σ vs π bonds — overlap, strength: Worked examples

This page trains you on every kind of question the σ/π distinction can throw at you. We first lay out a scenario matrix — a checklist of all the case-types — then work each one fully. If you can do all of these, you have seen every shape of exam question on this topic.

New here? Read the parent topic first. We lean on hybridisation, bond order and energy, and touch cis-trans isomerism and Molecular Orbital Theory.

Definition Notation used on this page (read this first)

==like this== just means "this term is a key point to remember" — it is emphasis, nothing mathematical.
$p$ orbital: a dumbbell-shaped region where a valence electron lives — two lobes ( $+$ on one side, $-$ on the other) pointing along one axis. An atom has three of them: $p_{x}, p_{y}, p_{z}$ , pointing along the $x$ , $y$ , $z$ directions.
Hybridisation: an atom can mix its one $s$ orbital (a sphere) with some of its $p$ orbitals to make new directional orbitals that point straight at neighbours. Mix $s$ with one $p$ → two == $s p$ == hybrids (linear); with two $p$ → three == $s p^{2}$ == hybrids (flat triangle); with three $p$ → four == $s p^{3}$ == hybrids (tetrahedron). Any $p$ orbital not used in mixing stays "pure" and is free to make a π bond. (Full detail in hybridisation.)
Rule of thumb: count the σ bonds + lone pairs on an atom → that number tells you the hybridisation (2 → $s p$ , 3 → $s p^{2}$ , 4 → $s p^{3}$ ), and $(3 - that number + 1)$ ... simpler: leftover pure $p$ orbitals = 3 minus (number of $p$ 's used in hybrids).

One number we keep using: the overlap integral $S$

The scenario matrix

Before working problems, here is the full set of "cells" a σ/π question can land in. Each worked example below is tagged with the cell it fills, and its steps directly tackle the "what makes it tricky" column.

Cell	Case class	What makes it tricky	Example
A	Single bond (baseline)	Only 1 σ, zero π — the "zero π" degenerate case	Ex 1
B	Double bond	1σ + 1π, leftover pure $p$	Ex 2
C	Triple bond	1σ + 2π, limiting max for a bond pair	Ex 3
D	Whole molecule count	Sum σ and π across a structure	Ex 4
E	Rotation / isomerism	Why π locks geometry	Ex 5
F	Numeric bond-energy reasoning	Extract π energy from data	Ex 6
G	Real-world word problem	Reactivity of π electrons	Ex 7
H	Exam twist — delocalised π	Resonance breaks the "1σ+1π" picture	Ex 8
I	Degenerate/limiting overlap	Head-on vs sideways at the extremes	Ex 9

Counting rule we will reuse

The figure below shows exactly this ladder. Read it left to right:

Figure — σ vs π bonds — overlap, strength

The two purple circles labelled A and B are the two bonded atoms. The coral line joining them straight through the centres is the σ bond (head-on, always present). In the double case a mint pair of lines above and below the axis appears — that is the 1 π (sideways overlap). In the triple case a second π (the butter-yellow pair) is added perpendicular to the first. So the picture literally is "0 π, then 1 π, then 2 π" — the ladder of the counting rule.

Worked examples

Worked example Ex 1 — Cell A: the single bond (zero-π degenerate case)

The tricky bit: recognising that a single bond is the degenerate end of the ladder — zero π, so nothing to lock rotation. Statement: In ethane, $H_{3} C - CH_{3}$ , describe the C–C bond in terms of σ and π, and state whether the two $CH_{3}$ groups can rotate relative to each other.

Forecast: Guess now — how many σ and how many π in that C–C bond, and rotation yes or no?

Identify hybridisation. Each C has 4 single bonds (3 H + 1 C) → four equivalent directions → == $s p^{3}$ == hybridised. Why this step? The number of σ + lone pairs around an atom sets its hybridisation; four σ means four $s p^{3}$ hybrids.
Overlap type of C–C. Two $s p^{3}$ hybrids point straight at each other → head-on overlap → this is a σ bond. Why this step? Hybrid orbitals are directional; head-on overlap is by definition σ.
Count π. There is no leftover parallel $p$ orbital (all four $p$ -character orbitals are used in hybrids) → $0 π$ . Why this step? π needs a spare parallel $p$ orbital; here there is none.
Rotation. A σ bond is cylindrically symmetric about the axis → rotating one end does not reduce overlap → free rotation: yes.

Answer: C–C in ethane = $1 σ + 0 π$ , free rotation allowed.

Verify: Bond count = 1 (a single line). Rule gives $1 σ + (1 - 1) π = 1 σ + 0 π$ . ✓ Ethane really does rotate freely (this is why it has no cis/trans forms).

Worked example Ex 2 — Cell B: the double bond

The tricky bit: not double-counting — a double bond is two different bonds (1σ + 1π), and you must locate the leftover pure $p$ that makes the π. Statement: For ethene $H_{2} C = CH_{2}$ , break the C=C into its σ and π components and name the orbitals involved.

Forecast: Which orbital does the σ come from, and which does the π come from?

Hybridisation of C. Each C has 3 σ bonds (2 H + 1 C) and no lone pair → three directions → == $s p^{2}$ == hybridised, leaving one unhybridised $p$ orbital. Why this step? Three σ groups → three hybrids → the fourth valence orbital (a $p$ ) stays pure.
Form the σ. One $s p^{2}$ hybrid on each C points at the other → head-on → 1 σ. Why this step? Directional hybrids overlap end-to-end.
Form the π. The leftover pure $p$ orbitals stand perpendicular to the molecular plane, parallel to each other → sideways overlap → 1 π. Why this step? Parallel $p$ lobes can only meet laterally, above and below the axis.

Answer: C=C = $1 σ$ ( $s p^{2}$ – $s p^{2}$ ) $+ 1 π$ ( $2 p$ – $2 p$ ).

Verify: Bond count = 2. Rule: $1 σ + (2 - 1) π = 1 σ + 1 π$ . ✓ Matches the parent note's ethene example.

Worked example Ex 3 — Cell C: the triple bond (limiting max)

The tricky bit: knowing that two is the ceiling for π between two atoms, and why (only two $p$ orbitals are perpendicular to the axis). Statement: In $N_{2}$ (N≡N), how many σ and π bonds are there, and which atomic orbitals form each?

Forecast: How many π bonds can two atoms maximally stack? Guess before reading.

Hybridisation of N. Each N has 1 σ bond (to the other N) + 1 lone pair → two directions → each N is == $s p$ == hybridised, leaving two unhybridised $p$ orbitals ( $p_{x}, p_{y}$ ). Why this step? One σ group + one lone pair → two $s p$ hybrids → the remaining two $p$ orbitals stay pure and are free for π.
Electron picture / form the σ. The $s p$ hybrid on each N points along the axis (call it $z$ ) → head-on → 1 σ; the other $s p$ holds the lone pair. Why this step? Only the orbital along the axis can overlap head-on.
The remaining two $p$ . $p_{x}$ and $p_{y}$ are perpendicular to the axis → each overlaps sideways → 2 π. Why this step? Perpendicular parallel lobes give lateral overlap only.

Answer: N≡N = $1 σ + 2 π$ . Two is the maximum number of π between two atoms, because only two $p$ orbitals ( $p_{x}, p_{y}$ ) are perpendicular to the axis.

Verify: Bond count = 3. Rule: $1 σ + (3 - 1) π = 1 σ + 2 π$ . ✓ Bond order $= 3$ , consistent with N₂'s huge inertness.

Worked example Ex 4 — Cell D: whole-molecule count

The tricky bit: summing across a whole structure without forgetting the C–H σ bonds — every line is a bond and every bonded pair has exactly one σ. Statement: Count the total number of σ and π bonds in propyne, $CH_{3} - C \equiv CH$ .

Forecast: Total σ = ?, total π = ? (Count every bond, including the C–H bonds.)

Note the hybridisations. The methyl C (4 single bonds) is $s p^{3}$ ; the two triple-bonded C's each have 1 σ to a neighbour + the triple, i.e. 2 σ directions → each is == $s p$ == hybridised, leaving two pure $p$ orbitals to make the two π. Why this step? Knowing the $s p$ carbons carry the pure $p$ orbitals tells us where the π bonds live.
List every bond. Bonds are: 3× (C–H on the methyl), 1× (C–C single, methyl to triple-carbon), 1× (C≡C triple), 1× (C–H on the terminal alkyne carbon). Why this step? Every line in the structure is a bond and contributes at least one σ.
Assign σ to each single line. Each single bond and the "first line" of the triple bond is a σ. Single-bond σ count: 3 C–H (methyl) + 1 C–C + 1 C–H (terminal) + 1 σ inside the triple = 6 σ. Why this step? Every bond, no matter its order, contributes exactly one σ.
Assign π. Only the triple bond has extra lines: 3 − 1 = 2 → 2 π (formed by the pure $p$ orbitals on the two $s p$ carbons). No other bond has a π. Why this step? Only multiple bonds carry π; the rest are single.

Answer: Propyne = 6 σ + 2 π.

Verify: Total lines drawn = 3 (C–H) + 1 (C–C) + 3 (C≡C) + 1 (C–H) = 8 bond-lines. σ = number of bonded pairs = 6. π = 8 − 6 = 2. ✓ ( $σ$ = one per bonded pair; $π$ = total lines − pairs.)

Worked example Ex 5 — Cell E: rotation and isomerism

The tricky bit: the π (not the σ) is what locks geometry — spotting that a double bond blocks rotation while a triple bond is symmetric again. Statement: But-2-ene $CH_{3} - CH = CH - CH_{3}$ exists as cis and trans forms, but but-2-yne does not exhibit this. Explain using σ/π.

Forecast: Which bond feature (σ or π) locks the geometry?

Find the locking bond. In but-2-ene the central bond is a C=C double bond = $1 σ + 1 π$ . Why this step? Isomerism about a bond depends on whether that bond can rotate.
Ask: can it rotate? Rotating one CH group by 90° would twist the two parallel $p$ orbitals out of alignment → the sideways π overlap is destroyed → costs energy → rotation blocked. Why this step? The π bond is not cylindrically symmetric, so twisting has an energy penalty (unlike a lone σ).
Fixed geometry → two arrangements. With rotation frozen, the two $CH_{3}$ groups are locked either on the same side (cis) or opposite sides (trans) → two distinct molecules. Why this step? A locked double bond makes the two spatial arrangements non-interconvertible.
Why the yne has none. But-2-yne has C≡C = $1 σ + 2 π$ ; the two π make the region cylindrically symmetric again (like a rod), and the substituents lie on the axis (linear) → no "sides" to be cis or trans. Why this step? Two perpendicular π bonds recover rotational symmetry and force linear geometry.

Answer: The single π of the C=C locks geometry → cis/trans in but-2-ene; the two π of C≡C give linear, symmetric geometry → no isomerism.

Verify: Consistent with cis-trans isomerism: isomerism requires a π (double bond) and two different groups on each carbon. But-2-ene qualifies; the linear yne cannot. ✓

Worked example Ex 6 — Cell F: numeric bond-energy reasoning

The tricky bit: you cannot read π energy off a table directly — you must subtract the σ from the double-bond total to isolate it. Statement: Given C–C $\approx 348$ kJ/mol and C=C $\approx 615$ kJ/mol, estimate the energy of the π contribution alone and show it is weaker than the σ.

Forecast: Will the π be more or less than 348 kJ/mol?

Model the double bond. C=C = 1σ + 1π. Assume the σ part of a C=C ≈ the C–C single-bond σ ≈ 348 kJ/mol. Why this step? Both are C–C σ bonds; treating them as roughly equal isolates the π's contribution.
Subtract to isolate π. $E_{π} \approx E_{C=C} - E_{C-C} = 615 - 348 = 267 kJ/mol$ . Why this step? Total double-bond energy minus the σ leaves the π energy.
Compare. $267 < 348$ , so π (267) < σ (348). The π adds less because its sideways overlap integral $S_{π}$ is smaller than the head-on $S_{σ}$ . Why this step? Smaller overlap $S$ → smaller stabilisation → weaker bond, exactly as defined at the top of this page.

Answer: $E_{π} \approx 267$ kJ/mol, clearly less than the σ's 348 kJ/mol.

Verify: If π equalled σ, C=C would be $2 \times 348 = 696$ kJ/mol, but it is only 615. The deficit $696 - 615 = 81$ kJ/mol is exactly $348 - 267$ . ✓ Two routes agree.

Worked example Ex 7 — Cell G: real-world word problem (reactivity)

The tricky bit: connecting a lab observation (colour vanishing) to the exposed position of π electrons — reactivity is about accessibility, not just strength. Statement: Ethene decolourises bromine water rapidly, but ethane does not react. Explain in σ/π language why the double-bonded molecule is the reactive one.

Forecast: Which electrons get attacked — the σ or the π?

Locate the vulnerable electrons. Ethene's C=C has a π bond whose electrons sit above and below the axis, exposed and loosely held. Why this step? Reactivity is governed by the most accessible, highest-energy electrons.
Why π and not σ. The σ electrons are buried between the nuclei (deep, tightly bound). The π cloud is on the outside → easy for an electrophile like $Br_{2}$ to reach. Why this step? Position of density decides accessibility.
The reaction. $Br_{2}$ attacks the π electrons → the π bond breaks (an addition reaction) → Br adds across the double bond → bromine colour vanishes. Why this step? Breaking the weak π (267 kJ/mol) is far easier than breaking a σ.
Why ethane is inert here. Ethane has only σ bonds (no exposed π) → nothing for $Br_{2}$ to grab under mild conditions → no reaction. Why this step? No π electrons = no easy electrophilic target.

Answer: The exposed, weaker π electrons of ethene make it reactive toward $Br_{2}$ ; ethane, all-σ, is unreactive.

Verify: Consistent with the energy numbers — π at 267 kJ/mol is the weakest bond present and breaks first; σ at 348 kJ/mol stays intact (addition, not substitution). ✓

Worked example Ex 8 — Cell H: exam twist — delocalised π (resonance)

The tricky bit: the "1σ+1π per double bond" bookkeeping gives the right count but the wrong picture — the π is spread over the whole ring, not locked into three double bonds. Statement: In benzene $C_{6} H_{6}$ , the naive "1σ + 1π per double bond" would give 3 localised π bonds. What actually happens, and how many σ and π systems are there? Relate to Resonance and conjugation.

Forecast: Are the 3 π bonds fixed in place, or something else?

Hybridisation. Each C is $s p^{2}$ (bonded to 2 C + 1 H) → each keeps one unhybridised $p$ orbital perpendicular to the ring. Why this step? $s p^{2}$ leaves a spare parallel $p$ on every carbon.
σ framework. 6 C–C σ + 6 C–H σ = 12 σ bonds forming the flat hexagon skeleton. Why this step? Every single line contributes exactly one σ.
Count the ring lines in a Kekulé structure. Draw benzene as alternating bonds: 3 single C–C (1 line each = 3) + 3 double C=C (2 lines each = 6) → total ring lines = 3 + 6 = 9. The 6 ring σ pairs use 6 of these, so π-bond count = $9 - 6 = 3$ . Why this step? We must spell out where "9" comes from (3 singles + 3 doubles × 2 lines) before subtracting the σ pairs.
The π twist. The six parallel $p$ orbitals all overlap sideways around the whole ring → the π electrons delocalise rather than lock into 3 separate double bonds. In localised bookkeeping this reads as 3 π bonds, but physically it is one delocalised π system holding 6 electrons. Why this step? Equivalent adjacent parallel $p$ orbitals merge into one continuous π system (a molecular-orbital picture), not three isolated π bonds.

Answer: Benzene = 12 σ bonds + a single delocalised π system holding 6 electrons (bookkept as 3 π bonds). The "1σ+1π" localised rule counts correctly but mislocates the π — the real twist.

Verify: σ = number of bonded pairs = 6 (ring) + 6 (C–H) = 12. π-bond count = ring lines (9) − ring σ pairs (6) = 3. Delocalisation redistributes these 3 without changing the count. ✓ All 6 C–C bond lengths are equal — the fingerprint of one shared π system.

Worked example Ex 9 — Cell I: degenerate / limiting overlap

The tricky bit: at the extremes of orientation the overlap $S$ hits its limits — maximum head-on, and exactly zero when orbitals are perpendicular because $+$ and $-$ lobes cancel. Statement: Two $p$ orbitals approach along the $z$ -axis. Compare the overlap $S$ when they meet (a) exactly head-on ( $p_{z}$ – $p_{z}$ ) versus (b) exactly perpendicular ( $p_{x}$ – $p_{z}$ ). What are the limiting overlap values?

Forecast: What is the overlap in the perpendicular (crossed) case — big, small, or exactly zero?

Head-on limit ( $p_{z}$ – $p_{z}$ ). Lobes point straight at each other → product $ψ_{A} ψ_{B}$ is large and same-sign in the overlap region → maximum overlap integral $S_{σ}$ (largest possible for two $p$ 's). Why this step? Aligned lobes maximise the shared region → strongest bond.
Perpendicular limit ( $p_{x}$ – $p_{z}$ ). One orbital's $+$ lobe overlaps the other's $+$ lobe on one side but its $-$ lobe on the other side → the two contributions cancel exactly → $S = 0$ . Why this step? By symmetry the positive and negative overlap regions are equal and opposite → net zero overlap, no bond.
Sideways limit (parallel $p_{x}$ – $p_{x}$ ). Between the two extremes: parallel same-axis lobes overlap laterally, giving a nonzero but smaller $S_{π}$ than $S_{σ}$ . Why this step? Off-axis overlap is real but reduced → weaker π bond.

Answer: Overlap ranges from maximum ( $S_{σ}$ , head-on) → intermediate ( $S_{π}$ , parallel sideways) → exactly zero ( $p_{x}$ – $p_{z}$ perpendicular, by cancellation). Ordering: $S_{σ} > S_{π} > 0 = S_{⊥}$ .

Verify: The perpendicular case gives $S = 0$ because the integrand is odd about the nodal plane → integral over symmetric limits = 0. This is why only matching-symmetry orbitals bond. ✓

Recall Self-test: which cell is each question?

"Total σ in $CH_{3} CN$ ?" is which cell type? ::: Cell D — whole-molecule count. "Why can't you rotate ethene's double bond?" is which cell? ::: Cell E — rotation/isomerism. "Estimate the π energy from C≡C and C–C data" is which cell? ::: Cell F — numeric bond-energy reasoning. "Overlap of a $p_{x}$ with a $p_{z}$ ?" is which cell? ::: Cell I — degenerate/limiting overlap (answer: zero). "How many π in benzene, really?" is which cell? ::: Cell H — delocalised π twist.

Recap

Recall Core takeaways from these examples

σ per bonded pair, π = extra lines? ::: Yes — every bond has exactly 1 σ; π count = (bond order − 1). Maximum π bonds between two atoms and why? ::: 2, because only two $p$ orbitals ( $p_{x}, p_{y}$ ) are perpendicular to the bond axis. Which electrons drive reactivity and why? ::: The π electrons — exposed above/below the axis, weaker, easily attacked. What does a delocalised π system do to the "3 π bonds" picture? ::: Keeps the count but spreads the electrons over the whole ring (benzene). Overlap of two perpendicular p orbitals? ::: Exactly zero — positive and negative overlaps cancel. What does the overlap integral $S$ measure? ::: How much two orbital clouds sit on top of each other; bigger $S$ = stronger bond.

One number we keep using: the overlap integral S

The scenario matrix

Counting rule we will reuse

Worked examples

Recap

One number we keep using: the overlap integral $S$