Covalent bonding — bond length, bond energy, bond order: Question bank

Quick vocabulary refresher so nothing below is unearned:

Bond length = the gap between two nuclei at the bottom of the energy well (the comfy resting distance).
Bond energy = the effort to rip that bond apart, always reported as a positive number.
Bond order (BO) = how many shared electron pairs sit between the two atoms (1, 2, 3, or fractions).

True or false — justify

Every answer explains why, never just "true/false".

True/false: A stronger bond is always a shorter bond.

False in general, true only for the same pair of atoms. Across different atoms, size dominates: H–F is stronger and shorter than H–I, but only because F is tiny — compare like with like.

True/false: Bond energy is negative because forming a bond releases energy.

False. Bond energy is defined as the energy to break the bond, so it is positive. Formation releases that same magnitude with a minus sign — don't swap the sign convention. See Hess's Law and Enthalpy.

True/false: A C=C double bond is exactly twice as strong as a C–C single bond.

False. C–C ≈ 348, C=C ≈ 614 — not 696. The first (

σ

) bond is the strongest; the added

π

bond has weaker side-on overlap, so it contributes less. See Sigma and Pi Bonds.

True/false: Bond order must be a whole number.

False. Delocalised or radical species give fractions: benzene C–C ≈ 1.5, and

O_{2}^{+}

has BO 2.5. Fractions come from spreading electrons or leaving one unpaired in an antibonding orbital. See Resonance and Delocalisation.

True/false: Removing an electron from a molecule always weakens its bond.

False. If the electron leaves an antibonding orbital,

N_{a}

drops and BO rises, strengthening and shortening the bond (NO → NO

^{+}

goes from BO 2.5 to 3). See Molecular Orbital Theory.

True/false: The bond energy from a data table is the exact value for any molecule containing that bond.

False. Table values are averages over many molecules. The C–H in methane differs slightly from the C–H in chloroform; averages are for estimating

Δ H

, not precision work.

True/false: At the equilibrium bond length the net force between the atoms is zero.

True. The energy curve is flat at its minimum (

d E / d r = 0

), and force is the slope of energy — zero slope means inward pull exactly balances outward push.

True/false: Two nuclei attract each other directly to form a covalent bond.

False. Two nuclei repel (both positive). What binds them is the shared electron cloud piled between them, pulling each nucleus toward the middle.

Spot the error

Each line contains a flawed statement; the reveal names and fixes the flaw.

_{2}

is unreactive because its bond is long." — find the error.

_{2}

is unreactive because its triple bond is short and extremely strong (~945 kJ mol

^{- 1}

), not long. Higher bond order means shorter, and the huge well depth is what resists breaking.

"Bond order of O

_{2}

is 3 because oxygen wants a full octet." — find the error.

Octet counting is a Lewis shortcut, not the MO answer. By MO theory BO =

(N_{b} - N_{a}) /2 = (10 - 6) /2 = 2

, and the two unpaired antibonding electrons even make O

_{2}

paramagnetic — something Lewis dots hide.

Δ H = \sum B E (formed) - \sum B E (broken)

." — find the error.

The order is reversed. Breaking costs energy (+), forming pays it back (−), so

Δ H = \sum B E (broken) - \sum B E (formed)

"Since C–F (135 pm) is shorter than C–C (154 pm), C–F must be a double bond." — find the error.

Both are single bonds. C–F is shorter because fluorine is a small, electronegative atom, not because of higher bond order. Atomic size sets length too. See Electronegativity and Bond Polarity.

"The Morse potential minimum gives the bond energy directly." — find the error.

The minimum's position gives bond length

r_{e}

; the bond energy is the well's depth (

D_{e}

), the vertical drop from

E (\infty) = 0

down to the minimum.

"Benzene has three single and three double bonds, so its C–C lengths alternate long-short." — find the error.

All six C–C bonds are identical at 139 pm. Delocalisation smears the electrons into a ring, giving each an effective bond order of ~1.5 — no alternation. See Resonance and Delocalisation.

"Electron–electron repulsion is what holds the two atoms together." — find the error.

Electron–electron repulsion is positive (destabilising). What binds is the nucleus–electron attraction to the shared cloud sitting between the nuclei.

Why questions

Why is the first bond between two atoms stronger than the second or third?

The first is a

σ

bond with efficient head-on orbital overlap; extra bonds are

π

bonds with weaker side-on

p

–

p

overlap, so each adds less energy than the one before.

Why does higher bond order simultaneously shorten the bond and deepen the well?

More shared pairs mean more negative glue between the nuclei, pulling them closer (shorter

r_{e}

) and making them harder to separate (deeper well = higher

B E

) — the same cause drives both. See Sigma and Pi Bonds.

Why must in-phase orbital overlap happen for a bond to form?

In-phase (constructive) overlap raises

∣ ψ ∣^{2}

in the internuclear region, piling electron density between the nuclei. Out-of-phase overlap creates a node there and gives an antibonding, destabilising arrangement. See Molecular Orbital Theory.

Why is bond energy always quoted as positive even though bonds form spontaneously?

It measures the breaking process, which always costs energy. Formation is just its negative; keeping bond energy positive avoids sign confusion in

Δ H

sums.

Why can we estimate reaction enthalpy from a table of average bond energies?

Because

Δ H

depends only on the net difference between bonds broken and bonds formed; averaged values are close enough for that bookkeeping even if each individual bond varies. See Hess's Law and Enthalpy.

Why does removing a bonding electron do the opposite of removing an antibonding one?

Bonding electrons raise BO, so losing one lowers it (weaker, longer bond); antibonding electrons lower BO, so losing one raises it (stronger, shorter bond).

Edge cases

What is the bond order and stability of He

_{2}

BO =

(2 - 2) /2 = 0

: equal bonding and antibonding electrons cancel, so no net bond forms and He stays monatomic.

r \to 0

in the Morse curve, what dominates and what happens to the energy?

Nucleus–nucleus repulsion (

\propto 1/ r

) blows up, so the energy shoots steeply upward — the atoms strongly resist being pushed together past

r_{e}

r \to \infty

, what is the energy and what does it represent?

E \to 0

, the reference of two separated non-interacting atoms. The bond energy is the drop from this zero down to the well minimum

- D_{e}

Can two different bonds have the same length but different bond energies?

Yes — length is set by both size and bond order, while energy also depends on overlap type and electronegativity, so the two need not track perfectly across unlike atom pairs.

What happens to bond order and length when O

_{2}

gains an electron to become O

_{2}^{-}

(superoxide)?

The extra electron enters an antibonding

π^{*}

orbital, raising

N_{a}

and lowering BO from 2 to 1.5, so the bond lengthens and weakens. See Molecular Orbital Theory.

Is a bond with BO = 0.5 (like H

_{2}^{+}

) real?

Yes — one bonding electron and none antibonding gives BO

= 1/2

, a genuine but weak, long half-bond, showing bonds don't need full pairs to exist.

Recall One-line escape hatches for the exam

"Break = positive" → bond energy sign never flips.
"Same two atoms only" → before using the BO-shortens-bonds rule.
"Antibonding out = stronger" → for the NO/NO $^{+}$ family.
"Averages, not exact" → whenever a table value tempts you toward false precision.