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ChemistryChemical Bonding
Consider the following statements regarding bond order and bond length:
Statement I: A molecule with bond order zero is stable.
Statement II: As bond order increases, bond length also increases.
Which of the above statements is/are INCORRECT?
Options
1
Only Statement I
2
Only Statement II
3
Both Statement I and II
4
Neither Statement I nor II
Correct Answer
Both Statement I and II are incorrect
Solution
1

Statement I: Bond order zero → equal bonding and antibonding electrons → NO net bonding → molecule is unstable/does not exist. Statement I is INCORRECT ✗

2

Statement II: Bond order ↑ → bond length ↓ (DECREASES, not increases).

N≡N: 110 pm (BO=3) < N=N: 124 pm (BO=2) < N-N: 145 pm (BO=1). Statement II is INCORRECT ✗

Both statements are incorrect.
Bond order 0 → unstable molecule | Bond order ↑ → bond length ↓, bond energy ↑
Theory: Chemical Bonding
1. Molecular Orbital Theory (MOT) — Fundamentals

Molecular Orbital Theory (MOT), developed by Hund and Mulliken, treats electrons as belonging to the entire molecule rather than individual atoms. Atomic orbitals combine to form molecular orbitals (MOs). Linear Combination of Atomic Orbitals (LCAO): when two AOs overlap, they form two MOs — a bonding MO (lower energy, constructive interference) and an antibonding MO (higher energy, destructive interference, marked with *). Rules for filling MOs: (1) Aufbau principle: fill lowest energy MO first. (2) Pauli exclusion: max 2 electrons per MO. (3) Hund rule: in degenerate MOs, fill one electron each before pairing. The energy order of MOs for diatomic molecules from Li₂ to N₂: σ1s < σ*1s < σ2s < σ*2s < π2p = π2p < σ2p < π*2p = π*2p < σ*2p. For O₂ and beyond: σ2p comes before π2p.

2. Bond Order and Its Significance

Bond order (BO) = (number of electrons in bonding MOs − number in antibonding MOs) / 2. Bond order indicates: (1) Stability: BO > 0 → stable molecule. BO = 0 → molecule does not exist. BO < 0 is not possible in practice. (2) Bond length: BO ↑ → bond length ↓. More bonding electrons pull nuclei together more tightly. (3) Bond energy: BO ↑ → bond energy ↑. Triple bonds are stronger than double bonds, which are stronger than single bonds. (4) Magnetic properties: molecules with unpaired electrons are paramagnetic; those with all electrons paired are diamagnetic. Examples: H₂ (BO=1, diamagnetic), O₂ (BO=2, paramagnetic — 2 unpaired π* electrons), N₂ (BO=3, diamagnetic), He₂ (BO=0, does not exist), He₂⁺ (BO=0.5, exists briefly).

3. MO Configurations of Important Molecules

H₂: σ1s² — BO=1, diamagnetic, stable. He₂: σ1s²σ*1s² — BO=0, does not exist. Li₂: σ1s²σ*1s²σ2s² — BO=1, diamagnetic. B₂: ...π2p¹π2p¹ — BO=1, paramagnetic (2 unpaired). C₂: ...π2p²π2p² — BO=2, diamagnetic. N₂: ...σ2p² — BO=3, diamagnetic, very stable (triple bond, 945 kJ/mol). O₂: ...π*2p¹π*2p¹ — BO=2, paramagnetic (2 unpaired electrons — explains why O₂ is attracted to a magnet). F₂: ...π*2p²π*2p² — BO=1, diamagnetic. Ne₂: BO=0, does not exist. Ions: O₂⁺ (BO=2.5), O₂⁻ (BO=1.5), N₂⁺ (BO=2.5), NO (BO=2.5). The prediction of paramagnetism of O₂ was a great triumph of MOT over VBT which incorrectly predicts O₂ to be diamagnetic.

4. Valence Bond Theory (VBT) vs MOT

VBT: bond forms when half-filled orbitals of two atoms overlap. Electron pair is localised between two bonded atoms. Hybridisation explains molecular geometry. Simple and intuitive. Limitation: cannot explain paramagnetism of O₂ (VBT predicts all electrons paired = diamagnetic), cannot explain fractional bond orders, fails for molecules like H₂⁺. MOT: electrons delocalized over entire molecule. Explains magnetic properties correctly. Explains bond orders including fractional ones. Predicts stability of molecular ions. Limitation: more complex, molecular orbital diagrams required. NEET requires both: VBT + hybridisation for structure/geometry; MOT for magnetic properties, bond order, stability of species. Resonance structures in VBT are an approximation of what MOT describes as electron delocalization.

5. VSEPR Theory and Molecular Geometry

VSEPR (Valence Shell Electron Pair Repulsion): electron pairs in valence shell repel each other and adopt geometry that minimises repulsion. Order of repulsion: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. This ordering explains why actual bond angles deviate from ideal. Examples: CH₄ (4 BP, 0 LP): tetrahedral, 109.5°. NH₃ (3 BP, 1 LP): trigonal pyramidal, 107° (one LP compresses angle). H₂O (2 BP, 2 LP): bent, 104.5° (two LPs compress further). PCl₅ (5 BP, 0 LP): trigonal bipyramidal. SF₆ (6 BP): octahedral. XeF₂ (2 BP, 3 LP): linear (3 LPs occupy equatorial positions in trigonal bipyramidal arrangement). XeF₄ (4 BP, 2 LP): square planar. The VSEPR model is remarkably successful at predicting geometry from Lewis structures alone, without complex quantum mechanical calculations.

6. Hybridisation and Molecular Shapes

sp: linear, 180°. Examples: BeCl₂, CO₂, C₂H₂ (acetylene). sp²: trigonal planar, 120°. Examples: BF₃, SO₃, C₂H₄ (ethylene), benzene. sp³: tetrahedral, 109.5°. Examples: CH₄, NH₄⁺, CCl₄, SiF₄. sp³d: trigonal bipyramidal, 90°/120°. Examples: PCl₅, SF₄, ClF₃. sp³d²: octahedral, 90°. Examples: SF₆, [Co(NH₃)₆]³⁺. sp³d³: pentagonal bipyramidal. Examples: IF₇. Hybridisation determines: molecular geometry, bond angles, existence of isomers (cis-trans), reactivity (sp³ less reactive toward electrophiles than sp²). In benzene: each C is sp² hybridised, unhybridised p orbitals form the π system (6 electrons delocalised in two ring-shaped MOs above and below the ring).

7. Hydrogen Bonding

Hydrogen bond forms when H is covalently bonded to a highly electronegative atom (F, O, N) — the H becomes partially positive and is attracted to a lone pair on another electronegative atom. Intermolecular H-bonding: between different molecules. Increases boiling point, melting point, viscosity, surface tension. Examples: H₂O (boiling point 100°C vs H₂S which has higher MW but no H-bonding at −60°C), HF (boiling point 20°C), NH₃, alcohols, carboxylic acids. Intramolecular H-bonding: within same molecule. Decreases boiling point. Examples: o-nitrophenol (lower bp than p-nitrophenol), o-chlorophenol, salicylaldehyde. Special case: water has 4 H-bonds per molecule (2 as donor, 2 as acceptor) → tetrahedral network → explains anomalous expansion on freezing (ice less dense than water), high specific heat, high surface tension.

8. Van der Waals Forces

Weak intermolecular forces that hold non-polar molecules together. Three types: (1) London dispersion (induced dipole-induced dipole): present in ALL molecules. Increases with molecular size (more electrons → larger temporary dipoles). Explains why larger noble gases (Xe, Kr) have higher boiling points than smaller ones (He, Ne). (2) Dipole-dipole: between polar molecules (HCl, SO₂). Stronger than London forces at same molecular size. (3) Dipole-induced dipole (Debye forces): permanent dipole of one molecule induces dipole in another. Weakest of the three directed forces. The total van der Waals force = London + dipole-dipole + dipole-induced. In addition: hydrogen bonding is much stronger (~20 kJ/mol) compared to van der Waals (~1-5 kJ/mol) and is responsible for many unique properties of water and biological molecules (DNA double helix, protein secondary structure).

Frequently Asked Questions
1. What happens to a molecule with bond order zero?
Bond order = 0 means bonding electrons = antibonding electrons. Net bonding = 0. This means: (1) No net attractive force between the two atoms. (2) The molecule is unstable and does not exist as a stable species. (3) The BO = 0 case does not mean atoms simply exist as separate atoms — rather, forming the "molecule" provides no energy advantage over being separate atoms. Example: He₂ has σ1s²σ*1s² giving BO = (2-2)/2 = 0. Experimentally, He₂ is not observed under normal conditions. Note: He₂⁺ has BO = 0.5 (one bonding, one antibonding electron → net of 0.5 bond) and does exist transiently. The statement "BO=0 → stable" is completely wrong.
2. What is the trend of bond lengths with bond order?
Bond order increases → bond length decreases. This is because: more electrons in bonding MOs means stronger attractive force between nuclei and electrons → nuclei pulled closer together → shorter bond. Quantitative examples for carbon compounds: C-C single bond: 154 pm (BO=1). C=C double bond: 134 pm (BO=2). C≡C triple bond: 120 pm (BO=3). For nitrogen: N-N: 145 pm (BO=1). N=N: 124 pm (BO=2). N≡N: 110 pm (BO=3). The decrease is not linear — each additional bond order decreases length by less than the previous one. The shortening from single to double (~20 pm) is more than from double to triple (~14 pm) for C-C bonds.
3. How does bond order relate to bond energy?
Bond energy (dissociation energy) increases with bond order: BO ↑ → bond energy ↑. More bonds hold atoms together more tightly → more energy needed to break them apart. Examples for C-C bonds: C-C: 347 kJ/mol. C=C: 614 kJ/mol. C≡C: 839 kJ/mol. Note: double bond energy (614) is not exactly twice single bond (347) — it is less. Triple is not three times single. This is because: (1) π bonds are weaker than σ bonds (less effective overlap). (2) σ bonds involve head-on overlap (maximum); π bonds involve side-on overlap (less effective). Bond order correlates with both short bonds (high BO → short) and strong bonds (high BO → high energy).
4. Why is O₂ paramagnetic according to MOT?
O₂ MO configuration: σ1s²σ*1s²σ2s²σ*2s²σ2p²π2p²π2p²π*2p¹π*2p¹σ*2p⁰. The last two electrons go into the two degenerate π*2p orbitals — by Hund rule, one in each with parallel spins. This gives 2 unpaired electrons → O₂ is paramagnetic. BO = (8 bonding - 4 antibonding)/2 = 2. VBT predicts O₂ as diamagnetic (all electrons paired in Lewis structure with one double bond) — this was a major failure of VBT. MOT's correct prediction of O₂ paramagnetism (confirmed experimentally — liquid O₂ is attracted to a magnet) was a major triumph and helped establish MOT as the more accurate theory.
5. What is the MO configuration and bond order of NO?
NO (11 electrons, one more than N₂): MO configuration: σ1s²σ*1s²σ2s²σ*2s²σ2p²π2p²π2p²π*2p¹. Bond order = (8-3)/2 = 2.5. The extra electron compared to N₂ goes into an antibonding π* MO, reducing bond order from 3 to 2.5. NO is paramagnetic (1 unpaired electron in π*). NO has fractional bond order — a key example of MOT's ability to handle species that VBT cannot describe simply. NO⁺ (nitrosonium, 10 electrons): same as N₂ configuration → BO=3, isoelectronic with N₂. NO⁻ (11+1=12 electrons): BO=2. So removing an electron from NO (→NO⁺) strengthens the bond; adding one (→NO⁻) weakens it.
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