sp³ (single bonds, e.g. methane, tetrahedral 109.5°). sp² (one double bond, e.g. ethene, trigonal planar 120°). sp (one triple bond, e.g. ethyne, linear 180°).
-- NCERT, p. 6Hybridization Organic
8 MCQs9-step worked example
Source: NCERT Alcohols, Phenols and EthersPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap you need to fix: Many aspirants mechanically assign sp³ to every carbon with four bonds without checking bond type, or forget that a triple bond means sp hybridization — not sp². This confusion costs easy marks because NEET regularly asks you to identify the hybridization state of specific carbons in organic molecules.
What hybridization actually means in organic chemistry. Carbon's ground-state electronic configuration (1s² 2s² 2p²) predicts only two bonds. To form four bonds, carbon promotes one 2s electron to the empty 2p orbital and then mixes (hybridizes) the resulting orbitals into equivalent hybrid sets. The type of hybridization depends on the number of sigma and pi bonds around that carbon (NCERT Class 11 Chemistry, Chapter 9 — Hybridisation section, page 6 of Part 2).
The three hybridization states of carbon:
- sp³ — four sigma bonds, zero pi bonds. Tetrahedral geometry (109.5°). Example: every carbon in ethane (CH₃–CH₃).
- sp² — three sigma bonds, one pi bond. Trigonal planar geometry (120°). Example: each carbon in ethene (CH₂=CH₂).
- sp — two sigma bonds, two pi bonds. Linear geometry (180°). Example: each carbon in ethyne (CH≡CH).
The quick-count rule: Count only the sigma bonds (and lone pairs, if relevant) around the carbon. Four sigma → sp³. Three sigma + one pi → sp². Two sigma + two pi → sp. Do not count pi bonds when assigning the hybrid orbital set — pi bonds use unhybridized p orbitals.
Watch out: In molecules like CH₂=C=CH₂ (allene), the central carbon is sp (two sigma bonds to adjacent carbons, two pi bonds) while the terminal carbons are sp² (three sigma bonds each, one pi bond each). Different carbons in the same molecule can have different hybridization states — NEET tests exactly this.
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
What is the hybridization of each carbon atom in ethane (C₂H₆)?
MCQ 2Easy RecallPractice
What is the bond angle in a molecule where the central carbon is sp² hybridized?
MCQ 3Easy RecallPractice
Which of the following correctly states the number of unhybridized p orbitals on an sp hybridized carbon atom?
MCQ 4Direct ApplicationPractice
In the molecule CH₃–CH=CH₂ (propene), what are the hybridization states of carbon-1 (CH₃), carbon-2 (CH=), and carbon-3 (=CH₂) respectively?
MCQ 5Direct ApplicationPractice
What is the hybridization of the central carbon atom in allene (CH₂=C=CH₂)?
MCQ 6Direct ApplicationPractice
How many sigma bonds and pi bonds are present in the molecule HC≡C–CH=CH₂?
MCQ 7CalculationPractice
In CH₃–C≡C–CH=CH₂, how many carbon atoms are sp hybridized, how many are sp² hybridized, and how many are sp³ hybridized?
MCQ 8Concept TrapPractice
A student claims that the carbon atom in formaldehyde (HCHO) is sp³ hybridized because it has four bonds (two C–H bonds, one C–O bond, and one extra bond from oxygen's lone pair). What is the error in this reasoning?
Worked Example
- 1
Given
A five-carbon chain: CH₃–CH=CH–C≡CH (pent-1-en-4-yne, numbered from the double-bond end).
- 2
Required
Hybridization state and geometry at each carbon (C1 through C5).
- 3
Concept
Hybridization is determined by the number of sigma bonds (and lone pairs) around the atom. Pi bonds use unhybridized p orbitals and do not count. Four sigma → sp³ (tetrahedral). Three sigma → sp² (trigonal planar). Two sigma → sp (linear). Reference: NCERT Class 11 Chemistry, Chapter 9, Hybridisation section.
- 4
Formula / rule
Count sigma bonds per carbon: - Each single bond = 1 sigma - Each double bond = 1 sigma + 1 pi - Each triple bond = 1 sigma + 2 pi
- 5
Substitution
| Carbon | Bonds | Sigma count | Pi count | |--------|-------|-------------|----------| | C1 (CH₃) | 3 C–H + 1 C–C | 4σ | 0π | | C2 (CH=) | 1 C–H + 1 C–C + 1 C=C | 3σ | 1π | | C3 (=CH–) | 1 C–H + 1 C=C + 1 C–C | 3σ | 1π | | C4 (–C≡) | 1 C–C + 1 C≡C | 2σ | 2π | | C5 (≡CH) | 1 C≡C + 1 C–H | 2σ | 2π |
- 6
Calculation
Apply the rule: - C1: 4σ → sp³, tetrahedral, ~109.5° - C2: 3σ → sp², trigonal planar, ~120° - C3: 3σ → sp², trigonal planar, ~120° - C4: 2σ → sp, linear, 180° - C5: 2σ → sp, linear, 180°
- 7
Final answer
CH₃–CH=CH–C≡CH contains: **1 sp³ carbon** (C1), **2 sp² carbons** (C2, C3), and **2 sp carbons** (C4, C5).
- 8
Common trap
Aspirants sometimes assign sp² to a triple-bonded carbon because they think "one bond type = one hybridization step down." The correct approach: count only sigma bonds. A triple bond contributes exactly one sigma bond to the count — the two pi bonds use unhybridized p orbitals.
- 9
Similar NEET-style question
"How many sp, sp², and sp³ hybridized carbon atoms are present in CH₂=CH–CH₂–C≡CH?" (Answer: 2 sp, 1 sp², 1 sp³ — but count carefully; the CH₂ at position 1 has three sigma bonds → sp².) ---
Before solving, remember these
Formulas
Carbocation stability order
Stability from hyperconjugation (more α-H) and inductive donation (alkyl groups). Resonance can elevate primary cations.
| Symbol | Quantity | SI Unit |
|---|---|---|
| stability | relative | - |
Valid when
- Gas phase or aprotic solvent
- Compare similar reaction conditions
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: Similar Terms
Student writes 1° > 2° > 3° (linear with substitution count, but inverted) or treats methyl as more stable.
When it triggers
Question gives multiple carbocations and asks for stability ranking.
How to avoid
Stability: 3° > 2° > 1° > methyl. Hyperconjugation (more α-H = more stable). Resonance can elevate (allyl, benzyl > 1°).
Category: Similar Terms
Student treats inductive (through sigma bonds, decreases with distance) like resonance (through pi system, often dominant).
When it triggers
Comparison of substituent effects (acidity, basicity, dipole moment).
How to avoid
Inductive: through-bond, weakens with distance, only sigma. Resonance: through-pi-system, often more powerful, requires conjugation.
Category: Similar Terms
Student numbers carbon chain from wrong end, giving higher locants to substituents than necessary.
When it triggers
IUPAC naming question with multiple substituents.
How to avoid
Choose end that gives the LOWEST set of locants for all substituents (compare set, not first-encountered). Functional group has priority for lowest locant.
Category: Similar Terms
Student conflates optical (chirality, R/S) with geometrical (cis/trans). They're different stereoisomerism types.
When it triggers
Question about isomerism of a specific compound.
How to avoid
Optical isomerism requires chiral center (sp³ with 4 different groups). Geometrical isomerism requires restricted rotation (C=C with two different groups on each carbon).
Root cause: concept gap
Correction
Stability follows hyperconjugation: more α-H → more stable. Order: 3° > 2° > 1° > methyl. Allyl/benzyl resonance-stabilised.
Root cause: concept gap
Correction
Hyperconjugation requires α-C-H bond. tert-Butyl carbocation: 9 α-H → very stable. No-α-H carbocation: no hyperconjugation.
Root cause: concept gap
Correction
Inductive: through sigma, weakens with distance, weak. Resonance: through pi, often dominant for activated systems.
Root cause: rushed under time pressure
Correction
Try numbering from BOTH ends; pick the one giving the lowest SET of locants. Functional group has priority for lowest locant.
Root cause: concept gap
Correction
Optical: chirality (sp³ with 4 different groups). Geometrical: restricted rotation (C=C). Different molecular features required.
Past Year Questions
11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
Which one of the following compounds can exist as cis-trans isomers?
11, 2-Dimethylcyclohexane
2Pent-1-ene
32-Methylhex-2-ene
41, 1-Dimethylcyclopropane
NTA Answer: Option 1(final)
NEET 2025
1II > III > I
2II > I > III
3I > II > III
4III > II > I
NTA Answer: Option 2(final)
NEET 2025
1A, C and D only
2B and E only
3A and C only
4D and E only
NTA Answer: Option 3(final)
NEET 2024Revised key
A compound with a molecular formula of C H has two tertiary carbons. Its IUPAC name is : 6 14
1n-hexane
22-methylpentane
32,3-dimethylbutane
42,2-dimethylbutane
NTA Answer: Option 3(revised_final)
NEET 2024Revised key
1Three resonance structures can be drawn for ozone
2BF has non-zero dipole moment 3
3Dipole moment of NF is greater than that of NH 3 3
4Three canonical forms can be drawn for CO2− ion 3
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
1propylamine
2butylamine
3butanamide
4–bromobutanoic acid
NTA Answer: Option 1(revised_final)
NEET 2023
1H = U + n g RT
2H – U = –nRT
3H + U = nR
4H = U – n RT g
NTA Answer: Option 1(final)
NEET 2022
The incorrect statement regarding chirality is
1A racemic mixture shows zero optical rotation
2S 1 reaction yields 1 : 1 mixture of both enantiomers N
3The product obtained by S 2 reaction of haloalkane having chirality at the reactive site shows inversion N of configuration
4Enantiomers are superimposable mirror images on each other
NTA Answer: Option 4(final)
NEET 2022
10.2303
21.3818
30.9212
40.4606
NTA Answer: Option 3(final)
NEET 2022
The correct IUPAC name of the following compound is
16-bromo-4-methyl-2-chlorohexan-4-ol
21-bromo-5-chloro-4-methylhexan-3-ol
36-bromo-2-chloro-4-methylhexan-4-ol
41-bromo-4-methyl-5-chlorohexan-3-ol
NTA Answer: Option 2(final)
NEET 2021
1(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
2(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
3(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
4(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
NTA Answer: Option 1(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Order carbocations by stability. 3° > 2° > 1° > methyl. Allyl/benzyl extra stabilised by resonance.
InterpretationEasy
Common distractors
ignores resonance
Compares only by alkyl substitution
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