O₂ has $8 + 8 = 16$ electrons total to fill into molecular orbitals.

For O₂ (and atoms with Z ≥ 8), the MO energy ordering is:

\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}

Note: For Z ≥ 8 (O, F, Ne), $\sigma_{2p}$ is lower in energy than the $\pi_{2p}$ orbitals. (For B, C, N, the order is reversed.)

Following Aufbau, Hund’s, and Pauli principles:

MO	Electrons
$\sigma_{1s}$	2
$\sigma^*_{1s}$	2
$\sigma_{2s}$	2
$\sigma^*_{2s}$	2
$\sigma_{2p_z}$	2
$\pi_{2p_x}$	2
$\pi_{2p_y}$	2
$\pi^*_{2p_x}$	1 ← Hund’s rule
$\pi^*_{2p_y}$	1 ← Hund’s rule

Total: 16 electrons ✓

\text{Bond Order} = \frac{N_b - N_a}{2}

where $N_b$ = electrons in bonding MOs, $N_a$ = electrons in antibonding MOs.

$N_b = 2 + 2 + 2 + 2 = 8$ (from $\sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}$ )

$N_a = 2 + 2 + 1 + 1 = 6$ (from $\sigma^*_{1s}, \sigma^*_{2s}, \pi^*_{2p_x}, \pi^*_{2p_y}$ )

\text{Bond Order} = \frac{8 - 6}{2} = \frac{2}{2} = \mathbf{2}

O₂ has a double bond — consistent with Lewis structure.

The last two electrons fill the degenerate $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ orbitals. By Hund’s rule, they enter separately with parallel spins — one up-spin in each orbital.

Paramagnetic substances have one or more unpaired electrons. O₂ has two unpaired electrons in its $\pi^*$ antibonding orbitals → it is paramagnetic.

This was the famous triumph of MOT over Lewis structures: Lewis structure of O₂ shows a double bond with all electrons paired, predicting it should be diamagnetic. But experimentally, liquid O₂ is attracted to magnets — it’s paramagnetic. MOT correctly predicts this.

Why This Works

The key insight is that molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). When degenerate MOs (same energy) are available, Hund’s rule applies just as it does for atomic orbitals: electrons prefer to occupy separate orbitals singly before pairing. This leads to the unpaired electrons in O₂.

The property of paramagnetism is directly linked to unpaired electrons — they have a net magnetic moment and are attracted to external magnetic fields. Paired electrons have opposite spins that cancel.

JEE Main 2023 asked about the bond order of O₂⁻ (superoxide ion). With one more electron, it goes into one of the $\pi^*$ orbitals: $N_a = 7$ , bond order = (8-7)/2 = 0.5. NEET asks about paramagnetic/diamagnetic predictions from MO configurations — know O₂ (paramagnetic), N₂ (diamagnetic), CO (diamagnetic).

Alternative Method

Without drawing the full MO diagram, you can use the bond order formula directly by knowing the electron configurations:

O₂ MO config: $(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^2$

The two $\pi^*_{2p}$ electrons occupy different orbitals (Hund’s rule) → paramagnetic with 2 unpaired electrons, bond order = 2.

Common Mistake

Students often place both last electrons in the same $\pi^*$ orbital (paired), predicting diamagnetism. This violates Hund’s rule — degenerate orbitals ( $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ have equal energy) must each receive one electron before any gets a second. The correct filling gives two unpaired electrons, confirming paramagnetism.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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