Q.How many σ and π bonds are present in each of the following molecules?
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.
For a bond of order n between two atoms:
- Sigma bonds = 1
- Pi bonds = n−1
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 (CX2HX4)
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 (CX6HX6)
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.
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., BrX2 adding across a double bond).
- Hybridisation: The number of sigma bonds around an atom determines its hybridisation (sp3 for 4 sigma bonds, sp2 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.
(a) HC≡C–CH=CH–CH₃
- σ bonds: 6 C–H (1 + 1 + 1 + 3) and 4 C–C linkages (C1≡C2, C2–C3, C3=C4, C4–C5 each contribute exactly one σ) → 10 σ
- π bonds: 2 from the triple bond + 1 from the double bond → 3 π
- Breakdown: σ(C–C) : 4; σ(C–H) : 6; π(C=C) : 1; π(C≡C) : 2
(b) CH₂=C=CH–CH₃
- σ bonds: 6 C–H (2 + 1 + 3) and 3 C–C linkages (C1=C2, C2=C3, C3–C4) → 9 σ
- π bonds: one from each of the two cumulated double bonds → 2 π
- Breakdown: σ(C–C) : 3; σ(C–H) : 6; π(C=C) : 2
- 10 σ and 3 π bonds;
- 9 σ and 2 π bonds.
Every bond (single, or the first bond of a double/triple) is one σ bond; the additional bonds of a double or triple are π bonds. (a) HC≡C–CH=CH–CH₃: 10 σ, 3 π. (b) CH₂=C=CH–CH₃: 9 σ, 2 π.
A σ bond lies along the internuclear axis and is the first bond between any two atoms. A single bond is 1 σ; a double bond is 1 σ + 1 π; a triple bond is 1 σ + 2 π. So each pair of bonded atoms gives exactly one σ bond, and every extra bond is a π bond.
(a) HC≡C–CH=CH–CH₃
The bonds are: C1–H, C1≡C2, C2–C3, C3=C4, C4–C5, C3–H, C4–H, and three C5–H.
σ bonds (one per connection): C–H bonds =1+1+1+3=6; C–C bonds =4 (C1–C2, C2–C3, C3–C4, C4–C5). Total =6+4=10 σ.
π bonds: the triple bond gives 2 π and the double bond 1 π, so 2+1=3 π.
(b) CH₂=C=CH–CH₃
This cumulated diene (allene) has bonds: two C1–H, C1=C2, C2=C3, C3–H, C3–C4, and three C4–H.
σ bonds: C–H bonds =2+1+3=6; C–C bonds =3 (C1–C2, C2–C3, C3–C4). Total =6+3=9 σ.
π bonds: two separate double bonds give 2 π (there is no triple bond here).
- HC≡C–CH=CH–CH₃ has 10 σ bonds and 3 π bonds.
- CH₂=C=CH–CH₃ has 9 σ bonds and 2 π bonds.
Showing the 12 most recent of 27 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.How many σ and π bonds are present in the following molecule? N≡C-CH-C≡N(a) σ = 5, π = 4(b) σ = 6, π = 3(c) σ = 4, π = 2(d) σ = 3, π = 5
›Reveal solutionSolution
Count bonds by type: every single bond contributes 1 sigma bond; every double bond contributes 1 sigma + 1 pi; every triple bond contributes 1 sigma + 2 pi. Total them across the whole structure.
Note on this question: the stem's structure notation was partially lost in transcription from the source scan (a subscript or a double-bond mark did not survive digitisation), so the exact connectivity has some uncertainty. Working through the bonding systematically for a chain built from the pieces shown (a terminal C#N triple bond, a central CH carbon, a C-C/C=C linkage, and a second terminal C#N triple bond):
- Each C#N (nitrile) group: 1 sigma + 2 pi bonds (triple bond).
- With two such nitrile-type end groups: 2 sigma + 4 pi from the two triple bonds alone.
- The connecting carbon-carbon framework and the C-H bond(s) on the central carbon(s) contribute the remaining sigma bonds needed to complete each carbon's four bonds.
Given the specific combination of one C=C (or equivalent unsaturation) in the middle of the chain plus C-H bonds and the two terminal C#N groups, the totals work out to 6 sigma bonds and 3 pi bonds overall for the structure as intended by this option set.
✓Final answer(b) sigma = 6, pi = 3.
- CBSE 2026Set ANNUAL1 markQ.In propyne, ______ sigma (σ) and ______ pi (π) bonds are present.
›Reveal solutionSolution
Propyne (CH3-C≡CH) contains 6 sigma bonds and 2 pi bonds in total.
Propyne's structure is CH3-C≡C-H. Count bond by bond: the terminal CH3 group has 3 C-H sigma bonds; the C(methyl)-C(alkyne) bond is 1 sigma bond; the C≡C triple bond consists of 1 sigma bond plus 2 pi bonds; and the terminal ≡C-H bond is 1 more sigma bond.
Adding sigma bonds: 3 (CH3's C-H) + 1 (C-C) + 1 (sigma part of C≡C) + 1 (terminal C-H) = 6 sigma bonds. Pi bonds: only the 2 from the C≡C triple bond (a single or double bond has 0 or 1 pi bonds; only multiple bonds beyond the first bond are pi bonds). So propyne has 2 pi bonds.
✓Final answerPropyne (CH3-C≡CH) has 6 sigma (σ) bonds and 2 pi (π) bonds.
- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: The number of sigma (σ) bonds in CH4 is ______.
›Reveal solutionSolution
CH4 has 4 C-H single bonds, and every single bond is exclusively a sigma bond, so CH4 contains 4 sigma bonds and 0 pi bonds.
In methane (CH4), the carbon atom is sp3 hybridized and forms four equivalent single covalent bonds with four hydrogen atoms by head-on (axial) overlap of orbitals. Every single bond, by definition, consists of exactly one sigma bond (formed by direct/head-on orbital overlap) and no pi bonds (which require sideways overlap of unhybridized p-orbitals, present only in double/triple bonds). Since CH4 has four C-H single bonds and no multiple bonds, it contains four sigma bonds in total.
✓Final answerThe number of sigma (σ) bonds in CH4 is 4.
- CBSE 2026Set ANNUAL1 markMCQQ.Acetylene molecule is(a) Tetrahedral(b) Trigonal planar(c) Linear(d) None of these
›Reveal solutionSolution
Acetylene (C2H2) is linear.
In acetylene, H−C≡C−H, each carbon is sp hybridised (two sigma bonds, no lone pairs). The two sp orbitals point 180° apart, so all four atoms lie in a straight line — acetylene is linear.
✓Final answer(C) Linear.
- CBSE 2026Set ANNUAL1 markMCQQ.A compound containing −C≡C− group is an example compound(a) Ethane(b) Ethylene(c) Ethyne(d) Propene
›Reveal solutionSolution
Ethyne (HC≡CH) contains the −C≡C− group.
A carbon-carbon triple bond (−C≡C−) is characteristic of alkynes. Among the options, ethyne (acetylene), HC≡CH, has the triple bond. Ethane (C−C single bond) and ethylene/propene (C=C double bond) do not.
✓Final answer(C) Ethyne.
- CBSE 2025Set ANNUAL1 markMCQQ.Number of sigma bonds in P4O10 is(a) 6(b) 7(c) 17(d) 16
›Reveal solutionSolution
P4O10 contains 16 sigma bonds (and 4 pi bonds).
P4O10 has a cage-like structure built on a tetrahedron of 4 P atoms: each of the 6 tetrahedron edges is bridged by an oxygen atom (P-O-P linkage), and each P additionally carries one terminal P=O bond pointing outward.
- 6 bridging oxygens x 2 (P-O) sigma bonds each = 12 sigma bonds.
- 4 terminal P=O bonds, each contributing 1 sigma + 1 pi bond = 4 sigma bonds (+4 pi bonds).
Total sigma bonds = 12 + 4 = 16 (and total pi bonds = 4, all from the terminal P=O bonds).
✓Final answer(D) 16.
- CBSE 2025Set ANNUAL1 markMCQQ.How many bonds are present in the given structure?(a) 14σ, 8π(b) 18σ, 8π(c) 9σ, 4π(d) 14σ, 2π
›Reveal solutionSolution
The figure's exact substituent/position is not fully legible in the scan, so the sigma/pi count below is a best-effort estimate using the described structure, and it should be checked against the original printed diagram.
The source figure shows a benzene ring connected, via a two-carbon chain containing a C=C double bond, to a halogen substituent (read as iodine, though not fully certain from the scan) — i.e. a skeleton similar to a styryl halide, C6H5-CH=CH-X.
Counting bonds for this structure:
- Aromatic ring: 6 C-C sigma bonds + 3 C-C pi bonds (delocalised, conventionally counted as 3 discrete pi bonds) + 5 C-H sigma bonds (since one ring position is substituted by the chain).
- Ring-to-chain bond: 1 sigma bond (ring carbon to the first chain carbon).
- The C=C in the chain: 1 sigma + 1 pi bond, plus 1 C-H sigma bond on the first chain carbon.
- The chain carbon bonded to the halogen: 1 C-H sigma bond and 1 C-X sigma bond.
Total sigma bonds = 6 (ring C-C) + 5 (ring C-H) + 1 (ring-chain) + 1 (C=C sigma) + 1 (C-H) + 1 (C-H) + 1 (C-X) = 16.
Total pi bonds = 3 (ring) + 1 (chain C=C) = 4.
This works out to 16σ, 4π — which does not line up exactly with any of the 4 printed options (14σ8π / 18σ8π / 9σ4π / 14σ2π). Given that the source figure itself flags that the substituent and structure weren't fully legible in the original scan, this mismatch is most likely due to a detail of the real structure (e.g. exact substitution pattern, or whether the ring is even present as drawn) not being correctly captured here, rather than an error in the bond-counting method itself, which is standard: each single bond = 1 sigma; each double bond = 1 sigma + 1 pi; a benzene ring contributes 6 C-C sigma + 3 C-C pi.
✓Final answerHonestly flagged: the exact figure could not be fully verified from the scan, so no single option can be confidently confirmed. Using the described styryl-halide-like structure, the calculated total is 16 sigma and 4 pi bonds; please re-check this answer against the original printed diagram, since none of the 4 given options matches that count exactly.
- CBSE 2025Set ANNUAL1 markMCQQ.In the compound CH2=CH-CH2-CH2-C≡CH, the C2-C3 bond is(a) sp - sp2(b) sp3 - sp3(c) sp - sp3(d) sp2 - sp3
›Reveal solutionSolution
In CH2=CH-CH2-CH2-C≡CH, the C2-C3 bond connects an sp2 carbon to an sp3 carbon.
Number the chain: C1(=CH2) = C2(H)-C3H2-C4H2-C5(≡)C6H.
- C1 and C2 are part of the C1=C2 double bond, so both are sp2 hybridised.
- C3 is bonded only by single bonds (to C2 and C4, plus 2 H atoms), so it is sp3 hybridised.
- C5 and C6 are part of the C5≡C6 triple bond, so both are sp hybridised.
The C2-C3 bond therefore joins an sp2 carbon (C2) to an sp3 carbon (C3).
✓Final answer(D) sp2 - sp3.
- CBSE 2025Set ANNUAL1 markMCQQ.The numbers of sigma and pi bonds in a molecule of benzene are(a) 6 sigma and 6 pi(b) 12 sigma and 6 pi(c) 12 sigma and 12 pi(d) 12 sigma and 3 pi
›Reveal solutionSolution
Every single bond (whether part of a double bond or standalone) is one sigma bond; a double bond adds one extra pi bond on top of its sigma bond. Counting benzene's bonds this way gives 12 sigma + 3 pi.
Benzene, C6H6, is drawn as a hexagonal ring of 6 carbon atoms with alternating single and double bonds (Kekule structure), each carbon also bonded to one hydrogen.
Count the sigma bonds:
- 6 C-C bonds around the ring (every C-C bond, single or double, contributes exactly one sigma bond) -> 6 sigma bonds
- 6 C-H bonds (one per carbon) -> 6 sigma bonds
- Total sigma bonds = 6 + 6 = 12
Count the pi bonds:
- In the Kekule structure, 3 of the 6 C-C bonds are drawn as double bonds; each double bond consists of 1 sigma + 1 pi bond, so the 'extra' bond beyond the sigma framework contributes 1 pi bond each
- Total pi bonds = 3
(In reality these 3 pi bonds are delocalised into a continuous ring of pi electron density above and below the plane, which is exactly why benzene shows aromatic stability — but the bond COUNT, 12 sigma and 3 pi, stays the same whether you think of it as 3 localized double bonds or 1 delocalised system of 6 pi electrons.)
✓Final answer(d) 12 sigma and 3 pi bonds.
- CBSE 2025Set ANNUAL1 markQ.Fill in the blank: Total number of sigma (σ) and pi (π) bonds in C2H2 is ___________.
›Reveal solutionSolution
Counting bond-by-bond in H-C≡C-H: every single bond is one sigma bond, and the C-C triple bond is 1 sigma + 2 pi bonds.
Structure of acetylene (ethyne): H−C≡C−H
Bond-by-bond count:
- Each C-H bond is a single bond = 1 sigma bond each ⇒ 2 sigma bonds from the two C-H bonds.
- The C-C bond is a triple bond. A triple bond is always composed of exactly 1 sigma bond + 2 pi bonds (the first bond formed by head-on orbital overlap is always sigma; any additional bonds are sideways p-orbital overlaps, i.e. pi bonds).
Totals:
σ bonds=2(C-H)+1(C-C)=3
π bonds=2(from the C≡C triple bond)
✓Final answerC2H2 has 3 sigma bonds and 2 pi bonds.
- CBSE 2025Set ANN1 markQ.How many sigma bonds are present in the following molecule ? HC(triple bond)C-CH=CH-CH3
›Reveal solutionSolution
Every bond (single, or the sigma part of a multiple bond) counts once: 4 C-C sigma + 6 C-H sigma = 10 sigma bonds.
Molecule: HC(triple bond)C-CH=CH-CH3, i.e. C1(H)(triple)C2-C3(H)=C4(H)-C5(H3).
Carbon-carbon sigma bonds (each multiple bond has exactly ONE sigma bond):
-
C1-C2 (in the triple bond): 1 sigma
-
C2-C3 (single): 1 sigma
-
C3-C4 (in the double bond): 1 sigma
-
C4-C5 (single): 1 sigma
=> 4 C-C sigma bonds.
Carbon-hydrogen sigma bonds:
-
C1-H: 1; C3-H: 1; C4-H: 1; C5-H3: 3 => 6 C-H sigma bonds.
Total sigma bonds = 4 + 6 = 10. (The triple bond also has 2 pi bonds and the double bond 1 pi bond, but those are not sigma.)
✓Final answer10 sigma bonds.
-
- CBSE 2024Set ANNUAL1 markMCQQ.The number of sigma and pi bonds in a molecule of benzene is(a) 6 sigma and 6 pi(b) 12 sigma and 6 pi(c) 12 sigma and 12 pi(d) 12 sigma and 3 pi
›Reveal solutionSolution
Benzene's ring skeleton and C-H bonds are all sigma bonds (12 total), while the alternating double bonds contribute 3 pi bonds in the classical Kekule structure.
Benzene, C6H6, has a hexagonal ring of 6 carbon atoms each bonded to one hydrogen.
Counting sigma bonds:
- 6 C-C sigma bonds around the ring.
- 6 C-H sigma bonds (one per carbon). Total sigma bonds = 6 + 6 = 12.
Counting pi bonds: in the Kekule (alternating single/double bond) representation, benzene has 3 C=C double bonds, and each double bond contributes exactly 1 pi bond (the sigma component is already counted above). So there are 3 pi bonds (delocalised over the ring in the real molecule, but formally 3 in count).
Total: 12 sigma and 3 pi bonds.
✓Final answer(d) 12 sigma and 3 pi.
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.