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. 129Bond Order Length Energy
8 MCQs4 revision cards9-step worked example
Source: NCERT Chemical Bonding and Molecular StructurePYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 26 Sep 2026
Bond Order Length Energy, explained for NEET
The trap that costs marks: students compute bond order by counting only bonding electrons, completely ignoring antibonding electrons. The formula BO = (N_b − N_a)/2 has a subtraction — skip the N_a term and you get a bond order that is too high, leading you to wrong predictions about bond length and bond energy.
Bond order is the number of net chemical bonds between two atoms. In molecular orbital (MO) theory, it is calculated as:
BO = (N_b − N_a) / 2
where N_b is the number of electrons in bonding molecular orbitals and N_a is the number in antibonding molecular orbitals (NCERT Class 11 Chemistry, Chapter 4, page 129).
The triad relationship — bond order, bond length, and bond energy — is tightly linked:
- Higher bond order → shorter bond length → greater bond energy. N₂ has BO = 3, bond length 1.10 Å, and bond dissociation energy 941 kJ/mol. O₂ has BO = 2, bond length 1.21 Å, energy 498 kJ/mol. F₂ has BO = 1, bond length 1.42 Å, energy 159 kJ/mol.
- Zero or negative bond order → molecule does not exist. He₂ gives BO = 0 (2 bonding, 2 antibonding). No stable He₂ molecule forms.
Watch-out for NEET: when a question asks you to rank molecules by bond length or bond strength, compute each bond order first using the full MO configuration. The common wrong answer comes from forgetting antibonding electrons — for F₂, counting only the 8 bonding electrons gives BO = 4, which is absurd for a single bond. The correct count (8 bonding, 6 antibonding) gives BO = 1, consistent with F₂ being a weak, long bond.
Can you answer these Bond Order Length Energy 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
Which of the following correctly states the relationship between bond order, bond length, and bond energy?
Show answer and why every option is right or wrong
Answer: C. C is correct. As bond order increases, more electron density holds the nuclei together — the bond becomes shorter and requires more energy to break (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why A is wrong: A is wrong because higher bond order produces a shorter bond, not a longer one. More bonding electrons pull nuclei closer.
Why B is wrong: B is wrong because shorter bonds (from higher bond order) are stronger, not weaker. Bond energy increases with bond order.
Why D is wrong: D is wrong because lower bond order means fewer net bonds, giving a longer and weaker bond — the opposite of what D claims.
MCQ 2Easy RecallPractice
The bond order of N₂ according to molecular orbital theory is:
Show answer and why every option is right or wrong
Answer: A. A is correct. N₂ has 10 bonding electrons and 4 antibonding electrons. BO = (10 − 4)/2 = 3 (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why B is wrong: B is wrong. BO = 2 corresponds to O₂ (10 bonding, 6 antibonding, counting all electrons as for N₂ above). N₂ has a higher bond order.
Why C is wrong: C is wrong. BO = 1 would mean only 2 more bonding than antibonding electrons — this describes F₂, not N₂.
Why D is wrong: D is wrong. BO = 4 would require (N_b − N_a) = 8, which does not occur in any homonuclear diatomic of period 2 elements.
MCQ 3Easy RecallPractice
A molecule with bond order zero implies that:
Show answer and why every option is right or wrong
Answer: C. C is correct. Bond order zero means bonding and antibonding contributions cancel completely — there is no net bond, so the molecule is not stable. He₂ (BO = 0) does not exist as a stable molecule (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why A is wrong: A is wrong. Zero bond order means no net bond at all — not a long bond, but no bond.
Why B is wrong: B is wrong. Paramagnetism relates to unpaired electrons, not to whether bond order is zero. A molecule must first exist to discuss its magnetic property.
Why D is wrong: D is wrong. The weakest single bond still has BO = 1 (e.g. F₂). BO = 0 means no bond exists.
MCQ 4Direct ApplicationPractice
The molecular orbital configuration of O₂ has 8 electrons in bonding MOs and 4 electrons in antibonding MOs. What is the bond order of O₂?
Show answer and why every option is right or wrong
Answer: D. D is correct. BO = (8 − 4)/2 = 2. This is consistent with O₂ being a double bond (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why A is wrong: A is wrong. BO = 1 would require (N_b − N_a) = 2. With 8 bonding and 4 antibonding, the difference is 4, not 2 (trap: forgetting to count antibonding electrons and then miscalculating).
Why B is wrong: B is wrong. BO = 4 would need (N_b − N_a) = 8, which means ignoring all antibonding electrons entirely — the classic mistake of using BO = N_b/2 instead of (N_b − N_a)/2.
Why C is wrong: C is wrong. BO = 3 applies to N₂, not O₂. Getting 3 here likely means confusing the MO configurations of the two molecules.
MCQ 5Direct ApplicationPractice
F₂ has 8 bonding electrons and 6 antibonding electrons. A student calculates bond order as 4 by using only bonding electrons. What is the correct bond order?
Show answer and why every option is right or wrong
Answer: D. D is correct. BO = (8 − 6)/2 = 1. The student's error was ignoring the 6 antibonding electrons. This is a common mistake — the bond order formula requires subtracting antibonding electrons before dividing (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why A is wrong: A is wrong. BO = 4 is exactly the student's incorrect answer — obtained by using 8/2 while ignoring the 6 antibonding electrons.
Why B is wrong: B is wrong. BO = 3 applies to N₂ (10 bonding, 4 antibonding). For F₂ the antibonding count is 6, not 2.
Why C is wrong: C is wrong. BO = 2 would require (N_b − N_a) = 4, i.e. only 4 antibonding electrons. F₂ has 6.
MCQ 6Direct ApplicationPractice
Among N₂ (BO = 3), O₂ (BO = 2), and F₂ (BO = 1), which has the longest bond length?
Show answer and why every option is right or wrong
Answer: B. B is correct. Bond length is inversely related to bond order. F₂ has the lowest bond order (1), so it has the longest bond (1.42 Å), compared to O₂ (1.21 Å) and N₂ (1.10 Å) (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why A is wrong: A is wrong. N₂ has the highest bond order (3), making it the shortest bond — the exact opposite of what the question asks.
Why C is wrong: C is wrong. O₂ has an intermediate bond order (2) and intermediate bond length (1.21 Å). It is not the longest.
Why D is wrong: D is wrong. Different bond orders produce different bond lengths. The inverse relationship between bond order and bond length is a core NCERT principle.
MCQ 7CalculationPractice
He₂⁺ has a total of 3 electrons. Using the MO configuration (σ1s)²(σ*1s)¹, calculate the bond order and predict whether the species is stable.
Show answer and why every option is right or wrong
Answer: B. B is correct. N_b = 2 (two electrons in σ1s), N_a = 1 (one electron in σ*1s). BO = (2 − 1)/2 = 0.5. A fractional but positive bond order means the species has some net bonding and can exist, though it is weak and transient (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why A is wrong: A is wrong. BO = 1 would require (N_b − N_a) = 2, meaning zero antibonding electrons. He₂⁺ has 1 antibonding electron — you cannot ignore it.
Why C is wrong: C is wrong. BO = 0 applies to neutral He₂ (2 bonding, 2 antibonding). He₂⁺ has only 1 antibonding electron, giving a positive bond order.
Why D is wrong: D is wrong. BO = 1.5 would require (N_b − N_a) = 3, which is impossible with only 3 total electrons and 2 in bonding MOs.
MCQ 8CalculationPractice
Two diatomic species X₂ and Y₂ have the following MO data: X₂ has 8 bonding and 4 antibonding electrons; Y₂ has 8 bonding and 6 antibonding electrons. Which species has the higher bond dissociation energy, and why?
Show answer and why every option is right or wrong
Answer: A. A is correct. X₂: BO = (8 − 4)/2 = 2. Y₂: BO = (8 − 6)/2 = 1. Higher bond order means greater bond dissociation energy, so X₂ has the stronger bond (NCERT Class 11 Chemistry, Chapter 4, page 129).
Why B is wrong: B is wrong. Total electron count does not determine bond strength — what matters is the net bonding, calculated as (N_b − N_a)/2. Y₂ has more electrons but a lower bond order.
Why C is wrong: C is wrong. This is exactly the mistake of ignoring antibonding electrons. Equal N_b does not mean equal bond order when N_a differs.
Why D is wrong: D is wrong. Antibonding electrons weaken the bond, not strengthen it. Each antibonding electron cancels part of the bonding contribution.
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Bond Order Length Energy: quick recall before you leave
How do you solve a Bond Order Length Energy question? A worked example
Pattern: Compute bond order from MO configuration and predict bond properties (pattern: MO bond order calculation, observed NEET 2021, 2022).
- 1
Given
The MO electronic configuration of O₂ 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
Total electrons: 16. - 2
Required
Calculate the bond order of O₂ and predict whether it is paramagnetic or diamagnetic.
- 3
Concept
Bond order from MO theory uses the count of electrons in bonding vs antibonding orbitals. Unpaired electrons in the MO configuration indicate paramagnetism (NCERT Class 11 Chemistry, Chapter 4, page 129).
- 4
Formula
BO = (N_b − N_a) / 2
- 5
Substitution
Count bonding electrons: σ1s(2) + σ2s(2) + σ2p_z(2) + π2p_x(2) + π2p_y(2) = 10
Count antibonding electrons: σ*1s(2) + σ*2s(2) + π\*2p_x(1) + π\*2p_y(1) = 6
BO = (10 − 6) / 2 - 6
Calculation
BO = 4 / 2 = 2
All numbers here are exact electron counts (integers), so no significant-figure considerations apply. - 7
Final answer
Bond order of O₂ = 2 (a double bond).
The two unpaired electrons in π\*2p_x and π\*2p_y make O₂ paramagnetic — it is attracted by a magnetic field. This is one of the key successes of MO theory; the Lewis structure of O₂ incorrectly predicts it as diamagnetic. - 8
Common trap
Ignoring the 6 antibonding electrons and computing BO = 10/2 = 5 — a nonsensical result. The subtraction in (N_b − N_a) is the entire point. For O₂, the mistake of counting only bonding electrons gives BO = 5, which no student should accept since even N₂ (a triple bond) has BO = 3.
- 9
Similar NEET-style question
"The species O₂⁺ has one fewer electron than O₂. Determine its bond order and predict whether it is more or less stable than O₂."
Hint: Remove one electron from the highest-energy antibonding MO. The new N_a = 5, so BO = (10 − 5)/2 = 2.5 — higher than O₂, meaning O₂⁺ has a shorter, stronger bond.
---
What to remember before solving Bond Order Length Energy questions
Which Bond Order Length Energy 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)
Where do students lose marks on Bond Order Length Energy?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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: 3 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.
Bond Order Length Energy questions from past NEET papers
2 questions from NEET 2025. Answers verified against NTA official keys.
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
1B, C only
2A, D only
3B, D only
4A, C only
NTA Answer: Option 2(final)
All 18 past-paper questions from Chemical Bonding and Molecular Structure →
How does NEET ask about Bond Order Length Energy?
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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