Chemical Bonding & Molecular Structure Formula Sheet — JEE Main Chemistry

Every key Chemical Bonding & Molecular Structure formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 15 sub-topics

Valence electrons
Ionic bond
Covalent bond
Bond parameters
Lewis structure
Polar character of covalent bond
Covalent character of ionic bond
Valence bond theory
Resonance
Geometry of covalent molecules
VSEPR theory
Concept of hybridization involving s, p and d orbitals
Shapes of simple molecules
Molecular orbital theory of homonuclear diatomic molecules (qualitative idea only)
Hydrogen bond

Kossel–Lewis, Octet Rule & Ionic Bonding

Why atoms bond — Kossel–Lewis approach

▸Atoms combine to attain the stable $noble-gas (octet) configuration$ of 8 electrons in the outermost shell (2 for H, He duplet).
▸ $Lewis symbol$ : element symbol surrounded by dots = its valence electrons (e.g. $\dot{\ddot{N}} :$ has 5).
▸Bonds form by $transfer$ of electrons (ionic) or $sharing$ of electron pairs (covalent).

Octet rule: In bonding, atoms gain, lose or share electrons so that each attains eight electrons in its valence shell. A shared pair is counted in the octet of

both

atoms.

Formal charge on an atom in a Lewis structure

FC = V - L - \frac{1}{2} B

V

= valence e of the free atom,

L

= lone-pair (non-bonding) e,

B

= bonding (shared) e. The lowest-energy Lewis structure has the smallest formal charges.

Drawing a Lewis structure

▸Total valence e $= \sum$ (group valence) $\pm$ ionic charge.
▸Place least electronegative atom in the centre; join atoms by single bonds.
▸Distribute remaining e as lone pairs to complete octets; form multiple bonds if the central atom is short (e.g. $CO$ , $N_{2}$ , $O_{3}$ ).

★ Remember · Three exceptions to the octet rule

Incomplete octet

: central atom

< 8

e —

LiCl, Be H_{2}, BC l_{3}, AlC l_{3}

Odd-electron

molecules —

NO, N O_{2}

(cannot pair all e).

Expanded octet

: 3rd-period onward use d-orbitals —

P F_{5} (10), S F_{6} (12), H_{2} S O_{4}

Sodium transfers one electron to chlorine forming Na+ and Cl-, which assemble into a 3-D rock-salt lattice stabilised by lattice enthalpy — Ionic bond: complete e-transfer; the solid is held by lattice energy, not by isolated ion pairs.

Ionic (electrovalent) bond: Electrostatic force of attraction between oppositely charged ions formed by complete transfer of electrons, typically from a low-IE metal to a high electron-affinity non-metal (e.g.

N a^{+} C l^{-}

Lattice enthalpy

Δ_{lattice} H : NaCl (s) \to N a^{+} (g) + C l^{-} (g) (+ 788 kJ mol^{- 1})

Energy to separate 1 mol of solid into gaseous ions. The large lattice energy (not the octet) is what makes the ionic solid stable.

Factors favouring ionic-bond formation

▸ $Low ionisation enthalpy$ of the metal (easy cation formation).
▸ $High negative electron-gain enthalpy$ of the non-metal (easy anion formation).
▸ $High lattice enthalpy$ — favoured by small, highly charged ions ( $\propto \frac{q _{+} q _{-}}{r _{+} + r _{-}}$ ).

🎯 Exam · Born–Haber cycle

Lattice enthalpy cannot be measured directly; it is obtained by applying Hess's law around the cycle:

Δ_{f} H = Δ_{s u b} H + \frac{1}{2} Δ_{d i ss} H + Δ_{i} H + Δ_{e g} H + (- Δ_{l a tt i ce} H)

🚫 Examiner Trap · Examiner traps

(1) Octet has THREE exception classes — incomplete (Be

H_{2}

, BC

l_{3}

), odd-electron (NO, N

O_{2}

), expanded (P

F_{5}

, S

F_{6}

). (2) Formal charge

= V - L - \frac{1}{2} B

— the best Lewis structure minimises it. (3) An ionic solid is stabilised by

lattice enthalpy

, not by 'reaching the octet'. (4) In Born–Haber, lattice enthalpy enters with a

negative

sign (formation releases it).

Covalent Bond — Bond Parameters & Resonance

Covalent bond: A bond formed by mutual sharing of one or more electron pairs between atoms (single, double or triple). Each shared pair contributes one bond; unshared pairs are lone pairs.

Bond length

R = r_{A} + r_{B}

(sum of covalent radii)

Bond angle

angle between two adjacent bond orbitals (e.g.

H - O - H = 104. 5^{\circ}

)

Bond enthalpy

energy to break 1 mol of bonds in gas phase (

H - H = 435.8

kJ mo

l^{- 1}

)

Bond order

number of bonds between two atoms

Bond	Order	Length (pm)	Enthalpy (kJ/mol)
$C - C$	1	154	348
$C = C$	2	134	615
$C \equiv C$	3	120	839

Bond-order trends

▸As bond order $↑$ : bond length $↓$ and bond enthalpy (strength) $↑$ .
▸ $Isoelectronic$ species have the same bond order — e.g. $F_{2}$ , $O_{2}^{2 -}$ (B.O. 1); $N_{2}$ , $CO$ , $N O^{+}$ (B.O. 3).
▸For polyatomics use $mean bond enthalpy$ (e.g. $H_{2} O = \frac{502 + 427}{2} = 464.5$ kJ mo $l^{- 1}$ ).

Bar chart showing bond length falling and bond enthalpy rising from single to triple bond, beside the three resonance canonical forms of the carbonate ion — Higher bond order ⇒ shorter, stronger bond. Resonance makes all C–O bonds in $C O_{3}^{2 -}$ equal.

Resonance: When a single Lewis structure cannot describe a molecule, it is represented as a hybrid of several canonical (contributing) structures that differ only in the position of electrons, not nuclei (shown with a

\leftrightarrow

arrow).

Key facts about resonance

▸The $resonance hybrid$ is more stable (lower energy) than any single canonical form — this lowering is the $resonance energy$ .
▸Resonance $averages$ bond lengths: all O–O bonds in $O_{3}$ are 128 pm (between single 148 and double 121).
▸More equivalent, low-formal-charge canonical structures ⇒ greater stability.

⚠️ Watch out · Resonance misconceptions

Canonical forms have

no real existence

; the molecule does

not

flip between them and there is no equilibrium. Only the single hybrid exists — resonance

\neq =

tautomerism.

✎ Example · Bond order in O₃

Why are both O–O bonds in ozone equal at 128 pm?

Two canonical forms: $O = O - O \leftrightarrow O - O = O$ .
Each O–O is a single bond in one form, double in the other.
Average bond order $= \frac{1 + 2}{2} = 1.5$ , so both bonds are identical and intermediate in length.

∴

B.O. = 1.5

; both bonds 128 pm (equal).

🚫 Examiner Trap · Examiner traps

(1) Bond order

↑

\Rightarrow

length

↓

, enthalpy

↑

— all three move together. (2) Resonance forms have

no real existence

; the molecule is ONE hybrid, not flipping between them (resonance

\neq =

tautomerism). (3) Isoelectronic species share bond order (

N_{2}

, CO, NO

^{+}

all B.O. 3). (4) The hybrid is MORE stable than any canonical form (resonance energy).

Bond Polarity — Electronegativity, Dipole Moment & Fajan's Rules

Polar covalent bond: When a covalent bond joins atoms of different electronegativity, the shared pair shifts toward the more electronegative atom, giving partial charges

δ^{+}

and

δ^{-}

(e.g.

H^{δ +} - F^{δ -}

Dipole moment

μ = Q \times r (1 D = 3.336 \times 1 0^{- 30} C m)

Q = magnitude of charge, r = separation.

μ

is a

vector

(points from

δ^{+}

δ^{-}

); net

μ

= vector sum of all bond dipoles.

Net dipole depends on shape

▸ $Symmetric$ molecules cancel: $C O_{2}$ (linear), $B F_{3}$ (trigonal planar), $C H_{4} / CC l_{4}$ (tetrahedral), $Be F_{2}$ — all $μ = 0$ .
▸ $Bent/pyramidal$ molecules don't cancel: $H_{2} O$ ( $1.85$ D), $N H_{3}$ ( $1.47$ D).
▸Greater $μ$ ⇒ more polar ⇒ generally higher boiling point and solubility in polar solvents.

Six molecules showing bond-dipole vectors: HCl polar, CO2 and BF3 cancelling to zero, H2O bent with net dipole, and NH3 versus NF3 where the lone pair adds or opposes — Net dipole = vector sum of bond dipoles; geometry decides whether they cancel.

🎯 Exam · NH₃ vs NF₃ — classic trap

Both are pyramidal, yet

μ_{N H_{3}} (1.47 D) > μ_{N F_{3}} (0.23 D)

. In

N H_{3}

the lone-pair dipole adds to the N–H bond dipoles; in

N F_{3}

it opposes the N–F dipoles, almost cancelling.

Fajan's rules — covalent character of an ionic bond

▸Covalent character $↑$ with $small cation$ and $large anion$ (more polarisation).
▸Covalent character $↑$ with $higher charge$ on cation/anion.
▸Pseudo-noble-gas cations ( $(n - 1) d^{10} n s^{0}$ , e.g. $C u^{+}, A g^{+}$ ) polarise more than noble-gas-type cations of the same size/charge.

Type	Example	$μ$ (D)	Shape
$A B$	HF / HCl	1.78 / 1.07	linear
$A B_{2}$	$H_{2} O$ / $C O_{2}$	1.85 / 0	bent / linear
$A B_{3}$	$N H_{3}$ / $B F_{3}$	1.47 / 0	pyramidal / planar
$A B_{4}$	$CHC l_{3}$ / $CC l_{4}$	1.04 / 0	tetrahedral

🚫 Examiner Trap · Examiner traps

(1) N

H_{3}

(

1.47

>

F_{3}

(

0.23

D): the lone-pair dipole

adds

in N

H_{3}

but

opposes

in N

F_{3}

— a classic trap. (2) Symmetric shapes (C

O_{2}

, B

F_{3}

, CC

l_{4}

) give

μ = 0

even with polar bonds. (3)

μ

is a

vector

— add bond dipoles head-to-tail. (4) Fajan: small cation

+

large anion

+

high charge

\Rightarrow

more covalent character.

VSEPR Theory & Molecular Shapes

VSEPR postulates

▸Shape is decided by the number of $valence-shell electron pairs$ (bonding + lone) around the central atom.
▸Electron pairs orient to be as far apart as possible (minimum repulsion).
▸A multiple bond is treated as a $single$ super-pair for geometry.
▸Lone pairs occupy more space than bonding pairs, so they $compress$ bond angles.

★ Remember · Repulsion order

lp-lp > lp-bp > bp-bp

. Each lone pair reduces the ideal bond angle (e.g.

109. 5^{\circ} \to 10 7^{\circ}

N H_{3} \to 104. 5^{\circ}

H_{2} O

e-pairs	Geometry	Angle	Example
2	Linear	$18 0^{\circ}$	$BeC l_{2}$
3	Trigonal planar	$12 0^{\circ}$	$B F_{3}$
4	Tetrahedral	$109. 5^{\circ}$	$C H_{4}$
5	Trig. bipyramidal	$12 0^{\circ} /9 0^{\circ}$	$PC l_{5}$
6	Octahedral	$9 0^{\circ}$	$S F_{6}$

Gallery of VSEPR shapes: linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, plus bent, pyramidal and see-saw shapes arising from lone pairs — Geometry from total electron pairs; lone pairs bend the shape away from the ideal.

Type	bp	lp	Shape	Example
$A B_{2} E$	2	1	Bent	$S O_{2}$
$A B_{3} E$	3	1	Pyramidal	$N H_{3}$
$A B_{2} E_{2}$	2	2	Bent	$H_{2} O$
$A B_{4} E$	4	1	See-saw	$S F_{4}$
$A B_{3} E_{2}$	3	2	T-shape	$Cl F_{3}$
$A B_{4} E_{2}$	4	2	Square planar	$Xe F_{4}$

🎯 Exam · Bond-angle reasoning

Same hybridisation, decreasing angle with more lone pairs:

C H_{4} (109. 5^{\circ}) > N H_{3} (10 7^{\circ}) > H_{2} O (104. 5^{\circ})

. In trigonal bipyramidal, lone pairs always take

equatorial

positions (fewer

9 0^{\circ}

repulsions).

🚫 Examiner Trap · Examiner traps

(1) Repulsion order lp–lp

>

lp–bp

>

bp–bp

\Rightarrow

lone pairs

compress

angles: C

H_{4} (109. 5^{\circ}

)

>

H_{3} (10 7^{\circ}

)

>

H_{2} O (104. 5^{\circ}

). (2) A multiple bond counts as

one

electron pair for shape. (3) In trigonal bipyramidal, lone pairs always go

equatorial

. (4) Count bonding

+

lone pairs on the CENTRAL atom only.

Valence Bond Theory, Hybridisation & σ/π Bonds

Valence Bond Theory (VBT)

▸A covalent bond forms by $overlap$ of half-filled atomic orbitals with electrons of opposite spin.
▸Greater overlap ⇒ stronger bond; the bond forms at the internuclear distance of $minimum potential energy$ (e.g. $H_{2}$ at 74 pm).
▸Explains bond directionality and shapes, which Lewis theory cannot.

σ (sigma) vs π (pi) bond:

σ

(sigma): end-to-end (head-on) overlap along the internuclear axis — strong, present in every bond.

π

(pi): sideways overlap of parallel p-orbitals above and below the axis — weaker, only in addition to a

σ

bond (in double/triple bonds).

sp, sp2 and sp3 hybrid orbital sets giving linear, trigonal and tetrahedral geometries, with an inset comparing sigma head-on overlap and pi sideways overlap — Hybridisation mixes s + p (+ d) into equivalent orbitals that fix the geometry.

Counting σ and π bonds

single

= 1 σ

; double

= 1 σ + 1 π

; triple

= 1 σ + 2 π

C_{2} H_{4}

: 5σ + 1π;

C_{2} H_{2}

: 3σ + 2π.

Hybrid.	Geometry	Angle	Example
$s p$	Linear	$18 0^{\circ}$	$BeC l_{2}, C_{2} H_{2}$
$s p^{2}$	Trigonal planar	$12 0^{\circ}$	$B F_{3}, C_{2} H_{4}$
$s p^{3}$	Tetrahedral	$109. 5^{\circ}$	$C H_{4}, N H_{3}$
$s p^{3} d$	Trig. bipyramidal	$9 0^{\circ}, 12 0^{\circ}$	$PC l_{5}$
$s p^{3} d^{2}$	Octahedral	$9 0^{\circ}$	$S F_{6}$
$dsp^{2}$	Square planar	$9 0^{\circ}$	$[Ni (CN)_{4}]^{2 -}$

🎯 Exam · Finding hybridisation fast

Steric number

= \frac{1}{2} (V + M - C + A)

where V = central-atom valence e, M = monovalent atoms,

C / A

= cation/anion charge. SN

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}

✎ Example · Hybridisation of C in ethene (C₂H₄)

Describe the bonding in

C_{2} H_{4}

Each C is bonded to 3 atoms (2 H + 1 C), no lone pair ⇒ SN $= 3$ .
So each C is $s p^{2}$ (trigonal planar, $12 0^{\circ}$ ).
The unhybridised $2 p$ orbitals overlap sideways to give one π bond.

∴

s p^{2}

carbon; C=C is

1 σ (s p^{2} - s p^{2}) + 1 π

, length 134 pm.

🚫 Examiner Trap · Examiner traps

(1) Every bond has

one

σ

;

π

bonds come only

in addition

(double

= 1 σ + 1 π

, triple

= 1 σ + 2 π

). (2)

σ

(head-on)

>

π

(sideways) in strength. (3) Get hybridisation from the steric number (SN

4 \to s p^{3}

), counting lone pairs. (4) Hybridisation decides shape;

π

bonds use

unhybridised

p-orbitals.

Molecular Orbital Theory

MOT — salient features (Hund & Mulliken)

▸Atomic orbitals of comparable energy & symmetry combine (LCAO) to give molecular orbitals spread over the whole molecule.
▸n atomic orbitals ⇒ n MOs: a low-energy $bonding$ MO ( $ψ_{A} + ψ_{B}$ ) and a high-energy $antibonding$ MO* ( $ψ_{A} - ψ_{B}$ ).
▸MOs fill by Aufbau, Pauli and Hund's rules — just like atomic orbitals.

Bond order (MOT)

B.O. = \frac{1}{2} (N_{b} - N_{a})

N_{b}, N_{a}

= electrons in bonding / antibonding MOs. B.O.

> 0

⇒ stable;

= 0

⇒ does not exist. B.O.

\propto

stability & bond strength,

\propto 1/

length.

Molecular orbital energy-level diagram for O2 showing sigma2s, sigma*2s, sigma2pz, pi2p, pi*2p with two unpaired electrons, giving bond order 2 and paramagnetism — $O_{2}$ : two unpaired e in $π^{*}$ ⇒ paramagnetic, B.O. $= 2$ — MOT's great success.

MO energy order (valence)

▸ $Up to N_{2}$ ( $L i_{2}, B_{2}, C_{2}, N_{2}$ , with $2 s - 2 p$ mixing): $σ 2 s < σ^{*} 2 s < π 2 p_{x} = π 2 p_{y} < σ 2 p_{z} < π^{*} 2 p < σ^{*} 2 p_{z}$ .
▸ $O_{2}$ and $F_{2}$ (no mixing): $σ 2 p_{z}$ drops $below$ the $π 2 p$ pair.
▸Magnetism: all paired ⇒ $diamagnetic$ ; unpaired e ⇒ $paramagnetic$ .

Molecule	B.O.	Magnetism
$H_{2}$	1	diamagnetic
$H e_{2}$	0	does not exist
$L i_{2}$	1	diamagnetic
$N_{2}$	3	diamagnetic
$O_{2}$	2	paramagnetic
$F_{2}$	1	diamagnetic

🎯 Exam · Why He₂ doesn't exist

H e_{2}

(σ 1 s)^{2} (σ^{*} 1 s)^{2}

, so

N_{b} = N_{a} = 2

and

B.O. = \frac{1}{2} (2 - 2) = 0

. No net bonding ⇒ no molecule. Likewise

B e_{2}

. For ions, removing an antibonding e

raises

B.O. (e.g.

O_{2}^{+}

B.O.

= 2.5 > O_{2}

🚫 Examiner Trap · Examiner traps

(1) B.O.

= \frac{1}{2} (N_{b} - N_{a})

; B.O.

= 0

\Rightarrow

molecule doesn't exist (H

e_{2}

, B

e_{2}

). (2)

O_{2}

paramagnetic

(2 unpaired in

π^{*}

) — MOT's triumph, VBT misses it. (3) Order changes at

O_{2}

: up to

N_{2}

the

σ 2 p_{z}

is ABOVE

π 2 p

; for

O_{2}

F_{2}

it drops below. (4) Removing an

antibonding

e raises B.O. (O

_{2}^{+}

>

O_{2}

Hydrogen Bonding & Intermolecular Forces

Hydrogen bond: The attractive force binding a covalently-bonded H atom (carrying

δ^{+}

) to a highly electronegative atom —

F, O or N

— of another or the same molecule. Written

X - H \dots Y

; weaker than a covalent bond but stronger than other van der Waals forces.

Conditions & strength

▸H must be bonded to a small, highly electronegative atom (F $>$ O $>$ N).
▸Strength order: $HF > H_{2} O$ (per bond) $> N H_{3}$ ; energy $\sim 10 - 40$ kJ mo $l^{- 1}$ .
▸Strongest in the solid state, weakest in the gas — it shapes structure and physical properties.

Left: water molecules linked by intermolecular hydrogen bonds; right: o-nitrophenol with an intramolecular hydrogen bond forming a six-membered ring — Intermolecular H-bonds raise boiling point; intramolecular H-bonds (o-nitrophenol) lower it.

Two types & their effects

▸ $Intermolecular$ (between molecules): associates molecules ⇒ high b.p., high viscosity, e.g. $H_{2} O$ , $HF$ , alcohols.
▸ $Intramolecular$ (within one molecule, forming a ring): e.g. o-nitrophenol — ties up the H so it can't bond between molecules ⇒ $lower$ b.p. and steam-volatility.

Force	Acts between	Strength
London (dispersion)	all molecules	weakest
Dipole–dipole	polar molecules	moderate
H-bond	H with F/O/N	strongest vdW

🎯 Exam · Anomalies explained by H-bonding

H_{2} O

has a far higher b.p. than

H_{2} S

;

HF > HCl

in b.p.;

ice is less dense than water

(open H-bonded cage). o-nitrophenol (intramolecular) is more volatile than p-nitrophenol (intermolecular).

✎ Example · Ordering boiling points

Arrange

H_{2} O, H_{2} S, H_{2} Se

by boiling point.

$H_{2} S, H_{2} Se$ have only weak dipole/London forces, rising with molar mass.
$H_{2} O$ has strong intermolecular hydrogen bonding (O is small & highly EN).
H-bonding outweighs the mass trend, lifting water far above the others.

∴

H_{2} O ≫ H_{2} Se > H_{2} S

🚫 Examiner Trap · Examiner traps

(1) H-bonding needs H bonded to

F, O or N

only — not Cl, S. (2)

Intermolecular

H-bonds RAISE boiling point;

intramolecular

(o-nitrophenol) LOWER it. (3) Ice is less dense than water (open H-bonded cage). (4) H-bond (

\sim

10–40 kJ/mol) is the strongest van der Waals force but far weaker than a covalent bond.

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