σ vs π bonds — overlap, strength: Exercises — Chemistry deep dive

Every symbol used here is built in the parent note; where a new idea (like counting bonds from a structure) enters, the solution re-derives it from scratch.

Definition Two words we lean on all page — build them now

Overlap integral $S$ . The parent note defines it as $S = \int ψ_{A} ψ_{B} d τ$ , where $ψ_{A}, ψ_{B}$ are the two orbital wavefunctions and the integral just adds up how much the two orbitals occupy the same space. Plain words: $S$ is a single number measuring how much two orbitals overlap — big when lobes sit on top of each other, small when they barely touch. Head-on overlap makes $S$ large; sideways overlap makes $S$ smaller. Full derivation lives in the parent σ vs π bonds note. Cylindrically symmetric. Picture spinning the bond like a rod on a lathe: a σ bond's electron cloud looks exactly the same at every twist angle — it wraps evenly around the axis like a sleeve on a pencil. That "same-at-every-angle-around-the-axis" property is what cylindrically symmetric means, and it is why a σ bond can rotate freely (nothing changes). A π cloud sits in two lobes above/below, so twisting does change it — not cylindrically symmetric.

Level 1 — Recognition

Problem L1.1

A bond forms by two orbitals overlapping head-on along the line joining the two nuclei. Name the bond type and say whether the electron density sits on or off that line.

Recall Solution L1.1

WHAT the question tests: can you read a geometry and name the bond? Head-on / axial / end-to-end overlap along the bond axis is the definition of a σ (sigma) bond. The shared electron density piles up ON the bond axis, symmetrically wrapped around it (cylindrically symmetric — same cloud at every twist angle, as built above). Answer: σ bond; density on the axis.

Problem L1.2

Two parallel $p$ orbitals overlap sideways, giving two lobes of electron density — one above and one below the bond axis — and an empty plane containing the axis. Name the bond type and name that empty plane.

Recall Solution L1.2

Sideways / lateral overlap of parallel $p$ orbitals is a π (pi) bond. The plane that contains the bond axis but has zero electron density is the nodal plane. "Nodal" just means "where the wavefunction is zero," so no electrons are found there. Answer: π bond; the empty plane is the nodal plane.

Figure s01 — what to look at. The picture below draws the same pair of $p$ orbitals twice. Top row: the $p$ lobes point straight at each other along the dashed bond axis; the red patch marks where they overlap — squarely ON the axis → that is a σ bond. Bottom row: the same lobes stand up vertically and sit side-by-side; the red patches show overlap above and below the axis, never on it → that is a π bond. Use it to see that orientation, not the letter " $p$ ", decides the bond.

Figure — σ vs π bonds — overlap, strength

Alt-text: two panels of p-orbital overlap; top panel shows head-on overlap on the bond axis (σ), bottom panel shows side-by-side parallel overlap above and below the axis (π), with the overlap regions highlighted in red.

Level 2 — Application

Problem L2.1

Classify the total σ and π bonds in each molecule: (a) H–Cl (single bond), (b) O=O (double bond), (c) N≡N (triple bond).

Recall Solution L2.1

The rule (re-derived): atoms first pull as close as possible, so the head-on σ forms first. Only after the σ fixes the geometry can leftover parallel $p$ orbitals overlap sideways to give π. Therefore:

a single line = 1σ
a double line = 1σ + 1π
a triple line = 1σ + 2π

Applying it:

(a) H–Cl single → 1σ, 0π
(b) O=O double → 1σ, 1π
(c) N≡N triple → 1σ, 2π

Problem L2.2

Ethyne (acetylene) is H–C≡C–H. Count the total number of σ bonds and the total number of π bonds in the whole molecule.

Recall Solution L2.2

Draw every connection:

Two C–H bonds → each is a single bond → 2 σ.
One C≡C triple bond → 1σ + 2π → adds 1 σ and 2 π.

Totals: σ = $2 + 1 = 3$ , π = $0 + 2 = 2$ . Answer: 3 σ bonds and 2 π bonds.

Level 3 — Analysis

Problem L3.1

Given bond energies C–C single $\approx 348 kJ/mol$ and C=C double $\approx 615 kJ/mol$ , estimate the energy contributed by the π component alone. Is your number smaller or larger than the σ contribution, and does that agree with "σ stronger than π"?

Recall Solution L3.1

WHAT we do: a C=C double bond is 1σ + 1π. If we assume the σ part of C=C is about the same as a lone C–C single σ ( $348$ ), then $E_{π} \approx E_{C=C} - E_{C-C} = 615 - 348 = 267 kJ/mol .$ WHY this works: subtracting the single-bond σ energy from the double-bond total leaves the extra stick, which is the π. Compare: $267 < 348$ , so the π contributes less than the σ. This matches the master principle from the parent note: sideways overlap gives a smaller overlap integral $S$ (the "how much do the orbitals share space" number defined at the top) than head-on overlap, so π is weaker than σ. ✓

Problem L3.2

Take N₂ with N≡N total bond energy $\approx 945 kJ/mol$ . Using the σ estimate for a N–N single bond $\approx 163 kJ/mol$ , find the average energy of one π bond in N₂. (A triple bond = 1σ + 2π.)

Recall Solution L3.2

N≡N = 1σ + 2π. Treat the σ part as the single-bond energy $163$ . Then the two π bonds together carry $E_{2 π} = 945 - 163 = 782 kJ/mol,$ WHY this subtraction works: the total triple-bond energy is the sum of its parts, $E_{N \equiv N} = E_{σ} + E_{2 π}$ . We know the whole ( $945$ ) and we estimate the σ piece as the lone single-bond energy ( $163$ ), so whatever remains after removing the σ must be the energy held by the two π bonds together. Rearranging the sum gives $E_{2 π} = 945 - 163$ . so each π averages $E_{π} = \frac{782}{2} = 391 kJ/mol .$ Reading it: here the π "number" looks larger than this particular σ estimate — a good reminder that these are rough model numbers (the σ inside a triple bond isn't identical to a lone N–N single bond). The reliable, direction-based conclusion stays: σ forms first, π is the add-on, and for the same pair of orbitals at the same distance head-on overlap beats sideways. Answer: ≈ 391 kJ/mol per π (with this estimate).

Level 4 — Synthesis

Problem L4.1

Ethene is $H_{2} C=CH_{2}$ ; each carbon is $s p^{2}$ hybridised. (a) State which orbitals make each σ and which make the π. (b) Explain, using orbital geometry, why the molecule is planar and cannot rotate about the C=C axis. Link this to a named phenomenon.

Recall Solution L4.1

(a) Building every bond from orbitals:

Each C has three $s p^{2}$ hybrids lying flat in a plane at $12 0^{\circ}$ (see Hybridisation (sp, sp2, sp3)). These are directional, so they overlap head-on → σ bonds: two go to H atoms, one goes to the other C. That gives the C–C σ and four C–H σ bonds.
Each C also has one leftover unhybridised $p$ orbital standing perpendicular to that flat plane. The two $p$ orbitals are parallel, so they overlap sideways → the single C=C π bond.

(b) Why planar + no rotation. The π overlap only survives while the two $p$ orbitals stay parallel. That forces both $CH_{2}$ groups into the same plane. If you twist one $CH_{2}$ by $9 0^{\circ}$ , the two $p$ orbitals become perpendicular → their sideways overlap drops to zero → the π bond breaks (costs ~267 kJ/mol from L3.1). Rotation is therefore locked. A σ bond, being cylindrically symmetric (same cloud at every twist angle), would allow free spin — but the π does not. This locked geometry is exactly why cis-trans isomerism exists across C=C double bonds.

Figure s02 — what to look at. The left panel shows the two carbon $p$ orbitals drawn vertical and parallel (in red): they overlap side-by-side, so the π bond is intact and the molecule stays flat. The right panel twists the right-hand carbon by $90°$ so its $p$ orbital now lies horizontal (red), perpendicular to its partner: the lobes no longer face each other, overlap collapses to zero, and the π bond snaps. Read the two panels as "before vs after twisting" to see why the double bond cannot rotate.

Alt-text: left panel shows two parallel vertical p orbitals on adjacent carbons with intact side-by-side π overlap (planar ethene); right panel shows one carbon twisted 90° so its p orbital is perpendicular, destroying the overlap and breaking the π bond.

Level 5 — Mastery

Problem L5.1

Benzene $C_{6} H_{6}$ has each carbon $s p^{2}$ hybridised in a flat hexagonal ring. (a) Count all σ bonds. (b) Count the electrons available for the π system, and explain why they are best described as delocalised rather than as three fixed double bonds.

Recall Solution L5.1

(a) σ framework: each C uses its three $s p^{2}$ hybrids for head-on overlap:

6 C–C σ bonds around the ring,
6 C–H σ bonds. Total = 12 σ bonds.

(b) π system: each of the 6 carbons keeps one unhybridised $p$ orbital perpendicular to the ring, each holding one electron → 6 π electrons. Every $p$ orbital is parallel to both of its ring neighbours, so sideways overlap happens all the way around — not just between three chosen pairs. The electrons therefore spread over the whole ring (Resonance and conjugation, Molecular Orbital Theory): they are delocalised. This lower-energy shared cloud is why benzene is unusually stable (aromatic) and why all six C–C bonds have the same length, intermediate between a single and a double bond (Bond order, length and energy). Answer: 12 σ bonds; 6 delocalised π electrons.

Problem L5.2

A student claims: "A π bond is weak, so it barely matters chemically." Give one quantitative and one behavioural reason this reasoning is wrong.

Recall Solution L5.2

Quantitative: although a single π (~267 kJ/mol, L3.1) is weaker than a σ (~348), it is still a large energy on the chemistry scale — comparable to many full single bonds. "Weaker than σ" ≠ "small." Behavioural: π electrons sit above and below the axis, exposed and loosely held, so they are the reactive site — they drive addition reactions, conjugation, and colour. Reactivity/importance is not the same as bond energy. So the parent-note point stands: small energy ≠ small importance.

Recall One-line self-check before you leave

Single = 1σ; double = 1σ + 1π; triple = 1σ + 2π; a triple bond is never 3σ; the π always comes from the unhybridised $p$ ; π overlap dies when the $p$ orbitals stop being parallel (no rotation).