The single most tested trap in hybridization questions: forgetting to count lone pairs on the central atom. Students see NH₃ with three bonds and write sp² — wrong. The lone pair on nitrogen makes the steric number 4, giving sp³ hybridization.
The steric number rule. Hybridization depends on the steric number (SN) of the central atom, defined as:
SN = (number of bond pairs around central atom) + (number of lone pairs on central atom)
The mapping is fixed:
Note: bond pairs here include single, double, and triple bonds equally — each counts as one steric unit regardless of bond multiplicity (NCERT Class 11 Chemistry Chapter 4, page 113).
Where students lose marks. The trap fires on molecules where lone pairs exist but are invisible in the molecular formula: H₂O has SN = 2 bonds + 2 lone pairs = 4 → sp³ (not sp). ClF₃ has SN = 3 bonds + 2 lone pairs = 5 → sp³d (not sp²). XeF₂ has SN = 2 bonds + 3 lone pairs = 5 → sp³d (not sp).
The NEET distractor pattern. When a question asks "What is the hybridization of the central atom in X?", the wrong options are always calculated by ignoring lone pairs. If you see an option that matches the bond-pair count alone, that is the trap distractor — skip it and count properly.
Quick audit before marking your answer: Write the Lewis structure. Count ALL electron pairs around the central atom. Bond pairs + lone pairs = steric number. Map to hybridization. Done.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
What is the hybridization of carbon in methane (CH₄)?
Answer: C. Carbon in CH₄ has 4 bond pairs and 0 lone pairs, giving SN = 4. SN = 4 corresponds to sp³ hybridization (NCERT Class 11 Chemistry Chapter 4, page 113).
Why A is wrong: A is wrong because sp hybridization requires SN = 2. Carbon in CH₄ has SN = 4 (4 bond pairs, 0 lone pairs) — not 2.
Why B is wrong: B is wrong because sp² hybridization requires SN = 3. Carbon in CH₄ has 4 bond pairs and 0 lone pairs, so SN = 4, not 3.
Why D is wrong: D is wrong because sp³d hybridization requires SN = 5, which needs d-orbital involvement. Carbon has no accessible d-orbitals, and CH₄ has SN = 4.
Which hybridization corresponds to a steric number of 5?
Answer: A. The steric number to hybridization mapping is fixed: SN = 5 → sp³d (NCERT Class 11 Chemistry Chapter 4, page 113).
Why B is wrong: B is wrong because sp³ corresponds to SN = 4, not SN = 5.
Why C is wrong: C is wrong because sp² corresponds to SN = 3, not SN = 5.
Why D is wrong: D is wrong because sp³d² corresponds to SN = 6, not SN = 5.
A double bond between two atoms counts as how many steric units when determining the hybridization of the central atom?
Answer: D. Each bond — single, double, or triple — counts as exactly one steric unit for hybridization purposes, because hybridization counts electron domains, not individual electron pairs within a bond (NCERT Class 11 Chemistry Chapter 4, page 113).
Why A is wrong: A is wrong because a double bond is an electron domain around the central atom and must be counted. It contributes 1 steric unit, not 0.
Why B is wrong: B is wrong because 4 has no basis in hybridization counting. A double bond, regardless of the number of electrons it contains, counts as 1 steric unit.
Why C is wrong: C is wrong because this counts the individual bond pairs within a double bond (σ + π). For hybridization, the entire double bond counts as one steric unit, not two (trap: confusing bond pairs with electron domains).
What is the hybridization of the central atom in NH₃?
Answer: A. Nitrogen in NH₃ has 3 bond pairs + 1 lone pair = SN of 4. SN = 4 → sp³ hybridization (NCERT Class 11 Chemistry Chapter 4, page 113).
Why B is wrong: B is wrong because sp² requires SN = 3. This is the classic lone-pair-ignoring trap — counting only the 3 N–H bonds gives SN = 3, but the lone pair on nitrogen must be included, making SN = 4 (trap: hybridization lone pair count).
Why C is wrong: C is wrong because sp requires SN = 2. NH₃ has SN = 4 (3 bonds + 1 lone pair), not 2.
Why D is wrong: D is wrong because sp³d requires SN = 5. NH₃ has SN = 4 (3 bonds + 1 lone pair), not 5.
The hybridisation of each carbon atom in ethyne (HC≡CH) is:
Answer: B. B is correct. Each carbon forms two σ bonds (one to H, one to the other C) and has no lone pairs, so it needs two hybrid orbitals: sp, which are 180° apart and make the molecule linear. The two unhybridised p orbitals on each carbon form the two π bonds of the triple bond.
Why A is wrong: A is wrong because sp² is the hybridisation in ethene, where each carbon forms three σ bonds. In ethyne each carbon forms only two.
Why C is wrong: C is wrong because sp³ is the hybridisation in ethane, with four σ bonds per carbon. Counting the three bonds of the triple bond as separate σ bonds leads here; only one of them is σ.
Why D is wrong: D is wrong because sp³d involves a d orbital and five electron domains. Carbon has no d orbitals in its valence shell.
What is the hybridization of the central atom in XeF₂?
Answer: D. Xenon in XeF₂ has 2 bond pairs + 3 lone pairs = SN of 5. SN = 5 → sp³d hybridization (NCERT Class 11 Chemistry Chapter 4, page 113).
Why A is wrong: A is wrong because sp requires SN = 2. This counts only the 2 Xe–F bonds and ignores all 3 lone pairs on xenon (trap: hybridization lone pair count). Actual SN = 5.
Why B is wrong: B is wrong because sp² requires SN = 3. XeF₂ has SN = 5 (2 bonds + 3 lone pairs), not 3.
Why C is wrong: C is wrong because sp³ requires SN = 4. XeF₂ has SN = 5 (2 bonds + 3 lone pairs), not 4.
Among BF₃, NF₃, and ClF₃, which molecule(s) have sp³ hybridization on the central atom?
Answer: B. BF₃: B has 3 bond pairs + 0 lone pairs = SN 3 → sp². NF₃: N has 3 bond pairs + 1 lone pair = SN 4 → sp³. ClF₃: Cl has 3 bond pairs + 2 lone pairs = SN 5 → sp³d. Only NF₃ is sp³ (NCERT Class 11 Chemistry Chapter 4, page 113).
Why A is wrong: A is wrong because BF₃ has SN = 3 (3 bonds, 0 lone pairs) → sp², not sp³. Boron in BF₃ is electron-deficient with no lone pairs.
Why C is wrong: C is wrong because BF₃ is sp² (SN = 3) and ClF₃ is sp³d (SN = 5). Neither is sp³. The trap is assuming all trifluorides share the same hybridization — they don't, because lone pair counts differ.
Why D is wrong: D is wrong because ClF₃ has SN = 5 (3 bonds + 2 lone pairs) → sp³d, not sp³. Only NF₃ (SN = 4) is sp³ (trap: hybridization lone pair count on ClF₃).
The central atom in a molecule has 2 bond pairs and 2 lone pairs. A student predicts sp² hybridization. What is the student's error, and what is the correct hybridization?
Answer: C. The student counted only the 2 bond pairs (SN = 2 → sp) or partially counted, arriving at SN = 3 → sp². The actual SN = 2 bonds + 2 lone pairs = 4 → sp³. The error is ignoring lone pairs in the steric number calculation (NCERT Class 11 Chemistry Chapter 4, page 113; trap: hybridization lone pair count).
Why A is wrong: A is wrong because counting lone pairs twice would give SN = 6, predicting sp³d², not sp². The student's sp² prediction (SN = 3) is consistent with partially or fully ignoring the 2 lone pairs, not double-counting them.
Why B is wrong: B is wrong because bond order is a molecular orbital concept irrelevant to hybridization assignment. The student's error is ignoring lone pairs, not confusing bond order with bond pairs. The correct SN is 4 → sp³, not 5 → sp³d.
Why D is wrong: D is wrong because SN = 2 bonds + 2 lone pairs = 4 → sp³. sp² requires SN = 3, which the student obtained by ignoring the lone pairs.
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Pattern: Determine hybridization of central atom from molecular formula (NEET pattern: hybridization from structure — observed in NEET 2023 and 2025).
Given
Molecule: ClF₃. Central atom: Cl (chlorine, Group 17, 7 valence electrons).
Required
Hybridization of the central chlorine atom.
Concept
Hybridization is determined by the steric number (SN) of the central atom. SN = bond pairs + lone pairs around the central atom. Each bond to a terminal atom counts as one steric unit regardless of bond multiplicity (NCERT Class 11 Chemistry Chapter 4, page 113).
Formula
SN = (bond pairs) + (lone pairs)
Mapping: SN = 2 → sp; SN = 3 → sp²; SN = 4 → sp³; SN = 5 → sp³d; SN = 6 → sp³d².
Substitution
Chlorine has 7 valence electrons. Three are used for 3 Cl–F bonds → 3 bond pairs. Remaining: 7 − 3 = 4 electrons = 2 lone pairs.
SN = 3 (bond pairs) + 2 (lone pairs) = 5.
Calculation
SN = 5 → sp³d hybridization.
Note: the integers 3, 2, and 5 are exact counting numbers and do not involve significant-figure considerations.
Final answer
The central chlorine atom in ClF₃ is sp³d hybridized (with trigonal bipyramidal electron geometry and T-shaped molecular geometry due to the 2 lone pairs occupying equatorial positions).
Common trap
A student who counts only the 3 Cl–F bond pairs would get SN = 3 → sp², which is the most common wrong answer on NEET. The 2 lone pairs on chlorine are invisible in the formula "ClF₃" but must be counted (trap: lone-pair omission in steric number).
Similar NEET-style question
What is the hybridization of the central atom in ICl₄⁻? (Answer: Iodine has 7 valence electrons + 1 from the negative charge = 8. Four bonds + 2 lone pairs → SN = 6 → sp³d².)
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Mixing of atomic orbitals to form equivalent hybrid orbitals. sp (linear, BeF₂), sp² (trigonal, BF₃), sp³ (tetrahedral, CH₄), sp³d (trigonal bipyramidal, PCl₅), sp³d² (octahedral, SF₆).
-- NCERT Class 11 Chemistry, Ch. 4, p. 121These 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.
1 question from NEET 2021. Answers verified against NTA official keys.
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 lone pairs
Counts only bonding pairs
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