Molecular Orbital TheoryJEE Advanced

Bond order from the MO ladder — interactive Chemistry simulation for IIT-JEE.

Concept

Atomic orbitals combine into delocalised molecular orbitals: constructive overlap → bonding (lower energy), destructive → antibonding (σ, π, higher). Fill them Aufbau-style; the bond order ½(bonding − antibonding) predicts existence, strength and length. MO theory's triumph: it predicts O₂'s two unpaired π* electrons — paramagnetic — which Lewis structures cannot.

Key formula

BO = \frac{N_b - N_a}{2}; \quad \text{Z} \le 7: \; \pi2p < \sigma2p \quad \text{(s–p mixing)}

Derivation

LCAO: two 1s AOs give σ1s and σ*1s. For 2p: head-on overlap → σ2p; sideways → two degenerate π2p.

Up to N₂, s–p mixing pushes σ2p ABOVE π2p; from O₂ on, the 'normal' order returns. He₂: BO = 0 → doesn't exist. N₂: BO = 3 (record strength). O₂: BO = 2 with π*² singly occupied → paramagnetic.

Scenarios to explore

MO Theory — Bond order from the MO ladder — why O₂ is paramagnetic.

Real-world applications

Liquid O₂ sticking to a magnet.
Bond-length/strength trends: O₂⁺ < O₂ < O₂⁻ < O₂²⁻ (BO 2.5, 2, 1.5, 1).
Conjugated π systems & band theory grow out of MO ideas.

JEE exam tips

Adding electrons to ANTIBONDING orbitals lowers BO: O₂ → O₂⁻ weakens the bond.
Removing an antibonding electron STRENGTHENS: O₂⁺ has BO 2.5 > O₂.
CO and NO: heteronuclear cousins — NO has BO 2.5 and one unpaired electron.

Common mistakes

Using the O₂ orbital order for N₂ (σ/π order swaps at Z = 8).
Forgetting Hund's rule in the degenerate π* set (O₂'s unpaired pair).
Counting core 1s electrons incorrectly (they cancel: σ1s² σ*1s²).

Exam traps to avoid

C₂ has BO = 2 from two π bonds and NO σ2p bond — famously weird.
Bond order ↑ ⇒ bond length ↓ — inverse relation, always paired in MCQs.