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 Class 11 Chemistry, Ch. 8, p. 257Hybridization Organic
8 MCQs9-step worked example
Source: NCERT Organic Chemistry — Some Basic PrinciplesPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 27 Sep 2026
Hybridization Organic, explained for NEET
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 8 — Hybridisation section, page 257 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.
Can you answer these Hybridization Organic 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)?
Show answer and why every option is right or wrong
Answer: A. Each carbon in ethane forms four sigma bonds (three C–H and one C–C) and zero pi bonds. Four sigma bonds → sp³ hybridization with tetrahedral geometry (NCERT Class 11 Chemistry, Chapter 8, page 257 of Part 2).
Why B is wrong: B is wrong because sp² hybridization requires three sigma bonds and one pi bond (as in alkenes). Ethane has no pi bonds.
Why C is wrong: C is wrong because sp hybridization requires two sigma bonds and two pi bonds (as in alkynes). Ethane carbons have four sigma bonds each.
Why D is wrong: D is wrong because sp³d hybridization involves d-orbital participation and is seen in expanded octets (e.g., PCl₅). Carbon in ethane does not use d orbitals.
MCQ 2Easy RecallPractice
What is the bond angle in a molecule where the central carbon is sp² hybridized?
Show answer and why every option is right or wrong
Answer: B. sp² hybridization produces three equivalent hybrid orbitals arranged in a trigonal planar geometry with bond angles of 120° (NCERT Class 11 Chemistry, Chapter 8, page 257 of Part 2).
Why A is wrong: A is wrong because 180° is the bond angle for sp hybridization (linear geometry), not sp².
Why C is wrong: C is wrong because 109.5° is the tetrahedral angle for sp³ hybridization, not sp².
Why D is wrong: D is wrong because 90° is associated with unhybridized p orbitals in octahedral geometry, not with sp² carbon.
MCQ 3Easy RecallPractice
Which of the following correctly states the number of unhybridized p orbitals on an sp hybridized carbon atom?
Show answer and why every option is right or wrong
Answer: B. In sp hybridization, one s orbital and one p orbital mix to form two sp hybrid orbitals. The remaining two p orbitals stay unhybridized and are available for pi bond formation (NCERT Class 11 Chemistry, Chapter 8, page 257 of Part 2).
Why A is wrong: A is wrong because sp hybridization uses only one of the three p orbitals for hybridization, leaving two unhybridized — not zero.
Why C is wrong: C is wrong because one unhybridized p orbital describes sp² hybridization (where two p orbitals are used in hybridization). sp uses only one p orbital, leaving two.
Why D is wrong: D is wrong because three unhybridized p orbitals would mean no p orbital participated in hybridization at all — that would be an unhybridized s orbital scenario, which does not describe sp carbon.
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?
Show answer and why every option is right or wrong
Answer: C. Carbon-1 (CH₃) has four sigma bonds → sp³. Carbon-2 (CH=) has three sigma bonds and one pi bond → sp². Carbon-3 (=CH₂) has three sigma bonds and one pi bond → sp². The count per carbon gives sp³, sp², sp² (sp³, sp² and sp carbon: NCERT Class 11 Chemistry, Chapter 8, page 257).
Why A is wrong: A is wrong because it assigns sp³ to carbon-2, which participates in the C=C double bond (one sigma + one pi). Three sigma bonds + one pi bond → sp², not sp³.
Why B is wrong: B is wrong because it assigns sp² to the CH₃ carbon (carbon-1), which has four sigma bonds and no pi bonds — that is sp³, not sp².
Why D is wrong: D is wrong because it assigns sp³ to carbon-2 (which is part of the double bond → sp²) and sp² to carbon-1 (which has four sigma bonds → sp³). Both assignments are inverted.
MCQ 5Direct ApplicationPractice
What is the hybridization of the central carbon atom in allene (CH₂=C=CH₂)?
Show answer and why every option is right or wrong
Answer: D. The central carbon in allene forms two sigma bonds (one to each terminal carbon) and two pi bonds (one to each side). Two sigma bonds + two pi bonds → sp hybridization with linear geometry at the central carbon (NCERT Class 11 Chemistry, Chapter 8, page 257 of Part 2).
Why A is wrong: A is wrong because sp³ requires four sigma bonds. The central carbon of allene has only two sigma bonds and two pi bonds — it is sp, not sp³.
Why B is wrong: B is wrong because sp² requires three sigma bonds and one pi bond. The central allene carbon has only two sigma bonds (to two terminal carbons) and participates in two pi bonds — that is sp.
Why C is wrong: C is wrong because sp³d hybridization involves d-orbital participation and expanded octets. Carbon does not use d orbitals in standard organic compounds.
MCQ 6Direct ApplicationPractice
How many sigma bonds and pi bonds are present in the molecule HC≡C–CH=CH₂?
Show answer and why every option is right or wrong
Answer: C. C is correct. The molecule is C₄H₄, so start from the formula: four C–H bonds, and three C–C skeletal links (C1–C2, C2–C3, C3–C4), each contributing one sigma — 7 sigma in all. Counting bond by bond gives the same: C1–H (1σ), C1≡C2 (1σ + 2π), C2–C3 (1σ), C3–H (1σ), C3=C4 (1σ + 1π), C4–H (1σ), C4–H (1σ). The pi count is 2 from the triple bond and 1 from the double bond, so 3 pi. Checking the hydrogens against the molecular formula is the fastest guard against miscounting sigma bonds here.
Why A is wrong: A is wrong because 8 sigma is one too many. A molecule of C₄H₄ has exactly four C–H bonds and three C–C bonds, which is seven sigma bonds — counting an eighth means a fifth hydrogen has been added somewhere that the formula does not allow.
Why B is wrong: B is wrong because it overcounts sigma bonds by two. C₄H₄ gives 4 C–H plus 3 C–C = 7 sigma, not 9.
Why D is wrong: D is wrong because it undercounts pi bonds. The triple bond contributes 2 pi bonds and the double bond contributes 1 pi bond, giving 3 pi bonds total — not 2.
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?
Show answer and why every option is right or wrong
Answer: D. Label the carbons C1 through C5. C1 (CH₃): four sigma bonds → sp³. C2 (≡C): two sigma bonds + two pi bonds → sp. C3 (C≡): two sigma bonds + two pi bonds → sp. C4 (CH=): three sigma bonds + one pi bond → sp². C5 (=CH₂): three sigma bonds + one pi bond → sp². Result: 2 sp, 2 sp², 1 sp³.
Why A is wrong: A is wrong because it counts three sp² carbons and only one sp. The triple bond means both C2 and C3 are sp (not just one), and only C4 and C5 are sp².
Why B is wrong: B is wrong because it assigns sp³ to two carbons. Only C1 (CH₃) is sp³. C5 (=CH₂) has three sigma bonds and one pi bond → sp², not sp³.
Why C is wrong: C is wrong because it counts only one sp carbon. Both C2 and C3 are part of the triple bond, and each has two sigma bonds + two pi bonds → both are sp.
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?
Show answer and why every option is right or wrong
Answer: C. The C=O double bond has one sigma bond and one pi bond. The pi bond uses an unhybridized p orbital and does not determine hybridization. Carbon in HCHO forms three sigma bonds (two C–H, one C–O sigma) and has one pi bond → sp² hybridization with trigonal planar geometry. The student's error is counting the pi component and oxygen's lone pair as sigma bonds to carbon.
Why A is wrong: A is wrong because the carbon in HCHO does form a total of four bonds (two C–H single bonds + one C=O double bond = four bonds). The issue is not the bond count — it is that the student counted the pi bond toward hybridization when only sigma bonds determine hybridization state.
Why B is wrong: B is wrong on two counts: while it is true that oxygen's lone pair does not bond to carbon, the conclusion that carbon is sp hybridized is incorrect. Carbon in HCHO has three sigma bonds → sp², not sp.
Why D is wrong: D is wrong because formaldehyde carbon is sp², not sp³. The student incorrectly counted the pi bond and a lone pair as sigma bonds.
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How do you solve a Hybridization Organic question? A worked example
- 1
Given
A five-carbon chain: CH₃–CH=CH–C≡CH (pent-3-en-1-yne). Number from the alkyne end: the locant set {1,3} beats {2,4} from the methyl end. Note that pent-1-en-4-yne is a different compound, CH₂=CH–CH₂–C≡CH — the one in Step 9.
- 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 8, page 257.
- 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, 2 sp², 1 sp³. Take the carbons in order: C1 (=CH₂) sp², C2 (–CH=) sp², C3 (–CH₂–) sp³, C4 (≡C–) sp, C5 (≡CH) sp. That is 2 + 2 + 1 = 5, which must equal the number of carbons — the count is the check. Both alkene carbons are sp², so answering 1 sp² leaves a carbon unaccounted for.)
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What to remember before solving Hybridization Organic questions
More in Organic Chemistry — Some Basic Principles: 9 exam traps and mistakes · 1 formula · 1 question pattern from its other lessons.
Hybridization Organic questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys.
NEET 2023
The number of σ bonds, π bonds and lone pair of electrons in pyridine, respectively are:
112, 3, 0
211, 3, 1
312, 2, 1
411, 2, 0
NTA Answer: Option 2(final)
All 12 past-paper questions from Organic Chemistry — Some Basic Principles →
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