Important Concepts Of Chemical Bonding

Q: Why does BeCl2 have a covalent character despite being formed by a metal and a non-metal?

According to Fajans' rules, the Be2+ cation is very small and highly charged, giving it strong polarising power. It distorts the Cl- electron cloud significantly, drawing electron density between the ions. This creates substantial covalent character.

Q: How do you decide the central atom in a Lewis structure?

The least electronegative atom (other than H) is usually placed at the centre. Hydrogen is never central because it can form only one bond.

Q: Why is the C-C bond in benzene shorter than a single bond but longer than a double bond?

Benzene is a resonance hybrid of two Kekule structures. The six pi-electrons are delocalised over all six carbons, giving each C-C bond an intermediate character with bond order 1.5, resulting in a bond length of 139 pm.

Q: Why does water have a higher dipole moment than hydrogen sulphide?

Oxygen is much more electronegative than sulphur, so the O-H bond is more polar than the S-H bond. The vector sum in H2O (1.85 D) is significantly greater than in H2S (0.95 D).

Q: Can bond order be a fraction?

Yes. In molecules with resonance or unpaired electrons in molecular orbitals, bond order can be fractional. Ozone has bond order 1.5, carbonate has 1.33, and superoxide O2- has 1.5.

Q: What decides whether a covalent bond is polar or non-polar?

The electronegativity difference between the bonded atoms. Identical atoms give non-polar bonds; increasing electronegativity difference makes the bond more polar, and very large differences give predominantly ionic bonds. Molecular geometry also matters for net molecular polarity.

ChemistryChemical Bonding And Molecular StructureFor NEET aspirants

A chemical bond is the attractive force that holds atoms, ions, or molecules together in a chemical species. Atoms bond to attain a lower-energy, more stable electron configuration - usually the octet of a noble gas. The three main bond types for JEE and NEET are the ionic (electrovalent) bond, formed by electron transfer between a metal and a non-metal; the covalent bond, formed by sharing electron pairs between non-metals; and the coordinate (dative) bond, a covalent bond where both shared electrons come from the same atom. This concept covers the Kössel–Lewis approach, Lewis structures, formal charge, ionic bond energetics (Born–Haber cycle, lattice enthalpy), Fajans' rules, resonance, dipole moment, percentage ionic character, and bond parameters (length, energy, angle).

Key Formulas - Quick Reference

Formal charge $= V - L - \frac{1}{2} B$ , where $V$ = valence electrons in free atom, $L$ = lone pair electrons, $B$ = bonding electrons
Bond order $= \frac{1}{2} (N_{b} - N_{a})$ (MOT) or number of shared electron pairs (Lewis)
Born–Haber cycle: $Δ H_{f} = S + \frac{1}{2} D + IE + EA + U$
Dipole moment: $μ = q \times d$ ; $1 D = 3.33564 \times 1 0^{- 30} C m$
Resultant of two dipoles: $μ_{R} = μ_{1}^{2} + μ_{2}^{2} + 2 μ_{1} μ_{2} cos θ$
Percent ionic character $= \frac{μ _{observed}}{μ _{100% ionic}} \times 100$
Bond length in covalent AB $= r_{A} + r_{B}$ ; in ionic $= r_{c}^{+} + r_{a}^{-}$

1. Kössel–Lewis Approach to Chemical Bonding

In 1916, Kössel and Lewis independently gave the first satisfactory electronic explanation of chemical bonding, based on the inertness of noble gases. Their combined view still forms the foundation of how we draw and interpret molecules today.

Lewis pictured the atom as a positively charged kernel (nucleus + inner electrons) surrounded by an outer shell that can accommodate at most eight electrons. He proposed that atoms achieve a stable octet by linking through chemical bonds - either by transferring electrons (ionic) or by sharing them (covalent).

Kössel drew attention to three key facts:

Highly electronegative halogens and highly electropositive alkali metals are separated by noble gases in the periodic table.
A halogen atom forms a negative ion by gaining an electron; an alkali metal forms a positive ion by losing one.
The ions so formed attain stable noble-gas configurations ( $n s^{2} n p^{6}$ ) and are held together by electrostatic attraction.

Lewis symbols and group valence

A Lewis symbol shows the valence electrons of an atom as dots around its chemical symbol. For example, sodium (1 valence electron) is written as $Na \cdot$ , and chlorine (7 valence electrons) has three lone pairs plus one unpaired dot. The group valence equals the number of dots, or $8 - (number of dots)$ , whichever is smaller.

Figure 1: Lewis dot symbols for second-period elements Li to Ne, showing valence electrons as dots around the atomic symbol.

The octet rule

Octet rule: Atoms combine by transferring or sharing valence electrons so that each attains an octet (eight electrons) in its outermost shell, mimicking the nearest noble gas configuration. Hydrogen and lithium aim for a duplet (two electrons, like helium).

2. Ionic (Electrovalent) Bond

An ionic bond forms when one atom transfers one or more electrons to another, producing oppositely charged ions held together by electrostatic (coulombic) attraction. The atom losing electrons must have a low ionization energy (typically an alkali or alkaline earth metal), and the atom gaining electrons must have a highly negative electron gain enthalpy (typically a halogen or chalcogen).

For NaCl, sodium's single 3s electron transfers to chlorine's half-filled 3p orbital:

$Na + Cl ⟶ Na^{+} + Cl^{-} ⟶ NaCl$

Figure 2: Sodium chloride crystal lattice: alternating

Na^{+}

(small orange) and

Cl^{-}

(large brown) ions in a face-centred cubic array. Each ion is surrounded by six neighbours of opposite charge.

Criteria for ionic bond formation

Low ionization energy of the metal (favours cation formation).
High negative electron gain enthalpy of the non-metal (favours anion formation).
High lattice enthalpy of the resulting crystal (drives the overall process to be exothermic).

Energy changes during ionic bond formation

The formation of an ionic solid proceeds in three conceptual steps:

Formation of gaseous cation: $A (g) + IE ⟶ A^{+} (g) + e^{-}$ (energy absorbed = ionization energy)
Formation of gaseous anion: $X (g) + e^{-} ⟶ X^{-} (g) + EA$ (energy released = electron gain enthalpy)
Packing of ions into the crystal lattice: $A^{+} (g) + X^{-} (g) ⟶ AX (s) + U$ (energy released = lattice enthalpy)

Lattice enthalpy is the energy released when one mole of an ionic solid is formed by the close packing of gaseous ions. Equivalently, it is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. For NaCl,

U = - 788 kJ mo l^{- 1}

Born–Haber cycle

Lattice enthalpy cannot be measured directly. The Born–Haber cycle uses Hess's law to calculate it from measurable quantities - sublimation enthalpy ( $S$ ), bond dissociation enthalpy ( $D$ ), ionization enthalpy (IE), electron gain enthalpy (EA), and the enthalpy of formation ( $Δ H_{f}$ ).

Figure 3: Born–Haber cycle for NaCl. Summing all steps around the cycle gives

Δ H_{f} = S + \frac{1}{2} D + IE + EA + U

Solved Example 1

Calculate the lattice enthalpy of

MgB r_{2}

given:

Δ H_{f} = - 524 kJ mo l^{- 1}

I E_{1} + I E_{2} = 2148 kJ mo l^{- 1}

, sublimation energy of Mg

= 148 kJ mo l^{- 1}

, vaporisation energy of

B r_{2} (l) = 31 kJ mo l^{- 1}

, dissociation energy of

B r_{2} (g) = 193 kJ mo l^{- 1}

, and electron gain enthalpy of Br

= - 331 kJ mo l^{- 1}

Solution:

From the Born–Haber cycle:

$Δ H_{f} = S + IE + Δ H_{vap} + D + 2 EA + U$

Solving for $U$ :

$U = Δ H_{f} - (S + IE + Δ H_{vap} + D + 2 EA)$

$U = - 524 - [148 + 2148 + 31 + 193 + 2 \times (- 331)]$

$U = - 524 - 1897 = - 2421 kJ mo l^{- 1}$

Characteristics of ionic compounds

Crystalline solids: Non-directional coulombic forces give three-dimensional lattices with high coordination numbers. Ionic compounds do not show stereoisomerism.
High melting and boiling points due to strong electrostatic attraction throughout the lattice.
Electrical conductivity: Do not conduct in the solid state (ions locked in place) but conduct in molten or aqueous form (ions become mobile).
Solubility: Soluble in polar solvents like water (high dielectric constant weakens ionic attraction; solvation energy compensates for lattice energy) and insoluble in non-polar solvents.
Ionic reactions are typically fast because they proceed through free ions in solution.

3. Covalent Bond

A covalent bond forms when two atoms share one or more pairs of electrons, each atom contributing at least one electron to the shared pair. This satisfies the octet (or duplet for H) of both atoms simultaneously. Covalent bonding is favoured when the two atoms have similar (usually high) electronegativity - typically two non-metals.

Depending on the number of shared pairs:

Single bond - one shared pair (e.g., $H_{2}$ , $C l_{2}$ )
Double bond - two shared pairs (e.g., $O_{2}$ , $C O_{2}$ , $C_{2} H_{4}$ )
Triple bond - three shared pairs (e.g., $N_{2}$ , $C_{2} H_{2}$ )

Sigma ( $σ$ ) and pi ( $π$ ) bonds

A $σ$ -bond is formed by head-on (axial) overlap of atomic orbitals along the internuclear axis. It has cylindrical symmetry about that axis and has the maximum possible electron density between the two nuclei. Any first bond between two atoms is always a $σ$ -bond.

A $π$ -bond is formed by sideways (lateral) overlap of parallel p-orbitals perpendicular to the internuclear axis. The electron density lies above and below the axis in two lobes, with a nodal plane containing the axis. $π$ -bonds occur only in multiple bonds - a double bond has one $σ$ + one $π$ , a triple bond has one $σ$ + two $π$ .

Figure 4: Formation of a

σ

-bond by axial overlap of two

s

-orbitals. The spherical clouds merge head-on along the internuclear axis, concentrating electron density between the nuclei.

Figure 5: Formation of a

σ

-bond by axial overlap of two

p

-orbitals. Lobes of the same phase overlap end-on, giving a cylindrically symmetric bond along the internuclear axis.

Figure 6:

π

-bond formation by lateral overlap of two parallel

p

-orbitals. The upper lobes overlap above the internuclear axis and the lower lobes overlap below it, giving two regions of electron density and a nodal plane containing the axis. A

π

-bond is weaker than a

σ

-bond because lateral overlap is less extensive than axial overlap.

$σ$ -bonds are stronger than $π$ -bonds because axial overlap is more extensive than lateral overlap. This is why a C≡C triple bond (1 $σ$ + 2 $π$ ≈ 837 kJ/mol) is not three times as strong as a C–C single bond (1 $σ$ ≈ 348 kJ/mol).

Coordinate (dative) bond

A coordinate bond is a covalent bond in which both shared electrons come from the same atom. The atom donating the pair is the donor (Lewis base), and the atom receiving is the acceptor (Lewis acid). Once formed, a coordinate bond is indistinguishable from a normal covalent bond. It is shown by an arrow from donor to acceptor: $D \to A$ .

Common examples: $N H_{4}^{+}$ ( $N H_{3}$ donates a lone pair to $H^{+}$ ), $S O_{3}$ , $H_{3} O^{+}$ , $PC l_{6}^{-}$ , $Sb F_{6}^{-}$ .

Figure 7: Formation of

N H_{4}^{+}

from

N H_{3}

and

H^{+}

. Nitrogen's lone pair forms a coordinate bond with the electron-deficient proton.

4. Lewis Structures and Formal Charge

A Lewis structure shows how bonded pairs and lone pairs of electrons are distributed among the atoms of a molecule or polyatomic ion. Follow this procedure:

Count total valence electrons of all atoms. Add one for each negative charge; subtract one for each positive charge.
Choose the least electronegative atom (except H) as the central atom.
Place a single bond (shared pair) between each pair of bonded atoms.
Distribute remaining electrons as lone pairs to complete octets on the terminal atoms first, then the central atom.
If the central atom lacks an octet, convert lone pairs on terminal atoms into double or triple bonds.

Figure 8: Lewis dot structure of methane (

C H_{4}

). Each pair of dots between C and an H represents one shared covalent bond. Carbon achieves an octet through four such bonds; each hydrogen achieves its duplet.

Formal charge

The formal charge of an atom in a Lewis structure equals the number of valence electrons in the free atom minus the number of lone-pair electrons and half the bonding electrons on it:

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

Formal charges help identify the most plausible Lewis structure. The best structure usually has the smallest formal charges, and negative formal charges on the most electronegative atoms.

Solved Example 2

Assign formal charges to each oxygen atom in the ozone (

O_{3}

) molecule.

Solution:

Ozone's Lewis structure has a central O double-bonded to one terminal O and single-bonded (with a lone pair) to the other. Label them $O_{1}$ (central), $O_{2}$ (double-bonded terminal), $O_{3}$ (single-bonded terminal).

Central $O_{1}$ : $V = 6$ , $L = 2$ (one lone pair), $B = 6$ (one double + one single); FC $= 6 - 2 - 3 = + 1$
Terminal $O_{2}$ (double bond): $V = 6$ , $L = 4$ , $B = 4$ ; FC $= 6 - 4 - 2 = 0$
Terminal $O_{3}$ (single bond): $V = 6$ , $L = 6$ , $B = 2$ ; FC $= 6 - 6 - 1 = - 1$

Sum of formal charges $= + 1 + 0 - 1 = 0$ , matching the neutral $O_{3}$ molecule.

Limitations of the octet rule

Incomplete octet: Li, Be, B, Al often form compounds with fewer than 8 electrons around the central atom, e.g., $LiCl$ , $Be H_{2}$ , $B F_{3}$ , $AlC l_{3}$ .
Odd-electron molecules: $NO$ , $N O_{2}$ , $Cl O_{2}$ have an odd total electron count, so at least one atom cannot have an octet.
Expanded octet: Elements of the third period and beyond can accommodate more than 8 electrons using vacant d-orbitals, e.g., $P F_{5}$ , $S F_{6}$ , $I F_{7}$ , $H_{2} S O_{4}$ .
Does not account for the shapes of molecules or their relative stabilities.

5. Fajans' Rules - Covalent Character in Ionic Bonds

No ionic bond is 100% ionic; every ionic bond has some covalent character because the cation polarises the anion's electron cloud, distorting it towards itself. Fajans' rules summarise the factors that increase this covalent (polarisation) character.

Small cation size - high charge density means strong polarising power. Down the group, ionic character increases: $BeC l_{2}$ (m.p. 405 °C) is more covalent than $BaC l_{2}$ (m.p. 960 °C).
High cation charge - greater positive charge means greater polarising power. $AlB r_{3}$ ( $A l^{3 +}$ ) is more covalent than $NaBr$ ( $N a^{+}$ ).
Large anion size - big, diffuse electron clouds are easily polarised. $Ca I_{2}$ (m.p. 575 °C) is more covalent than $Ca F_{2}$ (m.p. 1400 °C).
High anion charge - more electrons are more easily distorted. $M g_{3} N_{2}$ (with $N^{3 -}$ ) is more covalent than $Mg F_{2}$ .
Pseudo-noble-gas cation vs noble-gas cation - cations with $(n - 1) d^{10}$ configuration (like $C u^{+}$ , $A g^{+}$ ) polarise anions more strongly than noble-gas-core cations of similar size (like $N a^{+}$ , $K^{+}$ ), because d-electrons shield poorly. NaCl (m.p. 800 °C) is much more ionic than CuCl (m.p. 425 °C) despite similar ionic radii.

Solved Example 3

Explain why

Al F_{3}

is ionic while

AlC l_{3}

is largely covalent.

Solution:

Both compounds have the same cation ( $A l^{3 +}$ , small and highly charged, so strong polarising power). The difference lies in the anion. $F^{-}$ is small and holds its electrons tightly, so it is poorly polarised - Fajans' anion-size factor gives $Al F_{3}$ mostly ionic character. $C l^{-}$ is much larger and easily polarised by $A l^{3 +}$ , so $AlC l_{3}$ acquires significant covalent character (it even sublimes as covalent $A l_{2} C l_{6}$ dimers).

Solved Example 4

The melting point of KCl (776 °C) is higher than that of AgCl (455 °C), although

K^{+}

and

A g^{+}

have almost the same ionic radius. Explain.

Solution:

Cl⁻ is common to both, so the deciding factor is the cation's electronic configuration. $K^{+}$ has an [Ar] noble-gas core, while $A g^{+}$ has a [Kr]4d¹⁰ pseudo-noble-gas core. The 4d electrons in Ag⁺ shield the nucleus poorly, so its effective polarising power is much greater. This gives AgCl significant covalent character, weaker lattice attraction, and a lower melting point.

6. Resonance

When a single Lewis structure cannot adequately describe a molecule (typically because experimental bond lengths and bond orders are between those of single and multiple bonds), the true structure is described as a resonance hybrid of two or more canonical (resonating) structures. The canonical structures differ only in the positions of electrons, never atoms.

Consider ozone, $O_{3}$ . Two Lewis structures place the O=O double bond on either side of the central O. Experiment shows both O–O bonds are identical (128 pm), intermediate between O–O single (148 pm) and O=O double (121 pm). The real molecule is the resonance hybrid:

Figure 9: Resonance in the

O_{3}

molecule. Canonical forms I and II differ only in which side carries the O=O double bond. The real molecule (structure III, the resonance hybrid) is bent (bond angle 117°) with two equivalent O-O bonds of length 128 pm and bond order 1.5, intermediate between a single bond (148 pm) and a double bond (121 pm).

Rules for writing resonating structures

Only electrons (never atoms) are shifted between canonical forms. Nuclei stay in place.
The number of unpaired electrons must be the same in every canonical form.
Negative charge should reside on the more electronegative atom; positive charge on the less electronegative atom.
Like charges should not be placed on adjacent atoms.
The greater the number of equivalent (low-energy) canonical structures, the greater the resonance stabilisation.

Other important examples

Carbonate ion ( $C O_{3}^{2 -}$ ): three equivalent canonical structures; all three C–O bonds are identical with bond order $= 4/3 \approx 1.33$ .
Carboxylate ion (R–COO⁻): two equivalent canonical structures; both C–O bonds are equal.
Benzene ( $C_{6} H_{6}$ ): two Kekulé structures give a hybrid where all six C–C bonds are equivalent (139 pm), intermediate between single (154 pm) and double (134 pm).
Nitrate ion ( $N O_{3}^{-}$ ): three equivalent canonical structures; all three N–O bonds identical.

Resonance energy is the difference between the energy of any single canonical structure and the actual (lower) energy of the resonance hybrid. It provides additional stability to the molecule.

7. Polarity of Bonds and Dipole Moment

When a covalent bond joins two atoms of different electronegativity, the shared electron pair is displaced towards the more electronegative atom, creating partial charges: $δ^{+}$ on the less electronegative atom and $δ^{-}$ on the more electronegative. The bond becomes a polar covalent bond.

The dipole moment

μ

measures bond (or molecular) polarity. It is the product of the magnitude of the charge and the distance separating the positive and negative centres:

$μ = q \times d$

Unit: Debye (D); $1 D = 3.33564 \times 1 0^{- 30} C m$

Figure 10: Definition of the dipole moment

μ = q \times r

: a pair of equal and opposite partial charges separated by distance

r

, drawn as a vector pointing from the positive to the negative end.

Dipole moment is a vector quantity, drawn as an arrow pointing from the positive to the negative end. For a polyatomic molecule, the net dipole moment is the vector sum of all individual bond dipoles.

Resultant of two bond dipoles

For two dipole vectors of magnitudes $μ_{1}$ and $μ_{2}$ inclined at angle $θ$ :

$μ_{R} = μ_{1}^{2} + μ_{2}^{2} + 2 μ_{1} μ_{2} cos θ$

When $θ = 0°$ (same direction), $μ_{R} = μ_{1} + μ_{2}$ (maximum). When $θ = 180°$ (opposite), $μ_{R} = ∣ μ_{1} - μ_{2} ∣$ (minimum, zero if equal).

Symmetry and net dipole moment

Symmetrical molecules with polar bonds can have zero net dipole moment because bond dipoles cancel:

$C O_{2}$ (linear, O=C=O) - two C=O dipoles equal and opposite; $μ = 0$
$B F_{3}$ (trigonal planar) - three B–F dipoles at 120° cancel; $μ = 0$
$CC l_{4}$ (tetrahedral) - four C–Cl dipoles arranged symmetrically cancel; $μ = 0$
$Be F_{2}$ (linear) - $μ = 0$

Unsymmetrical molecules have non-zero dipole moments:

$H_{2} O$ (bent, 104.5°) - $μ = 1.85 D$
$N H_{3}$ (pyramidal) - $μ = 1.47 D$
$CHC l_{3}$ (tetrahedral but with different atoms) - $μ \neq = 0$

Figure 11: Zero net dipole in

C O_{2}

. The two

C = O

bond dipoles are equal in magnitude and point in exactly opposite directions along the 180° axis, so they cancel:

μ_{net} = 0

Figure 12: Net dipole moment in water. Because

H_{2} O

is bent (104.5°), the two

O - H

bond dipoles do not cancel. Their vector sum bisects the H-O-H angle, giving

μ = 1.85 D

Solved Example 5

N H_{3}

has

μ = 1.47 D

, while

N F_{3}

has only

μ = 0.23 D

, even though F is much more electronegative than H. Explain.

Solution:

Both molecules are pyramidal with a lone pair on N. In $N H_{3}$ , N is more electronegative than H, so the three N–H bond dipoles point towards N. The lone pair dipole on N also points away from the hydrogens, i.e., in the same direction as the bond dipole resultant. They add up, giving a large net $μ$ .

In $N F_{3}$ , F is more electronegative than N, so the three N–F bond dipoles point away from N. The lone pair dipole still points away from the F atoms, i.e., in the opposite direction to the bond dipole resultant. They partially cancel, giving a very small net $μ$ .

Solved Example 6

Compare the dipole moments of cis- and trans-1,2-dichloroethene.

Solution:

In the trans isomer, the two C–Cl bond dipoles are equal, oppositely oriented across the C=C double bond, and cancel exactly: $μ = 0$ . In the cis isomer, the two C–Cl dipoles point in similar directions (on the same side of the double bond) and add vectorially to give a non-zero net dipole. Generally, trans isomers of symmetrically substituted alkenes have lower (often zero) dipole moments than the corresponding cis isomers.

8. Percentage Ionic Character

Every real bond has both ionic and covalent character. The percentage ionic character of a bond can be estimated by comparing the observed dipole moment with the value calculated assuming 100% ionic bonding (i.e., a full electronic charge separated by the bond length):

$% ionic character = \frac{μ _{observed}}{μ _{100% ionic}} \times 100$

where $μ_{100% ionic} = e \times d$ with $e = 1.602 \times 1 0^{- 19} C$ and $d$ = interatomic distance.

Solved Example 7

The observed dipole moment of KCl is

3.336 \times 1 0^{- 29} C m

and the K–Cl interatomic distance is

2.6 \times 1 0^{- 10} m

. Calculate the percentage ionic character.

Solution:

For 100% ionic bond: $μ_{ionic} = e \times d = 1.602 \times 1 0^{- 19} \times 2.6 \times 1 0^{- 10} = 4.165 \times 1 0^{- 29} C m$

$% ionic character = \frac{3.336 \times 1 0 ^{- 29}}{4.165 \times 1 0 ^{- 29}} \times 100 = 80.1%$

KCl is highly ionic, but still has about 20% covalent character.

Even "covalent" molecules like HF and HCl have significant ionic character - about 43% and 17% respectively - because of the electronegativity difference between the bonded atoms. There is no such thing as a purely ionic or purely covalent bond in real molecules.

9. Bond Parameters

Bond length

Bond length is the equilibrium distance between the nuclei of two bonded atoms. It is measured in picometres (pm) or angstroms (Å), where $1 \overset{˚}{A} = 1 0^{- 10} m = 100 pm$ .

Bond length = sum of covalent radii of the two atoms (for covalent bonds), or sum of ionic radii (for ionic bonds).

Key trends:

Bond order: Higher bond order gives shorter bond. C–C (154 pm) $>$ C=C (134 pm) $>$ C≡C (120 pm).
s-character: Greater s-character in the hybrid orbital gives a shorter bond. $s p^{3}$ C–H (112 pm) $>$ $s p^{2}$ C–H (110 pm) $>$ $s p$ C–H (108 pm).
Polar bonds are shorter than the theoretical non-polar bond length because of partial ionic contraction.
Homonuclear diatomic bond length $= 2 \times$ covalent radius.

Bond energy (bond dissociation enthalpy)

Bond energy is the amount of energy required to break one mole of a particular bond in the gaseous state. Most covalent bond energies fall between 200 and 400 kJ/mol.

Key trends:

$σ$ -bonds are stronger than $π$ -bonds.
Bond energy increases with bond order: C–C $<$ C=C $<$ C≡C.
Bond energy decreases as bond length increases: HF $>$ HCl $>$ HBr $>$ HI.
Bond energy decreases with the number of lone pairs on bonded atoms due to lone-pair repulsion: C–C $>$ N–N $>$ O–O.
Homolytic fission requires less energy than heterolytic fission.

Bond angle

Bond angle is the angle between two adjacent bonds at a central atom. It depends on three factors:

Hybridisation: $s p$ gives 180°, $s p^{2}$ gives 120°, $s p^{3}$ gives 109.5°. Higher s-character means larger bond angle.
Lone pairs on the central atom repel bond pairs more strongly, compressing the bond angle. $N H_{3}$ (107°) $<$ $C H_{4}$ (109.5°); $H_{2} O$ (104.5°) $<$ $N H_{3}$ (107°).
Electronegativity of surrounding atoms: more electronegative surrounding atoms pull bond pairs away from the central atom, reducing bond-pair repulsion and decreasing bond angle. Hence $∠$ H–N–H (107°) $>$ $∠$ F–N–F (102°) in $N F_{3}$ .

Common Mistakes to Avoid

Watch out

Assuming ionic bonds are 100% ionic and covalent bonds are 100% covalent. Both are extremes; real bonds always have mixed character.
Confusing formal charge with oxidation state. Formal charge assumes equal electron sharing; oxidation state assumes complete transfer to the more electronegative atom.
Thinking resonance means the molecule oscillates between structures. Canonical forms have no real existence; the molecule has one, unchanging hybrid structure.
Forgetting that dipole moment is a vector. Symmetrical molecules like $C O_{2}$ , $B F_{3}$ , and $CC l_{4}$ have zero net dipole moment despite having polar bonds.
Miscounting valence electrons in charged species. Add electrons for negative charges, subtract for positive - students often reverse the sign.
Ignoring the lone-pair contribution to dipole moment. Lone pairs contribute significantly (as in $N H_{3}$ vs $N F_{3}$ ) and can either add to or oppose the bond-dipole resultant.
Applying the octet rule rigidly. Elements from period 3 onwards can expand their octet using d-orbitals (e.g., PF₅, SF₆).

Frequently Asked Questions

Q1. What is the difference between an ionic bond and a covalent bond?

An ionic bond is formed by the complete transfer of electrons from a metal (low ionization energy) to a non-metal (high electron affinity), producing oppositely charged ions held by electrostatic attraction. A covalent bond is formed by the sharing of electron pairs between two atoms of similar (usually high) electronegativity. Ionic compounds are solids with high melting points and conduct electricity when molten or dissolved; covalent compounds are often gases or liquids and are poor conductors.

Q2. Why does $BeC l_{2}$ have a covalent character despite being formed by a metal and a non-metal?

According to Fajans' rules, the $B e^{2 +}$ cation is very small and highly charged, giving it strong polarising power. It distorts the $C l^{-}$ electron cloud significantly, drawing electron density between the ions. This creates substantial covalent character. As predicted, $BeC l_{2}$ has a relatively low melting point (405 °C) and dissolves in organic solvents - behaviour more typical of covalent compounds.

Q3. How do you decide the central atom in a Lewis structure?

The least electronegative atom (other than H, which forms only one bond) is usually placed at the centre. For example, in $N F_{3}$ , N is central; in $C O_{3}^{2 -}$ , C is central; in $H_{2} S O_{4}$ , S is central. Hydrogen is never central because it can form only one bond.

Q4. Why is the C–C bond in benzene shorter than a single bond but longer than a double bond?

Benzene is a resonance hybrid of two Kekulé structures. The six $π$ -electrons are delocalised over all six carbons, giving each C–C bond an intermediate character with bond order $\approx 1.5$ . This gives a bond length of 139 pm, between C–C single (154 pm) and C=C double (134 pm).

Q5. Why does water have a higher dipole moment than hydrogen sulphide?

Both molecules are bent with two lone pairs, but oxygen is much more electronegative than sulphur. The O–H bond is therefore much more polar than the S–H bond, giving each bond a larger dipole. The vector sum in $H_{2} O$ (1.85 D) is significantly greater than in $H_{2} S$ (0.95 D).

Q6. What is the significance of the Born–Haber cycle?

The Born–Haber cycle applies Hess's law to calculate the lattice enthalpy of an ionic compound, which cannot be measured directly. By combining measurable quantities - sublimation enthalpy, dissociation energy, ionization energy, electron gain enthalpy, and enthalpy of formation - the cycle indirectly gives the lattice enthalpy, allowing us to compare and predict the stability of ionic compounds.

Q7. Can bond order be a fraction?

Yes. In molecules or ions with resonance or with unpaired electrons in molecular orbitals, bond order can be a fraction. Ozone has O–O bond order 1.5. The carbonate ion has C–O bond order 4/3 ≈ 1.33. In MOT, $O_{2}^{-}$ (superoxide) has bond order 1.5 and $H_{2}^{+}$ has bond order 0.5.

Q8. Why does $PbC l_{4}$ exist but $Pb I_{4}$ does not?

Pb(+IV) is a strong oxidising state due to the inert pair effect (the $6 s^{2}$ electrons are reluctant to bond). Iodide, $I^{-}$ , is a good reducing agent - it reduces $P b^{4 +}$ to $P b^{2 +}$ while itself being oxidised to $I_{2}$ . So $Pb I_{4}$ is unstable and spontaneously decomposes to $Pb I_{2} + I_{2}$ . $C l^{-}$ is not a strong enough reducing agent to reduce $P b^{4 +}$ , so $PbC l_{4}$ is stable at low temperatures.

Q9. What decides whether a covalent bond is polar or non-polar?

The electronegativity difference ( $Δ$ EN) between the bonded atoms. If $Δ$ EN $= 0$ (identical atoms, as in $H_{2}$ , $C l_{2}$ ), the bond is purely covalent (non-polar). If $0 < Δ EN < 1.7$ (approx.), the bond is polar covalent. If $Δ$ EN $> 1.7$ (approx.), the bond is predominantly ionic. However, molecular geometry also matters: a molecule with polar bonds can still have zero net dipole moment if symmetrical (e.g., $C O_{2}$ , $CC l_{4}$ ).

Previous year questions on Important Concepts Of Chemical Bonding

20 questions from past papers, each with a step-by-step solution.

Show all 20 questions

Ready to master Chemical Bonding And Molecular Structure?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. Kössel–Lewis Approach to Chemical Bonding

Lewis symbols and group valence

The octet rule

2. Ionic (Electrovalent) Bond

Criteria for ionic bond formation

Energy changes during ionic bond formation

Born–Haber cycle

Characteristics of ionic compounds

3. Covalent Bond

Sigma (σ) and pi (π) bonds

Coordinate (dative) bond

4. Lewis Structures and Formal Charge

Formal charge

Limitations of the octet rule

5. Fajans' Rules - Covalent Character in Ionic Bonds

6. Resonance

Rules for writing resonating structures

Other important examples

7. Polarity of Bonds and Dipole Moment

Resultant of two bond dipoles

Symmetry and net dipole moment

8. Percentage Ionic Character

9. Bond Parameters

Bond length

Bond energy (bond dissociation enthalpy)

Bond angle

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between an ionic bond and a covalent bond?

Q2. Why does BeCl2​ have a covalent character despite being formed by a metal and a non-metal?

Q3. How do you decide the central atom in a Lewis structure?

Q4. Why is the C–C bond in benzene shorter than a single bond but longer than a double bond?

Q5. Why does water have a higher dipole moment than hydrogen sulphide?

Q6. What is the significance of the Born–Haber cycle?

Q7. Can bond order be a fraction?

Q8. Why does PbCl4​ exist but PbI4​ does not?

Q9. What decides whether a covalent bond is polar or non-polar?

Previous year questions on Important Concepts Of Chemical Bonding

Ready to master Chemical Bonding And Molecular Structure?

Sigma ( $σ$ ) and pi ( $π$ ) bonds

Q2. Why does $BeC l_{2}$ have a covalent character despite being formed by a metal and a non-metal?

Q8. Why does $PbC l_{4}$ exist but $Pb I_{4}$ does not?