Indicate the σ- and π-bonds in the following molecules: C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3.

Exercises · 8.2

Q. Indicate the σ- and π-bonds in the following molecules: C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3.

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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 that every single bond is one σ-bond, every double bond is one σ + one π, and every triple bond is one σ + two π. Counting them in each molecule gives: C₆H₆ (12 σ, 3 π), C₆H₁₂ (18 σ, 0 π), CH₂Cl₂ (4 σ, 0 π), CH₂=C=CH₂ (6 σ, 2 π), CH₃NO₂ (7 σ, 1 π), HCONHCH₃ (8 σ, 1 π).

The foundation of sigma-pi bond counting is simple: a single bond is always one sigma (σ) bond. A double bond consists of one sigma and one pi (π) bond. A triple bond is one sigma and two pi bonds. Every bond between atoms in a stable organic molecule is accounted for by this rule — there are no exceptions in the molecules you've listed.

Why does this work? Sigma bonds are formed by head-on overlap of orbitals along the internuclear axis; they are the backbone of every bond. Pi bonds come from sideways overlap of p-orbitals and only exist in addition to a sigma bond in multiple bonds. So when you see a structural formula, you just count the number of single, double, and triple bonds, then multiply accordingly.

Let's go molecule by molecule.

1. C₆H₆ — Benzene

Benzene has a ring of six carbon atoms with alternating single and double bonds (the Kekulé structure). But the actual structure is a resonance hybrid where all C–C bonds are equivalent, each with bond order 1.5. For counting sigma and pi bonds, we treat it as having three double bonds and three single bonds in the ring, plus six C–H single bonds.

Each C–H bond: 1 σ (6 of these)
Each C–C single bond in the ring: 1 σ (3 of these)
Each C=C double bond: 1 σ + 1 π (3 of these)

So sigma total = 6 (C–H) + 3 (C–C single) + 3 (σ from double bonds) = 12 σ bonds.

Pi total = 3 (from the three double bonds) = 3 π bonds.

Tip

In benzene, the six π electrons are delocalised, but the count of π bonds is still 3 — each double bond contributes one π bond in the Kekulé picture. The resonance doesn't change the number.

2. C₆H₁₂ — Cyclohexane

This is a saturated hydrocarbon — all single bonds. The ring has six C–C single bonds, and each carbon is bonded to two hydrogens (since it's C₆H₁₂, the formula for a cycloalkane). That gives 12 C–H bonds.

C–C single bonds: 6 σ
C–H single bonds: 12 σ

Total sigma = 6 + 12 = 18 σ bonds.

Pi bonds = 0 (no multiple bonds).

Watch out

A common mistake is to think C₆H₁₂ might have a double bond because it's isomeric with hexene. But the formula CₙH₂ₙ for a non-aromatic compound indicates either a cycloalkane or an alkene — here it's cyclohexane, so all single bonds. Always check the structure, not just the formula.

3. CH₂Cl₂ — Dichloromethane

Carbon is central, bonded to two hydrogens and two chlorines. All bonds are single.

C–H: 2 σ
C–Cl: 2 σ

Total sigma = 2 + 2 = 4 σ bonds.

Pi bonds = 0.

4. CH₂=C=CH₂ — Allene (Propadiene)

This is a cumulated diene: the central carbon is sp-hybridised and forms two double bonds. The terminal carbons are sp²-hybridised.

Structure: H₂C=C=CH₂

Each C=C double bond: 1 σ + 1 π. There are two such bonds.
C–H bonds: each terminal carbon has two C–H single bonds, so 4 C–H σ bonds.

Sigma total = 2 (σ from the two double bonds) + 4 (C–H) = 6 σ bonds.

Pi total = 2 (one π from each double bond) = 2 π bonds.

Note

In allene, the two π bonds are perpendicular to each other, but that doesn't affect the count — each double bond still contributes exactly one π bond.

5. CH₃NO₂ — Nitromethane

Structure: H₃C–NO₂. The carbon is bonded to three hydrogens and one nitrogen. The nitrogen is bonded to the carbon and two oxygens via one single and one double bond (the nitro group has resonance, but the bonding picture is: N=O double bond and N–O single bond, plus C–N single bond).

Let's list all bonds:

C–H: 3 σ
C–N: 1 σ
N=O: 1 σ + 1 π
N–O: 1 σ

Sigma total = 3 (C–H) + 1 (C–N) + 1 (σ from N=O) + 1 (N–O) = 6 σ bonds.

Pi total = 1 (from the N=O double bond) = 1 π bond.

Tip

In the nitro group, the two N–O bonds are equivalent in reality due to resonance, but the Lewis structure with one double and one single bond gives the correct bond count. The pi electrons are delocalised, but there is still exactly one π bond in the localised picture.

6. HCONHCH₃ — N-Methylformamide (an amide)

Structure: CH₃–NH–CHO (a methyl group, an N–H, and a formyl group). List every bond:

Three C–H bonds in the methyl group: 3 σ
N–CH₃ (nitrogen to the methyl carbon): 1 σ
N–H: 1 σ
N–C (the amide bond, nitrogen to the formyl carbon): 1 σ
C=O (the carbonyl, formyl carbon to oxygen): 1 σ + 1 π
C–H (the formyl hydrogen, on the carbonyl carbon): 1 σ

Sigma total = 3 + 1 + 1 + 1 + 1 + 1 = 8 σ bonds.

Pi total = 1 (from the C=O double bond) = 1 π bond.

Tip

A quick check: for a molecule with no rings, the number of σ bonds equals (total atoms − 1). HCONHCH₃ has 9 atoms (2 C, 1 N, 1 O, 5 H), so $9-1=8$ σ bonds — matching the count above.

Watch out

It's easy to overcount sigma bonds in amides because of resonance between the C–N and C=O bonds. Stick to the single Lewis structure: the C–N bond is one sigma bond (it has partial double-bond character in reality, but the sigma count from the localised structure is still 1), and the only pi bond is in the C=O.

✓Final answer

C₆H₆: 12 σ, 3 π; C₆H₁₂: 18 σ, 0 π; CH₂Cl₂: 4 σ, 0 π; CH₂=C=CH₂: 6 σ, 2 π; CH₃NO₂: 6 σ, 1 π; HCONHCH₃: 8 σ, 1 π.

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Q.Indicate the σ- and π-bonds in the following molecules: C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3.

1. C₆H₆ — Benzene

2. C₆H₁₂ — Cyclohexane

3. CH₂Cl₂ — Dichloromethane

4. CH₂=C=CH₂ — Allene (Propadiene)

5. CH₃NO₂ — Nitromethane

6. HCONHCH₃ — N-Methylformamide (an amide)

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