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 20).
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)?
Which hybridization corresponds to a steric number of 5?
A double bond between two atoms counts as how many steric units when determining the hybridization of the central atom?
What is the hybridization of the central atom in NH₃?
What is the hybridization of the central atom in H₂O?
What is the hybridization of the central atom in XeF₂?
Among BF₃, NF₃, and ClF₃, which molecule(s) have sp³ hybridization on the central atom?
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?
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 20).
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.) ---
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. 20Higher 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 | - |
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 |
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.
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).
Root cause: concept gap
Memorise the ordering switch at N2/O2 boundary. Affects bond order and magnetic property predictions.
Root cause: formula misuse
BO = (Nb - Na)/2. Antibonding electrons subtract. F2: 8 bonding, 6 antibonding → BO = 1.
14 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
The element expected to form largest ion to achieve the nearest noble gas configuration is
Which statement regarding polymers is not correct?
Amongst the following which one will have maximum ‘lone pair - lone pair’ electron repulsions?
In one molal solution that contains 0.5 mole of a solute, there is
Which of the following molecules is non-polar in nature?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
ignores lone pairs
Counts only bonding pairs
forgets anti bonding
Counts only bonding electrons
ignores lp bp distortion
Doesn't adjust for lone-pair repulsion
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