Electronic effects, substitution and elimination mechanisms, electrophilic addition, aromaticity, VSEPR, crystal field theory, and molecular orbital theory for NEET.
1. Inductive Effect & Mesomeric Effect
💡 Why this matters in real life
Why is acetic acid (CH₃COOH) a weak acid while trichloroacetic acid (CCl₃COOH) is much stronger? Why is phenol more acidic than ethanol? Why is aniline less basic than ammonia? The answer lies in electronic effects — the inductive and mesomeric effects that govern how electron density moves through molecules. Understanding these effects is essential for predicting reactivity, acidity, basicity, and stability in organic chemistry.
Fig 1.1: Inductive effect (left) — CCl₃ group pulls electron density. Mesomeric effect (right) — delocalized π electrons in benzene.
🧠 The Big Idea — In Plain Words
Inductive effect (I-effect): The polarization of σ-bonds due to electronegativity differences. It's a permanent effect that decreases with distance. Electron-withdrawing groups (−I: Cl, F, NO₂, CN, COOH) pull electron density. Electron-donating groups (+I: CH₃, C₂H₅, other alkyl groups) push electron density.
Mesomeric effect (M-effect) or Resonance effect: The delocalization of π-electrons through conjugated systems. It's stronger than the inductive effect but requires the system to be planar with overlapping p-orbitals. −M groups withdraw electrons via resonance (NO₂, CN, COOH). +M groups donate via resonance (OH, NH₂, OR).
🔍 Step-by-Step: Comparing Inductive and Mesomeric Effects
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Inductive effect operates through σ-bonds. It is distance-dependent and falls off rapidly after 3-4 bonds. The magnitude: F > Cl > Br > I (halogen electronegativity order). Alkyl groups show +I effect: (CH₃)₃C− > (CH₃)₂CH− > CH₃CH₂− > CH₃−
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Mesomeric effect operates through π-bonds and resonance. It requires parallel p-orbitals. The −M groups (electron withdrawing by resonance): NO₂ > CN > COOH > CHO > COCH₃. The +M groups (electron donating by resonance): OH > NH₂ > OCH₃ > Halogens
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Relative strength: M-effect dominates over I-effect. For example, in phenol, the +M effect of OH (donating electrons into the ring) outweighs its −I effect (withdrawing through σ bond). This makes phenol more reactive than benzene toward electrophilic substitution.
Halogens have both −I (strong) and +M (weak) effects. The −I effect dominates in inductive withdrawal, but the +M effect also operates. In electrophilic aromatic substitution, halogens are deactivating but ortho/para-directing — the +M effect directs to ortho/para, but net electron withdrawal deactivates the ring.
🎯 NEET Pattern
🎯 NEET Pattern — Electronic Effects
• "Arrange in increasing acidity: CH₃COOH, ClCH₂COOH, Cl₃CCOOH" → CH₃COOH < ClCH₂COOH < Cl₃CCOOH (more Cl = more −I = stronger acid).
• "Which is more basic: aniline or ammonia?" → Ammonia (NH₃) is more basic because the lone pair on N in aniline is delocalized into the benzene ring (+M effect).
• "Why is phenol more acidic than ethanol?" → Phenoxide ion is stabilized by resonance (delocalization of negative charge into the ring).
✍️ Worked Example
Example 1: Inductive Effect on Acidity
Arrange in order of increasing acid strength: CH₃COOH, ClCH₂COOH, FCH₂COOH, BrCH₂COOH.
Solution: Electronegativity: F > Cl > Br. So −I effect: F > Cl > Br. Acid strength: FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH. Key insight: More electronegative substituent stabilizes the conjugate base more → stronger acid.
2. SN1 & SN2 Mechanisms
💡 Why this matters in real life
Nucleophilic substitution reactions are among the most important in organic chemistry. They're used to synthesize pharmaceuticals, pesticides, and polymers. Understanding whether a reaction follows SN1 or SN2 determines the product's stereochemistry, which can be critical for drug activity — one enantiomer may be therapeutic while the other is toxic.
🧠 The Big Idea — In Plain Words
SN2: Bimolecular nucleophilic substitution. It's a concerted process where the nucleophile attacks from the back side as the leaving group departs. This causes inversion of configuration (Walden inversion). Favored by primary substrates, strong nucleophiles, and aprotic solvents. Rate = k[Nu][R-X].
SN1: Unimolecular nucleophilic substitution. It's a two-step process. First, the leaving group departs, forming a carbocation. Then, the nucleophile attacks the planar carbocation from either side, giving racemization. Favored by tertiary substrates, weak nucleophiles, and protic solvents. Rate = k[R-X].
🔍 Step-by-Step: SN2 Mechanism
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One-step concerted process. The nucleophile (Nu⁻) approaches the electrophilic carbon from the back side (opposite to the leaving group). As the Nu-C bond forms, the C-LG bond breaks simultaneously. Nu:⁻ + R₃C−LG → [Nu⋯C⋯LG]‡ → R₃C−Nu + :LG⁻
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Stereochemistry: complete inversion. The transition state has a trigonal bipyramidal geometry. The nucleophile and leaving group are at 180°. The other three groups are in a plane. After the reaction, the configuration is inverted.
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Factors favoring SN2. Primary alkyl halide > secondary ≫ tertiary (steric hindrance blocks backside attack). Strong nucleophile (OH⁻, CN⁻, CH₃O⁻). Aprotic solvent (DMSO, acetone, DMF) because protic solvents solvate the nucleophile and slow it down.
🔍 Step-by-Step: SN1 Mechanism
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Step 1 — Slow (rate-determining): Carbocation formation. The C-LG bond breaks heterolytically. The leaving group departs with its bonding electrons: R₃C−LG → R₃C⁺ + :LG⁻. Rate = k[R-X]. This is the slow step.
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Step 2 — Fast: Nucleophilic attack. The nucleophile attacks the planar carbocation. Since the carbocation is sp² hybridized and planar, the nucleophile can attack from either face with equal probability → racemic mixture (50:50 enantiomers).
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Factors favoring SN1. Tertiary alkyl halide > secondary ≫ primary (stability of carbocation: 3° > 2° > 1° > methyl). Weak nucleophile (H₂O, ROH). Protic solvent (H₂O, CH₃OH) stabilizes the carbocation and leaving group through solvation.
SN2 at a tertiary carbon is essentially impossible due to steric hindrance — the three alkyl groups block backside attack. Conversely, SN1 at a primary carbon is very slow because the primary carbocation is highly unstable. For secondary substrates, both pathways can compete depending on conditions.
🎯 NEET Pattern
🎯 NEET Pattern — SN1 vs SN2
• "What is the product when (R)-2-bromobutane reacts with NaOH in aqueous ethanol?" → SN1 (aq. EtOH is protic) → racemic mixture.
• "Which alkyl halide reacts fastest with CN⁻ in DMSO?" → 1-bromobutane (primary, SN2, strong Nu, aprotic solvent).
• "What is the stereochemistry of product when CH₃CH₂CHBrCH₃ reacts with CH₃O⁻ in DMSO?" → SN2 → inversion.
✍️ Worked Example
Example 2: SN2 Reaction
Predict the product and stereochemistry when (R)-2-bromooctane reacts with NaOH in DMSO.
Solution: DMSO is aprotic, OH⁻ is a strong nucleophile, and the substrate is secondary. SN2 is favored. The product is (S)-2-octanol (inversion of configuration). Key insight: The solvent choice (aprotic vs protic) often determines SN1 vs SN2 pathway.
3. E1 & E2 Elimination Mechanisms
💡 Why this matters in real life
Elimination reactions are used to synthesize alkenes, which are building blocks for polymers, plastics, and many industrial chemicals. The competition between substitution (SN1/SN2) and elimination (E1/E2) is a classic challenge in organic synthesis that depends on reaction conditions.
🧠 The Big Idea — In Plain Words
E2: Bimolecular elimination. Concerted one-step process where a base abstracts a β-hydrogen while the leaving group departs, forming a double bond. Requires anti-periplanar geometry. Favored by strong bases (OH⁻, RO⁻, NH₂⁻) and gives the more substituted alkene (Zaitsev's rule) unless a bulky base is used (Hofmann product). Rate = k[base][R-X].
E1: Unimolecular elimination. Two-step process. First, the leaving group departs forming a carbocation. Then, a base abstracts a β-proton to form the double bond. Favored by tertiary substrates and weak bases. Rate = k[R-X]. Always gives the more substituted alkene (Zaitsev).
🔍 Step-by-Step: E2 Mechanism
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Concerted anti-periplanar elimination. The base abstracts a β-hydrogen while the leaving group departs from the opposite side. The C-H and C-LG bonds must be in the same plane and anti to each other. B:⁻ + R₂CH−CR₂−LG → R₂C=CR₂ + BH + LG⁻
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Zaitsev's rule (thermodynamic control). The more substituted alkene (more alkyl groups on the double bond) is the major product. This is because alkyl groups stabilize the double bond through hyperconjugation. R₂C=CR₂ (tetrasubstituted) > R₂C=CHR (trisubstituted) > RHC=CHR (disubstituted)
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Hofmann product with bulky bases. When using bulky bases like potassium tert-butoxide (t-BuOK), the less substituted alkene (Hofmann product) is favored due to steric hindrance preventing access to the more substituted β-hydrogen.
🔍 Step-by-Step: E1 Mechanism
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Step 1 — Carbocation formation (rate-determining). Same as SN1: R₃C−LG → R₃C⁺ + LG⁻. The rate depends only on substrate concentration.
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Step 2 — β-Hydrogen abstraction by base. A weak base (often the solvent: H₂O, ROH) abstracts a β-proton. The more substituted alkene (Zaitsev) is formed because the transition state resembles the more stable alkene.
E2: Rate = k[base][R-X], Zaitsev product (or Hofmann with bulky base) E1: Rate = k[R-X], Zaitsev only
🔍 Common Mistake
Common Mistake — E2 vs SN2 Competition
Strong nucleophiles that are also strong bases (OH⁻, RO⁻) can promote both SN2 and E2. Higher temperature favors elimination (E2) over substitution. Primary substrates favor SN2. Tertiary substrates favor E2 (since SN2 is blocked by steric hindrance). Also: E2 requires a β-hydrogen — if no β-H exists, elimination cannot occur.
🎯 NEET Pattern
🎯 NEET Pattern — Elimination
• "What is the major product when 2-bromo-2-methylbutane reacts with KOH in ethanol?" → E2 (strong base) → 2-methyl-2-butene (Zaitsev, more substituted).
• "Which base gives Hofmann product?" → t-BuOK (bulky base).
• "What happens when 2-chloro-2-methylbutane is heated in water?" → E1 (weak base, heat) → 2-methyl-2-butene + rearranged products.
✍️ Worked Example
Example 3: E2 Elimination
Predict the major product when 2-bromopentane reacts with sodium ethoxide in ethanol.
Solution: Strong base (EtO⁻) → E2. Possible products: 1-pentene (less substituted) and 2-pentene (more substituted, Zaitsev). Zaitsev product (2-pentene) is major. Key insight: 2-pentene can be cis or trans, with trans being more stable and major.
4. Markownikoff's Rule & Electrophilic Addition
💡 Why this matters in real life
Electrophilic addition to alkenes is how many important industrial chemicals are made, including ethanol (from ethylene), isopropanol, and ethylene glycol (antifreeze). Markownikoff's rule predictably guides which product forms when an unsymmetrical reagent adds to an unsymmetrical alkene.
🧠 The Big Idea — In Plain Words
Markownikoff's rule: When HX adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already (the less substituted carbon), and the halogen attaches to the carbon with fewer hydrogens (the more substituted carbon). This is explained by carbocation stability — the reaction proceeds through the more stable (more substituted) carbocation intermediate.
Anti-Markownikoff addition occurs in the presence of peroxides (ROOR) for HBr addition. It follows a free radical mechanism where the bromine radical attacks the less substituted carbon, giving the opposite regioselectivity.
🔍 Step-by-Step: Markownikoff's Rule
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Electrophilic addition proceeds via carbocation intermediate. For CH₃−CH=CH₂ + H⁺: The proton adds to either C1 or C2. If C1 gets H⁺: CH₃−C⁺H−CH₃ (2° carbocation). If C2 gets H⁺: CH₃−CH₂−C⁺H₂ (1° carbocation). The 2° carbocation is more stable.
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Carbocation stability determines the pathway. Stability: 3° > 2° > 1° > CH₃⁺. The reaction follows the more stable carbocation path. This means H⁺ adds to the carbon that gives the more stable carbocation — which is the carbon with more alkyl substituents (more hydrogens originally).
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The halide ion then attacks the carbocation.R₃C⁺ + X⁻ → R₃C−X. The overall result: H adds to the less substituted C, X adds to the more substituted C. For propene: CH₃−CH=CH₂ + HBr → CH₃−CHBr−CH₃ (2-bromopropane, Markovnikov product).
Markownikoff: H adds to C with more H's Anti-Markownikoff: H adds to C with fewer H's (HBr + peroxide only)
🔍 Common Mistake
Common Mistake — Peroxide Effect Only for HBr
The anti-Markownikoff (peroxide) effect works only with HBr. HCl and HI do not show the peroxide effect. For HCl, the H-Cl bond is too strong for radical initiation. For HI, the I radical is too stable and tends to recombine. Also: in the absence of peroxides, HBr also follows Markownikoff's rule.
🎯 NEET Pattern
🎯 NEET Pattern — Markownikoff
• "Product of CH₃−CH=CH₂ + HCl?" → CH₃−CHCl−CH₃ (Markownikoff).
• "Product of CH₃−CH=CH₂ + HBr in presence of peroxide?" → CH₃−CH₂−CH₂Br (anti-Markownikoff).
• "Which alkene gives only one product with HBr?" → Symmetrical alkene like CH₃−CH=CH−CH₃ (either way gives same product).
✍️ Worked Example
Example 4: Markownikoff's Rule
Predict the major product when 2-methyl-2-butene reacts with HBr (no peroxides).
Solution: 2-methyl-2-butene: CH₃−C(CH₃)=CH−CH₃. H⁺ adds to C3 (CH=, more H's) → forms 3° carbocation at C2: (CH₃)₂C⁺−CH₂−CH₃. Br⁻ attacks → 2-bromo-2-methylbutane. Key insight: The 3° carbocation intermediate is highly stable.
5. Aromaticity & Hückel's Rule
💡 Why this matters in real life
Aromatic compounds are everywhere — from the benzene in gasoline to the DNA bases in your cells (purines and pyrimidines are aromatic). Many drugs, dyes, and plastics contain aromatic rings. Aromaticity explains why benzene is unusually stable and undergoes substitution rather than addition reactions.
🧠 The Big Idea — In Plain Words
A compound is aromatic if it satisfies four conditions: (1) cyclic, (2) planar, (3) fully conjugated (every atom in the ring has a p-orbital), and (4) follows Hückel's rule: 4n+2 π electrons (n = 0, 1, 2, ...). Benzene (C₆H₆) with 6 π electrons is the classic example. Anti-aromatic compounds have 4n π electrons and are destabilized. Non-aromatic compounds lack full conjugation or planarity.
🔍 Step-by-Step: Hückel's Rule
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Benzene: the paradigm of aromaticity. 6-membered carbon ring, sp² hybridized, planar. Each carbon contributes one p-orbital with one electron. Total: 6 π electrons (4n+2, n=1). The π electrons are completely delocalized, giving benzene exceptional stability (150 kJ/mol resonance energy).
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Other aromatic compounds. Pyridine (6 π e⁻, N replaces one CH), pyrrole (6 π e⁻, N contributes 2 e⁻), furan (6 π e⁻, O contributes 2 e⁻), thiophene (6 π e⁻). Cyclopentadienyl anion (C₅H₅⁻) is aromatic (6 π e⁻). Cycloheptatrienyl cation (C₇H₇⁺, tropylium) is aromatic (6 π e⁻).
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Anti-aromatic compounds. Cyclobutadiene (4 π e⁻, 4n with n=1) is anti-aromatic and highly unstable. Cyclooctatetraene is non-aromatic because it adopts a tub shape (not planar) to avoid anti-aromaticity. Planar cyclooctatetraene would have 8 π e⁻ (4n, n=2) and be anti-aromatic.
Only count π electrons in the conjugated system. For heterocycles: N in pyridine contributes 1 π e⁻ (its lone pair is in sp², not in p-orbital). N in pyrrole contributes 2 π e⁻ (its lone pair is in p-orbital). Also: ions can be aromatic — the cyclopentadienyl anion becomes aromatic by gaining one electron.
🎯 NEET Pattern
🎯 NEET Pattern — Aromaticity
• "Which of the following is aromatic: cyclopropenyl cation, cyclopropenyl anion?" → Cation: 2 π e⁻ (4n+2, n=0) → aromatic. Anion: 4 π e⁻ → anti-aromatic.
• "Is pyridine aromatic?" → Yes, 6 π e⁻, planar, cyclic, conjugated. The N lone pair is in sp² (not part of π system).
• "Which has higher resonance energy: benzene or naphthalene?" → Naphthalene (2 fused benzene rings, ~255 kJ/mol vs 150 kJ/mol).
✍️ Worked Example
Example 5: Aromaticity
Determine whether the following are aromatic, anti-aromatic, or non-aromatic: (a) cyclopentadiene, (b) cyclopentadienyl anion, (c) cyclopentadienyl cation.
Solution: (a) Cyclopentadiene: not fully conjugated (one sp³ C) → non-aromatic. (b) Cyclopentadienyl anion: 6 π e⁻, planar, cyclic, conjugated → aromatic. (c) Cyclopentadienyl cation: 4 π e⁻ → anti-aromatic. Key insight: Losing or gaining a proton at the sp³ carbon makes cyclopentadiene aromatic.
6. VSEPR Theory — Molecular Shapes
💡 Why this matters in real life
The shape of a molecule determines its properties — whether it's polar or nonpolar, how it interacts with other molecules, and its biological activity. The bent shape of water gives it its unique solvent properties essential for life. The tetrahedral geometry of carbon is the basis of organic chemistry.
🧠 The Big Idea — In Plain Words
VSEPR (Valence Shell Electron Pair Repulsion) theory states that electron pairs around a central atom repel each other and arrange to minimize repulsion. Lone pairs repel more strongly than bonding pairs. The steric number (number of atoms bonded + number of lone pairs) determines the electron domain geometry.
🔍 Step-by-Step: VSEPR Classification
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Steric number = 2: Linear. 2 electron domains, 180° angle. Example: CO₂, BeCl₂, C₂H₂. No lone pairs on central atom.
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Steric number = 3: Trigonal planar. 3 domains, 120° angle. With 0 lone pairs (AX₃): BF₃, SO₃. With 1 lone pair (AX₂E): SO₂, O₃ (bent, <120°).
Common Mistake — Molecular Shape vs Electron Domain
The shape (molecular geometry) considers only atoms, not lone pairs. The electron domain geometry considers both. For NH₃: electron domain = tetrahedral (4 domains), molecular shape = trigonal pyramidal (3 atoms + 1 lone pair).
🎯 NEET Pattern
🎯 NEET Pattern — VSEPR
• "Shape of SF₄?" → See-saw (AX₄E, SN = 5).
• "Shape of XeF₂?" → Linear (AX₂E₃, SN = 5, lone pairs occupy equatorial positions).
• "Which has maximum bond angle: H₂O, NH₃, CH₄?" → CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°) — fewer lone pairs → less repulsion → larger angle.
✍️ Worked Example
Example 6: VSEPR
Predict the shape and bond angles of ClF₃.
Solution: Cl has 7 valence e⁻. 3 F atoms use 3 e⁻ for bonding, leaving 4 e⁻ (2 lone pairs). Steric number = 5 (3 bonds + 2 LPs). Electron domain: trigonal bipyramidal. LP-LP repulsion places them in equatorial positions. Molecular shape: T-shaped (AX₃E₂). Key insight: Lone pairs occupy equatorial positions to minimize 90° LP-LP repulsions.
7. Crystal Field Theory — d-Orbital Splitting
💡 Why this matters in real life
Crystal field theory explains the colors of gemstones and transition metal complexes. Ruby is red because Cr³⁺ in Al₂O₃ absorbs specific wavelengths. The blue color of CuSO₄ solution, the green of NiCl₂, and the purple of [Ti(H₂O)₆]³⁺ are all explained by d-orbital splitting. CFT also explains magnetic properties (paramagnetism vs diamagnetism) of coordination compounds.
🧠 The Big Idea — In Plain Words
In a transition metal ion, the five d-orbitals are degenerate (same energy) in free space. When ligands approach, the d-orbitals split into different energy levels. In octahedral complexes, the d_xy, d_yz, d_zx orbitals (t₂g set) are lower in energy, and the d_x²−y², d_z² orbitals (e_g set) are higher. The splitting energy Δ₀ depends on the ligand: spectrochemical series orders ligands by their field strength.
🔍 Step-by-Step: Octahedral Splitting
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Ligands approach along the x, y, and z axes in an octahedral complex. The d_x²−y² and d_z² orbitals have electron density along these axes → experience strong repulsion → higher energy (e_g set). The d_xy, d_yz, d_zx orbitals have electron density between the axes → less repulsion → lower energy (t₂g set).
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The energy difference Δ₀ = 10 Dq. The t₂g set is stabilized by 4 Dq (0.4Δ₀) and the e_g set is destabilized by 6 Dq (0.6Δ₀). The total energy change is zero (barycenter is conserved).
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Spectrochemical series (increasing Δ₀). I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < F⁻ < OH⁻ < oxalate < H₂O < NCS⁻ < CH₃CN < NH₃ < en < bipy < phen < NO₂⁻ < CO, CN⁻. Strong field ligands (CN⁻, CO) cause large splitting → low spin. Weak field ligands (I⁻, Br⁻) → high spin.
🔍 Step-by-Step: High Spin vs Low Spin
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For d⁴, d⁵, d⁶, d⁷ configurations, two arrangements are possible. If Δ₀ > pairing energy (P): electrons pair in t₂g first → low spin. If Δ₀ < P: electrons occupy t₂g and e_g following Hund's rule → high spin.
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Example: [Fe(CN)₆]⁴⁻ vs [Fe(H₂O)₆]²⁺. Fe²⁺ is d⁶. CN⁻ is strong field → low spin: t₂g⁶ e_g⁰ (all paired, diamagnetic). H₂O is weak field → high spin: t₂g⁴ e_g² (4 unpaired, paramagnetic).
Octahedral: t₂g (lower, 3 orbitals) e_g (higher, 2 orbitals) Δ₀ = 10 Dq Weak field → high spin Strong field → low spin CFSE = (−0.4n_t₂g + 0.6n_e_g)Δ₀
🔍 Common Mistake
Common Mistake — Tetrahedral vs Octahedral
In tetrahedral complexes, the splitting is reversed and smaller: e set (lower, d_x²−y², d_z²) and t₂ set (higher, d_xy, d_yz, d_zx). Also, Δ_t ≈ (4/9)Δ₀, so tetrahedral complexes are always high spin (Δ_t < P).
🎯 NEET Pattern
🎯 NEET Pattern — CFT
• "Which complex is paramagnetic: [Co(NH₃)₆]³⁺ or [CoF₆]³⁻?" → Co³⁺ is d⁶. NH₃ strong → low spin → diamagnetic. F⁻ weak → high spin → paramagnetic (4 unpaired).
• "Color of [Ti(H₂O)₆]³⁺?" → Ti³⁺ is d¹. One electron in t₂g. Absorbs green light → transmits purple (complementary).
• "Which has higher CFSE: [Co(en)₃]³⁺ or [Co(NH₃)₆]³⁺?" → en > NH₃ in spectrochemical series → higher Δ₀ → higher CFSE.
✍️ Worked Example
Example 7: Crystal Field Theory
Calculate the CFSE for [Fe(H₂O)₆]²⁺ (high spin d⁶) and [Fe(CN)₆]⁴⁻ (low spin d⁶) in terms of Δ₀.
Solution: High spin: t₂g⁴ e_g². CFSE = 4×(−0.4) + 2×(0.6) = −1.6 + 1.2 = −0.4Δ₀. Low spin: t₂g⁶ e_g⁰. CFSE = 6×(−0.4) = −2.4Δ₀. Key insight: Low spin configuration has much higher CFSE, which is why strong field ligands favor low spin.
8. Molecular Orbital Theory
💡 Why this matters in real life
MO theory explains why O₂ is paramagnetic (has unpaired electrons) — something that VBT cannot explain. It also explains bond order, bond length, and magnetic properties of diatomic molecules. The concept of bonding and antibonding orbitals is fundamental to understanding chemical bonding.
🧠 The Big Idea — In Plain Words
When atomic orbitals combine, they form molecular orbitals that extend over the entire molecule. Bonding MOs are lower in energy, antibonding MOs are higher. The number of MOs equals the number of atomic orbitals combined. Bond order = (bonding e⁻ − antibonding e⁻)/2. A higher bond order means a stronger, shorter bond.
🔍 Step-by-Step: MO Diagram for Homonuclear Diatomics
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For elements Li₂ to N₂ (Z ≤ 7): The order of MOs is: σ1s < σ*1s < σ2s < σ*2s < π2p_x = π2p_y < σ2p_z < π*2p_x = π*2p_y < σ*2p_z. The π2p orbitals are lower than σ2p due to s-p mixing.
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For O₂, F₂, Ne₂ (Z > 7): No significant s-p mixing. Order: σ1s < σ*1s < σ2s < σ*2s < σ2p_z < π2p_x = π2p_y < π*2p_x = π*2p_y < σ*2p_z. The σ2p is lower than π2p.
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O₂ is paramagnetic. O₂ has 12 valence e⁻. Filling MOs: σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹. Two unpaired e⁻ in π* orbitals → paramagnetic. Bond order = (10 − 6)/2 = 2.
Bond order = (Bonding e⁻ − Antibonding e⁻)/2 | O₂: BO = 2, paramagnetic N₂: BO = 3, diamagnetic
🔍 Common Mistake
Common Mistake — S-p Mixing
Students use the O₂/F₂ MO order for all molecules. For B₂, C₂, N₂, the π2p orbitals are lower than σ2p. This changes the electron configuration. For C₂: (σ2s)²(σ*2s)²(π2p)⁴ — bond order = 2, and all electrons paired.
Calculate the bond order of N₂⁺ and determine if it's paramagnetic or diamagnetic.
Solution: N₂ has 10 e⁻. N₂⁺ removes one from σ2p_z highest orbital. Configuration: σ2s² σ*2s² π2p⁴ σ2p_z¹. BO = (9 − 4)/2 = 2.5. One unpaired e⁻ → paramagnetic. Key insight: N₂⁺ has higher bond order than N₂²⁺ but lower than N₂.