JEE Mains Chemistry · Chemical Bonding and Molecular Structure

Bond Length, Bond Angle and Resonance

Bond length shrinks as bond order rises, resonance averages the bond order over equivalent positions, and bond angles open or close with the repulsion between electron pairs.

All Chemical Bonding and Molecular Structure notes JEE Mains Chemistry strategy

Why this matters

Fifteen PYQs, all of them multiple choice, and five from 2026. Four test resonance: ozone, carbonate and the carbon-oxygen bonds of esters and ketones; five rank bond lengths or ask which species has unequal bonds; six compare bond angles. Three ideas cover the page.

Concept 1 of 3: Resonance and fractional bond order

Some species need more than one Lewis structure because the double bond could sit in several places. The real species is a single hybrid of all of them, not a mixture flipping between them. Every equivalent bond is the same length, with a bond order between single and double.

Definition

Canonical forms are drawings only; they cannot be isolated and are not in equilibrium.
The hybrid is more stable than any one canonical form.
Bond order of each equivalent bond $= \dfrac{\text{total bonds between the central atom and those atoms}}{\text{number of equivalent positions}}$ .
$\mathrm{O_3}$ : both O–O bonds 128 pm, between O=O (121 pm) and O–O (148 pm); bond order 1.5.
$\mathrm{CO_3^{2-}}$ and $\mathrm{NO_3^-}$ : three equal bonds of order $4/3$ . Carboxylate ions $\mathrm{RCOO^-}$ : two equal C–O bonds of order 1.5.
In an ester $\mathrm{R{-}C(=O){-}O{-}R'}$ : C=O is shortest, the C(=O)–O bond is next (it shares some double-bond character), and O–R' is longest.

Bond order in a resonance hybrid

\text{bond order} = \frac{\text{total bonds to the equivalent atoms}}{\text{number of equivalent atoms}}

Worked example

Nitrate

\mathrm{NO_3^-}

has three canonical forms. Find the N–O bond order and say how the N–O bonds compare in length.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q107Moderate

Example 1 · Chemical Bonding and Molecular Structure · Bond Length, Bond Angle and Resonance

Given below are two statements: Statement(I) : Experimentally determined oxygen-oxygen bond lengths in the

O_{3}

are found to be same and the bond length is greater than that of a

O = O

(double bond) but less than that of a single

(O - O)

bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond

(O = O)

but more than that of a single bond

(O - O)

. In the light of the above statements, choose the correct answer from the options given below:

The hybrid does not flip between forms

Options such as 'the structures are in dynamic equilibrium' or 'each structure exists for an equal time' are always wrong. Resonance forms are not real species; there is one structure, the hybrid.

Resonance, not repulsion, sets ozone's bond length

Ozone's two O–O bonds are equal at 128 pm because the double bond is spread over both positions. A statement that lone-pair repulsion alone causes the intermediate length is false.

Drill 3 more on resonance and fractional bond order

Concept 2 of 3: Bond length and what sets it

More shared pairs pull two nuclei closer together, so a triple bond is shorter than a double, and a double shorter than a single. Bigger atoms have bigger radii, so their bonds are longer. A bond to hydrogen is short because hydrogen is so small.

Definition

Bond length falls as bond order rises between the same two atoms.
Bond length rises with atomic size.
$\mathrm{O_2^+ < O_2 < O_2^- < O_2^{2-}}$ in length, because the bond order falls 2.5, 2, 1.5, 1.
Isoelectronic species have the same bond order, as $\mathrm{N_2}$ , CO and $\mathrm{CN^-}$ (all 3) do.
In a trigonal bipyramid ( $\mathrm{PCl_5}$ ) the axial bonds are longer and weaker than the equatorial ones: they feel three bond pairs at 90°. $\mathrm{SF_4}$ also has two longer axial bonds. $\mathrm{SiF_4}$ , $\mathrm{BF_4^-}$ and $\mathrm{XeF_4}$ have all bonds equal.

Bond	Typical length (pm)	Note
C–H	109	Shortest here: hydrogen is tiny
C≡C	120	Triple bond
C=C	134	Double bond
C–C	154	Single bond
C≡N	116	Shorter than C=O despite N being larger than C
C=O	122	Carbonyl
C–O	143	Alcohols and ethers
O=O	121	In O₂
O–O	148	In H₂O₂
P–Cl in PCl₅	219 axial, 204 equatorial	Axial bonds are the longer, weaker pair Calling the axial bonds of PCl₅ stronger is a standard wrong statement.

For the same pair of atoms: triple shorter than double, double shorter than single.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q39Moderate

Example 2 · Chemical Bonding and Molecular Structure · Bond Length, Bond Angle and Resonance

The correct increasing order of

C - H(A)

C - O(B),C = O(C)

and

C \equiv N(D)

bonds in terms of covalent bond length is:

Bond length is not set by bond order alone

C≡N (116 pm) is shorter than C=O (122 pm), and C–H (109 pm) is shorter than both, though it is a single bond. Compare orders only between the same two atoms; otherwise size matters too.

See-saw and trigonal bipyramid give unequal bonds

Four bonds do not mean four equal bonds. Tetrahedral

\mathrm{SiF_4}

and square planar

\mathrm{XeF_4}

have equal bonds, but see-saw

\mathrm{SF_4}

has two long axial and two short equatorial bonds.

Drill 4 more on bond length and what sets it

Concept 3 of 3: Bond angles and lone-pair repulsion

Electron pairs push each other apart. A lone pair sits closer to the central atom than a bond pair and spreads wider, so it pushes harder and squeezes the bonds together. The more lone pairs on the centre, the smaller the bond angle.

Definition

Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
$\mathrm{CH_4}$ 109.5°, $\mathrm{NH_3}$ 107°, $\mathrm{H_2O}$ 104.5°: 0, 1 and 2 lone pairs on an $sp^3$ centre.
An electronegative outer atom pulls the bond pairs away from the centre, so they repel less and the angle closes: $\mathrm{OF_2}$ (103°) < $\mathrm{H_2O}$ (104.5°) < $\mathrm{Cl_2O}$ (about 111°, the large Cl atoms also crowd each other).
Down a group the angle closes: $\mathrm{H_2O}$ 104.5° > $\mathrm{H_2S}$ 92°; $\mathrm{NH_3}$ 107° > $\mathrm{PH_3}$ 93.5°.
In $\mathrm{OF_2}$ oxygen is in the +2 oxidation state, because F is more electronegative.

Species	Pairs on the centre	Bond angle
$\mathrm{BF_3}$	3 bond, 0 lone	120°
$\mathrm{SO_2}$	2 bond (plus π), 1 lone	about 119°
$\mathrm{CH_4}$	4 bond, 0 lone	109.5°
$\mathrm{NH_3}$	3 bond, 1 lone	107°
$\mathrm{H_2O}$	2 bond, 2 lone	104.5°
$\mathrm{NF_3}$	3 bond, 1 lone	102°
$\mathrm{PF_3}$	3 bond, 1 lone	about 98°
$\mathrm{ClF_3}$	3 bond, 2 lone	about 87.5° (axial F–Cl–equatorial F) The two lone pairs bend the axial F atoms back below 90°.

More lone pairs on the centre, or more electronegative outer atoms, means a smaller angle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q124Moderate

Example 3 · Chemical Bonding and Molecular Structure · Bond Length, Bond Angle and Resonance

The correct increasing order for bond angles among

BF_{3},PF_{3}

and

ClF_{3}

is :

Both SO₂ and H₂O are bent, at very different angles

S in

\mathrm{SO_2}

has three electron domains (

sp^2

) and one lone pair, so its angle is near 119°. O in water has four domains (

sp^3

) and two lone pairs, so 104.5°. Same shape, larger angle for

\mathrm{SO_2}

Fluorine closes the angle; chlorine opens it

\mathrm{OF_2}

(103°) is smaller than

\mathrm{H_2O}

, but

\mathrm{Cl_2O}

(about 111°) is larger. Electronegative F draws the bond pairs outward; the big Cl atoms push each other apart.

Drill 5 more on bond angles and lone-pair repulsion

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Resonance and fractional bond order
Bond order in a resonance hybrid
$\text{bond order} = \frac{\text{total bonds to the equivalent atoms}}{\text{number of equivalent atoms}}$

Reference tables (2)

Bond length and what sets it10 rows

Bond	Typical length (pm)	Note
C–H	109	Shortest here: hydrogen is tiny
C≡C	120	Triple bond
C=C	134	Double bond
C–C	154	Single bond
C≡N	116	Shorter than C=O despite N being larger than C
C=O	122	Carbonyl
C–O	143	Alcohols and ethers
O=O	121	In O₂
O–O	148	In H₂O₂
P–Cl in PCl₅	219 axial, 204 equatorial	Axial bonds are the longer, weaker pair Calling the axial bonds of PCl₅ stronger is a standard wrong statement.

For the same pair of atoms: triple shorter than double, double shorter than single.

Bond angles and lone-pair repulsion8 rows

Species	Pairs on the centre	Bond angle
$\mathrm{BF_3}$	3 bond, 0 lone	120°
$\mathrm{SO_2}$	2 bond (plus π), 1 lone	about 119°
$\mathrm{CH_4}$	4 bond, 0 lone	109.5°
$\mathrm{NH_3}$	3 bond, 1 lone	107°
$\mathrm{H_2O}$	2 bond, 2 lone	104.5°
$\mathrm{NF_3}$	3 bond, 1 lone	102°
$\mathrm{PF_3}$	3 bond, 1 lone	about 98°
$\mathrm{ClF_3}$	3 bond, 2 lone	about 87.5° (axial F–Cl–equatorial F) The two lone pairs bend the axial F atoms back below 90°.

More lone pairs on the centre, or more electronegative outer atoms, means a smaller angle.

Watch out for (6)

The hybrid does not flip between forms
→ Resonance and fractional bond order
Resonance, not repulsion, sets ozone's bond length
→ Resonance and fractional bond order
Bond length is not set by bond order alone
→ Bond length and what sets it
See-saw and trigonal bipyramid give unequal bonds
→ Bond length and what sets it
Both SO₂ and H₂O are bent, at very different angles
→ Bond angles and lone-pair repulsion
Fluorine closes the angle; chlorine opens it
→ Bond angles and lone-pair repulsion

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