What are the hybridisation states of each carbon atom in the following compounds? CH2=C=O, CH3CH=CH2, (CH3)2CO

Exercises · 8.1

Q. What are the hybridisation states of each carbon atom in the following compounds? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6.

Tripura TbseTextbookSubjective· 3mImportance★★★★★est

19% · 25/130 Questions

✓ Free question

Concept understanding — Sigma Pi Bond Counting

Sigma Pi Bond Counting: From Intuition to Precision

Imagine you're building a molecular model with sticks and balls. Every single bond you see — a single line between two atoms — is made of one sigma bond. That's the backbone. A double bond? That's one sigma plus one pi bond. A triple bond? One sigma plus two pi bonds.

This is the core idea: sigma bonds are the first bond formed between any two atoms; any additional bonds are pi bonds.

Why sigma comes first

When two atoms approach each other, their orbitals overlap end-to-end along the line joining the nuclei. That head-on overlap creates a sigma bond — strong, cylindrically symmetric, and free to rotate. If the atoms need to share more electrons (to satisfy octets, for example), they can't form another sigma bond because the orbitals are already used up in that direction. Instead, they use sideways overlap of p-orbitals above and below the internuclear axis. That sideways overlap is a pi bond — weaker, and it locks the molecule into a plane (no free rotation).

So the rule is simple: between any two bonded atoms, exactly one bond is sigma; the rest are pi.

The precise counting method

For any molecule, you can count sigma and pi bonds systematically:

Count sigma bonds: Every single bond is one sigma. Every double bond contributes one sigma (and one pi). Every triple bond contributes one sigma (and two pi). Also, every bond to hydrogen is sigma.
Count pi bonds: For each multiple bond, subtract 1 from the bond order. That remainder is the number of pi bonds.

n

So:

Single bond ( $n=1$ ): 1 sigma, 0 pi
Double bond ( $n=2$ ): 1 sigma, 1 pi
Triple bond ( $n=3$ ): 1 sigma, 2 pi

A worked example: ethene ( $\ce{C2H4}$ )

Draw the structure: each carbon is double-bonded to the other, and each carbon has two single bonds to hydrogen.

The C=C double bond: 1 sigma + 1 pi
Each C–H single bond: 1 sigma (4 such bonds)
Total: 5 sigma bonds, 1 pi bond

Check: The molecule has 5 sigma bonds holding the skeleton together, and 1 pi bond in the double bond region.

A trickier case: benzene ( $\ce{C6H6}$ )

Benzene has six C–C bonds that are all equivalent — each is 1.5 bonds (resonance hybrid). But for counting purposes, treat each ring bond as a single bond (sigma) plus a delocalised pi system.

6 C–H bonds: all sigma
6 C–C ring bonds: each is sigma
The pi system: 3 pi bonds (delocalised over the ring)

Total: 12 sigma bonds, 3 pi bonds.

Watch out

Do not count each C–C bond in benzene as 1.5 sigma bonds. Sigma bonds are always whole numbers. The fractional bond order comes from pi electrons being shared across multiple bonds.

Why this matters

Sigma-pi counting is not just a classification exercise. It explains:

Rotation barriers: Single bonds (pure sigma) rotate freely; double bonds (sigma + pi) do not.
Reactivity: Pi bonds are weaker and more exposed — they're where addition reactions happen (e.g., $\ce{Br2}$ adding across a double bond).
Hybridisation: The number of sigma bonds around an atom determines its hybridisation ( $sp^3$ for 4 sigma bonds, $sp^2$ for 3, $sp$ for 2).

The one-sentence summary

Every bond has exactly one sigma bond; any additional bond order comes from pi bonds.

This topic is commonly searched as "Sigma Pi Bond Counting 11 chemistry important questions" or "Sigma Pi Bond Counting formula and examples", and it maps cleanly onto the Class 11 Chemistry portion of the NCERT/CBSE syllabus. Because sigma pi bond counting shows up repeatedly in JEE Main, NEET and state CET Chemistry papers, mastering the underlying idea (not just the formula) is genuinely worth the extra time.

The key idea is to count sigma bonds and lone pairs around each carbon to determine its steric number, which directly gives the hybridisation. For these compounds: CH₂=C=O → C1 sp², C2 sp; CH₃CH=CH₂ → C1 sp³, C2 sp², C3 sp²; (CH₃)₂CO → all three carbons sp²; CH₂=CHCN → C1 sp², C2 sp², C3 sp; C₆H₆ → all six carbons sp².

The Concept: Sigma-Pi Bond Counting

Hybridisation is a model that explains how atomic orbitals mix to form bonds. The rule is simple: the number of sigma bonds plus the number of lone pairs around an atom equals its steric number, which tells you the hybridisation.

Steric number 2 → sp (linear, 180°)
Steric number 3 → sp² (trigonal planar, 120°)
Steric number 4 → sp³ (tetrahedral, ~109.5°)

A double bond consists of one sigma bond and one pi bond. A triple bond consists of one sigma bond and two pi bonds. Pi bonds do not count toward the steric number — they use unhybridised p-orbitals. So when you see a carbon with a double bond, it has 3 sigma bonds (including the one in the double bond) and no lone pairs → steric number 3 → sp².

Watch out

A common mistake is to count both bonds in a double bond as sigma bonds. Only one bond in any multiple bond is sigma; the rest are pi. Always count sigma bonds explicitly.

Let's work through each compound.

1. CH₂=C=O (Ketene)

Draw the structure: H₂C=C=O. The central carbon (C2) is doubly bonded to both C1 and O.

Carbon 1 (CH₂): It forms two sigma bonds with hydrogens and one sigma bond with C2 (the double bond contributes one sigma). That's 3 sigma bonds, no lone pairs. Steric number = 3 → sp².

Carbon 2 (the central C): It forms one sigma bond with C1 and one sigma bond with O (each double bond gives one sigma). That's 2 sigma bonds, no lone pairs. Steric number = 2 → sp.

Oxygen: Not asked, but for completeness — it has one sigma bond to C2, two lone pairs → steric number 3 → sp².

Tip

In cumulated double bonds (like C=C=C or C=C=O), the central atom is always sp hybridised because it makes only two sigma bonds.

2. CH₃CH=CH₂ (Propene)

Structure: H₃C–CH=CH₂. Number the carbons from left to right.

Carbon 1 (CH₃–): Three sigma bonds to hydrogens, one sigma bond to C2. Total 4 sigma bonds, no lone pairs → steric number 4 → sp³.

Carbon 2 (–CH=): One sigma bond to C1, one sigma bond to C3, one sigma bond to H. That's 3 sigma bonds. The double bond to C3 contributes one sigma and one pi. No lone pairs → steric number 3 → sp².

Carbon 3 (=CH₂): Two sigma bonds to hydrogens, one sigma bond to C2. Total 3 sigma bonds → steric number 3 → sp².

3. (CH₃)₂CO (Acetone)

Structure: (H₃C)₂C=O. The central carbon is doubly bonded to oxygen and singly bonded to two methyl groups.

Central carbon (C=O): Two sigma bonds to the two methyl carbons, one sigma bond to oxygen (from the double bond). That's 3 sigma bonds, no lone pairs → sp².

Each methyl carbon (CH₃–): Three sigma bonds to hydrogens, one sigma bond to the central carbon. Total 4 sigma bonds → sp³.

Note

In acetone, the carbonyl carbon is sp² hybridised, making the molecule planar around that carbon. The methyl groups are tetrahedral.

4. CH₂=CH–CN (Acrylonitrile)

Structure: H₂C=CH–C≡N. Number carbons: C1 = CH₂, C2 = CH, C3 = CN.

C1 (CH₂=): Two sigma bonds to H, one sigma bond to C2 → 3 sigma bonds → sp².

C2 (=CH–): One sigma bond to C1, one sigma bond to C3, one sigma bond to H → 3 sigma bonds → sp².

C3 (–C≡N): One sigma bond to C2, one sigma bond to N (from the triple bond). That's 2 sigma bonds. The triple bond contributes one sigma and two pi bonds. No lone pairs → steric number 2 → sp.

Nitrogen: One sigma bond to C3, two lone pairs → steric number 3 → sp².

Watch out

In a triple bond, only one bond is sigma. The carbon in –C≡N is sp, not sp² or sp³. Students often mistakenly count all three bonds as sigma.

5. C₆H₆ (Benzene)

Benzene is a planar ring of six carbons, each bonded to one hydrogen and two neighbouring carbons via alternating single and double bonds (resonance). Every carbon forms:

One sigma bond to H
Two sigma bonds to adjacent carbons (one from each C–C bond, whether single or double)

That's 3 sigma bonds per carbon. No lone pairs. Steric number = 3 → sp².

The pi electrons are delocalised above and below the ring, but each carbon still uses sp² hybrid orbitals for its sigma framework.

For any carbon with a double bond (or in a conjugated system like benzene): Steric number = number of sigma bonds + lone pairs = 3 \to sp².

✓Final answer

The hybridisation states are: CH₂=C=O \to C1 sp², C2 sp; CH₃CH=CH₂ \to C1 sp³, C2 sp², C3 sp²; (CH₃)₂CO \to central C sp², methyl C sp³; CH₂=CHCN \to C1 sp², C2 sp², C3 sp; C₆H₆ \to all carbons sp².

←Previous section · exercisesExercises →Next question · exercises8.2 — Indicate the σ- and π-bonds in the following molecules: C6H6, C6H12, C…

Q.What are the hybridisation states of each carbon atom in the following compounds? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6.

Concept understanding — Sigma Pi Bond Counting

Sigma Pi Bond Counting: From Intuition to Precision

Why sigma comes first

The precise counting method

A worked example: ethene ( $\ce{C2H4}$ )

A trickier case: benzene ( $\ce{C6H6}$ )

Why this matters

The one-sentence summary

Key idea: Sigma-Pi Bond Counting

The Concept: Sigma-Pi Bond Counting

1. CH₂=C=O (Ketene)

2. CH₃CH=CH₂ (Propene)

3. (CH₃)₂CO (Acetone)

4. CH₂=CH–CN (Acrylonitrile)

5. C₆H₆ (Benzene)

Q.What are the hybridisation states of each carbon atom in the following compounds? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6.

The Concept: Sigma-Pi Bond Counting

1. CH₂=C=O (Ketene)

2. CH₃CH=CH₂ (Propene)

3. (CH₃)₂CO (Acetone)

4. CH₂=CH–CN (Acrylonitrile)

5. C₆H₆ (Benzene)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days