This MCQ module is based on: Molecular Orbital Theory Hbond
TOPIC 17 OF 28
Molecular Orbital Theory Hbond
🎓 Class 11
Chemistry
CBSE
Theory
Ch 4 – Chemical Bonding and Molecular Structure
⏱ ~14 min
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This assessment will be based on: Molecular Orbital Theory Hbond
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Molecular Orbital Theory and Hydrogen Bonding
4.11 Molecular Orbital Theory (MOT)
Molecular Orbital Theory (MOT) was developed by F. Hund and R. S. Mulliken in 1932. It treats the entire molecule as a single unit and the bonding electrons as occupying molecular orbitals (MOs) that are spread over the whole molecule, rather than belonging to a specific bond.
4.11.1 Key Postulates
- Atomic orbitals (AOs) of comparable energy and proper symmetry combine to form MOs.
- The number of MOs formed = number of AOs combined.
- Combination is described by the Linear Combination of Atomic Orbitals (LCAO) method.
- Adding wave functions (in-phase) gives a bonding MO (lower energy); subtracting (out-of-phase) gives an antibonding MO (higher energy, marked with *).
- MOs are filled following the Aufbau principle, Pauli exclusion and Hund's rule.
Bonding MO: energy lower than the constituent AOs; electron density concentrated between the nuclei → stabilises the molecule.
Antibonding MO: energy higher than the constituent AOs; a node between the nuclei → destabilises the molecule.
Antibonding MO: energy higher than the constituent AOs; a node between the nuclei → destabilises the molecule.
4.11.2 Conditions for Linear Combination
- Combining atomic orbitals must have comparable energies (e.g., 1s of one H with 1s of another, not 1s of H with 2s of Na).
- They must have proper symmetry with respect to the inter-nuclear axis.
- They must have significant overlap.
4.11.3 Types of Molecular Orbitals from p-orbitals
- 2px + 2px (axial) → σ₂px (bonding) and σ*₂px (antibonding).
- 2py + 2py (lateral) → π₂py (bonding) and π*₂py (antibonding).
- 2pz + 2pz (lateral) → π₂pz (bonding) and π*₂pz (antibonding).
4.11.4 Energy Order of MOs
For homonuclear diatomic molecules of period 2:
For Li₂, Be₂, B₂, C₂, N₂ (s–p mixing operative):
σ1s < σ*1s < σ2s < σ*2s < π2py = π2pz < σ2px < π*2py = π*2pz < σ*2px
For O₂, F₂, Ne₂ (no s–p mixing):
σ1s < σ*1s < σ2s < σ*2s < σ2px < π2py = π2pz < π*2py = π*2pz < σ*2px
σ1s < σ*1s < σ2s < σ*2s < π2py = π2pz < σ2px < π*2py = π*2pz < σ*2px
For O₂, F₂, Ne₂ (no s–p mixing):
σ1s < σ*1s < σ2s < σ*2s < σ2px < π2py = π2pz < π*2py = π*2pz < σ*2px
4.11.5 Bond Order (BO)
Bond Order = (Nb − Na)/2, where Nb = number of bonding electrons and Na = number of antibonding electrons.
Implications: BO > 0 → stable molecule; BO = 0 → molecule does not exist; higher BO → shorter, stronger bond. An odd BO (½, 3⁄2) means an unpaired electron and paramagnetism.
4.12 Bonding in Some Homonuclear Diatomic Molecules
H₂ Molecule
Total electrons = 2; configuration: (σ1s)². Nb = 2, Na = 0 → BO = 1. Diamagnetic. Bond enthalpy 435.8 kJ mol⁻¹, bond length 74 pm.
He₂ Molecule (does not exist)
4 electrons: (σ1s)² (σ*1s)². BO = (2−2)/2 = 0 → He₂ does not form. Same applies to Be₂.
Li₂
6 electrons: KK (σ2s)². BO = 1. Diamagnetic. Detected in vapour phase.
N₂ Molecule
14 electrons: KK (σ2s)² (σ*2s)² (π2py)² (π2pz)² (σ2px)². Nb = 10, Na = 4 → BO = 3. Diamagnetic. Bond enthalpy 946 kJ mol⁻¹, bond length 110 pm.
O₂ Molecule (Paramagnetism Explained)
16 electrons: KK (σ2s)² (σ*2s)² (σ2px)² (π2py)² (π2pz)² (π*2py)¹ (π*2pz)¹.
Nb = 10, Na = 6 → BO = 2. The two unpaired electrons in degenerate π* orbitals (Hund's rule) make O₂ paramagnetic — a triumph of MOT over VBT.
Summary of Period-2 Diatomics
| Molecule | Total e⁻ | Configuration (after KK = (σ1s)²(σ*1s)²) | BO | Magnetic | Bond enthalpy (kJ/mol) | Bond length (pm) |
|---|---|---|---|---|---|---|
| H₂ | 2 | (σ1s)² | 1 | Dia | 436 | 74 |
| He₂ | 4 | (σ1s)²(σ*1s)² | 0 | — | not formed | — |
| Li₂ | 6 | KK (σ2s)² | 1 | Dia | 105 | 267 |
| Be₂ | 8 | KK (σ2s)²(σ*2s)² | 0 | — | — | — |
| B₂ | 10 | KK (σ2s)²(σ*2s)²(π2p)² | 1 | Para | 289 | 159 |
| C₂ | 12 | KK (σ2s)²(σ*2s)²(π2p)⁴ | 2 | Dia | 620 | 131 |
| N₂ | 14 | … (π2p)⁴(σ2p)² | 3 | Dia | 946 | 110 |
| O₂ | 16 | … (σ2p)²(π2p)⁴(π*2p)² | 2 | Para | 498 | 121 |
| F₂ | 18 | … (π*2p)⁴ | 1 | Dia | 158 | 142 |
| Ne₂ | 20 | … (σ*2p)² | 0 | — | — | — |
Interactive: MO Bond Order Calculator
Pick a homonuclear diatomic species; the simulator returns the MO configuration, bond order, magnetic behaviour and stability.
Total electrons: 14
Bond order: 3
Magnetic behaviour: Diamagnetic
Triple bond, very stable.
4.13 Hydrogen Bonding
Nitrogen, oxygen and fluorine are very electronegative. When such an atom is bonded to a hydrogen atom, the H–X bond is highly polar. The slightly positive H of one molecule is attracted by the lone pair of the electronegative atom of a neighbouring molecule — this attraction is called the hydrogen bond.
Hydrogen bond: An electrostatic attraction X–H···Y where X = F/O/N (donor) and Y carries a lone pair (acceptor). Bond enthalpy 10–40 kJ mol⁻¹.
4.13.1 Cause of H-Bond Formation
Two conditions must be met:
- H must be bonded to a small, highly electronegative atom (F, O, N).
- The acceptor atom must have at least one lone pair.
4.13.2 Types of Hydrogen Bonds
(a) Intermolecular — between two different molecules of the same or different substances. Examples: H₂O···H₂O (water), H–F···H–F (HF), CH₃OH···CH₃OH.
(b) Intramolecular — within the same molecule between two groups bearing H and the lone pair. Examples: o-nitrophenol, salicylaldehyde — these have lower b.p. than their para isomers because the H-bond is internal and does not connect different molecules.
Why Ice Floats on Water
In ice, every water molecule is H-bonded tetrahedrally to four neighbours, building a low-density hexagonal cage. On melting, ~15% of the H-bonds break and the molecules pack more efficiently → liquid water at 0–4 °C is denser than ice. Density maximum at 4 °C (1.000 g cm⁻³) versus ice (0.917 g cm⁻³). This anomaly is critical for aquatic life: ice forms on top, insulating the water below.
Hydrogen Bonding in Biology
The two strands of DNA are held together by H-bonds: A=T (two H-bonds) and G≡C (three H-bonds). This makes the genetic code stable yet readable. Proteins also fold into α-helices and β-sheets via networks of N–H···O hydrogen bonds.
Activity 4.4 — H-Bond Detective
Setup: Compare the boiling points (in °C): HF (19), HCl (−85), HBr (−67), HI (−35); H₂O (100), H₂S (−60), H₂Se (−42), H₂Te (−2); NH₃ (−33), PH₃ (−87), AsH₃ (−55).
Predict: Why are HF, H₂O and NH₃ anomalously high compared to their group neighbours?
F, O and N are small and very electronegative. They form strong intermolecular H-bonds in HF, H₂O and NH₃. Extra energy is needed to break these H-bonds in addition to van der Waals forces → much higher boiling points. Cl, S, P etc. are bigger and less electronegative, so no significant H-bonding → boiling points follow the normal trend (rise with molar mass).
Strength of H-bonds in the order F···H–F > O···H–O > N···H–N (matches electronegativity of the donor atom).
Worked Example 4.8 — Bond order, magnetic behaviour of O₂⁻
O₂⁻ has 17 electrons. Configuration: KK (σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)³.
Nb = 10, Na = 7 → BO = 1.5. One unpaired electron in π* → paramagnetic. Bond order intermediate between O₂ (2) and O₂²⁻ (1) — consistent with the longer O–O bond in superoxide.
Nb = 10, Na = 7 → BO = 1.5. One unpaired electron in π* → paramagnetic. Bond order intermediate between O₂ (2) and O₂²⁻ (1) — consistent with the longer O–O bond in superoxide.
Worked Example 4.9 — Why is N₂ more stable than O₂?
N₂ BO = 3 (triple bond, 946 kJ mol⁻¹) — all bonding electrons, no electrons in antibonding π*. O₂ BO = 2 with 2 unpaired electrons in π* (498 kJ mol⁻¹). The antibonding electrons in O₂ weaken the bond and make oxygen reactive (combustion!), while N₂ is famously inert.
Competency-Based Questions
Q1. The bond order of O₂²⁻ (peroxide ion) is:L1 Remember
Answer: (b) 1. 18 electrons; π* orbitals filled with 4 e⁻; BO = (10−8)/2 = 1.
Q2. Why does He₂ not exist?L2 Understand
4 electrons fill (σ1s)²(σ*1s)². Bonding and antibonding contributions cancel → BO = 0 → no net bond → no molecule.
Q3. Apply MOT to predict the magnetic property of B₂.L3 Apply
B₂ has 10 e⁻. Owing to s–p mixing, the order is σ2s < σ*2s < π2p < σ2p. Configuration: KK (σ2s)²(σ*2s)²(π2py)¹(π2pz)¹. Two unpaired electrons → paramagnetic; BO = 1.
Q4. Why is the boiling point of H₂O (100 °C) much higher than that of H₂S (−60 °C)?L4 Analyse
O is much smaller and far more electronegative than S → strong O–H···O hydrogen bonds in liquid water. H₂S has only weak van der Waals forces (S is too large/less electronegative for effective H-bonding). Breaking the extensive H-bond network in water requires much more energy.
Q5. Compare the bond orders and bond lengths of O₂, O₂⁺, O₂⁻ and O₂²⁻ and arrange in order of bond strength.L5 Evaluate
O₂⁺ (15 e⁻): BO = 2.5 (shortest, strongest). O₂ (16): BO = 2. O₂⁻ (17): BO = 1.5. O₂²⁻ (18): BO = 1 (longest, weakest). Order of bond strength: O₂⁺ > O₂ > O₂⁻ > O₂²⁻. Removing an antibonding electron strengthens the bond; adding one weakens it.
Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Liquid oxygen is paramagnetic.
R: O₂ has two unpaired electrons in degenerate π*2p orbitals.
Answer: (A). Both true; R correctly explains A. This was a major MOT victory over VBT, which predicted O₂ to be diamagnetic.
A: Ice is less dense than liquid water.
R: In ice each water molecule is tetrahedrally H-bonded to four neighbours forming an open hexagonal cage with empty spaces.
Answer: (A). Both true; R correctly explains A. The empty cages collapse on melting → density increases.
A: The boiling point of o-nitrophenol is lower than that of p-nitrophenol.
R: o-Nitrophenol has intramolecular H-bonding while p-nitrophenol has intermolecular H-bonding.
Answer: (A). Both true; R correctly explains A. Intramolecular H-bond stays inside the molecule → no association → low boiling point. Intermolecular H-bonding links many molecules → much higher boiling point.
Frequently Asked Questions — Molecular Orbital Theory and Hydrogen Bonding
What is molecular orbital theory and how does it differ from VBT?
Molecular orbital theory (MOT), developed by Hund and Mulliken, treats electrons in molecules as moving in molecular orbitals (MOs) that extend over the entire molecule, formed by linear combination of atomic orbitals (LCAO). Two AOs combine to form one bonding MO (lower energy) and one antibonding MO (higher energy). MOT differs from VBT in that VBT views bonds as localised between atom pairs, whereas MOT views electrons as delocalised. NCERT Class 11 Chemistry Chapter 4 highlights MOT's strength in explaining the paramagnetism of O₂ and the existence of stable He₂⁺ — properties VBT cannot explain easily.
How is bond order calculated using MOT?
Bond order in MOT is calculated as: Bond order = ½ (N_b − N_a), where N_b is the number of electrons in bonding MOs and N_a is in antibonding MOs. Bond order indicates bond stability: 1 = single bond, 2 = double bond, 3 = triple bond, 0 = no bond. Higher bond order means stronger and shorter bonds. NCERT Class 11 Chemistry Chapter 4 examples: H₂ has bond order 1 (2 electrons in σ1s, 0 in σ*1s), He₂ has bond order 0 (does not exist), N₂ has bond order 3, O₂ has bond order 2, and F₂ has bond order 1.
Why is O₂ paramagnetic according to MOT?
O₂ is paramagnetic because its molecular orbital configuration places two unpaired electrons in the degenerate π*2p_x and π*2p_y antibonding orbitals. The configuration is (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)¹. According to Hund's rule, these two electrons occupy separate orbitals with parallel spins, making O₂ paramagnetic and attracted by a magnetic field. VBT predicts O₂ should be diamagnetic, contradicting experiment. This is a major triumph of MOT and is emphasised in NCERT Class 11 Chemistry Chapter 4.
What is hydrogen bonding?
Hydrogen bonding is a special dipole-dipole attractive force between a hydrogen atom covalently bonded to a highly electronegative atom (F, O or N) and another electronegative atom with a lone pair in the same or different molecule. The H atom carries a partial positive charge and is attracted to the lone pair of F, O or N. Hydrogen bonds are weaker than covalent bonds (~5–40 kJ/mol vs ~300–600 kJ/mol) but stronger than other intermolecular forces. NCERT Class 11 Chemistry Chapter 4 covers hydrogen bonding as the reason for unusually high boiling points of H₂O, HF and NH₃.
What are intermolecular and intramolecular hydrogen bonds?
Intermolecular hydrogen bonds form between H of one molecule and the electronegative atom of another molecule. Examples: H₂O, NH₃, alcohols, carboxylic acids — these have high boiling points and water solubility. Intramolecular hydrogen bonds form within the same molecule when geometry permits. Example: o-nitrophenol has an intramolecular H-bond between the -OH and -NO₂ groups giving a 6-membered ring; p-nitrophenol has only intermolecular H-bonds. NCERT Class 11 Chemistry uses this distinction to explain why o-nitrophenol has a lower boiling point and is more volatile than p-nitrophenol.
Why is ice less dense than water?
Ice is less dense than liquid water because of extensive intermolecular hydrogen bonding. In ice, each water molecule forms four H-bonds with its neighbours in a perfectly tetrahedral arrangement, creating an open cage-like crystal structure with empty hexagonal channels. When ice melts, the rigid H-bond network partially collapses, packing molecules closer and increasing density. Density of liquid water reaches a maximum at 4°C. NCERT Class 11 Chemistry Chapter 4 notes this anomaly is critical for aquatic life — ice floats on water, insulating fish during winter, while water below stays liquid at ~4°C.
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