Why does the molecular orbital energy order change between nitrogen and oxygen?

Because of s-p mixing. The $\sigma 2s$ and $\sigma 2p$ orbitals have the same symmetry, so they interact and push each other apart — the $\sigma 2p$ being pushed up. That interaction is strong when the 2s and 2p levels are close in energy, as they are in boron, carbon and nitrogen, and it lifts $\sigma 2p$ above the $\pi 2p$ pair. Moving right across the period the 2s level drops away from 2p, the mixing becomes negligible, and by oxygen the $\sigma 2p$ has settled back below the $\pi$ orbitals. It is a continuous effect that happens to cross over between N$_2$ and O$_2$.

Chapter 4 of 18

22%

JEE Main Chemistry★ ~2 questions (8 marks) in JEE Main, the largest and most consistently examined Chemistry chapter, and the one that decides whether Organic Chemistry makes sense later

Chemical Bonding and Molecular Structure

Atoms bond because the bonded arrangement has lower energy. Every model here — ionic, Lewis, VSEPR, hybridisation, molecular orbital — is a different tool for predicting how far it falls and what shape results.

⏱ 25 min readHard difficulty✓ Free · exam-pattern aligned

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Choose the cheapest model that answers a bonding question, and say what each model can and cannot predict
2Use lattice enthalpy and a Born-Haber cycle to explain why ionic compounds form, and apply Fajans' rules to covalent character
3Assign formal charges to select between competing Lewis structures, and check the choice against bond length
4Predict shape and bond angles from the steric number, including why lone pairs take equatorial sites in a trigonal bipyramid
5Assign hybridisation and count sigma and pi bonds in any structure, including chains with multiple bonds
6Compute bond order from a molecular orbital diagram and predict magnetism, including the ordering switch after nitrogen

💡

Why this chapter matters in JEE Main

This is the largest and most consistently examined chapter in JEE Main Chemistry, and the one that decides whether Organic Chemistry makes sense later. Students who treat it as five unrelated theories to memorise find it enormous; it is not. Atoms bond because the bonded arrangement is lower in energy than the separated one, and every model here is a different tool for predicting how far that energy falls and what shape results. The models form a ladder of increasing power and increasing cost: Lewis structures are fastest and weakest, VSEPR predicts shape but not energy, valence bond theory explains bonding through overlap, and molecular orbital theory is the most accurate and slowest to apply. Knowing which tool a question wants is half the skill — use the cheapest model that answers it, and reach for molecular orbital theory only when magnetic behaviour or an odd bond order is involved. JEE Main returns every year to VSEPR shapes with lone pairs, hybridisation from steric number, bond order of diatomic ions, and dipole moment comparisons.

Before you start — revise these

🔗

Electronic configuration and orbital shapes from Atomic Structure

🔗

Electronegativity, ionisation enthalpy and electron gain enthalpy from Periodicity

🔗

Vector addition, for dipole moment reasoning

🔗

Hess's law, for the Born-Haber cycle

Chemical Bonding and Molecular Structure

Oxygen's Lewis structure shows a double bond with every electron paired, so it should be diamagnetic. Pour liquid oxygen between the poles of a magnet.

It sticks.

Two electrons sit unpaired in degenerate $π^{*}$ orbitals — something no Lewis structure, no VSEPR diagram and no hybridisation scheme can show. Only molecular orbital theory gets it right, and that is the clearest statement of why this chapter contains five models rather than one.

This is the largest and most consistently examined chapter in JEE Main Chemistry, and the one that decides whether Organic Chemistry makes sense later.

The organising principle: atoms bond because the bonded arrangement is lower in energy, and each model is a different tool for predicting how far it falls and what shape results. They form a ladder of increasing power and increasing cost:

Model	Gives you	Cost
Lewis	Connectivity, formal charge	Seconds
VSEPR	Shape and angles	Seconds
Valence bond	Sigma/pi count, hybridisation	Fast
Molecular orbital	Bond order, magnetism, why He $_{2}$ fails	Slow

Use the cheapest model that answers the question. Reach for molecular orbital theory only when magnetism or an odd bond order is involved.

1. Why Atoms Bond

The Kossel-Lewis approach: atoms combine to reach a noble gas configuration, by transferring electrons (ionic) or sharing them (covalent). The octet rule captures this — and fails often enough that its exceptions are exam favourites.

Failure	Examples
Incomplete octet	BeCl $_{2}$ , BCl $_{3}$ , AlCl $_{3}$
Expanded octet	PCl $_{5}$ , SF $_{6}$ , IF $_{7}$
Odd electron	NO, NO $_{2}$ , ClO $_{2}$

Expanded octets need accessible d orbitals, which is why they appear from period 3 downwards and never in period 2.

2. The Ionic Bond

Formation is favoured by a low ionisation enthalpy for the metal, a highly negative electron gain enthalpy for the non-metal, and above all by a high lattice enthalpy.

Lattice enthalpy is the decisive term

Lattice enthalpy is the energy released when one mole of ionic solid forms from gaseous ions. It is what makes ionic bonding worthwhile at all, because ionisation enthalpy almost always exceeds the energy released on electron gain.

$Na: + 496 Cl: - 349 \Rightarrow transfer alone costs + 147 kJ mol^{- 1}$

It is the 788 kJ mol⁻¹ of lattice enthalpy that pays for it.

Lattice enthalpy rises with higher charges and smaller radii. MgO, with doubly charged ions, reaches near 3800 kJ mol⁻¹ — which is why it melts above 2800 °C while NaCl melts at 801 °C.

Illustration 1

Use a Born-Haber cycle to find $Δ H_{f}$ of NaCl(s), given: sublimation of Na $= + 108$ , ionisation of Na $= + 496$ , half the Cl–Cl bond enthalpy $= + 121$ , electron gain by Cl $= - 349$ , lattice enthalpy $= - 788$ kJ mol⁻¹.

Hess's law round the loop — every step is measurable except the last:

$Δ H_{f} = 108 + 496 + 121 - 349 - 788 \Rightarrow Δ H_{f} = - 412 kJ mol^{- 1}$

The measured value is $- 411$ kJ mol⁻¹. Note the arithmetic: the first four terms sum to $+ 376$ , so without the lattice term the compound would not form at all. Lattice enthalpy is not one contribution among several — it is the whole reason ionic solids exist.

Fajans' rules

No bond is purely ionic. A small, highly charged cation distorts a large anion's electron cloud, pulling density between the nuclei and giving covalent character.

Covalent character increases with: smaller cation, larger anion, higher charge on either ion, and a cation with a pseudo noble gas configuration such as Ag $^{+}$ or Cu $^{+}$ .

Illustration 2

Which is more covalent, LiF or LiI? And why does AlCl $_{3}$ sublime while NaCl melts at 801 °C?

The cation is identical in the first pair, so the anion decides: iodide is far larger than fluoride and much more easily polarised, so LiI is more covalent.

For the second, Al $^{3 +}$ is small and triply charged, giving enormous polarising power, while Na $^{+}$ is larger and singly charged. Aluminium chloride is therefore substantially covalent, existing as discrete Al $_{2}$ Cl $_{6}$ dimers held only by weak intermolecular forces. Sodium chloride is a genuine ionic lattice, and melting it means breaking electrostatic attraction throughout the crystal.

3. Lewis Structures and Formal Charge

Draw the skeleton, count total valence electrons, place lone pairs to complete octets, and convert lone pairs to multiple bonds if any atom is short.

$F C = V - L - \frac{B}{2}$

with $V$ the free atom's valence count, $L$ lone pair electrons, $B$ bonding electrons.

The best structure has formal charges closest to zero, with any negative charge on the most electronegative atom.

Illustration 3

Nitrous oxide, N $_{2}$ O, has 16 valence electrons. Two structures satisfy the octet rule. Use formal charge to choose.

Applying $F C = V - L - B /2$ atom by atom gives the values shown above. Both structures carry a total of $- 1$ , as they must. The tie-break is where the negative charge sits, and oxygen is the more electronegative atom — so the triple-bonded structure is preferred.

The prediction is testable: it demands a short, strong N–N bond, and the measured N–N distance in N $_{2}$ O is 113 pm, close to the 110 pm of a genuine triple bond. Formal charge is not bookkeeping; it makes claims about geometry.

4. Bond Parameters and Polarity

Bond	Order	Length (pm)	Enthalpy (kJ mol⁻¹)
C–C	1	154	348
C=C	2	134	614
C≡C	3	120	839

Trap. A double bond is not twice as strong as a single bond — 614 against 348, not 696. The second bond is a pi bond from sideways overlap, which is weaker than the sigma. That single fact drives most of alkene chemistry: the pi bond is the one that breaks.

$μ = q \times d, 1 D = 3.336 \times 1 0^{- 30} C m$

The molecular dipole is the vector sum of the bond dipoles, so a molecule with strongly polar bonds can be entirely non-polar if symmetry cancels them — CO $_{2}$ , BF $_{3}$ and CCl $_{4}$ all have zero dipole moment.

Illustration 4

Rank CH $_{3}$ Cl, CH $_{2}$ Cl $_{2}$ , CHCl $_{3}$ and CCl $_{4}$ by dipole moment.

Molecule	$μ$ (D)
CH $_{3}$ Cl	1.87
CH $_{2}$ Cl $_{2}$	1.60
CHCl $_{3}$	1.04
CCl $_{4}$	0

More chlorine does not mean more polar. Each added C–Cl dipole points outward from a different tetrahedral vertex, so successive additions increasingly cancel one another, and the fourth cancels the resultant completely by symmetry. Adding polar bonds to a molecule can reduce its polarity — a result that only vector reasoning predicts.

Illustration 5

The dipole moment of HCl is 1.03 D and its bond length is 1.27 Å. Find the percentage ionic character of the bond.

First ask what the dipole moment would be if the bond were completely ionic, with a full electronic charge separated across that distance. In the units chemists use, a unit charge one angstrom apart gives 4.8 D:

$μ_{i o ni c} = 4.8 \times 1.27 = 6.10 D$

The measured value is a fraction of that:

$% ionic character = \frac{1.03}{6.10} \times 100 = 16.9 %$

So the H–Cl bond is about one-sixth ionic and five-sixths covalent, which matches its behaviour: HCl is a gas, not a salt.

The number is a measure of charge separation, not a classification. There is no threshold at which a bond stops being covalent and starts being ionic; the two extremes are idealisations and every real bond sits somewhere between them. That is the same point Fajans' rules made from the opposite direction, working out how much covalent character an ionic compound picks up.

5. VSEPR Theory

One idea: electron pairs around a central atom get as far apart as possible. The repulsion order is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, because a lone pair is held by one nucleus only and spreads out more.

Count the steric number (bond pairs + lone pairs) for the electron geometry, then ignore the lone pairs to read the molecular shape.

Steric no.	Electron geometry	Lone pairs	Shape	Example
2	Linear	0	Linear	BeCl $_{2}$
3	Trigonal planar	0	Trigonal planar	BF $_{3}$
3	Trigonal planar	1	Bent	SO $_{2}$
4	Tetrahedral	0	Tetrahedral	CH $_{4}$
4	Tetrahedral	1	Trigonal pyramidal	NH $_{3}$
4	Tetrahedral	2	Bent	H $_{2}$ O
5	Trigonal bipyramidal	0	Trigonal bipyramidal	PCl $_{5}$
5	Trigonal bipyramidal	1	See-saw	SF $_{4}$
5	Trigonal bipyramidal	2	T-shaped	ClF $_{3}$
5	Trigonal bipyramidal	3	Linear	XeF $_{2}$
6	Octahedral	0	Octahedral	SF $_{6}$
6	Octahedral	1	Square pyramidal	BrF $_{5}$
6	Octahedral	2	Square planar	XeF $_{4}$

Bond angle compression. CH $_{4}$ , NH $_{3}$ and H $_{2}$ O all have steric number 4, yet their angles fall 109.5 → 107 → 104.5°. Each added lone pair pushes the bonding pairs closer, roughly 2.5° per lone pair. Explaining the ordering is worth more than remembering the numbers.

Illustration 6

Predict the shape and hybridisation of ClF $_{3}$ .

Chlorine has seven valence electrons; three go into bonds, leaving four as two lone pairs.

$steric number = 3 + 2 = 5 \Rightarrow trigonal bipyramidal, s p^{3} d$

Both lone pairs take equatorial sites, by the rule above. Removing them from the picture leaves three atoms in a T-shape, with the F–Cl–F angles slightly under 90° because the lone pairs squeeze them.

Illustration 7

Account for the bond angle order $C H_{4} (109. 5^{\circ}) > N H_{3} (10 7^{\circ}) > H_{2} O (104. 5^{\circ})$ , and then explain why $N H_{3}$ has a larger angle than $N F_{3}$ .

All three of the first set have four electron pairs and are built on the same tetrahedral framework. What differs is how many of those pairs are lone pairs.

A lone pair is held by one nucleus rather than shared between two, so it spreads wider and pushes harder. Methane has none, ammonia one, water two, and each addition squeezes the bonding pairs closer:

$lone pair-lone pair > lone pair-bond pair > bond pair-bond pair$

The second comparison uses a different lever. NH₃ and NF₃ have the same shape and the same single lone pair, so repulsion between pairs cannot separate them.

Fluorine is more electronegative than nitrogen, so it drags each bonding pair away from the central atom. Those pairs now sit further out and repel each other less, letting the angle close to 102° against ammonia's 107°. In NH₃ the nitrogen is the more electronegative partner and holds the pairs close, where they push back harder.

So two separate effects set a bond angle: how many lone pairs there are, and where the bonding pairs sit, which the substituent's electronegativity controls.

Illustration 8

For XeOF₄, determine the steric number, hybridisation, shape and whether the molecule is polar.

Xenon has 8 valence electrons. Four go into single bonds with fluorine and two into a double bond with oxygen, leaving 2 as one lone pair.

$steric number = 5 sigma bonds + 1 lone pair = 6$

$\Rightarrow octahedral electron geometry, s p^{3} d^{2}$

With one position occupied by a lone pair, the shape is square pyramidal: four fluorines in a plane, oxygen at the apex, lone pair opposite it.

Polarity: the four Xe–F dipoles lie in a plane and cancel by symmetry, but the Xe=O dipole and the lone pair both lie along the unique axis with nothing to oppose them. The molecule is polar.

Notice the double bond contributed one sigma to the steric number, not two. Pi bonds occupy no separate position in the electron geometry — a point that decides this question and many like it.

6. Valence Bond Theory and Hybridisation

A covalent bond is the overlap of two half-filled atomic orbitals, with greater overlap giving a stronger bond.

Sigma bond — head-on overlap along the internuclear axis; permits free rotation.
Pi bond — sideways overlap of parallel p orbitals; weaker, and locks rotation, which is why alkenes show geometrical isomerism.

Single = 1 sigma. Double = 1 sigma + 1 pi. Triple = 1 sigma + 2 pi.

Hybridisation	Mixed	Geometry	Angle	Example
$s p$	1s, 1p	Linear	180°	BeCl $_{2}$ , C $_{2}$ H $_{2}$
$s p^{2}$	1s, 2p	Trigonal planar	120°	BF $_{3}$ , C $_{2}$ H $_{4}$
$s p^{3}$	1s, 3p	Tetrahedral	109.5°	CH $_{4}$ , NH $_{3}$
$s p^{3} d$	1s, 3p, 1d	Trigonal bipyramidal	120°, 90°	PCl $_{5}$
$s p^{3} d^{2}$	1s, 3p, 2d	Octahedral	90°	SF $_{6}$

Hybridisation follows the steric number directly — 2 gives $s p$ , 3 gives $s p^{2}$ , 4 gives $s p^{3}$ , and so on. Lone pairs count.

More s character means a shorter, stronger bond and a wider angle, because s orbitals hold electrons closer to the nucleus. That is why an $s p$ carbon is more electronegative than an $s p^{3}$ one — and why terminal alkynes are acidic.

Illustration 9

For $C H_{3} - CH = CH - C \equiv CH$ , give the hybridisation of each carbon and count sigma and pi bonds.

Carbon	Environment	Hybridisation
C1	4 single bonds	$s p^{3}$
C2, C3	one double bond each	$s p^{2}$
C4, C5	triple bond	$s p$

Sigma bonds: 4 carbon-carbon plus 6 carbon-hydrogen $= 10$ . Pi bonds: 1 from the double plus 2 from the triple $= 3$ .

Count sigma bonds as one per connection, whatever its order, and then add one pi for each order above the first. Counting the double bond as two sigmas is the standard error.

Resonance

When one Lewis structure cannot represent a molecule, the true structure is a resonance hybrid. Ozone has two equivalent structures, and its two bonds are found identical at 128 pm — between a single and a double bond.

Trap. Resonance is not oscillation. The molecule does not flip between structures; it exists permanently in one state that none of the drawn structures represents alone. Delocalisation lowers the energy, and more equivalent contributors means more stability.

7. Molecular Orbital Theory

Molecular orbital theory abandons the idea that electrons belong to particular bonds. Atomic orbitals combine by linear combination into molecular orbitals spread over the whole molecule: a bonding orbital lower in energy with density between the nuclei, and an antibonding orbital higher, with a node between them.

Electrons fill them by the same rules as atomic orbitals.

The energy ordering, and why it switches

For molecules up to and including N $_{2}$ (14 electrons or fewer):

$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < π 2 p_{x} = π 2 p_{y} < σ 2 p_{z} < π^{*} 2 p < σ^{*} 2 p_{z}$

For O $_{2}$ and F $_{2}$ , $σ 2 p_{z}$ drops below the $π 2 p$ pair. The cause is s-p mixing, significant while 2s and 2p are close in energy and negligible once the gap widens. Getting this switch right is essential, because the magnetic prediction depends on it.

$Bond order = \frac{N _{b} - N _{a}}{2}$

Species	Electrons	Bond order	Magnetism
H $_{2}$	2	1	Diamagnetic
He $_{2}$	4	0	Does not exist
B $_{2}$	10	1	Paramagnetic
C $_{2}$	12	2	Diamagnetic
N $_{2}$	14	3	Diamagnetic
O $_{2}$	16	2	Paramagnetic
F $_{2}$	18	1	Diamagnetic

A bond order of zero means the molecule does not exist — exactly why He $_{2}$ and Be $_{2}$ have never been isolated.

Illustration 10

Compare O $_{2}$ , O $_{2}^{+}$ and O $_{2}^{-}$ for bond order, bond length and magnetism.

Species	Electrons	$N_{b}$	$N_{a}$	Bond order	Unpaired
O $_{2}^{+}$	15	10	5	2.5	1
O $_{2}$	16	10	6	2.0	2
O $_{2}^{-}$	17	10	7	1.5	1

Bond length runs inversely to bond order, so O $_{2}^{+}$ is shortest and O $_{2}^{-}$ longest. Every one of the three is paramagnetic — a common distractor, since it is tempting to assume the ions must differ from the neutral molecule in that respect too.

Removing an antibonding electron strengthens the bond. That is counterintuitive until you notice which orbital the electron came out of.

Illustration 11

Show that CN⁻, CO, N₂ and NO⁺ are isoelectronic, and predict the bond order and magnetism of each.

Count valence electrons:

$C N^{-} : 4 + 5 + 1 = 10 CO : 4 + 6 = 10 N_{2} : 5 + 5 = 10 N O^{+} : 5 + 6 - 1 = 10$

All four have ten valence electrons, so all four fill the same molecular orbitals to the same level — the N $_{2}$ ordering, since none exceeds 14 electrons in total.

$N_{b} = 8, N_{a} = 2 \Rightarrow bond order = \frac{8 - 2}{2} = 3$

All are diamagnetic, all have a triple bond, and all have very similar bond lengths near 110 pm.

This is why CO and CN $^{-}$ turn up together as strong-field ligands in Coordination Compounds. They are the same electronic structure wearing different atoms, and a metal ion binding one behaves much as it does binding the other.

8. Hydrogen Bonding and Metallic Bonding

A hydrogen bond forms when hydrogen bonded to a small, highly electronegative atom — in practice only F, O or N — is attracted to a lone pair on another such atom. At 10 to 40 kJ mol⁻¹ it is far weaker than a covalent bond, but there are many of them and the consequences are enormous.

Water boils at 100 °C while the heavier H $_{2}$ S boils at $- 60$ °C. Ice floats because hydrogen bonding forces an open tetrahedral lattice less dense than the liquid.

Trap. Intramolecular hydrogen bonding has the opposite effect on boiling point. Ortho-nitrophenol bonds to itself and so cannot bond to neighbours, making it more volatile than para-nitrophenol, which bonds intermolecularly.

Metallic bonding is a lattice of positive ions in a sea of delocalised electrons. Mobile electrons explain conductivity; non-directional bonding explains why metals are malleable rather than brittle.

Illustration 12

Explain the boiling point order HF > HI > HBr > HCl.

	HF	HCl	HBr	HI
bp (°C)	+20	−85	−67	−35

Two effects run in opposite directions. Down the group, molecules get larger and more polarisable, so van der Waals forces strengthen: HCl < HBr < HI. But HF alone has hydrogen bonding, worth far more than the size difference, and it jumps to the top.

The order is not a single trend broken by an exception — it is two trends, one of which applies to one member only. The same shape appears in NH $_{3}$ and H $_{2}$ O against their group neighbours.

Summary

Atoms bond because the bonded state is lower in energy. Five models, increasing power and cost — use the cheapest that answers the question.
Octet failures: incomplete (BCl $_{3}$ ), expanded (SF $_{6}$ , period 3 down only), odd-electron (NO).
Ionic bonding is driven by lattice enthalpy — without it NaCl would not form at all.
Lattice enthalpy rises with higher charges and smaller radii; MgO melts above 2800 °C.
Fajans: small cation, large anion, high charge, pseudo noble gas core all raise covalent character.
$F C = V - L - B /2$ ; the best structure has charges near zero, negative on the most electronegative atom.
A double bond is not twice a single bond — the pi component is weaker, and it is the one that breaks.
Molecular dipole is the vector sum; adding polar bonds can reduce polarity, as CH $_{3}$ Cl to CCl $_{4}$ shows.
VSEPR: steric number gives electron geometry, then delete lone pairs for the shape.
Lone pairs always go equatorial in a trigonal bipyramid — 2 neighbours at 90°, not 3.
Angles compress about 2.5° per lone pair: 109.5 → 107 → 104.5.
Hybridisation follows the steric number, and lone pairs count. More s character means shorter, stronger, wider.
Sigma per connection, plus one pi per order above the first.
Resonance is delocalisation, not oscillation.
Bond order $= (N_{b} - N_{a}) /2$ ; zero means the molecule cannot exist.
The $σ$ / $π$ order switches after N $_{2}$ because s-p mixing weakens.
O $_{2}$ is paramagnetic with two unpaired $π^{*}$ electrons — the theory's decisive success.
Removing an antibonding electron raises bond order: O $_{2}^{+}$ is stronger than O $_{2}$ .
Hydrogen bonding at 10–40 kJ mol⁻¹ explains water's boiling point and floating ice; intramolecular H-bonding lowers boiling point.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Formal charge

F C = V - L - \frac{B}{2}

$V$ is the free atom's valence count, $L$ its lone pair electrons, $B$ its bonding electrons. The best structure has charges closest to zero, with any negative charge on the most electronegative atom — and that prediction is testable against bond length.

Lattice enthalpy trend

U \propto \frac{q _{+} \times q _{-}}{r _{+} + r _{-}}

Rises with higher charges and smaller radii. MgO reaches near 3800 kJ mol$^{-1}$ against NaCl's 788, which is why it melts above 2800 °C. Without the lattice term NaCl would not form at all — the transfer step alone costs $+147$ kJ mol$^{-1}$.

Born-Haber cycle

Δ H_{f} = Δ H_{s u b} + I E + \frac{1}{2} B E + E A + U

Hess's law round a closed loop of measurable steps, used to obtain the lattice enthalpy that cannot be measured directly. For NaCl the terms are $108 + 496 + 121 - 349 - 788 = -412$ kJ mol$^{-1}$, against a measured $-411$.

Fajans' rules

covalent character ↑ with smaller cation, larger anion, higher charge, d^{10} core

No bond is purely ionic. A small highly charged cation distorts a large anion's cloud, pulling density between the nuclei. This is why AlCl$_3$ sublimes as a dimer while NaCl is a high-melting lattice, and why AgCl is far more covalent than NaCl.

Dipole moment

μ = q \times d, 1 D = 3.336 \times 1 0^{- 30} C m

The molecular dipole is the vector sum of the bond dipoles, so symmetry can cancel strongly polar bonds entirely — CO$_2$, BF$_3$ and CCl$_4$ all read zero. Adding polar bonds can even lower polarity, as CH$_3$Cl through to CCl$_4$ shows.

Steric number and shape

S N = bond pairs + lone pairs

Gives the electron geometry; delete the lone pairs to read the molecular shape. Repulsion runs lone-lone > lone-bond > bond-bond, so each lone pair compresses the angles by roughly 2.5 degrees.

Lone pairs in a trigonal bipyramid

equatorial: 2 neighbours at 9 0^{\circ}; axial: 3 at 9 0^{\circ}

So lone pairs always take equatorial positions, and that one rule generates see-saw for one lone pair, T-shaped for two and linear for three. In an octahedron two lone pairs instead go opposite each other, giving square planar.

Hybridisation from steric number

2 \to s p, 3 \to s p^{2}, 4 \to s p^{3}, 5 \to s p^{3} d, 6 \to s p^{3} d^{2}

Lone pairs count towards the steric number. More s character gives a shorter, stronger bond and a wider angle, which is why an $sp$ carbon is more electronegative than an $sp^{3}$ one and terminal alkynes are acidic.

Sigma and pi counting

single = 1 σ; double = 1 σ + 1 π; triple = 1 σ + 2 π

One sigma per connection whatever its order, then one pi for each order above the first. A double bond is not twice as strong as a single one — 614 against 348 kJ mol$^{-1}$ — because the pi component is the weaker half, and the one that breaks.

Bond order and MO ordering

BO = \frac{N _{b} - N _{a}}{2}

Zero bond order means the molecule cannot exist, which is why He$_2$ and Be$_2$ have never been isolated. For O$_2$ and F$_2$ the $\sigma 2p$ level drops below the $\pi 2p$ pair, because s-p mixing weakens once the 2s-2p gap widens — and the magnetic prediction depends on getting that switch right.

Percentage ionic character

% ionic = \frac{μ _{o b ser v e d}}{μ _{i o ni c}} \times 100, μ_{i o ni c} [D] = 4.8 \times d [\overset{˚}{A}]

The 4.8 is one electronic charge separated by one angstrom, expressed in debye. HCl comes out about 17 per cent ionic. The figure measures charge separation and does not classify the bond — there is no threshold at which covalent becomes ionic.

What sets a bond angle

lp-lp > lp-bp > bp-bp

Two separate levers act. More lone pairs squeeze the angle down, giving $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$. A more electronegative substituent pulls bonding pairs away so they repel less, giving $\mathrm{NH_3}\ (107^\circ) > \mathrm{NF_3}\ (102^\circ)$ at identical shape and lone pair count.

Isoelectronic species

same total electrons \Rightarrow same MO filling, bond order and magnetism

CN⁻, CO, N₂ and NO⁺ all carry 14 electrons, so all four have bond order 3 and are diamagnetic. Counting electrons and charge together is faster than rebuilding the diagram, and it is how these questions are meant to be answered.

JEE Main test series

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⚠️

Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Predicting oxygen to be diamagnetic from its Lewis structure

✓ The Lewis model cannot represent degenerate orbitals. O

_{2}

's last two electrons occupy the two

π^{*}

orbitals singly with parallel spins, leaving two unpaired electrons. Liquid oxygen sticks to a magnet, and only molecular orbital theory predicts it.

Why it happens: The Lewis picture shows a double bond with every electron neatly paired.

WATCH OUT

✗ Using the same molecular orbital ordering for N

_{2}

and O

_{2}

✓ Up to and including N

_{2}

, the

π 2 p

pair lies below

σ 2 p

. For O

_{2}

and F

_{2}

they swap, because s-p mixing weakens as the 2s-2p gap widens. The magnetic prediction turns entirely on this.

Why it happens: One diagram is usually memorised and applied to the whole period.

WATCH OUT

✗ Assuming NF

_{3}

has a larger dipole moment than NH

_{3}

because fluorine is more electronegative

✓ The molecular dipole is a vector sum. In NH

_{3}

the bond dipoles point towards nitrogen, reinforcing the lone pair dipole; in NF

_{3}

they point away and largely cancel it. The measured values are 1.47 D against 0.24 D.

Why it happens: Electronegativity difference is the usual guide to bond polarity, and it does favour NF

_{3}

WATCH OUT

✗ Placing a lone pair axially in a trigonal bipyramidal arrangement

✓ An axial site has three neighbours at 90 degrees; an equatorial site has only two, the rest at 120. Lone pairs, which repel most strongly, always go equatorial — which is what produces see-saw, T-shaped and linear geometries.

Why it happens: Axial looks like the roomier position on a two-dimensional drawing.

WATCH OUT

✗ Treating resonance as the molecule flipping between structures

✓ The molecule exists permanently in a single delocalised state that none of the drawn structures represents. Ozone's two bonds are identical at 128 pm at all times, not alternating. Delocalisation lowers the energy; oscillation would not.

Why it happens: The double-headed arrow looks like an equilibrium arrow, and the structures are drawn one after another.

WATCH OUT

✗ Forgetting to count lone pairs when assigning hybridisation

✓ The steric number is bond pairs plus lone pairs. Water is

s p^{3}

, not

s p^{2}

, because its two lone pairs occupy hybrid orbitals just as bonding pairs do. Count every region of electron density round the central atom.

Why it happens: Hybridisation is introduced through methane, where there are none.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Chemical Bonding and Molecular Structure?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~12 marks in JEE Main exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Five models of increasing power and cost — use the cheapest that answers the question
•Octet fails three ways: incomplete, expanded (period 3 down only), and odd-electron
•Lattice enthalpy is why ionic solids form at all; it rises with higher charge and smaller radius
•Fajans: small cation, large anion, high charge, $d^{10}$ core all raise covalent character
• $F C = V - L - B /2$ ; negative formal charge belongs on the most electronegative atom
•Molecular dipole is a vector sum, so symmetry can cancel strongly polar bonds entirely
•Steric number gives electron geometry; delete lone pairs for the shape; about 2.5° compression each
•Lone pairs always go equatorial in a trigonal bipyramid — 2 neighbours at 90°, not 3
•Hybridisation follows steric number and lone pairs count; sigma per connection, pi per extra order
•BO $= (N_{b} - N_{a}) /2$ ; the $σ$ / $π$ order switches after N $_{2}$ ; O $_{2}$ is paramagnetic with two $π^{*}$ electrons
•Symmetry decides polarity, not bond polarity: CO₂ and BF₃ cancel while H₂O and NH₃ do not, despite polar bonds in all four
•NH₃ beats NF₃ on bond angle because fluorine pulls the bonding pairs away, not because the shapes differ

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: ~3 questions (12 marks) of the 100-mark Chemistry section

Question style	Marks each	Typical count	What it tests
VSEPR shapes and hybridisation	3	1	Steric number from bonding pairs plus lone pairs, lone pair placement in a trigonal bipyramid, bond angles set by both lone pair count and substituent electronegativity, and hybridisation with sigma and pi counting
Molecular orbital theory and bond order	2	1	The 2p and 2s ordering switch before and after nitrogen, bond order from $(N_b-N_a)/2$, magnetism from unpaired electrons, and isoelectronic species treated by electron count
Dipole moment and polarity	1	1	Vector addition of bond dipoles and why symmetric molecules cancel, ranking a substituted series, and percentage ionic character from $\mu$ and bond length
Ionic bonding and Fajans' rules	1	1	Born-Haber cycles solved for a missing term, lattice enthalpy trends with charge and size, and Fajans' rules predicting covalent character in an ionic compound
Hydrogen bonding and metallic bonding	1	1	Boiling point anomalies from hydrogen bonding, intramolecular against intermolecular cases, and the electron sea explaining conductivity and malleability

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Read what the question actually asks for — shape, polarity, bond order, magnetism — and pick the cheapest model that supplies it. Building a molecular orbital diagram for a shape question wastes minutes you do not have.
Count the steric number before anything else in a structure question. It gives the electron geometry, the hybridisation and the starting bond angle in one step, and lone pairs count towards it.
For polarity, draw the shape and look for symmetry rather than comparing electronegativities. Symmetric arrangements of identical outer atoms cancel, and any lone pair on the central atom usually breaks that.
Check whether a diatomic species has more or fewer than 14 electrons before drawing a molecular orbital diagram, since the ordering switches. Then remember that removing an antibonding electron raises the bond order.
In hydrogen bonding questions, ask whether the bond is intramolecular or intermolecular. Only the intermolecular kind raises boiling point; an internal bond lowers it by using up both donor and acceptor.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Drug design turns on hydrogen bonding

Drug design turns on hydrogen bonding, since a molecule binds its target protein through a specific pattern of donors and acceptors, and moving one hydroxyl group can destroy the fit entirely

Liquid oxygen's paramagnetism is used to measure oxygen c…

Liquid oxygen's paramagnetism is used to measure oxygen concentration in medical gas analysers, exploiting the one property no other atmospheric gas shares

Kevlar owes its strength to sheets of polymer chains held…

Kevlar owes its strength to sheets of polymer chains held by dense hydrogen bonding between amide groups, which is why a bond worth only 20 kJ per mole can stop a bullet when there are enough of them

Where else this topic is tested

Prepare once, score in every exam that asks it.

JEE Main

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NEET UG

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CBSE Class 11 Chemistry

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because accuracy is not the only thing that matters in an exam or in practice. Building a molecular orbital diagram for a molecule with five different atoms is slow and, for most questions, tells you nothing you needed. VSEPR gives the shape of XeF $_{4}$ in about ten seconds; molecular orbital theory would take a page and still not give it more directly. The right habit is to ask what the question wants — shape, polarity, sigma count, magnetism — and pick the cheapest model that supplies it. Molecular orbital theory earns its cost only for bond order, magnetism, and species like He $_{2}^{+}$ that the other models cannot describe at all.

No. Hybridisation is a mathematical device, not a physical process — there is no moment at which a carbon atom's orbitals mix. What is real is the observed geometry: methane genuinely has four identical bonds at 109.5 degrees, and pure s and p orbitals cannot describe that. Hybridisation is the bookkeeping that makes valence bond theory reproduce the observed shape. Treat it as a language for describing geometry, and never as a cause of it — the geometry comes first, the label second.

Draw the shape first, then look for symmetry. If every position around the central atom is occupied by an identical atom and there are no lone pairs, the bond dipoles cancel and the molecule is non-polar — that covers CO $_{2}$ , BF $_{3}$ , CCl $_{4}$ , PCl $_{5}$ and SF $_{6}$ . Any lone pair on the central atom, or any mixture of different outer atoms, breaks the symmetry and usually makes it polar. The two exceptions worth knowing are XeF $_{2}$ and XeF $_{4}$ , where the lone pairs themselves sit symmetrically and cancel, leaving both non-polar.

Because of s-p mixing. The $σ 2 s$ and $σ 2 p$ orbitals have the same symmetry, so they interact and push each other apart — the $σ 2 p$ being pushed up. That interaction is strong when the 2s and 2p levels are close in energy, as they are in boron, carbon and nitrogen, and it lifts $σ 2 p$ above the $π 2 p$ pair. Moving right across the period the 2s level drops away from 2p, the mixing becomes negligible, and by oxygen the $σ 2 p$ has settled back below the $π$ orbitals. It is a continuous effect that happens to cross over between N $_{2}$ and O $_{2}$ .

Because there are four regions of electron density round the oxygen, not two. Two are bonding pairs and two are lone pairs, and all four repel one another, so they spread into a roughly tetrahedral arrangement. The lone pairs are invisible in a structural formula but entirely real in deciding geometry, and putting the hydrogens opposite each other would force the two lone pairs together at 180 degrees — the worst possible arrangement. The measured angle of 104.5 degrees is the tetrahedral 109.5 compressed by the two lone pairs, which repel more strongly than bonding pairs do.

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Sources and How This Chapter Was CheckedSyllabus scope, what was derived rather than quoted, and how every answer here was checked.

Scope follows the NTA JEE Main syllabus (Unit 4, Chemical Bonding and Molecular Structure): the Kossel-Lewis approach, ionic and covalent bonds, the octet rule and its limitations, factors favouring ionic bonds, lattice enthalpy, bond parameters, and Fajans' rules.

It also covers the dipole moment as a vector quantity, VSEPR theory and the shapes of simple molecules, valence bond theory with orbital overlap and hybridisation involving s, p and d orbitals, resonance, and molecular orbital theory with LCAO, bonding and antibonding orbitals, bond order and the electronic configurations of homonuclear diatomic species. Hydrogen bonding and elementary metallic bonding complete the unit.

Results were derived rather than quoted: the Born-Haber sum computed term by term to show the first four steps are endothermic overall; formal charges evaluated atom by atom for both N $_{2}$ O structures and then checked against the measured N–N bond length; the trigonal bipyramidal preference argued by counting 90-degree neighbours; and the isoelectronic series established by counting valence electrons rather than by assertion.

Every illustration was checked. The Born-Haber result was compared against the measured enthalpy of formation of NaCl and agrees to 1 kJ mol⁻¹. The chloromethane dipole ordering was checked against tabulated values, since the non-monotonic result is the point of the question. The O $_{2}$ ion series was verified to leave all three species paramagnetic, which is the distractor the question exists to catch.

The illustrations are teaching problems written for this chapter, not previous-year questions, and are not labelled as such.

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Chemical Bonding and Molecular Structure

By the end of this chapter you'll be able to…

Before you start — revise these

Chemical Bonding and Molecular Structure

1. Why Atoms Bond

2. The Ionic Bond

Lattice enthalpy is the decisive term

Illustration 1

Fajans' rules

Illustration 2

3. Lewis Structures and Formal Charge

Illustration 3

4. Bond Parameters and Polarity

Illustration 4

Illustration 5

5. VSEPR Theory

Illustration 6

Illustration 7

Illustration 8

6. Valence Bond Theory and Hybridisation

Illustration 9

Resonance

7. Molecular Orbital Theory

The energy ordering, and why it switches

Illustration 10

Illustration 11

8. Hydrogen Bonding and Metallic Bonding

Illustration 12

Summary

Key formulas & results

Traps JEE Main sets — and how to dodge them

Exam-pattern practice

Are you exam-ready for Chemical Bonding and Molecular Structure?

5-minute revision

JEE Main question blueprint

Exam-hall strategy

Beyond the exam

Where else this topic is tested

📝 My notes

Questions aspirants ask

If molecular orbital theory is the most accurate model, why bother with the others?

Is hybridisation something atoms actually do?

How do I decide quickly whether a molecule is polar?

Why does the molecular orbital energy order change between nitrogen and oxygen?

Why is water bent when the oxygen could just as easily hold the hydrogens opposite each other?

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