The trap that costs marks: You see NH₃ and count three bonds, so you call the geometry "trigonal planar." But that lone pair on nitrogen is invisible in the formula and very visible in the shape. VSEPR counts all electron pairs around the central atom — bonding and lone — to determine the electron-pair geometry first, and only then names the molecular shape from the atom positions alone.
VSEPR core idea. Valence Shell Electron Pair Repulsion theory states that electron pairs around a central atom arrange themselves to minimise repulsion (NCERT Class 11 Chemistry Chapter 4, page 113). The number of electron pairs — the steric number (SN) — fixes the electron-pair geometry:
| SN | Electron-pair geometry | Example with 0 lone pairs |
|---|---|---|
| 2 | Linear | BeCl₂ |
| 3 | Trigonal planar | BF₃ |
| 4 | Tetrahedral | CH₄ |
| 5 | Trigonal bipyramidal | PCl₅ |
| 6 | Octahedral | SF₆ |
Lone pairs distort. Lone-pair–bond-pair repulsion is stronger than bond-pair–bond-pair repulsion (NCERT Class 11 Chemistry Chapter 4, page 113). This compresses bond angles below the ideal values. So while NH₃ has SN = 4 (tetrahedral electron geometry), its molecular shape is trigonal pyramidal with bond angle ~107° instead of 109.5°. H₂O with SN = 4 but two lone pairs is bent at ~104.5°.
NEET relevance. Questions typically give you a molecular formula and ask for the shape or bond angle comparison. The high-frequency trap: forgetting to include lone pairs in the steric number. Always compute SN = (bonding pairs) + (lone pairs) before naming the geometry.
Watch-out: Electron-pair geometry ≠ molecular shape when lone pairs are present. NEET distractors routinely offer the electron-pair geometry name as a wrong option for the molecular shape.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
According to VSEPR theory, electron pairs around a central atom arrange themselves to:
Answer: B. VSEPR theory is based on the principle that electron pairs (both bonding and lone) around a central atom position themselves to minimise repulsion (NCERT Class 11 Chemistry Chapter 4, page 113).
Why A is wrong: A describes orbital overlap, which is the basis of Valence Bond Theory, not VSEPR.
Why C is wrong: C describes a bonding tendency but not the geometric arrangement principle that VSEPR addresses.
Why D is wrong: D refers to antibonding orbitals from MO theory, which is a different bonding model entirely.
In VSEPR theory, which type of repulsion is the strongest?
Answer: C. The order of repulsion strength is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, because lone pairs are held closer to the nucleus and spread over a larger volume (NCERT Class 11 Chemistry Chapter 4, page 113).
Why A is wrong: A is the weakest of the three repulsion types since bonding electrons are localised between nuclei and occupy less angular space.
Why B is wrong: B is intermediate in strength — stronger than bp–bp but weaker than lp–lp repulsion.
Why D is wrong: D is incorrect; the three types of repulsion are explicitly ranked in VSEPR theory, with lone pair–lone pair being the strongest.
What is the electron-pair geometry of a molecule with a steric number of 4 and no lone pairs on the central atom?
Answer: C. SN = 4 with zero lone pairs gives tetrahedral geometry with bond angles of 109.5° (e.g., CH₄). Both the electron-pair geometry and molecular shape are tetrahedral when no lone pairs are present.
Why A is wrong: A (square planar) corresponds to SN = 6 with 2 lone pairs (e.g., XeF₄), not SN = 4 with 0 lone pairs.
Why B is wrong: B (trigonal pyramidal) is the molecular shape for SN = 4 with 1 lone pair (e.g., NH₃), not 0 lone pairs.
Why D is wrong: D (trigonal planar) corresponds to SN = 3 with 0 lone pairs (e.g., BF₃), not SN = 4.
The steric number of the central atom in H₂O is:
Answer: C. Oxygen in H₂O has 2 bonding pairs + 2 lone pairs = SN of 4. The electron-pair geometry is tetrahedral, though the molecular shape is bent due to the two lone pairs. The common trap is counting only the 2 bonds and ignoring the 2 lone pairs.
Why A is wrong: A counts only the 2 O–H bonds and ignores the 2 lone pairs on oxygen — a common mistake (trap: forgetting lone pairs in steric number count).
Why B is wrong: B would imply 3 electron pairs, which doesn't match oxygen's electron configuration in water (2 bp + 2 lp = 4).
Why D is wrong: D (SN = 6) would require 6 electron pairs around the central atom, which is far more than oxygen has in H₂O.
NH₃ has a bond angle of approximately 107° instead of the ideal tetrahedral angle of 109.5°. This compression is because:
Answer: B. The lone pair on nitrogen occupies more angular space than a bonding pair, exerting stronger repulsion on the three N–H bond pairs and compressing the H–N–H angle below 109.5° (NCERT Class 11 Chemistry Chapter 4, page 113).
Why A is wrong: A is incorrect — bond length affects bond strength but does not explain why bond angles deviate from ideal tetrahedral values in VSEPR.
Why C is wrong: C confuses electronegativity with steric repulsion. Electronegativity affects bond polarity, not the geometric distortion caused by lone pairs in VSEPR.
Why D is wrong: D is a common mistake: NH₃ has SN = 4 (3 bp + 1 lp), so it is sp³ hybridized, not sp². Forgetting the lone pair leads to this wrong assignment (trap: counting only bonded atoms for hybridization).
Which of the following molecules has a bent molecular shape?
Answer: B. H₂O has SN = 4 (2 bp + 2 lp), giving tetrahedral electron-pair geometry but a bent molecular shape with a bond angle of ~104.5°.
Why A is wrong: A (CO₂) has SN = 2 with 0 lone pairs on carbon, making it linear (180°), not bent.
Why C is wrong: C (BeCl₂) has SN = 2 with 0 lone pairs, giving a linear shape — same reasoning as CO₂.
Why D is wrong: D (BF₃) has SN = 3 with 0 lone pairs on boron, giving trigonal planar geometry (120°), not bent.
A student predicts that NH₃ is trigonal planar because nitrogen forms three bonds. The error in this reasoning is:
Answer: B. The student counted only the 3 N–H bond pairs (SN = 3 → trigonal planar) but failed to include the lone pair. The correct steric number is 4 (3 bp + 1 lp), giving tetrahedral electron-pair geometry and trigonal pyramidal molecular shape.
Why A is wrong: A is incorrect — NH₃ does form exactly three N–H bonds. The issue is not the bond count but the omission of the lone pair from the steric number (trap: forgetting lone pairs in steric number).
Why C is wrong: C is incorrect — trigonal planar geometry does not require double bonds. BF₃ is trigonal planar with all single bonds. The error is about lone-pair omission, not bond type.
Why D is wrong: D is incorrect — VSEPR theory applies to all covalent molecules including nitrogen compounds. There is no such restriction.
Arrange the following in order of decreasing bond angle: CH₄, NH₃, H₂O.
Answer: B. All three have SN = 4, but CH₄ (0 lp) has 109.5°, NH₃ (1 lp) has ~107°, and H₂O (2 lp) has ~104.5°. Each additional lone pair compresses the bond angle further due to stronger lp–bp repulsion. So CH₄ > NH₃ > H₂O.
Why A is wrong: A reverses the trend — more lone pairs means more compression, so H₂O has the smallest angle, not the largest (trap: confusing the direction of lone-pair compression).
Why C is wrong: C places NH₃ first, but CH₄ with zero lone pairs has the ideal tetrahedral angle (109.5°), which is the largest of the three.
Why D is wrong: D incorrectly places H₂O above NH₃. H₂O has 2 lone pairs and therefore greater compression (~104.5°) than NH₃ with 1 lone pair (~107°).
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Pattern: Predict molecular geometry from VSEPR steric number and lone pairs (P.CHE.U03.VSEPR_GEOMETRY; observed in NEET 2022, 2024, 2025).
Given
• Molecule: ClF₃• Chlorine is the central atom (less electronegative)• Cl has 7 valence electrons; each F has 7 valence electrons
Required
• Molecular shape and approximate bond angle of ClF₃
Concept
VSEPR theory: compute steric number (SN = bonding pairs + lone pairs on central atom), determine electron-pair geometry, then identify molecular shape from atom positions only.
Formula
SN = (number of bonding pairs) + (number of lone pairs on central atom)
Substitution
Total valence electrons = 7 + 3(7) = 28.
Three Cl–F bonds use 6 electrons. Remaining on Cl: 7 − 3 = 4 electrons (after formal assignment) → but let's count systematically.
Place 3 bond pairs (6 e⁻). Remaining: 28 − 6 = 22 e⁻.
Each F gets 3 lone pairs: 3 × 6 = 18 e⁻ used. Remaining on Cl: 22 − 18 = 4 e⁻ = 2 lone pairs.
SN = 3 (bond pairs) + 2 (lone pairs) = 5.
Calculation
SN = 5 → electron-pair geometry is trigonal bipyramidal.
With 2 lone pairs: lone pairs occupy equatorial positions (to minimise 90° lp–bp repulsions).
Molecular shape: T-shaped.
Bond angle: approximately 87.5° (less than 90° due to lp–bp compression).
Final answer
ClF₃ has a T-shaped molecular geometry with bond angles of approximately 87.5°.
Note: the counting numbers (3 bond pairs, 2 lone pairs, 28 total electrons) are exact integers and do not affect significant-figure considerations.
Common trap
Forgetting the 2 lone pairs on Cl and assigning SN = 3 → trigonal planar. This is the same lone-pair omission trap seen in NH₃ and H₂O. Always count ALL electron pairs on the central atom.
Similar NEET-style question
"Predict the molecular shape of XeF₂. Given: Xe has 8 valence electrons."
(Answer: SN = 5, 3 lone pairs + 2 bond pairs → linear.)
---
Electron pairs (bonding + lone) around central atom arrange to minimize repulsion. Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. Geometry from steric number = bond pairs + lone pairs.
-- NCERT Class 11 Chemistry, Ch. 4, p. 113AB₂: linear (180°). AB₃: trigonal planar (120°). AB₄: tetrahedral (109.5°). AB₅: trigonal bipyramidal. AB₆: octahedral (90°). Lone pairs distort: AB₃E pyramidal, AB₂E₂ bent.
-- NCERT Class 11 Chemistry, Ch. 4, p. 114These 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.
Molecule with central atom having lone pairs (e.g., NH₃: 3 bonds + 1 lp = 4 = sp³; H₂O: 2+2 = 4 = sp³).
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
Steric number = bond pairs + lone pairs. NH3: SN=4 → sp³ (despite trigonal pyramidal shape).
More in Chemical Bonding and Molecular Structure: 2 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
5 questions from NEET 2021, 2022, 2024, 2025, 2026. Answers verified against NTA official keys.
Amongst the following which one will have maximum ‘lone pair - lone pair’ electron repulsions?
All 18 past-paper questions from Chemical Bonding and Molecular Structure →
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
ignores lp bp distortion
Doesn't adjust for lone-pair repulsion
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →