Prerequisites
Before studying MOT, review:
- Atomic Orbitals — s, p, d orbital shapes
- Electronic Configuration — Aufbau principle
- Covalent Bonding — Basic bonding concepts
- Hybridization — Alternative bonding theory (VBT)
The Hook: The Oxygen Mystery VBT Couldn’t Solve
The year is 1920. A problem haunts chemists worldwide:
Liquid oxygen is BLUE and sticks to magnets!
But according to Lewis structures and Valence Bond Theory:
- O₂ should have all electrons paired → diamagnetic
- Reality: O₂ has unpaired electrons → paramagnetic!
VBT predicts: O=O with no unpaired electrons ✗ Reality: O₂ attracted to magnets ✓
How did scientists solve this? Enter Molecular Orbital Theory — the theory that treats molecules as unified quantum systems, not just atoms holding hands!
In Oppenheimer (2023), understanding quantum mechanics (including MO theory) was crucial for predicting nuclear reactions. The atomic bomb worked because physicists understood how electrons behave in molecules — not as separate atoms, but as quantum wavefunctions!
JEE Reality: 2-3 questions every year worth 8-12 marks. MOT explains what VBT can’t — magnetic properties, bond order variations, and stability of unusual molecules!
The Core Concept
What is Molecular Orbital Theory?
Molecular Orbital Theory (MOT): Electrons in molecules occupy molecular orbitals (MOs) that belong to the entire molecule, not individual atoms.
$$\boxed{\text{Atomic Orbitals combine} \rightarrow \text{Molecular Orbitals (spread over whole molecule)}}$$In simple terms:
VBT (old way): Two atoms shake hands (overlap orbitals) but keep their identities.
MOT (quantum way): Two atoms merge completely. Their orbitals create new molecular orbitals that belong to BOTH atoms. Electrons move freely in these new orbitals!
Key Principles of MOT
1. Atomic orbitals combine to form molecular orbitals
- When two atoms approach, their atomic orbitals (AOs) interact
- Linear combination creates new molecular orbitals (MOs)
2. Number of MOs = Number of AOs combined
- 2 atomic orbitals → 2 molecular orbitals
- Conservation of orbitals!
3. Two types of MOs form:
- Bonding MO (σ, π): Lower energy, stabilizing (electron density between nuclei)
- Antibonding MO (σ, π):** Higher energy, destabilizing (electron density outside bonding region)
4. Electrons fill MOs following:
- Aufbau principle: Lowest energy first
- Pauli exclusion: Max 2 electrons per MO (opposite spins)
- Hund’s rule: Equal energy MOs fill singly first
5. Bond order determines stability:
$$\text{Bond Order} = \frac{1}{2}(N_b - N_a)$$Where $N_b$ = electrons in bonding MOs, $N_a$ = electrons in antibonding MOs
Formation of Molecular Orbitals
Bonding Molecular Orbital (σ)
Formation: Atomic orbitals combine in-phase (constructive interference)
AO₁ + AO₂ → Bonding MO (σ)
Wave: ∼ + ∼ → ∼∼ (amplified in middle)
Characteristics:
- Lower energy than parent AOs
- Electron density concentrated between nuclei
- Stabilizes the molecule
- Designated: σ (sigma) or π (pi)
Example: H₂
1s (H₁) + 1s (H₂) → σ₁s (bonding)
Energy decreases by Δ
Stability increases
Antibonding Molecular Orbital (σ*)
Formation: Atomic orbitals combine out-of-phase (destructive interference)
AO₁ - AO₂ → Antibonding MO (σ*)
Wave: ∼ + ∿ → |∼∿| (node in middle)
Characteristics:
- Higher energy than parent AOs
- Electron density outside bonding region
- Destabilizes the molecule (weakens bond)
- Designated: σ* (sigma star) or π* (pi star)
- Has a node (zero electron density) between nuclei
Think of it like waves:
Bonding MO: Two waves add up (peaks align) → Constructive interference → Electron density builds up between nuclei → Attractive → Stabilizing
Antibonding MO: Two waves cancel (peak + trough) → Destructive interference → Node between nuclei → Electron density outside → Repulsive → Destabilizing
Energy diagram:
Energy ↑
σ* (antibonding) — Higher energy
1s 1s (atomic orbitals) — Original energy
σ (bonding) — Lower energy
Key fact: Energy decrease in bonding MO > Energy increase in antibonding MO That’s why bonds form!
Types of Molecular Orbitals
σ (Sigma) Molecular Orbitals
Formation: Head-on (axial) overlap of atomic orbitals
Types:
- σ(s-s): s orbital + s orbital (e.g., H₂)
- σ(s-p): s orbital + p orbital (e.g., HCl)
- σ(p-p): p orbital + p orbital (end-to-end) (e.g., pz + pz)
Characteristics:
- Cylindrically symmetrical around bond axis
- Electron density along internuclear axis
- Stronger than π bonds (better overlap)
Example: In O₂, σ₂pz is formed by head-on overlap of 2pz orbitals.
π (Pi) Molecular Orbitals
Formation: Lateral (sideways) overlap of p orbitals
Types:
- π(p-p): px + px or py + py (sideways overlap)
Characteristics:
- Electron density above and below bond axis (not on axis!)
- Weaker than σ bonds (poorer overlap)
- Zero electron density on internuclear axis
- Two lobes (one above, one below bonding axis)
Example: In O₂, π₂px and π₂py are formed by sideways overlap of 2px and 2py orbitals.
MO Energy Order for Homonuclear Diatomic Molecules
For O₂ and F₂ (Z > 7)
$$\boxed{\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}}$$Key feature: σ₂pz below π₂px, π₂py
For N₂, C₂, B₂ (Z ≤ 7)
$$\boxed{\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}}$$Key feature: π₂px, π₂py below σ₂pz (order swapped!)
The energy order of σ₂pz and π₂p orbitals DEPENDS on the element!
For B₂, C₂, N₂ (Z ≤ 7): π₂p comes before σ₂pz (due to s-p mixing)
For O₂, F₂ (Z > 7): σ₂pz comes before π₂p (less s-p mixing)
Why? As Z increases, energy gap between 2s and 2p increases → less orbital mixing → order changes!
JEE Trick:
- Elements Li to N: π below σ
- Elements O to Ne: σ below π
Memory trick: “Nitrogen and Lighter: π comes Lower”
Drawing MO Diagrams: Step-by-Step
Example 1: H₂ Molecule
Step 1: Count total electrons = 1 + 1 = 2 electrons
Step 2: Combine atomic orbitals
- H: 1s
- Two 1s orbitals → σ₁s (bonding) + σ*₁s (antibonding)
Step 3: Fill MOs with electrons (Aufbau principle)
- 2 electrons → both in σ₁s (lowest energy)
Step 4: Calculate bond order
$$\text{BO} = \frac{1}{2}(2 - 0) = 1$$MO Diagram:
Energy ↑
H atom H₂ molecule H atom
σ*₁s __
1s ↑↓ ↑↓ 1s
σ₁s ↑↓
Result:
- Bond order = 1 (single bond)
- All electrons paired → Diamagnetic
- Stable molecule ✓
Example 2: O₂ Molecule (The Famous One!)
Step 1: Count total electrons = 8 + 8 = 16 electrons
Step 2: Energy order (O has Z = 8 > 7)
$$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}$$Step 3: Fill 16 electrons
- σ₁s: 2e⁻
- σ*₁s: 2e⁻
- σ₂s: 2e⁻
- σ*₂s: 2e⁻
- σ₂pz: 2e⁻
- π₂px: 2e⁻
- π₂py: 2e⁻
- π*₂px: 1e⁻ ← unpaired!
- π*₂py: 1e⁻ ← unpaired!
Step 4: Calculate bond order
- Bonding electrons: σ₁s(2) + σ₂s(2) + σ₂pz(2) + π₂px(2) + π₂py(2) = 10
- Antibonding electrons: σ₁s(2) + σ₂s(2) + π₂px(1) + π₂py(1) = 6 $$\text{BO} = \frac{1}{2}(10 - 6) = 2$$
MO Diagram:
Energy ↑
σ*₂pz __
π*₂px ↑ π*₂py ↑ ← UNPAIRED!
π₂px ↑↓ π₂py ↑↓
σ₂pz ↑↓
σ*₂s ↑↓
σ₂s ↑↓
σ*₁s ↑↓
σ₁s ↑↓
Result:
- Bond order = 2 (double bond) ✓
- 2 unpaired electrons → Paramagnetic ✓✓✓
- This explains why O₂ is attracted to magnets!
Interactive Demo: Visualize Molecular Orbital Formation
See how atomic orbitals combine to form bonding and antibonding molecular orbitals. Explore electron filling patterns and understand why O₂ is paramagnetic!
VBT prediction:
- Lewis structure: O=O
- All electrons paired
- Diamagnetic ✗
MOT prediction:
- Bond order = 2 (double bond) ✓
- 2 unpaired electrons in π*
- Paramagnetic ✓
Reality: O₂ is paramagnetic (proven experimentally)
Winner: MOT! This is why MOT is superior for predicting magnetic properties.
Example 3: N₂ Molecule
Step 1: Count total electrons = 7 + 7 = 14 electrons
Step 2: Energy order (N has Z = 7 ≤ 7)
$$\pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z}$$(Note the order!)
Step 3: Fill 14 electrons
- σ₁s: 2e⁻
- σ*₁s: 2e⁻
- σ₂s: 2e⁻
- σ*₂s: 2e⁻
- π₂px: 2e⁻
- π₂py: 2e⁻
- σ₂pz: 2e⁻
Step 4: Calculate bond order
- Bonding: 2 + 2 + 2 + 2 + 2 = 10
- Antibonding: 2 + 2 = 4 $$\text{BO} = \frac{1}{2}(10 - 4) = 3$$
Result:
- Bond order = 3 (triple bond!)
- All electrons paired → Diamagnetic
- Extremely stable (highest bond order = strongest bond)
This is why N₂ is so unreactive — triple bond is very strong!
Bond Order: The Stability Predictor
Formula
$$\boxed{\text{Bond Order (BO)} = \frac{1}{2}(N_b - N_a)}$$Where:
- $N_b$ = Number of electrons in bonding MOs
- $N_a$ = Number of electrons in antibonding MOs
Interpretation
| Bond Order | Meaning | Example |
|---|---|---|
| 0 | No bond, molecule doesn’t exist | He₂ |
| 0.5 | Very weak bond, unstable | He₂⁺ |
| 1 | Single bond | H₂, F₂ |
| 1.5 | Between single and double | O₂⁺ |
| 2 | Double bond | O₂ |
| 2.5 | Between double and triple | — |
| 3 | Triple bond | N₂ |
Bond Order Correlations
Higher bond order means:
✓ Stronger bond (more energy to break) ✓ Shorter bond length (atoms pulled closer) ✓ Higher bond energy ✓ More stable molecule ✓ Lower reactivity (harder to break)
Lower bond order means:
✗ Weaker bond ✗ Longer bond length ✗ Lower bond energy ✗ Less stable ✗ Higher reactivity
Example: O₂ series
| Species | Electrons | Bond Order | Bond Length | Magnetic |
|---|---|---|---|---|
| O₂²⁻ | 18 | 1 | Longest | Diamagnetic |
| O₂⁻ | 17 | 1.5 | Long | Paramagnetic |
| O₂ | 16 | 2 | Medium | Paramagnetic |
| O₂⁺ | 15 | 2.5 | Short | Paramagnetic |
| O₂²⁺ | 14 | 3 | Shortest | Diamagnetic |
Pattern: As electrons are removed, bond order increases → bond gets shorter and stronger!
Magnetic Properties from MOT
Diamagnetic vs Paramagnetic
Diamagnetic:
- All electrons paired
- Weakly repelled by magnetic field
- Examples: H₂, N₂, F₂, O₂²⁻
Paramagnetic:
- Has unpaired electrons
- Attracted to magnetic field
- Degree of attraction ∝ number of unpaired electrons
- Examples: O₂, B₂, O₂⁻, O₂⁺
After filling MO diagram:
Count unpaired electrons:
- 0 unpaired → Diamagnetic
- 1+ unpaired → Paramagnetic
Quick check without full diagram:
- Even total electrons + no half-filled orbitals → Usually diamagnetic
- Odd total electrons → Always paramagnetic (must have at least 1 unpaired)
- O₂ (16e⁻) → Exception! Even electrons but paramagnetic due to MO ordering
For O₂ family:
- O₂²⁺, O₂²⁻: Diamagnetic (all paired)
- O₂, O₂⁺, O₂⁻: Paramagnetic (unpaired in π*)
Stability Comparisons Using MOT
Example: H₂ vs He₂
H₂:
- Total electrons: 2
- Configuration: σ₁s²
- BO = ½(2 - 0) = 1
- Stable, exists ✓
He₂:
- Total electrons: 4
- Configuration: σ₁s² σ*₁s²
- BO = ½(2 - 2) = 0
- No bond, doesn’t exist ✗
Conclusion: He₂ cannot exist because bonding and antibonding effects cancel!
Example: O₂ vs O₂⁺ vs O₂⁻
| Species | Electrons | BO | Bond Length (pm) | Stability | Magnetic |
|---|---|---|---|---|---|
| O₂⁺ | 15 | 2.5 | 112 | Most stable | Paramagnetic (1 unpaired) |
| O₂ | 16 | 2.0 | 121 | Stable | Paramagnetic (2 unpaired) |
| O₂⁻ | 17 | 1.5 | 126 | Less stable | Paramagnetic (3 unpaired) |
| O₂²⁻ | 18 | 1.0 | 149 | Least stable | Diamagnetic |
Trend: BO decreases as electrons are added → bond weakens and lengthens
Common question: “Which has the shortest bond length: O₂, O₂⁺, or O₂⁻?”
Solution:
- Calculate bond orders: O₂⁺ (2.5), O₂ (2.0), O₂⁻ (1.5)
- Higher BO → Shorter bond
- Answer: O₂⁺
Takes 10 seconds if you know the pattern!
Common Molecules: Quick Reference
H₂ (Hydrogen)
- Electrons: 2
- Configuration: σ₁s²
- BO: 1, Diamagnetic, Stable
He₂ (Helium — doesn’t exist!)
- Electrons: 4
- Configuration: σ₁s² σ*₁s²
- BO: 0, Cannot exist
N₂ (Nitrogen)
- Electrons: 14
- BO: 3 (triple bond!)
- Diamagnetic, Very stable (unreactive)
O₂ (Oxygen)
- Electrons: 16
- BO: 2 (double bond)
- Paramagnetic (2 unpaired), Blue liquid
F₂ (Fluorine)
- Electrons: 18
- BO: 1 (single bond)
- Diamagnetic, Reactive (weak F-F bond)
B₂ (Boron)
- Electrons: 10
- BO: 1
- Paramagnetic (2 unpaired)
C₂ (Carbon)
- Electrons: 12
- BO: 2
- Diamagnetic, Exists in vapor at high temp
Memory Tricks & Patterns
MO Energy Order Mnemonic
For Z ≤ 7 (B, C, N): “Silly Students See People Playing Before Singing”
- Sigma 1s → Sigma* 1s → Sigma 2s → Sigma* 2s → Pi 2p → Pi* 2p (wait, wrong!)
Actually: “π before σ” for lighter elements
For Z > 7 (O, F): “σ before π” (normal order)
Bond Order Quick Calculations
For homonuclear diatomics:
Total electrons → Bond Order:
- 2 (H₂): BO = 1
- 4 (He₂): BO = 0
- 10 (B₂): BO = 1
- 12 (C₂): BO = 2
- 14 (N₂): BO = 3
- 16 (O₂): BO = 2
- 18 (F₂): BO = 1
Pattern recognition saves time!
Magnetic Properties
“Odd = Always Paramagnetic”
- Any species with odd total electrons must be paramagnetic
- Even electrons can be either (check MO diagram)
Common Mistakes to Avoid
Mistake: Using σ₂pz before π₂p for ALL elements
Correct:
- Z ≤ 7 (B to N): π₂p comes before σ₂pz
- Z > 7 (O, F): σ₂pz comes before π₂p
JEE Trap: If you use wrong order for N₂, you’ll get wrong BO!
Always check: What element am I dealing with?
Mistake: “O₂ has 6+6 = 12 valence electrons”
Correct: Count ALL electrons for MO diagram!
- O: 8 electrons total (not just valence)
- O₂: 8 + 8 = 16 electrons
Core electrons (1s) also form MOs (σ₁s, σ*₁s)!
JEE Tip: For bond order calculations, you can ignore core electrons if they’re fully paired (σ₁s² σ*₁s² cancel out). But for complete MO diagram, include them!
Mistake: “Diamagnetic molecules are more stable than paramagnetic”
Wrong! Magnetism ≠ Stability
Example:
- N₂: Diamagnetic, BO = 3, very stable ✓
- O₂: Paramagnetic, BO = 2, stable ✓
- F₂: Diamagnetic, BO = 1, reactive ✗
Stability depends on bond order, not magnetism!
Practice Problems
Level 1: Foundation (NCERT Style)
Question: Calculate the bond order of H₂⁺ and predict if it’s stable.
Solution:
Step 1: Count electrons
- H: 1 electron
- H⁺: 0 electrons (lost one)
- Total: 1 electron
Step 2: Fill MO diagram
- σ₁s: ↑ (1 electron)
Step 3: Calculate BO
$$\text{BO} = \frac{1}{2}(1 - 0) = 0.5$$Step 4: Interpret
- BO = 0.5 > 0 → Bond exists (though weak)
- 1 unpaired electron → Paramagnetic
Answer: H₂⁺ has bond order 0.5, is paramagnetic, and exists as a weak bond (bond energy ≈ 60% of H₂).
Question: Why does He₂ not exist?
Solution:
MO diagram for He₂:
- Total electrons: 2 + 2 = 4
- Configuration: σ₁s² σ*₁s²
Bond order:
$$\text{BO} = \frac{1}{2}(2 - 2) = 0$$Interpretation:
- BO = 0 → No net bonding
- Bonding stabilization = Antibonding destabilization
- No energy benefit from bond formation
Answer: He₂ does not exist because its bond order is zero — the antibonding MO cancels the bonding MO completely.
Level 2: JEE Main Type
Question: Arrange the following in order of increasing bond length: O₂, O₂⁺, O₂⁻, O₂²⁻
Solution:
Calculate bond orders:
O₂²⁻: 18 electrons
- π₂px: ↑↓, π₂py: ↑↓ (all antibonding filled)
- BO = ½(10 - 8) = 1
O₂⁻: 17 electrons
- π₂px: ↑↓, π₂py: ↑ (one antibonding half-filled)
- BO = ½(10 - 7) = 1.5
O₂: 16 electrons
- π₂px: ↑, π₂py: ↑ (two antibonding half-filled)
- BO = ½(10 - 6) = 2
O₂⁺: 15 electrons
- π₂px: ↑, π₂py: __ (one antibonding half-filled)
- BO = ½(10 - 5) = 2.5
Bond length trend: Higher BO → Shorter bond
Order: O₂⁺ (2.5) < O₂ (2.0) < O₂⁻ (1.5) < O₂²⁻ (1.0)
Answer (increasing bond length): O₂⁺ < O₂ < O₂⁻ < O₂²⁻
Question: Which of the following is paramagnetic? (A) N₂ (B) F₂ (C) O₂ (D) C₂
Solution:
(A) N₂: 14 electrons
- All paired in MO diagram → Diamagnetic
(B) F₂: 18 electrons
- All paired → Diamagnetic
(C) O₂: 16 electrons
- 2 unpaired electrons in π*₂p → Paramagnetic ✓
(D) C₂: 12 electrons
- All paired → Diamagnetic
Answer: (C) O₂
JEE Shortcut: Among common diatomics, only O₂ and B₂ are paramagnetic. Memorize this!
Level 3: JEE Advanced Type
Question: (a) Draw the MO diagram for N₂ (b) Calculate bond order (c) Predict magnetic nature (d) Compare with N₂⁺ — which is more stable?
Solution:
(a) MO diagram for N₂ (14 electrons):
Energy order (Z = 7 ≤ 7): π₂p comes before σ₂pz
Configuration:
- σ₁s²
- σ*₁s²
- σ₂s²
- σ*₂s²
- π₂px²
- π₂py²
- σ₂pz²
(b) Bond order:
- Bonding: σ₁s(2) + σ₂s(2) + π₂px(2) + π₂py(2) + σ₂pz(2) = 10
- Antibonding: σ₁s(2) + σ₂s(2) = 4 $$\text{BO} = \frac{1}{2}(10 - 4) = 3$$
(c) Magnetic nature:
- All electrons paired → Diamagnetic
(d) Compare with N₂⁺:
N₂⁺: 13 electrons
- Remove one electron from σ₂pz (highest occupied)
- Configuration: …π₂px² π₂py² σ₂pz¹
- BO = ½(9 - 4) = 2.5
- 1 unpaired → Paramagnetic
Stability comparison:
- N₂: BO = 3 → More stable
- N₂⁺: BO = 2.5 → Less stable
Answer: (a) MO diagram as above (b) BO = 3 (c) Diamagnetic (d) N₂ is more stable than N₂⁺ (higher BO)
Question: Using MOT, explain why O₂ is paramagnetic but O₂²⁻ is diamagnetic. Which has a longer bond?
Solution:
O₂ (16 electrons):
- Configuration: …σ₂pz² π₂px² π₂py² π₂px¹ π₂py¹
- 2 unpaired electrons in π* orbitals
- Paramagnetic
- BO = ½(10 - 6) = 2
O₂²⁻ (18 electrons):
- Configuration: …σ₂pz² π₂px² π₂py² π₂px² π₂py²
- All electrons paired (both π* orbitals filled)
- Diamagnetic
- BO = ½(10 - 8) = 1
Bond length:
- O₂: BO = 2 → shorter
- O₂²⁻: BO = 1 → longer
Explanation: Adding 2 electrons to O₂ fills the antibonding π* orbitals completely:
- Pairs all electrons → diamagnetic
- Reduces bond order from 2 to 1
- Weakens and lengthens the bond
Answer: O₂ is paramagnetic due to 2 unpaired e⁻ in π*. O₂²⁻ is diamagnetic because those π* orbitals are filled (all paired). O₂²⁻ has longer bond (BO = 1 vs 2).
Quick Revision Box
MO Energy Orders
For Z ≤ 7 (B, C, N):
σ1s < σ*1s < σ2s < σ*2s < π2p < σ2pz < π*2p < σ*2pz
For Z > 7 (O, F):
σ1s < σ*1s < σ2s < σ*2s < σ2pz < π2p < π*2p < σ*2pz
Bond Order Formula
$$\boxed{\text{BO} = \frac{1}{2}(N_b - N_a)}$$Key Predictions
| Property | Determination |
|---|---|
| Bond strength | Higher BO → Stronger |
| Bond length | Higher BO → Shorter |
| Stability | Higher BO → More stable |
| Magnetism | Unpaired electrons → Paramagnetic |
Common Bond Orders
| Molecule | Electrons | BO | Magnetic |
|---|---|---|---|
| H₂ | 2 | 1 | Dia |
| N₂ | 14 | 3 | Dia |
| O₂ | 16 | 2 | Para |
| F₂ | 18 | 1 | Dia |
| B₂ | 10 | 1 | Para |
Teacher’s Summary
1. MOT treats molecules as unified quantum systems
- Electrons occupy molecular orbitals (not atomic orbitals)
- MOs belong to entire molecule
- More accurate than VBT for magnetic properties
2. AOs combine to form bonding and antibonding MOs
- 2 AOs → 2 MOs (conservation)
- Bonding MO: lower energy, stabilizing
- Antibonding MO: higher energy, destabilizing (marked with *)
3. Energy order changes with atomic number
- Z ≤ 7: π₂p before σ₂pz
- Z > 7: σ₂pz before π₂p
- Critical for JEE — memorize this!
4. Bond order predicts everything
$$\text{BO} = \frac{1}{2}(N_b - N_a)$$- Higher BO = Stronger, shorter, more stable
- BO = 0 → molecule doesn’t exist
- Fractional BO is possible!
5. Magnetic properties from unpaired electrons
- All paired → Diamagnetic
- Any unpaired → Paramagnetic
- O₂ is paramagnetic (VBT fails, MOT succeeds!)
6. MOT explains what VBT cannot
- ✓ Paramagnetism of O₂
- ✓ Stability of ions (O₂⁺, O₂⁻)
- ✓ Bond order variations
- ✓ Why He₂ doesn’t exist
7. For JEE:
- Master energy order for Z ≤ 7 vs Z > 7
- Practice bond order calculations (fastest marks!)
- Remember O₂ family trends (common question!)
- Use MO diagrams for magnetic predictions
“Molecular Orbital Theory: Where atoms lose their individuality and electrons go quantum!”
Related Topics
Within Chemical Bonding
- Covalent Bonding — Lewis structures (simpler, less accurate)
- Hybridization — VBT approach (localized orbitals)
- VSEPR Theory — Predicts shapes (complementary to MOT)
- Hydrogen Bonding — Weak bonds between molecules
Cross-Chapter Links
- Atomic Orbitals — s, p, d shapes used in MOs
- Electronic Configuration — Aufbau, Pauli, Hund’s
- Quantum Model — Quantum mechanics foundation
- Coordination Compounds — Crystal field vs MOT
Cross-Subject: Physics
- Quantum Mechanics — Wave functions, interference
- Magnetism — Para vs diamagnetic materials
Cross-Subject: Math
- Vectors — MO as vector combinations
- Wave Interference — Constructive vs destructive