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. 131MO Homonuclear Diatomics
8 MCQs5 revision cards9-step worked example
Source: NCERT Chemical Bonding and Molecular StructurePYQ coverage: NEET 2020, 2022, 2023Official key: NTA-verifiedLast updated: 26 Sep 2026
MO Homonuclear Diatomics, explained for NEET
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 129):
- 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.
Can you answer these MO Homonuclear Diatomics 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?
Show answer and why every option is right or wrong
Answer: D. For N₂ and lighter homonuclear diatomics (Li₂ to N₂), s-p mixing raises the energy of σ2p above the π2p orbitals. Hence π2p_x = π2p_y < σ2p is the correct bonding-MO ordering (NCERT Class 11 Chemistry Chapter 4, page 129).
Why A is wrong: A places σ2p below π2p, which is the ordering for O₂ and F₂ (no s-p mixing), not for N₂.
Why B is wrong: B describes antibonding orbitals relative to σ2p. The question asks about the bonding-level ordering.
Why C is wrong: C describes antibonding orbitals (π*), not the bonding orbital ordering asked in the question.
MCQ 2Easy RecallPractice
The MO energy ordering switches between which pair of consecutive homonuclear diatomic molecules?
Show answer and why every option is right or wrong
Answer: A. The s-p mixing effect is significant for Li₂ through N₂ but negligible from O₂ onward. The ordering switch — σ2p moving below π2p — occurs at the N₂/O₂ boundary (NCERT Class 11 Chemistry Chapter 4, page 129).
Why B is wrong: B is wrong. Be₂ and B₂ both use the same ordering (with s-p mixing). The switch is later in the period.
Why C is wrong: C is wrong. Both Li₂ and Be₂ fall in the s-p mixing regime; no ordering change occurs between them.
Why D is wrong: D is wrong. O₂ and F₂ both use the same ordering (without s-p mixing). The switch has already happened before O₂.
MCQ 3Easy RecallPractice
Which homonuclear diatomic molecule is paramagnetic?
Show answer and why every option is right or wrong
Answer: C. O₂ has the electronic configuration (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)². The two electrons in the degenerate π*2p orbitals occupy them singly with parallel spins (Hund's rule), making O₂ paramagnetic (NCERT Class 11 Chemistry Chapter 4, page 130).
Why A is wrong: A is wrong. N₂ has all electrons paired in its MO configuration: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(π2p)⁴(σ2p)². Bond order 3, diamagnetic.
Why B is wrong: B is wrong. F₂ has the configuration ending in (π2p)⁴ — both π orbitals fully occupied, all electrons paired. Diamagnetic.
Why D is wrong: D is wrong. C₂ has 8 valence electrons filling through (π2p)⁴. All electrons paired. Diamagnetic.
MCQ 4Direct ApplicationPractice
The bond order of O₂ using MO theory is:
Show answer and why every option is right or wrong
Answer: B. O₂ has 16 electrons. MO configuration: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)². Bonding electrons = 10, antibonding electrons = 6. BO = (10 − 6)/2 = 2 (NCERT Class 11 Chemistry Chapter 4, page 130).
Why A is wrong: A gives BO = 1, which is the bond order of F₂ (not O₂). This results from miscounting electrons or confusing the two configurations.
Why C is wrong: C gives BO = 3, which is the bond order of N₂. Applying N₂'s configuration to O₂ (wrong ordering) could lead to this error.
Why D is wrong: D gives BO = 2.5, which would require an odd electron count (like NO or O₂⁺). Neutral O₂ has an even electron count and integer bond order.
MCQ 5Direct ApplicationPractice
What is the bond order of F₂? Given: F has atomic number 9.
Show answer and why every option is right or wrong
Answer: B. F₂ has 18 electrons. MO filling (O₂/F₂ ordering): (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)⁴. Bonding = 10, antibonding = 8. BO = (10 − 8)/2 = 1.
Why A is wrong: A gives BO = 0, which would mean no bond exists. That describes Ne₂, not F₂. F₂ is a stable molecule.
Why C is wrong: C gives BO = 2. That is the bond order of O₂. It results from leaving out one filled π*2p orbital: (10 − 6)/2 = 2. F₂ has two more electrons than O₂, which complete the π*2p set (8 antibonding electrons), giving (10 − 8)/2 = 1.
Why D is wrong: D gives BO = 3. This is the bond order of N₂, not F₂. F₂ has many more antibonding electrons filled.
MCQ 6Direct ApplicationPractice
Among O₂, O₂⁺, O₂⁻, and O₂²⁻, which has the highest bond order?
Show answer and why every option is right or wrong
Answer: A. O₂⁺ has 15 electrons. Removing one electron from O₂ takes it from a π*2p orbital: antibonding electrons drop from 6 to 5. BO = (10 − 5)/2 = 2.5. Compare: O₂ = 2, O₂⁻ = 1.5, O₂²⁻ = 1. O₂⁺ has the highest bond order.
Why B is wrong: B gives O₂ with BO = 2. While stable, adding a positive charge removes an antibonding electron, increasing BO to 2.5 in O₂⁺.
Why C is wrong: C gives O₂⁻ with BO = (10 − 7)/2 = 1.5. Adding an electron increases antibonding count, weakening the bond relative to O₂.
Why D is wrong: D gives O₂²⁻ with BO = (10 − 8)/2 = 1. Two extra antibonding electrons make this the weakest bond in the series.
MCQ 7CalculationPractice
B₂ has 10 electrons. Using the correct MO energy ordering for B₂, determine its bond order and magnetic character.
Show answer and why every option is right or wrong
Answer: D. B₂ uses the s-p mixing ordering (like N₂). Filling 10 electrons: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(π2p_x)¹(π2p_y)¹. Bonding = 6, antibonding = 4. BO = (6 − 4)/2 = 1. The two electrons in the degenerate π2p orbitals are unpaired (Hund's rule) → paramagnetic.
Why A is wrong: A correctly identifies BO = 1 but claims diamagnetic. In the s-p mixing ordering, the last two electrons enter degenerate π2p orbitals singly (Hund's rule), giving two unpaired electrons — paramagnetic, not diamagnetic.
Why B is wrong: B gives BO = 0, implying no bond. This would require equal bonding and antibonding electrons, which is the case for Be₂ (8 electrons), not B₂ (10 electrons).
Why C is wrong: C gives BO = 2, which would require 8 bonding and 4 antibonding electrons. The actual count with 10 total electrons gives only 6 bonding electrons. BO = 1.
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?
Show answer and why every option is right or wrong
Answer: C. N₂ has 14 electrons. MO configuration (s-p mixing ordering): (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(π2p)⁴(σ2p)². Bonding = 10, antibonding = 4. BO = (10 − 4)/2 = 3. The student's formula BO = N_b/2 omits the antibonding subtraction — a documented common mistake.
Why A is wrong: A validates the student's incorrect method. BO = N_b/2 ignores antibonding electrons entirely, violating the definition BO = (N_b − N_a)/2.
Why B is wrong: B claims BO = 2, which is the bond order of O₂, not N₂. The student's error is about formula misuse, not MO ordering. Correct BO for N₂ is 3.
Why D is wrong: D claims BO = 4, which is not achievable with s and p orbitals for a second-period diatomic. The maximum bond order in this series is 3 (N₂).
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MO Homonuclear Diatomics: quick recall before you leave
How do you solve a MO Homonuclear Diatomics question? A 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.)
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What to remember before solving MO Homonuclear Diatomics questions
Where do students lose marks on MO Homonuclear Diatomics?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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.
More in Chemical Bonding and Molecular Structure: 2 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
MO Homonuclear Diatomics questions from past NEET papers
3 questions from NEET 2020, 2022, 2023. Answers verified against NTA official keys.
NEET 2023
The correct order of energies of molecular orbitals of N₂ molecule, is
1σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz
2σ1s < σ*1s < σ2s < σ*2s < σ2pz < σ*2pz < (π2px = π2py) < (π*2px = π*2py)
3σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < (π*2px = π*2py) < σ2pz < σ*2pz
4σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz
NTA Answer: Option 4(final)
NEET 2022
Which amongst the following is incorrect statement?
1O₂⁺ ion is diamagnetic
2The bond orders of O₂⁺, O₂, O₂⁻ and O₂²⁻ are 2.5, 2, 1.5 and 1, respectively
3C₂ molecule has four electrons in its two degenerate π molecular orbitals
4H₂⁺ ion has one electron
NTA Answer: Option 1(final)
NEET 2020
Identify a molecule which does not exist.
1O2
2He2
3Li2
4C2
NTA Answer: Option 2(final)
All 18 past-paper questions from Chemical Bonding and Molecular Structure →
How does NEET ask about MO Homonuclear Diatomics?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
From MO configuration of homonuclear diatomic, compute bond order and predict magnetic property.
Multi StepMedium
Common distractors
forgets anti bonding
Counts only bonding electrons
Sources
NCERT refs: Class 11 Chemistry Chapter 4, p.129
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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