Atoms tend to have 8 electrons in valence shell (octet). Lewis dot structures represent valence electrons as dots. Exceptions: incomplete octet (BF₃), expanded octet (PCl₅, SF₆).
-- NCERT Class 11 Chemistry, Ch. 4, p. 4Kossel Lewis
8 MCQs9-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 Kossel-Lewis approach is the foundational framework NEET uses to test whether you understand why atoms bond and how to represent bonding electrons. Getting this wrong cascades into errors across the entire Chemical Bonding chapter.
The core idea. Kossel (1916) observed that noble gases are exceptionally stable due to their complete outer shells — 2 electrons for helium, 8 for the rest. He proposed that atoms form ions by losing or gaining electrons to achieve this configuration. Lewis, working independently, introduced the electron-dot structure: valence electrons are shown as dots around the atomic symbol, and a shared pair of electrons between two atoms constitutes a covalent bond.
What NCERT says. NCERT Class 11 Chemistry Chapter 4, page 4, establishes the octet rule as the central principle: atoms combine to achieve 8 electrons in their valence shell (or 2 for the first period). Ionic bonding arises when electronegativity difference is large enough for complete electron transfer; covalent bonding arises when atoms share pairs.
Where aspirants lose marks. The high-frequency confusion at this topic level is conflating the octet rule with a universal law. It is a useful guideline with well-documented exceptions: incomplete octets (BF₃ — boron has only 6 electrons), expanded octets (PCl₅, SF₆ — elements in period 3+ use d-orbitals), and odd-electron species (NO, NO₂). NEET distractors exploit the assumption that every stable molecule obeys the octet rule.
A second common error is drawing Lewis structures with incorrect formal charges — placing more bonds than necessary on an atom or failing to minimise formal charge. The formal charge formula (FC = valence electrons − lone pair electrons − ½ bonding electrons) is the checkpoint.
Watch-out. When a question asks "which of the following violates the octet rule," scan for period-3+ central atoms (expanded octet) and group-13 elements (incomplete octet) before committing to an answer.
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
According to Kossel's theory, atoms form ionic bonds to achieve the electronic configuration of which group of elements?
MCQ 2Easy RecallPractice
In a Lewis dot structure, a lone pair refers to:
MCQ 3Easy RecallPractice
Which of the following is the correct Lewis dot structure representation of a double bond?
MCQ 4Direct ApplicationPractice
Which of the following species does NOT obey the octet rule?
MCQ 5Direct ApplicationPractice
The formal charge on the nitrogen atom in the ammonium ion (NH₄⁺) is:
MCQ 6Direct ApplicationPractice
SF₆ is a stable molecule despite sulfur having 12 electrons in its valence shell. This is possible because sulfur:
MCQ 7Concept TrapPractice
NO₂ is a paramagnetic molecule with an odd number of electrons. Within the Kossel-Lewis framework, this species is best described as:
MCQ 8Concept TrapPractice
A student draws the Lewis structure of CO₂ with single bonds between C and each O, placing 3 lone pairs on each oxygen and 2 lone pairs on carbon. The total electron count in this structure is 16, matching the actual valence electron count. Why is this structure still considered incorrect?
Worked Example
- 1
Given
- Carbon: 4 valence electrons - Nitrogen: 5 valence electrons - Charge on ion: −1 (one extra electron) - Total valence electrons: 4 + 5 + 1 = 10
- 2
Required
Lewis dot structure of CN⁻ with formal charges on C and N. Bond multiplicity.
- 3
Concept
The best Lewis structure minimises formal charges on all atoms. We try single, double, and triple bond options and compare formal charges.
- 4
Formula
Formal charge: FC = (valence electrons) − (lone pair electrons) − ½(bonding electrons)
- 5
Substitution (testing the triple bond structure)
Triple bond = 6 bonding electrons. Remaining electrons = 10 − 6 = 4, distributed as 1 lone pair on C and 1 lone pair on N. - FC on C = 4 − 2 − ½(6) = 4 − 2 − 3 = −1 - FC on N = 5 − 2 − ½(6) = 5 − 2 − 3 = 0
- 6
Calculation (compare with alternatives)
**Single bond attempt:** C−N with 8 remaining electrons → C gets 3 lone pairs (6e), N gets 1 lone pair (2e). FC on C = 4 − 6 − ½(2) = −3. FC on N = 5 − 2 − ½(2) = +2. Large formal charges — rejected. **Double bond attempt:** C=N with 6 remaining electrons → C gets 2 lone pairs (4e), N gets 1 lone pair (2e). FC on C = 4 − 4 − ½(4) = −2. FC on N = 5 − 2 − ½(4) = +1. Better, but still non-zero on both atoms — not optimal. **Triple bond:** FC on C = −1, FC on N = 0. Sum = −1 = charge on the ion. Formal charges are minimised. Note: the integers 4, 5, 6, 2, 3 used in formal charge arithmetic are exact counting numbers and do not affect any significant-figure consideration.
- 7
Final answer
The correct Lewis structure of CN⁻ has a **triple bond** between C and N, with one lone pair on each atom. Formal charges: C = −1, N = 0. The negative charge resides on carbon.
- 8
Common trap
Aspirants often stop at the double bond structure because "C=N looks familiar from organic chemistry." But in CN⁻, the double bond gives FC = −2 on C and +1 on N — violating formal charge minimisation. Always check all bond orders systematically.
- 9
Similar NEET-style question
"Calculate the formal charge on oxygen in the ozone molecule (O₃), given that the central oxygen forms one single bond and one double bond with the terminal oxygens." (Answer: central O has FC = +1; the single-bonded terminal O has FC = −1; the double-bonded terminal O has FC = 0.) ---
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.4
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