Question bank — σ vs π bonds — overlap, strength

True — head-on overlap piles electron density directly between the nuclei, giving a larger overlap integral $S$ and more binding than the off-axis π lobes.

False — a double bond is 1σ + 1π, and the π (sideways overlap) is weaker than the σ, so its energy is less than double (e.g. C=C ≈ 615, not 2×348 = 696 kJ/mol).

False — atoms first pull close and overlap head-on (σ); only the leftover parallel p orbitals can then overlap sideways, so π is always the "plus-one".

False — rotating twists the two parallel p orbitals out of alignment and destroys the sideways π overlap; the σ alone is cylindrically symmetric so it would rotate, but the π locks it.

True — hybrids are directional and point straight at the neighbour, so their overlap is head-on (axial) → σ; the unhybridised leftover p orbitals are what make π.

False — that describes π. A σ bond is cylindrically symmetric about the axis with electron density on the axis; π has the nodal plane through the axis.

False — weak energy ≠ low importance; π electrons are exposed above/below the axis and loosely held, so they drive most organic reactivity (additions, colour, conjugation).

False — a triple bond is 1σ + 2π: one head-on overlap plus two perpendicular sideways overlaps of the two remaining p orbitals.

False — more shared electron density between the nuclei means shorter, stronger binding; total bond energy rises single < double < triple.

False — bond order counts all shared electron pairs (σ + π); a triple bond has bond order 3 but only 1 σ. See Bond order, length and energy.

The three bonds are not identical: 1 σ (from $p_z$–$p_z$ head-on) + 2 π (from $p_x$ and $p_y$ sideways). Only the total is very strong; the bonds differ in type and strength.

No — one bond is a σ from the directional $s{p}^2$ hybrids (head-on), the other is a single π from the leftover unhybridised p orbitals (sideways). One σ + one π, not two π.

A σ bond is cylindrically symmetric about the axis, so rotation leaves the overlap unchanged — single bonds rotate freely. It's the π (in double bonds) that rotation would break.

An $s$ orbital is spherical with no lobes to lay parallel, so it cannot overlap sideways; $s$–$s$ overlap is inherently head-on → σ only. π needs lobed ($p$/$d$) orbitals.

They add density, but off-axis where binding is least effective, and the overlap integral is smaller than head-on. Adding density ≠ adding it in the useful place — π is weaker than σ.

$S=\int {\psi }_A{\psi }_{B}d\tau$ measures how much orbitals overlap; bonding energy depends on both $S$ and the interaction element $H_{ab}$. Qualitatively larger $|S|$ ⇒ stronger, but it is not the energy itself.

Head-on lobes point directly at each other, so the product ${\psi }_A{\psi }_B$ is large along the whole axis; sideways lobes are parallel and displaced, so their product is smaller and spread off-axis.

Atoms approach to minimise energy, and head-on overlap maximises electron density between the nuclei (lowest energy first); only once the σ fixes the geometry can the remaining parallel p orbitals meet sideways.

Twisting steadily misaligns the parallel p orbitals, weakening then destroying the sideways overlap — a real energy barrier, high enough to fix the geometry and give cis-trans isomerism, but finite.

The extra π electrons sit exposed above and below the plane, loosely held, so they are easily attacked (addition) and, when extended through Resonance and conjugation, absorb visible light.

Hybrids are built to point directionally at the neighbour, so they overlap head-on → σ; forming π requires an orbital lying perpendicular to the bond axis, which only an unhybridised p orbital provides.

Each added π contributes extra stabilisation on top of the σ; the sum grows even though each successive π adds less than a σ would (348 → 615 → still higher for C≡C).

No — one lobe of the along-axis orbital points at the node between the two lobes of the crosswise one, so positive and negative overlap cancel; net $S=0$, no bond.

The two p orbitals become perpendicular, their sideways overlap goes to zero, and the π bond is fully broken — this defines the top of the rotation barrier and is the origin of cis-trans isomers.

In normal ground-state covalent bonding, no — π always accompanies a prior σ; only in excited or exotic species can the ordering break, which is why the "σ first" rule is the reliable default.

Yes — $p$–$d$ or $d$–$d$ sideways overlap gives π-type bonds (e.g. in some phosphorus/sulfur species), since π only requires two lobed orbitals lying parallel, not specifically two p's.

N₂ (1σ+2π) has more π bonds than O₂ (1σ+1π), so N≡N has higher bond order, hence shorter and stronger bonding — consistent with Bond order, length and energy.

No — a bond needs shared overlap between two atoms; a lone pair is localised on one atom. It can, however, donate into a π system via Resonance and conjugation.