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. 24Bond Order Length Energy
8 MCQs1 revision card9-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: 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 24).
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.
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
Which of the following correctly states the relationship between bond order, bond length, and bond energy?
MCQ 2Easy RecallPractice
The bond order of N₂ according to molecular orbital theory is:
MCQ 3Easy RecallPractice
A molecule with bond order zero implies that:
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?
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?
MCQ 6Direct ApplicationPractice
Among N₂ (BO = 3), O₂ (BO = 2), and F₂ (BO = 1), which has the longest bond length?
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.
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?
Quick recall before you leave
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 24).
- 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. ---
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.24
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