O₂ is paramagnetic. This is explained by:

Two unpaired electrons in degenerate π* orbitals

The bond angle order NH₃ > PH₃ > AsH₃ is mainly because:

The central atom becomes less electronegative, so bond pairs sit further away and repel less

The number of lone pairs increases down the group

The central atom becomes less electronegative, so bond pairs sit further away and repel less

Hydrogen becomes larger down the group

The molecules change from pyramidal to planar

Why is ortho-nitrophenol more volatile than para-nitrophenol?

It forms intramolecular hydrogen bonds instead of intermolecular ones

It cannot form hydrogen bonds at all

AlCl₃ sublimes readily while NaCl melts at 801 °C. This is explained by:

Fajans' rules — the small, highly charged Al³⁺ polarises chloride, giving covalent character

Aluminium being a heavier atom than sodium

Chloride being larger in AlCl₃

NaCl having a lower lattice enthalpy

Chemical Bonding and Molecular Structure

Bonding is the single most productive chapter in the chemistry syllabus, because it is asked directly and it underlies coordination compounds, organic reactivity and almost every "explain the observation" question in inorganic. The chapter also carries the syllabus's most common wrong answer, and it is worth naming now: shape is decided by all electron pairs, but named after only the bonded ones. Water is tetrahedral in its electron geometry and bent in its molecular shape, and a candidate who does not hold those two ideas apart will get every VSEPR question with lone pairs wrong. Learn to count first and name second.

Ionic and Covalent Character

No bond is purely ionic or purely covalent. Fajans' rules predict how much covalent character an "ionic" compound has, and all four reduce to how easily the cation distorts the anion's electron cloud: covalent character rises with a small, highly charged cation, a large, highly charged anion, and a cation with a non-noble-gas (pseudo-inert) configuration such as Cu⁺ or Ag⁺. That is why AlCl₃ is covalent enough to sublime while NaCl is a high-melting solid, and why AgI is much more covalent than AgF.

🎯 Why lattice enthalpy explains melting points better than "ionic bonds are strong"

Lattice enthalpy scales as the product of the ionic charges divided by the sum of the radii, so it answers comparison questions instantly. MgO melts at 2852 °C against NaCl's 801 °C, and the reason is charge: 2 × 2 = 4 for MgO against 1 × 1 = 1, a fourfold advantage that dwarfs any size difference. Within a group the charges are equal so size decides, which is why LiF melts higher than LiI. This single ratio also predicts solubility trends, thermal stability of carbonates, and which of two salts is harder — questions that look unrelated all reduce to charge over size.

A covalent bond is polar when the two atoms differ in electronegativity, and the molecule has a dipole moment only if those bond dipoles do not cancel by symmetry. This is why CO₂ is non-polar despite two strongly polar bonds (linear, so they cancel) while H₂O is strongly polar (bent, so they add). BF₃ is non-polar because trigonal planar symmetry cancels three dipoles, but NF₃ is polar because the lone pair breaks the symmetry — and NF₃ is less polar than NH₃, because in NF₃ the bond dipoles point away from the lone pair and partly oppose it, while in NH₃ they reinforce it.

VSEPR — Shape Versus Geometry

Count the steric number = bonded atoms + lone pairs on the central atom. That number fixes the electron geometry: 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral. Then subtract the lone pairs to name the shape. Repulsion follows the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, which is why each lone pair squeezes the bond angles inward.

Steric number 4 to 6 with lone pairs
Example	Bond pairs + lone pairs	Shape and angle
CH₄	4 + 0	Tetrahedral, 109.5°
NH₃	3 + 1	Pyramidal, 107°
H₂O	2 + 2	Bent, 104.5°
PCl₅	5 + 0	Trigonal bipyramidal, 90° and 120°
SF₄	4 + 1	See-saw — the lone pair takes an equatorial site
XeF₄	4 + 2	Square planar — the two lone pairs go opposite each other

🧠 In a trigonal bipyramid, lone pairs always take equatorial positions

An axial position has three neighbours at 90°, while an equatorial position has only two, so a lone pair — which repels most strongly — always chooses equatorial to minimise 90° contacts. This one rule generates all three shapes of the family without further thought: SF₄ (4 + 1) becomes see-saw, ClF₃ (3 + 2) becomes T-shaped, and XeF₂ (2 + 3) becomes linear, because three equatorial lone pairs leave the two bonds pointing straight up and down. Candidates who put the lone pair axially get bent shapes for all three and lose every mark in the set.

🎯 Why the bond angle falls from NH₃ to PH₃ to AsH₃

The angles are 107°, 93.6° and 91.8°, dropping much faster than lone-pair repulsion alone could explain. The reason is electronegativity: as the central atom becomes less electronegative down the group, the bonding electron pairs are held further from it and closer to hydrogen, so they repel each other less and the angle collapses towards 90° — that is, towards nearly pure p orbitals with almost no s contribution. The same reasoning explains H₂O (104.5°) > H₂S (92°) and, in the other direction, why replacing H with the more electronegative F narrows the angle further: NF₃ is 102° against NH₃'s 107°, because fluorine pulls the bond pairs away from nitrogen.

Hybridisation and Molecular Orbital Theory

Hybridisation mixes atomic orbitals to give equivalent bonding orbitals: sp linear, sp² trigonal planar, sp³ tetrahedral, sp³d trigonal bipyramidal, sp³d² octahedral. The steric number gives the hybridisation directly — 2 is sp, 3 is sp², 4 is sp³, and so on — so the two ideas are the same count seen from different angles. A useful consequence: s character raises electronegativity and shortens bonds, which is why an sp carbon in ethyne is more electronegative than an sp³ carbon in ethane, and why ethyne is acidic enough to lose a proton to sodamide while ethane is not.

Molecular orbital theory succeeds precisely where VSEPR and hybridisation fail — on magnetic properties and on bond order in odd-electron species. Atomic orbitals combine into bonding and antibonding molecular orbitals, and bond order = (N_bonding − N_antibonding)/2. A species is paramagnetic if any MO holds an unpaired electron.

📖 Liquid oxygen sticks to a magnet, and only MO theory predicted it

Pour liquid oxygen between the poles of a strong magnet and it clings there — a striking demonstration that O₂ is paramagnetic. The Lewis structure O=O shows every electron neatly paired and therefore predicts the opposite. MO theory gets it right: the last two electrons of O₂ go into two degenerate π* antibonding orbitals and, by Hund's rule, occupy them singly with parallel spins. So O₂ has two unpaired electrons, a bond order of 2, and genuine paramagnetism. This is the standard illustration of why MO theory is needed at all, and questions on it usually ask you to rank O₂, O₂⁺, O₂⁻ and O₂²⁻ — bond orders 2, 2.5, 1.5 and 1, so O₂⁺ has the shortest and strongest bond while the peroxide ion has the longest.

Bond order and magnetism for the diatomics the paper names
Species	Bond order	Magnetic behaviour
N₂	3 — the strongest of the set	Diamagnetic
O₂	2	Paramagnetic, two unpaired electrons
O₂⁺ (dioxygenyl)	2.5 — shorter than O₂	Paramagnetic, one unpaired
O₂²⁻ (peroxide)	1 — the longest bond	Diamagnetic
Be₂	0 — does not exist	Not applicable
NO	2.5, odd electron count	Paramagnetic

ℹ️ Hydrogen bonding is why the anomalies of water exist

A hydrogen bond forms when H is attached to N, O or F and approaches another such atom's lone pair. It is only about 5–10 % of a covalent bond in strength, yet it accounts for a startling list: water boils 160 degrees higher than H₂S, ice floats because its open hydrogen-bonded lattice is less dense than liquid water, HF is a weaker acid than HCl because H–F is stabilised by association, and ortho-nitrophenol is more volatile than the para isomer because it hydrogen bonds internally rather than to its neighbours. That last contrast — intramolecular versus intermolecular — is a recurring exam point, since internal bonding lowers boiling point while external bonding raises it.

Key takeaways

Fajans: covalent character rises with a small highly charged cation and a large highly charged anion.
Lattice enthalpy ∝ charge product / sum of radii — that one ratio settles melting-point comparisons.
A molecule is polar only if the bond dipoles fail to cancel: CO₂ and BF₃ are non-polar, H₂O and NF₃ are polar.
Steric number fixes the geometry; subtract lone pairs to get the shape. Repulsion: lp–lp > lp–bp > bp–bp.
In a trigonal bipyramid lone pairs go equatorial, giving see-saw, T-shaped and linear for 1, 2 and 3 lone pairs.
Bond angle falls down a group (NH₃ > PH₃ > AsH₃) as the central atom becomes less electronegative.
Bond order = (bonding − antibonding)/2. O₂ is paramagnetic with two unpaired electrons — only MO theory predicts it.
Intermolecular hydrogen bonding raises boiling point; intramolecular hydrogen bonding lowers it.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The shape of XeF₂ is:
1. Bent
2. Linear
3. T-shaped
4. Trigonal planar
Show answer
Answer: B — Linear
Xenon has two bond pairs and three lone pairs, giving a steric number of 5 and trigonal bipyramidal electron geometry. All three lone pairs occupy equatorial positions, since equatorial sites have only two neighbours at 90° rather than three, leaving the two fluorines axial and the molecule linear. The bent answer is what you get by placing a lone pair axially, which is exactly the error the rule is there to prevent.
Which molecule has a zero dipole moment?
1. H₂O
2. NH₃
3. BF₃
4. NF₃
Show answer
Answer: C — BF₃
BF₃ is trigonal planar with three identical bonds at 120°, so the three bond dipoles cancel exactly by symmetry and the net moment is zero, despite each B–F bond being strongly polar. The other three all have a lone pair on the central atom that destroys the symmetry, so their dipoles add to a non-zero resultant — which is why symmetry, not bond polarity, is the thing to check first.
O₂ is paramagnetic. This is explained by:
1. Its Lewis structure O=O
2. Two unpaired electrons in degenerate π* orbitals
3. sp hybridisation of oxygen
4. VSEPR theory
Show answer
Answer: B — Two unpaired electrons in degenerate π* orbitals
The last two electrons of O₂ enter two degenerate π* antibonding orbitals and by Hund's rule occupy them singly with parallel spins, leaving two unpaired electrons. The Lewis structure shows all electrons paired and so predicts diamagnetism — it is wrong, and this failure is the classic argument for molecular orbital theory. VSEPR addresses shape and says nothing about magnetism.
The species with the shortest bond length among O₂, O₂⁺, O₂⁻ and O₂²⁻ is:
1. O₂
2. O₂⁺
3. O₂⁻
4. O₂²⁻
Show answer
Answer: B — O₂⁺
Bond orders are 2, 2.5, 1.5 and 1 respectively, and a higher bond order means a shorter, stronger bond, so O₂⁺ wins. Removing an electron from O₂ takes it out of an antibonding π* orbital, which is why ionisation strengthens the bond here rather than weakening it — an outcome that is counter-intuitive until you notice which orbital the electron left.
The bond angle order NH₃ > PH₃ > AsH₃ is mainly because:
1. The number of lone pairs increases down the group
2. The central atom becomes less electronegative, so bond pairs sit further away and repel less
3. Hydrogen becomes larger down the group
4. The molecules change from pyramidal to planar
Show answer
Answer: B — The central atom becomes less electronegative, so bond pairs sit further away and repel less
All three have one lone pair and three bond pairs, so the lone-pair count cannot explain the difference. As the central atom becomes less electronegative from N to As, it holds the bonding pairs less tightly and they drift towards hydrogen, so mutual repulsion between them weakens and the angle collapses towards 90° — the value expected for almost pure p orbitals. Hydrogen is the same atom throughout, and all three remain pyramidal.
Why is ortho-nitrophenol more volatile than para-nitrophenol?
1. It has a lower molecular mass
2. It forms intramolecular hydrogen bonds instead of intermolecular ones
3. It is more polar
4. It cannot form hydrogen bonds at all
Show answer
Answer: B — It forms intramolecular hydrogen bonds instead of intermolecular ones
In the ortho isomer the OH and NO₂ groups are adjacent, so the hydrogen bond forms within the same molecule and is unavailable for linking molecules together. The para isomer must hydrogen bond to its neighbours instead, which holds the molecules in the liquid and raises the boiling point. Both isomers have identical molecular masses, so the first option cannot be the discriminator.
AlCl₃ sublimes readily while NaCl melts at 801 °C. This is explained by:
1. Fajans' rules — the small, highly charged Al³⁺ polarises chloride, giving covalent character
2. Aluminium being a heavier atom than sodium
3. Chloride being larger in AlCl₃
4. NaCl having a lower lattice enthalpy
Show answer
Answer: A — Fajans' rules — the small, highly charged Al³⁺ polarises chloride, giving covalent character
Al³⁺ is small and triply charged, so it polarises the chloride electron cloud strongly and pulls charge into the bonding region, giving AlCl₃ substantial covalent character and a molecular rather than ionic lattice. NaCl's singly charged, larger Na⁺ polarises far less, so it remains a genuine ionic lattice with high lattice enthalpy — the opposite of what the last option claims.
The hybridisation of sulphur in SF₄ is:
1. sp³
2. sp³d
3. sp³d²
4. sp²
Show answer
Answer: B — sp³d
Sulphur has four bond pairs and one lone pair, so the steric number is 5 and the hybridisation is sp³d with trigonal bipyramidal electron geometry; the lone pair takes an equatorial site and the shape is see-saw. The answer sp³ comes from counting only the four bonded atoms and forgetting the lone pair, which is the standard slip — lone pairs occupy hybrid orbitals just as bonds do.