The trap that costs marks: You see NH₃ and count three bonds — so you write sp². Wrong. You forgot the lone pair. Hybridization depends on the steric number (bond pairs + lone pairs), not just the number of atoms bonded to the central atom.
Valence Bond Theory (VBT) essentials for NEET:
VBT explains covalent bond formation through the overlap of half-filled (or lone-pair) atomic orbitals. Key principles from NCERT Class 11 Chemistry Chapter 4, page 119:
Hybridization within VBT:
The central atom's orbitals mix (hybridize) to form equivalent hybrid orbitals that minimize repulsion. The steric number dictates hybridization:
Steric number = bond pairs + lone pairs on the central atom.
The high-frequency NEET trap: Lone pairs occupy hybrid orbitals but are invisible in the molecular formula. H₂O has SN = 4 (2 bonds + 2 lone pairs) → sp³, not sp. NH₃ has SN = 4 (3 bonds + 1 lone pair) → sp³, not sp².
Watch-out: When NEET gives you a structure and asks "hybridization of the central atom," count ALL electron domains — bonded pairs AND lone pairs — before assigning hybridization.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
According to Valence Bond Theory, which condition is essential for the formation of a covalent bond?
Answer: B. VBT states that a covalent bond forms by the overlap of half-filled (singly occupied) atomic orbitals on two atoms, resulting in pairing of electrons with opposite spins (NCERT Class 11 Chemistry, Chapter 4, page 119).
Why A is wrong: A describes ionic bonding (Kossel theory), not covalent bonding under VBT. (trap: confusing ionic and covalent mechanisms)
Why C is wrong: C is wrong because completely filled orbitals have no unpaired electrons available for bonding under standard VBT. Overlap requires half-filled orbitals. (trap: conflating filled orbitals with bonding capacity)
Why D is wrong: D is wrong because coordinate bonds are a special case (both electrons from one atom); the general VBT condition requires half-filled orbitals on each atom. (trap: overgeneralizing coordinate bonding)
The numbers of σ and π bonds in a molecule of ethene (CH₂=CH₂) are:
Answer: A. A is correct. There are four C–H bonds, each a σ bond, and one C=C double bond, which is one σ bond plus one π bond. Total: 4 + 1 = 5 σ and 1 π. Every single bond is σ; each additional bond in a multiple bond is π.
Why B is wrong: B is wrong because 4 σ counts the C–H bonds and forgets the σ component of the C=C double bond.
Why C is wrong: C is wrong because 4 σ and 2 π treats both bonds of the double bond as π. A double bond is one σ, from head-on overlap, plus one π, from sideways overlap.
Why D is wrong: D is wrong because 6 σ and 0 π counts both bonds of the double bond as σ. Two atoms can share only one σ bond, because there is only one axis between them for head-on overlap.
A π (pi) bond is formed by:
Answer: C. C is correct. In a π bond the electron density lies above and below the internuclear axis, which comes from p orbitals overlapping side by side. Sideways overlap is smaller than head-on overlap, which is why a π bond is weaker than a σ bond and why it forms only alongside one.
Why A is wrong: A is wrong because head-on s–s overlap gives a σ bond, as in H₂. Its electron density lies along the internuclear axis.
Why B is wrong: B is wrong because head-on overlap of p orbitals along the internuclear axis gives a σ bond, as in F₂. What makes a bond π is the sideways orientation, not the use of p orbitals.
Why D is wrong: D is wrong because head-on s–p overlap gives a σ bond, as in HF.
Which of the following molecules has sp² hybridization on the central atom?
Answer: C. Boron in BF₃ has 3 bond pairs + 0 lone pairs → steric number = 3 → sp² hybridization. It is electron-deficient with an empty p-orbital.
Why A is wrong: A is wrong: H₂O has SN = 4 (2 bonds + 2 lone pairs) → sp³. The two lone pairs push the steric number to 4. (trap: hybridization lone pair count)
Why B is wrong: B is wrong: NH₃ has SN = 4 (3 bonds + 1 lone pair) → sp³, not sp². Forgetting the lone pair leads to the incorrect sp² assignment. (trap: hybridization lone pair count)
Why D is wrong: D is wrong: CH₄ has SN = 4 (4 bonds + 0 lone pairs) → sp³, not sp². (trap: confusing number of bonds with hybridization type)
According to VBT, a sigma (σ) bond is formed by:
Answer: D. In VBT, a σ bond results from head-on (axial) overlap of orbitals along the line joining the two nuclei. This includes s-s, s-p, and p-p head-on overlaps (NCERT Class 11 Chemistry, Chapter 4).
Why A is wrong: A describes π-bond formation (lateral overlap of p-orbitals), not σ-bond. (trap: reversing σ and π overlap definitions)
Why B is wrong: B describes ionic bonding, not covalent bond formation under VBT. (trap: confusing bonding models)
Why C is wrong: C describes δ-bond type overlap seen in transition metal complexes — not the σ-bond mechanism taught in NEET-level VBT. (trap: importing d-orbital concepts beyond syllabus scope)
The hybridization of carbon in ethylene (C₂H₄) is:
Answer: A. Each carbon in C₂H₄ forms 2 C–H σ bonds + 1 C=C bond (σ + π). The σ framework uses 3 hybrid orbitals → sp² hybridization. The unhybridized p-orbital forms the π bond.
Why B is wrong: B (sp) would mean only 2 σ bonds and a linear geometry (180°). Carbon in ethylene has 3 σ bonds (trigonal planar, 120°). (trap: confusing ethylene with acetylene)
Why C is wrong: C (sp³) implies 4 equivalent σ bonds with tetrahedral geometry, which describes methane, not ethylene. (trap: defaulting to sp³ for all carbon compounds)
Why D is wrong: D (sp³d) requires d-orbital participation and steric number 5 — impossible for second-period carbon. (trap: expanded octet confusion)
PCl₅ has trigonal bipyramidal geometry. The hybridization of phosphorus in PCl₅ is:
Answer: B. Phosphorus in PCl₅ has 5 bond pairs + 0 lone pairs → steric number = 5 → sp³d hybridization, giving trigonal bipyramidal geometry.
Why A is wrong: A (sp²) corresponds to SN = 3. PCl₅ has 5 bonds to chlorine, not 3. (trap: undercounting bonds)
Why C is wrong: C (sp³) corresponds to SN = 4. PCl₅ has 5 bond pairs, exceeding the octet via d-orbital participation. (trap: ignoring expanded octet for Period 3 elements)
Why D is wrong: D (sp³d²) corresponds to SN = 6 (octahedral). PCl₅ has only 5 electron domains. (trap: confusing PCl₅ with SF₆)
A molecule XY₃ has one lone pair on X and trigonal pyramidal shape. The steric number and hybridization of X are:
Answer: D. XY₃ with one lone pair on X has 3 bond pairs + 1 lone pair = steric number 4 → sp³ hybridization. The shape is trigonal pyramidal (not tetrahedral) because the lone pair distorts the geometry, but hybridization depends on steric number, not shape.
Why A is wrong: A counts only the 3 bonded atoms (SN = 3 → sp²), ignoring the lone pair. This is exactly the trap: hybridization is decided by steric number, which INCLUDES lone pairs. (trap: hybridization lone pair count)
Why B is wrong: B assigns SN = 4 to sp³d, but sp³d corresponds to SN = 5. SN = 4 maps to sp³. (trap: confusing hybridization-SN correspondence table)
Why C is wrong: C has the right SN (3) paired with wrong hybridization (sp³). SN = 3 gives sp², and SN = 4 gives sp³. This option is internally inconsistent. (trap: mixing up SN-to-hybridization mapping)
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Pattern: Determine hybridization of a central atom from molecular formula (NEET pattern: hybridization from structure)
Given
Molecule: ClF₃ (chlorine trifluoride). Asked: hybridization of the central chlorine atom.
Required
Hybridization of Cl in ClF₃.
Concept
Hybridization is determined by the steric number (total electron domains around the central atom). We need the Lewis structure to identify bond pairs and lone pairs on Cl.
Formula
Steric number (SN) = number of bond pairs + number of lone pairs on the central atom.
SN → hybridization: 2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d².
Substitution
Chlorine: 7 valence electrons. Three Cl–F bonds use 3 electrons for bonding. Remaining electrons on Cl: 7 − 3 = 4 → 2 lone pairs.
SN = 3 (bond pairs) + 2 (lone pairs) = 5.
Calculation
SN = 5 → sp³d hybridization.
Final answer
Chlorine in ClF₃ is sp³d hybridized. The molecular shape is T-shaped (trigonal bipyramidal electron geometry with 2 equatorial lone pairs).
Common trap
Students who count only the 3 Cl–F bonds get SN = 3 and incorrectly assign sp². The two lone pairs on Cl are invisible in the formula but essential for the steric number. Always draw the Lewis structure first.
Similar NEET-style question
"What is the hybridization of the central atom in XeF₂?" (Answer: Xe has 3 lone pairs + 2 bond pairs → SN = 5 → sp³d; shape is linear.)
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. 121More in Chemical Bonding and Molecular Structure: 4 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
1 question from NEET 2024. 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
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.
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