VKC Notes · Chemical bonding · Civil Services P-I · IFoS P-I

2.11 Molecular Orbital Theory: LCAO, Homonuclear and Heteronuclear Diatomics (O₂, N₂, CO, NO, CN⁻, HF), Bond Order and Magnetism

★★★ 2021–2025: 9 Mains parts, 100 marks (Civil Services 4 parts / 50 m · IFoS 5 / 50 m), including a 20-mark NO question in 2025. Almost every part asks you to draw an MO diagram and read off bond order and magnetism: learn to draw four diagrams (N₂, O₂, CO, NO) cleanly and the rest is arithmetic.

Syllabus lines. Civil Services: “Molecular orbital theory (LCAO method); bonding in H₂⁺, H₂, He₂⁺ to Ne₂, NO, CO, HF, and CN⁻; comparison of valence bond and molecular orbital theories, bond order, bond strength and bond length.” IFoS: “Molecular orbital theory (LCAO method); bonding in homo- and heteronuclear diatomic molecules … bond order, bond strength and bond length.”

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

2.11.1 LCAO: bonding and antibonding orbitals ★★★

(2.11.1)

A molecular orbital is a linear combination of atomic orbitals. In-phase addition piles up electron density between the nuclei: bonding, lower energy. Out-of-phase gives a nodal plane between the nuclei: antibonding (*), higher energy. α = Coulomb integral (energy of the AO), β = resonance integral (negative), S = overlap.SN
Because of the 1 ± S denominators, the antibonding MO is raised more than the bonding MO is lowered: He₂ (2 bonding + 2 antibonding electrons) is unbound. Number of MOs formed = number of AOs used.SN
Three conditions for two AOs to combine: (i) similar energy, (ii) good overlap, (iii) same symmetry about the bond axis (σ with σ, π with π). If the symmetry differs, + overlap on one side cancels − overlap on the other: S = 0 at every distance, and the orbital stays non-bonding.SN

Fig. 2.11.1 Which atomic orbitals combine (top) and which cannot (bottom); z is the bond axis.

Draw it in 60 s: top: two + circles making an oval (σ); two p dumbbells end to end with like signs facing (σ); two upright p dumbbells side by side (π). Bottom: an s next to an upright p, a horizontal p next to an upright p, an upright p next to a dotted-circle p: write S = 0.

Solved PYQ CSE 2022 Q1(b) 10m: atom A offers 2s, 2p_x; atom B offers 2p_y, 2p_z (z = bond axis). Which interactions are non-bonding?

2s(A)–2p_z(B): both symmetric about the axis (σ): net overlap, bonding/antibonding pair.
2s(A)–2p_y(B): the s lobe meets the + and − halves of p_y equally: S = 0, non-bonding. 2p_x(A)–2p_z(B): the same cancellation with roles exchanged: non-bonding.SN
2p_x(A)–2p_y(B): both π type but in perpendicular planes; the product is + in two quadrants and − in two: S = 0, non-bonding. The cancellation comes from symmetry, so it holds at every A–B distance.SN

2.11.2 Homonuclear diatomics and s–p mixing ★★★

(2.11.2)

s–p mixing: σ_g2s and σ_g2p have the same symmetry, so they interact when the 2s–2p gap is small (B, C, N: gap 5.8–12.4 eV). σ_g2s is pushed down and σ_g2p pushed above π_u2p. In O and F the gap is large (16.5, 21.5 eV): little mixing, σ_g2p stays below π_u2p (Fig. 2.11.2). Photoelectron spectra confirm both orders.SN
Each π level is doubly degenerate; Hund’s rule fills them singly first, which is why B₂ and O₂ are paramagnetic.SN

Fig. 2.11.2 N₂ (s–p mixing: π below σ) and O₂ (σ below π). Left and right: the atomic levels; centre: the MOs. Atomic energies from Miessler–Tarr Table 5.2.

Draw it in 60 s: two AO columns, MOs between, dashed tie lines; for N₂ draw π (two lines) under σ2p and fill 10 e; for O₂ σ2p under π, fill 12 e: the last two singly into the two π* lines.

Molecule	Valence e	Configuration (beyond σ2s² σ*2s²)	BO	Unpaired e	r / pm
B₂	6	π2p²	1	2 (para)	159.0
C₂	8	π2p⁴	2	0	124.2
N₂	10	π2p⁴ σ2p²	3	0	109.8
O₂	12	σ2p² π2p⁴ π*2p²	2	2 (para)	120.8
F₂	14	σ2p² π2p⁴ π*2p⁴	1	0	141.2
Ne₂	16	… σ*2p²	0	0	not bound

Bond lengths: Miessler–Tarr Table 5.1. The higher the bond order, the shorter and stronger the bond.

Solved PYQ IFoS 2025 Q2(c)(ii) 5m: “B₂ has a bond order of one and is paramagnetic.” Substantiate.

B₂ has 6 valence electrons. For B, C, N, s–p mixing puts π_u2p below σ_g2p, so the configuration is σ2s² σ*2s² π2p_x¹ π2p_y¹.SN
BO = ½(4 − 2) = 1; the two π electrons sit singly in degenerate orbitals with parallel spins (Hund): paramagnetic. Without mixing, σ2p² would give a diamagnetic B₂, which experiment rules out: the magnetism itself proves s–p mixing.SN

2.11.3 The O₂ family: bond order, length, energy, magnetism ★★★

Species	Valence e	π*2p electrons	Bond order	r(O–O) / pm	D / kJ mol⁻¹	Magnetism
O₂⁺ (dioxygenyl)	11	1	2.5	111.6	642	para, 1 unpaired
O₂	12	2	2	120.8	494	para, 2 unpaired
O₂⁻ (superoxide)	13	3	1.5	135	394	para, 1 unpaired
O₂²⁻ (peroxide)	14	4	1	150.4	210	dia

r from Miessler–Tarr Table 5.1; D as given in IFoS 2025 Q2(b)(ii).

Each electron added to π*2p lowers the bond order by ½; each removed raises it by ½. Bond order ↑ ⇒ bond length ↓, dissociation energy ↑, stretching frequency ↑ CSE 2024 Q2(b)(i) 10m IFoS 2025 Q2(b)(ii) 10m IFoS 2023 Q2(c) 10m.SN
Stability order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻. Valence-bond theory (O=O, all paired) cannot explain why O₂ is paramagnetic (liquid O₂ hangs between the poles of a magnet); MO theory gets it at once: two unpaired π* electrons, a triplet ground state.SN

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