Question bank — Resonance — delocalization, resonance energy (benzene, ozone, carbonate)

False. $\leftrightarrow$ is not the equilibrium arrow $\rightleftharpoons$; the molecule is permanently one blended hybrid and is never structure A nor structure B at any instant.

False. They are imaginary drawings, artifacts of our limited Lewis notation; only their weighted average (the hybrid) physically exists.

False. Nuclei must stay put; only $\pi$ electrons and lone pairs may shift. Moving an atom gives an isomer, not a resonance form.

False. Only when the structures are equivalent by symmetry (ozone, benzene, carbonate). Otherwise more stable forms (more bonds, less charge, negative charge on the electronegative atom) contribute more.

True. Delocalization lowers energy, so the hybrid lies below the best single structure — that energy gap is the resonance energy.

False. Nothing oscillates; RE is a static comparison between the real hybrid and a hypothetical un-delocalized structure — no motion is involved.

False. Larger RE means the real molecule departs further from any one structure, so the single-structure picture is a worse description and the hybrid a much better one.

False. Resonance needs an alternative legitimate placement for the $\pi$ electrons; an isolated C=C with nowhere for electrons to delocalize (e.g. ethene) has only one structure.

True. A localized Kekulé structure predicts alternating short (134 pm) and long (154 pm) bonds; the observed uniform 139 pm can only come from ==spreading the $\pi$ electrons== over all six carbons.

False. It can — as long as every contributing structure keeps the same one unpaired electron. NO₂ has two equivalent forms with the lone radical electron and the double bond swapped between the two N–O bonds, giving equal bond lengths.

The flipping model is wrong. Both bonds are simultaneously and permanently order 1.5; there is no time-sharing, the $\pi$ electron is smeared over both O–O links at once.

The number $\frac{4}{3}$ is right but the reasoning is wrong: it is not time-sharing. One $\pi$ bond is delocalized over three equivalent C–O positions at all times, giving each a permanent order of $\frac{4}{3}$.

Wrong. Either single structure predicts two different bond lengths, contradicting experiment. The equivalence is exactly why the true structure must be their symmetric hybrid.

The interpretation is off. RE is stability held, not released. Benzene starts 152 kJ/mol lower than fake cyclohexatriene, so it releases less on hydrogenation — the gap is stored stability, not an emitted energy.

Wrong for period-2 atoms. Carbon, nitrogen, oxygen cannot exceed an octet; an octet violation makes the structure invalid (not merely minor), so it cannot contribute at all.

Wrong. Delocalization requires the p-orbitals to be parallel (atoms roughly coplanar) so they overlap side-to-side; twisting them out of plane destroys the overlap and kills delocalization.

Overstated. Hyperconjugation (a σ C–H bond donating into an adjacent empty or π orbital) is a weak, small delocalization and needs an adjacent unsaturation or cation to donate into — it is nothing like benzene's full ring delocalization and does not make ethane "aromatic."

Quantum mechanically a bigger "box" for an electron lowers its ground-state kinetic energy (particle-in-a-box: bigger box ⇒ lower energy). Delocalization enlarges the box, so the electrons — and the molecule — settle to lower energy.

A single Lewis structure can only lock electrons between two atoms; it physically cannot depict a smear over many atoms. We draw several imperfect sketches and declare the truth to be their average.

A σ bond is the head-on skeleton holding two nuclei together, so moving it would move atoms — forbidden. π electrons sit sideways off the bond axis and are loosely held, so they can be shuffled without disturbing the nuclei.

They describe the same molecule in different electron arrangements. Changing net charge or spin multiplicity would describe a different chemical species, not a resonance form of the original.

An electronegative atom is more comfortable holding extra electron density, so that arrangement is lower in energy and hence a bigger, more faithful contributor to the hybrid.

Because benzene is already stabilized by 152 kJ/mol before the reaction; sitting lower means it has less energy to give up when hydrogenated, so less heat is released.

$s{p}^2$ leaves exactly one unhybridized p-orbital per carbon, perpendicular to the ring; six such parallel p-orbitals overlap continuously to form the delocalized $\pi$ cloud that resonance describes.

MO theory builds genuinely delocalized $\pi$ molecular orbitals spanning all the atoms at once — it directly represents the smear that resonance can only approximate with several localized drawings.

No. Resonance requires two or more legitimate structures differing only in electron position; with a single valid arrangement there is nothing to average, so no delocalization from resonance.

The delocalized bond order and charge are evenly shared: each C–O bond is order $\frac{4}{3}$ and each oxygen carries exactly $-\frac{2}{3}$ of the $2-$ charge — perfect symmetry.

No. Resonance moves only $\pi$ electrons and lone pairs; breaking a $\sigma$ (single) bond moves nuclei-holding framework electrons, producing a different species rather than a resonance form.

The side-to-side p-orbital overlap is lost, so delocalization collapses — the molecule behaves like a single localized structure with essentially zero resonance stabilization.

Effectively zero. With no alternative electron arrangement there is no delocalization, so there is no extra stability to measure as resonance energy.

Very little. The hybrid is dominated by the low-energy structure; the high-energy form barely nudges the average, so the molecule looks almost like the good single structure alone.

Less. Separating opposite charges costs energy, raising that structure's energy, so it contributes less than a form where opposite charges sit close together.

Yes, provided net charge, spin, and valence limits are all obeyed. But the structure with more bonds is lower energy and contributes more heavily to the hybrid.

Yes. As long as the single unpaired electron stays put in every contributing form, the π bond delocalizes and both N–O bonds become equal, intermediate in length — resonance is not blocked by having a radical.

No — not always. In antiaromatic rings (a cyclic, planar, conjugated system with $4n$ π electrons, e.g. cyclobutadiene with 4), the delocalized arrangement is actually destabilizing; such molecules distort or avoid planarity. See Aromaticity & Hückel's Rule for the $4n+2$ vs $4n$ dividing line.

Weakly, yes. In hyperconjugation a filled σ (usually C–H) bond partially overlaps an adjacent empty p-orbital or π system, spreading a little electron density — a minor, secondary effect, far smaller than genuine π resonance and only present when a suitable neighbouring orbital exists.