Molecular orbital theory — bonding, antibonding, bond order calculation

Question

How does Molecular Orbital Theory (MOT) explain bonding? How do we fill molecular orbitals, and how do we calculate bond order to predict stability and magnetic behaviour?

(JEE Main, JEE Advanced, CBSE 11 — MOT bond order and magnetic properties are high-frequency JEE questions)

Solution — Step by Step

When two atomic orbitals (AOs) combine, they form two molecular orbitals (MOs):

Bonding MO ( $\sigma$ , $\pi$ ): lower energy than parent AOs, electron density concentrated between nuclei
Antibonding MO ( $\sigma^*$ , $\pi^*$ ): higher energy, electron density pushed away from the internuclear region, has a nodal plane

The number of MOs formed always equals the number of AOs combined. Two AOs give one bonding + one antibonding MO.

For homonuclear diatomics, the filling order differs based on atomic number:

For $O_2$ , $F_2$ , $Ne_2$ (Z > 7):

\sigma 1s \lt \sigma^* 1s \lt \sigma 2s \lt \sigma^* 2s \lt \sigma 2p_z \lt \pi 2p_x = \pi 2p_y \lt \pi^* 2p_x = \pi^* 2p_y \lt \sigma^* 2p_z

For $Li_2$ to $N_2$ (Z $\leq$ 7):

\sigma 1s \lt \sigma^* 1s \lt \sigma 2s \lt \sigma^* 2s \lt \pi 2p_x = \pi 2p_y \lt \sigma 2p_z \lt \pi^* 2p_x = \pi^* 2p_y \lt \sigma^* 2p_z

The key difference: for lighter elements, $\sigma 2p_z$ is ABOVE $\pi 2p$ (due to significant 2s-2p mixing). For heavier elements ( $O_2$ onwards), $\sigma 2p_z$ drops below $\pi 2p$ .

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

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

Rules:

Bond order > 0 means the molecule is stable
Higher bond order = shorter bond length = greater bond energy
Bond order = 0 means the molecule does not exist (e.g., $He_2$ , $Ne_2$ )

$O_2$ has 16 electrons. Using the Z > 7 ordering:

$(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$

$N_b = 10$ (bonding electrons), $N_a = 6$ (antibonding electrons)

\text{Bond Order} = \frac{10 - 6}{2} = 2

The two unpaired electrons in $\pi^*$ orbitals make $O_2$ paramagnetic — this is the triumph of MOT. VBT predicted $O_2$ should be diamagnetic, which contradicts experimental evidence.

flowchart TD
    A["Count total electrons"] --> B{"Z ≤ 7 for both atoms?"}
    B -->|"Yes"| C["Use σ2p above π2p ordering"]
    B -->|"No"| D["Use σ2p below π2p ordering"]
    C --> E["Fill MOs following Aufbau + Hund's rule"]
    D --> E
    E --> F["Count Nb and Na"]
    F --> G["Bond Order = (Nb - Na)/2"]
    G --> H{"Any unpaired electrons?"}
    H -->|"Yes"| I["Paramagnetic"]
    H -->|"No"| J["Diamagnetic"]

Why This Works

MOT treats electrons as belonging to the molecule as a whole, not to individual atoms. By constructive and destructive interference of wavefunctions, bonding and antibonding MOs arise naturally. Bond order quantifies the net bonding — every antibonding electron cancels one bonding electron. The beauty of MOT is that it correctly predicts magnetic properties, fractional bond orders, and the stability of species like $He_2^+$ (bond order = 0.5, which exists) that VBT cannot explain.

Common Mistake

The most frequent error: using the wrong MO ordering. Students apply the Z > 7 order for $N_2$ or $C_2$ , which gives the wrong electronic configuration and wrong magnetic prediction. Remember: the switch happens between $N_2$ and $O_2$ . For $N_2$ and lighter molecules, $\sigma 2p_z$ is ABOVE the $\pi 2p$ pair. JEE Advanced 2021 directly tested this distinction.

Quick bond orders to memorise: $H_2 = 1$ , $He_2 = 0$ (does not exist), $N_2 = 3$ (triple bond), $O_2 = 2$ (double bond, paramagnetic), $F_2 = 1$ , $Ne_2 = 0$ . For ions, add/remove electrons from the highest occupied MO and recalculate.

Question

Solution — Step by Step

Why This Works

Common Mistake

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