Key Fact
Diatomic MO configurations
H₂: σ1s² (BO=1, diamagnetic). N₂: σ1s²σ*1s²σ2s²σ*2s²π2p⁴σ2p² (BO=3). O₂: π*2p² unpaired (BO=2, paramagnetic). F₂: π*2p⁴ (BO=1).
-- NCERT Class 11 Chemistry, Ch. 4, p. 28Mo Homonuclear Diatomics
8 MCQs2 revision cards9-step worked example
Source: NCERT Chemical Bonding and Molecular StructurePYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap that costs marks in MO theory questions is not the bond order formula — it is the MO energy ordering switch between N₂ and O₂.
For homonuclear diatomic molecules, molecular orbitals form by linear combination of atomic orbitals. The 2p set produces three bonding MOs (σ2p, π2p_x, π2p_y) and three antibonding MOs (σ2p, π2p_x, π*2p_y). The critical detail: s-p mixing changes the energy ordering.
Two orderings exist (NCERT Class 11 Chemistry Chapter 4, page 28):
- For Li₂ through N₂ (s-p mixing active): π2p_x = π2p_y sit below σ2p. The filling order is σ1s < σ1s < σ2s < σ2s < π2p_x = π2p_y < σ2p < π2p_x = π2p_y < σ*2p.
- For O₂, F₂, Ne₂ (negligible s-p mixing): σ2p drops below π2p_x = π2p_y.
A common mistake: applying one ordering universally. Students who use the O₂ ordering for N₂ misplace electrons and get the wrong bond order or wrong magnetic prediction.
Bond order tells you stability: BO = (N_b − N_a)/2, where N_b = bonding electrons, N_a = antibonding electrons. Another frequent error: forgetting antibonding electrons entirely and computing BO = N_b/2, which inflates the result. F₂ has 8 bonding and 6 antibonding electrons → BO = (8−6)/2 = 1, not 4.
Magnetic behaviour follows directly from the configuration. Unpaired electrons in π*2p orbitals make O₂ paramagnetic — a fact VBT cannot explain but MOT predicts cleanly.
Watch out: if a question asks "which species is paramagnetic," write the full MO configuration first, then check for unpaired electrons. Do not guess from Lewis structures.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the MO energy level diagram for N₂, which of the following orderings is correct for the 2p molecular orbitals?
MCQ 2Easy RecallPractice
The MO energy ordering switches between which pair of consecutive homonuclear diatomic molecules?
MCQ 3Easy RecallPractice
Which homonuclear diatomic molecule is paramagnetic?
MCQ 4Direct ApplicationPractice
The bond order of O₂ using MO theory is:
MCQ 5Direct ApplicationPractice
What is the bond order of F₂? Given: F has atomic number 9.
MCQ 6Direct ApplicationPractice
Among O₂, O₂⁺, O₂⁻, and O₂²⁻, which has the highest bond order?
MCQ 7CalculationPractice
B₂ has 10 electrons. Using the correct MO energy ordering for B₂, determine its bond order and magnetic character.
MCQ 8CalculationPractice
A student calculates the bond order of N₂ by counting only bonding electrons and writes BO = N_b/2 = 10/2 = 5. What is the correct bond order, and what error did the student make?
Quick recall before you leave
Worked Example
Pattern: Compute bond order and predict magnetic property from MO configuration (pattern NEET pattern: mo bond order).
- 1
Given
O₂ is a homonuclear diatomic molecule. Oxygen has atomic number 8, so each atom contributes 8 electrons. Total electrons in O₂ = 16.
- 2
Required
Bond order of O₂ and whether it is paramagnetic or diamagnetic.
- 3
Concept
Molecular orbital theory fills electrons into MOs formed by LCAO. For O₂, the ordering WITHOUT s-p mixing applies (σ2p is lower than π2p). Bond order is calculated from bonding vs. antibonding electron count. Magnetic character depends on the presence of unpaired electrons.
- 4
Formula
BO = (N_b − N_a) / 2
- 5
Substitution
MO filling for 16 electrons (O₂ ordering): | MO | Electrons | Type | |---|---|---| | σ1s | 2 | bonding | | σ*1s | 2 | antibonding | | σ2s | 2 | bonding | | σ*2s | 2 | antibonding | | σ2p | 2 | bonding | | π2p_x | 2 | bonding | | π2p_y | 2 | bonding | | π*2p_x | 1 | antibonding | | π*2p_y | 1 | antibonding | N_b = 2 + 2 + 2 + 2 + 2 = 10 N_a = 2 + 2 + 1 + 1 = 6
- 6
Calculation
BO = (10 − 6) / 2 = 4/2 = 2 Note: N_b and N_a are exact counting integers. They do not limit significant figures.
- 7
Final answer
Bond order of O₂ = **2**. The two electrons in the degenerate π*2p orbitals are unpaired (one in π*2p_x, one in π*2p_y, by Hund's rule). Therefore O₂ is **paramagnetic**.
- 8
Common trap
Forgetting antibonding electrons entirely: a student who computes BO = 10/2 = 5 has used the wrong formula (N_b/2 instead of (N_b − N_a)/2). This is a documented common mistake (mistake: mo bond order no antibonding in natural terms: "counts only bonding electrons in bond order calculation").
- 9
Similar NEET-style question
"Write the MO configuration of N₂ and calculate its bond order. Predict whether N₂ or O₂ has a stronger bond." (Requires applying the s-p mixing ordering for N₂ and comparing BO = 3 vs. BO = 2.) ---
Before solving, remember these
Formulas
Bond order from MO theory
Higher bond order: shorter, stronger bond. N₂: BO=3, O₂: BO=2, F₂: BO=1.
| Symbol | Quantity | SI Unit |
|---|---|---|
| N_b | bonding electrons | - |
| N_a | antibonding electrons | - |
| BO | bond order | - |
Valid when
- MO theory framework
- Closed-shell molecule (or with appropriate treatment)
Dipole moment
Product of charge magnitude and bond length. SI: C·m. Common: Debye (1 D = 3.336e-30 C·m).
| Symbol | Quantity | SI Unit |
|---|---|---|
| q | charge | C |
| d | bond length | m |
| mu | dipole moment | C*m or D |
Valid when
- Diatomic or vector-summed for polyatomic
- Polar bond
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student counts only bonded atoms when assigning hybridization. Lone pairs count toward steric number too. Steric number = bond pairs + lone pairs → hybridization.
When it triggers
Molecule with central atom having lone pairs (e.g., NH₃: 3 bonds + 1 lp = 4 = sp³; H₂O: 2+2 = 4 = sp³).
How to avoid
Steric number formula: SN = (bond pairs) + (lone pairs). SN=2: sp; SN=3: sp²; SN=4: sp³; SN=5: sp³d; SN=6: sp³d². Lone pairs distort but still count.
Root cause: concept gap
Correction
Steric number = bond pairs + lone pairs. NH3: SN=4 → sp³ (despite trigonal pyramidal shape).
Root cause: concept gap
Correction
Memorise the ordering switch at N2/O2 boundary. Affects bond order and magnetic property predictions.
Root cause: formula misuse
Correction
BO = (Nb - Na)/2. Antibonding electrons subtract. F2: 8 bonding, 6 antibonding → BO = 1.
Past Year Questions
14 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
10.215
20.115
30.125
40.225
NTA Answer: Option 4(final)
NEET 2025
1Statement I is false but Statement II is true
2Both Statement I and Statement II are true
3Both Statement I and Statement II are false
4Statement I is true but Statement II is false
NTA Answer: Option 3(final)
NEET 2025
1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2025
1A-IV, B-II, C-I, D-III
2A-II, B-I, C-IV, D-III
3A-II, B-I, C-III, D-IV
4A-IV, B-II, C-III, D-I
NTA Answer: Option 2(final)
NEET 2025
1B, C only
2A, D only
3B, D only
4A, C only
NTA Answer: Option 2(final)
NEET 2024Revised key
1A-I, B-IV, C-II, D-III
2A-II, B-IV, C-III, D-I
3A-III, B-IV, C-I, D-II
4A-II, B-III, C-IV, D-I
NTA Answer: Option 1(revised_final)
NEET 2023
The element expected to form largest ion to achieve the nearest noble gas configuration is
1F
2N
3Na
4O
NTA Answer: Option 2(final)
NEET 2023
12
24
31
43
NTA Answer: Option 4(final)
NEET 2022
Which statement regarding polymers is not correct?
1Thermosetting polymers are reusable
2Elastomers have polymer chains held together by weak intermolecular forces
3Fibers possess high tensile strength
4Thermoplastic polymers are capable of repeatedly softening and hardening on heating and cooling respectively
NTA Answer: Option 1(final)
NEET 2022
Amongst the following which one will have maximum ‘lone pair - lone pair’ electron repulsions?
1XeF
2CIF 2 3
3IF
4SF 5 4
NTA Answer: Option 1(final)
NEET 2022
In one molal solution that contains 0.5 mole of a solute, there is
11000 g of solvent
2500 mL of solvent
3500 g of solvent
4100 mL of solvent
NTA Answer: Option 3(final)
NEET 2021
13
27
35
42
NTA Answer: Option 1(final)
NEET 2021
1(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
2(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
3(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
4(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
NTA Answer: Option 2(final)
NEET 2021
Which of the following molecules is non-polar in nature?
1NO 2
2POCl 3
3CH O 2
4SbCl 5
NTA Answer: Option 4(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Determine hybridization of central atom from molecular formula or geometry.
InterpretationMedium
Common distractors
ignores lone pairs
Counts only bonding pairs
From MO configuration of homonuclear diatomic, compute bond order and predict magnetic property.
Multi StepMedium
Common distractors
forgets anti bonding
Counts only bonding electrons
Predict molecular geometry (linear, bent, trigonal pyramidal, etc.) from VSEPR steric number and lone pairs.
InterpretationEasy
Common distractors
ignores lp bp distortion
Doesn't adjust for lone-pair repulsion
Sources
NCERT refs: Class 11 Chemistry Chapter 4, p.28
Test yourself on this topic with real past-paper questions:
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