Atomic orbitals combine to form molecular orbitals (bonding lower energy, antibonding higher). Bond order = (N_b - N_a)/2. Higher bond order: shorter, stronger bond. Magnetic property: paramagnetic if unpaired electrons.
-- NCERT Class 11 Chemistry, Ch. 4, p. 129Molecular Orbital Theory
8 MCQs2 revision cards9-step worked example
Source: NCERT Chemical Bonding and Molecular StructurePYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Molecular Orbital Theory, explained for NEET
The trap that costs marks: students fill molecular orbital diagrams correctly but then count only bonding electrons when calculating bond order — forgetting that antibonding electrons subtract. For F₂, this gives bond order 4 instead of the correct 1. A second common confusion is the MO energy ordering: for B₂, C₂, and N₂, the σ2p_z orbital sits above the π2p orbitals (due to s–p mixing), but for O₂, F₂, and Ne₂ the σ2p_z drops below the π2p orbitals. Using the wrong ordering leads to incorrect electron configurations, wrong bond orders, and wrong magnetic-property predictions.
What MOT actually says. When two atomic orbitals combine (LCAO — Linear Combination of Atomic Orbitals), they produce two molecular orbitals: one bonding (lower energy, constructive overlap) and one antibonding (higher energy, destructive overlap, marked with an asterisk *). Electrons fill MOs following Aufbau, Pauli exclusion, and Hund's rule — exactly as in atoms. Bond order is then calculated as:
BO = (N_b − N_a) / 2
where N_b = total bonding electrons and N_a = total antibonding electrons (NCERT Class 11 Chemistry Chapter 4, page 129).
The ordering switch you must memorise. For homonuclear diatomics up to and including N₂ (Z ≤ 7): σ1s < σ1s < σ2s < σ2s < π2p_x = π2p_y < σ2p_z < π2p_x = π2p_y < σ2p_z. From O₂ onward (Z ≥ 8): σ2p_z drops below the π2p pair. The boundary is at the N₂/O₂ transition. Getting this wrong flips paramagnetism predictions — O₂ is paramagnetic (two unpaired electrons in π), and this is a frequent NEET question.
Watch out: if BO = 0, the molecule does not exist (e.g., He₂ with BO = 0; He₂⁺ with BO = 0.5 does exist). A fractional bond order is valid and physically meaningful.
Can you answer these Molecular Orbital Theory 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 molecular orbital theory, how many molecular orbitals are formed when two atomic orbitals combine by LCAO?
Show answer and why every option is right or wrong
Answer: B. LCAO of two atomic orbitals always produces exactly two MOs — one bonding (constructive overlap) and one antibonding (destructive overlap). This is a fundamental postulate of MOT (NCERT Class 11 Chemistry Chapter 4, page 126).
Why A is wrong: A is wrong because LCAO always produces two MOs, not one. The number of MOs must equal the number of combining AOs.
Why C is wrong: C is wrong because one of the two MOs formed is antibonding (higher energy, destructive interference), not both bonding.
Why D is wrong: D is wrong because two AOs produce exactly two MOs. Non-bonding MOs arise only in polyatomic systems where symmetry prevents certain AOs from mixing.
MCQ 2Direct ApplicationPractice
What is the bond order of the F₂ molecule according to MOT?
Show answer and why every option is right or wrong
Answer: A. F₂ has 18 electrons. MO configuration (Z ≥ 8 ordering): σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x² π*2p_y². N_b = 10, N_a = 8. BO = (10 − 8)/2 = 1 (NCERT Class 11 Chemistry Chapter 4).
Why B is wrong: B is wrong because this likely results from miscounting antibonding electrons (e.g., forgetting π* electrons), giving (10 − 6)/2 = 2.
Why C is wrong: C is wrong because this results from counting only bonding electrons (10/2 = 5, or miscounting as 8/2 = 4) and ignoring antibonding electrons entirely — a common mistake (trap: forgetting antibonding subtraction).
Why D is wrong: D is wrong because BO = 0 would mean F₂ doesn't exist. F₂ is a stable diatomic halogen with BO = 1.
MCQ 3Easy RecallPractice
Which of the following homonuclear diatomic molecules is paramagnetic?
Show answer and why every option is right or wrong
Answer: C. O₂ has two unpaired electrons in the degenerate π*2p orbitals, making it paramagnetic. This is a key success of MOT over VBT, which incorrectly predicts O₂ as diamagnetic (NCERT Class 11 Chemistry Chapter 4, page 130).
Why A is wrong: A is wrong because N₂ has all electrons paired in its MO configuration (BO = 3, no unpaired electrons) and is diamagnetic.
Why B is wrong: B is wrong because F₂ has all electrons paired (including the π* electrons, which are fully filled) and is diamagnetic.
Why D is wrong: D is wrong because C₂ has two electrons in the degenerate π2p orbitals following Hund's rule, but in the Z ≤ 7 ordering these π electrons are paired across the two π orbitals giving BO = 2 and diamagnetic character.
MCQ 4Easy RecallPractice
For which of the following diatomics does the σ2p_z molecular orbital lie higher in energy than the π2p_x and π2p_y orbitals?
Show answer and why every option is right or wrong
Answer: D. For homonuclear diatomics with Z ≤ 7 (Li₂ through N₂), s–p mixing raises the σ2p_z above the π2p pair. From O₂ onward (Z ≥ 8), s–p mixing is negligible and σ2p_z falls below π2p. N₂ (Z = 7) follows the "anomalous" ordering (NCERT Class 11 Chemistry Chapter 4).
Why A is wrong: A is wrong because O₂ (Z = 8) follows the normal ordering where σ2p_z is below π2p — the s–p mixing effect is negligible at this atomic number.
Why B is wrong: B is wrong because F₂ (Z = 9) also follows the normal (Z ≥ 8) ordering with σ2p_z below π2p.
Why C is wrong: C is wrong because Ne₂ (Z = 10) follows the Z ≥ 8 ordering. Additionally, Ne₂ has BO = 0 and does not exist as a stable molecule.
MCQ 5Direct ApplicationPractice
The bond order of O₂⁺ is:
Show answer and why every option is right or wrong
Answer: A. O₂ has 16 electrons with BO = 2. Removing one electron from the highest occupied MO (π*2p) to form O₂⁺ (15 electrons) gives N_b = 10, N_a = 5, so BO = (10 − 5)/2 = 2.5. Fewer antibonding electrons → higher bond order than O₂ (NCERT Class 11 Chemistry Chapter 4).
Why B is wrong: B is wrong because 2 is the bond order of neutral O₂. Removing an antibonding electron increases bond order, not maintains it.
Why C is wrong: C is wrong because 1.5 would be the bond order of O₂⁻ (17 electrons, one extra antibonding electron), not O₂⁺. Confusing cation with anion reverses the direction of bond order change.
Why D is wrong: D is wrong because bond order 3 belongs to N₂. O₂⁺ with 15 electrons cannot achieve BO = 3 — that would require only 4 antibonding electrons, but it has 5.
MCQ 6Direct ApplicationPractice
He₂ does not exist as a stable molecule. What is its bond order from MOT?
Show answer and why every option is right or wrong
Answer: D. He₂ has 4 electrons: σ1s² σ*1s². N_b = 2, N_a = 2, so BO = (2 − 2)/2 = 0. Zero bond order means no net bonding and the molecule is unstable (NCERT Class 11 Chemistry Chapter 4).
Why A is wrong: A is wrong because BO = 0.5 applies to He₂⁺ (3 electrons: σ1s² σ*1s¹), which does exist transiently. He₂ with 4 electrons has equal bonding and antibonding, giving BO = 0.
Why B is wrong: B is wrong because bond order cannot be negative in the standard MO treatment of homonuclear diatomics. When N_a > N_b, we simply say BO < 0 is physically meaningless (molecule doesn't form), but for He₂ specifically, BO = 0 exactly.
Why C is wrong: C is wrong because BO = 1 would require N_a = 0 (only bonding electrons), which is not the case — He₂ has 2 antibonding electrons cancelling the 2 bonding ones.
MCQ 7Concept TrapPractice
Among N₂, O₂, and F₂, which has the shortest bond length?
Show answer and why every option is right or wrong
Answer: C. Bond order correlates inversely with bond length: higher BO → shorter bond. N₂ (BO = 3) > O₂ (BO = 2) > F₂ (BO = 1). Therefore N₂ has the shortest bond length. This is a direct consequence of the bond-order formula BO = (N_b − N_a)/2 (NCERT Class 11 Chemistry Chapter 4).
Why A is wrong: A is wrong because F₂ has the lowest bond order (BO = 1) among the three, meaning it has the longest and weakest bond, not the shortest.
Why B is wrong: B is wrong because O₂ has BO = 2, which is intermediate. Its bond is shorter than F₂ but longer than N₂.
Why D is wrong: D is wrong because these molecules have different bond orders (3, 2, 1 respectively), so their bond lengths differ significantly.
MCQ 8CalculationPractice
A student writes the MO configuration of O₂ using the Z ≤ 7 energy ordering (σ2p_z above π2p). What error will this introduce?
Show answer and why every option is right or wrong
Answer: B. For O₂ (16 electrons), even if you use the wrong ordering (σ2p_z above π2p), the electron filling still places 2 electrons in π*2p orbitals — giving the same N_b = 10, N_a = 6, BO = 2, and paramagnetic prediction. The ordering error matters for molecules like B₂ or C₂ where it changes which orbitals get the last few electrons, but for O₂ (where all bonding MOs are fully filled), it happens to give the same result.
Why A is wrong: A is wrong because the bond order remains (10 − 6)/2 = 2 regardless of which ordering is used — all bonding and antibonding orbitals have the same occupancy either way for 16 electrons.
Why C is wrong: C is wrong because the ordering changes only which orbitals fill first, not the total electron count. O₂ always has 16 electrons.
Why D is wrong: D is wrong because the two unpaired electrons still end up in the π*2p orbitals under either ordering, so paramagnetism is correctly predicted.
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Molecular Orbital Theory: quick recall before you leave
How do you solve a Molecular Orbital Theory question? A worked example
Pattern: MO bond order calculation and magnetic property prediction (P.CHE.U03.MO_BOND_ORDER, observed in NEET 2021, 2022).
- 1
Given
O₂⁻ ion. Oxygen has atomic number 8, so neutral O₂ has 16 electrons. The superoxide ion O₂⁻ has 17 electrons (one extra electron added). Use the Z ≥ 8 MO energy ordering (σ2p_z below π2p).
- 2
Required
Bond order of O₂⁻ and whether it is paramagnetic or diamagnetic.
- 3
Concept
In MOT, electrons fill molecular orbitals in order of increasing energy. Bond order = (N_b − N_a)/2. A species with unpaired electrons is paramagnetic.
- 4
Formula
BO = (N_b − N_a) / 2
- 5
Substitution
MO configuration of O₂⁻ (17 electrons, Z ≥ 8 ordering):
σ1s² σ\*1s² σ2s² σ\*2s² σ2p_z² π2p_x² π2p_y² π\*2p_x² π\*2p_y¹
Bonding electrons (σ1s², σ2s², σ2p_z², π2p_x², π2p_y²): N_b = 10
Antibonding electrons (σ\*1s², σ\*2s², π\*2p_x², π\*2p_y¹): N_a = 7 - 6
Calculation
BO = (10 − 7) / 2 = 3/2 = 1.5
All values here are exact counting integers (number of electrons), so no significant-figure limitation applies. - 7
Final answer
Bond order of O₂⁻ = 1.5. The ion has one unpaired electron (in π\*2p_y), so O₂⁻ is paramagnetic.
- 8
Common trap
Forgetting to include the antibonding electrons. If a student counts only bonding electrons: 10/2 = 5 (nonsensical). Another trap: using the Z ≤ 7 ordering for O₂⁻ — for this particular case the result happens to be the same (see MCQ 8 discussion), but it reflects a conceptual error that will cause failures in B₂ or C₂ problems.
- 9
Similar NEET-style question
"Arrange O₂, O₂⁺, O₂⁻, and O₂²⁻ in order of increasing bond length." (Answer: O₂⁺ < O₂ < O₂⁻ < O₂²⁻, following decreasing bond order 2.5 > 2 > 1.5 > 1.)
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What to remember before solving Molecular Orbital Theory questions
Which Molecular Orbital Theory formulas do you need for NEET?
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)
More in Chemical Bonding and Molecular Structure: 4 exam traps and mistakes · 1 formula · 3 question patterns from its other lessons.
Molecular Orbital Theory questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys.
NEET 2026
The highest occupied molecular orbital for Ne₂ is
1π₂p
2σ₂p
3π*₂p
4σ*₂p
NTA Answer: Option 4(final)
All 18 past-paper questions from Chemical Bonding and Molecular Structure →
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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