Classly AILog in

Chemical Bonding and Molecular StructureClass 11 Chemistry Part I, Chapter 4

This chapter explains why atoms bond and how that decides the shape and properties of a molecule. You move from Lewis structures and the octet rule to resonance, dipole moments, VSEPR shapes, hybridisation, molecular orbitals and hydrogen bonding. After revising here, you can predict shapes, bond orders and polarity.

Follows the Class 11 Chemistry Part I textbook, Chapter 4. Made with Classly and checked by our team.

Try Flashcards free

Cheat sheet

Chemical Bonding and Molecular Structure on one page, for a last look before a test.

Chemical Bonding and Molecular Structure

Cheat sheetClass 11 · Chemistry

⚑ Panic box

VSEPR repulsion: lp–lp > lp–bp > bp–bp; more lone pairs usually compress bond angles.

Bond order: Lewis count = shared pairs; MO value = $\tfrac12(N_b-N_a)$.

Resonance: the molecule is one hybrid, not a structure switching between canonical forms.

Multiple bond: one $\sigma$ bond plus additional $\pi$ bond(s), except the unusual MO description of $\mathrm{C_2}$.

MO magnetism: unpaired electrons → paramagnetic; all paired → diamagnetic.

1. Lewis structures and the octet idea

  • Chemical bond — attraction holding atoms or ions together; bonding lowers a system’s energy.
  • Valence electrons — outer-shell electrons available for bonding; Lewis-symbol dots count them.
  • Octet rule — atoms commonly gain, lose or share electrons to reach eight outer-shell electrons; H aims for a duplet.
  • Ionic bond — electrostatic attraction between oppositely charged ions; electron transfer creates the ions.
  • Covalent bond — shared electron pair(s) between atoms; one pair = single, two = double, three = triple bond.
  • Lone pair — valence-electron pair not shared in a bond.
  • Electrovalence — number of unit charges on an ion.

Lewis structure routine

  1. Add all atoms’ valence electrons; add one per negative charge, subtract one per positive charge.
  2. Choose the skeletal arrangement; usually put the least electronegative atom centrally (never H).
  3. Join atoms with single bonds; complete terminal atoms’ shells, then the central atom’s.
  4. If the central atom lacks an octet, make multiple bonds where possible; check total electrons and charge.

Octet-rule exceptions

CaseRecognise itExamples
Incomplete octetCentral atom has fewer than 8 electrons$\mathrm{BeH_2, BCl_3, BF_3}$
Odd-electron speciesTotal electron count is odd$\mathrm{NO, NO_2}$
Expanded octetCentral atom has more than 8 electrons$\mathrm{PF_5, SF_6, H_2SO_4}$

Useful for many molecules, especially second-period atoms; it does not predict shape or relative molecular stability, and noble gases can form compounds.

Formal charge — bookkeeping charge on an atom in a Lewis structure:

$$\mathrm{FC}=V-L-\tfrac12B$$

$V$ = free-atom valence electrons; $L$ = nonbonding electrons; $B$ = bonding electrons. Formal charge is not necessarily the atom’s real charge; among alternatives, a structure with smaller formal charges is generally favoured.

2. Ionic bonding and bond parameters

Ionic formation: $\mathrm{M(g)\rightarrow M^+(g)+e^-}$ (ionisation; endothermic) · $\mathrm{X(g)+e^-\rightarrow X^-(g)}$ (electron gain; may release or absorb energy) · gaseous ions assemble into a crystal lattice.

  • Favourable ion formation — relatively low ionisation enthalpy for the cation-former and strongly negative electron-gain enthalpy for the anion-former.
  • Lattice enthalpy — energy needed to separate one mole of ionic solid into gaseous ions; lattice formation releases this energy.
  • Ionic-solid stability — lattice formation can outweigh the energy cost of forming gaseous ions; octets alone do not decide stability.
ParameterMeaning / relationRecall point
Bond lengthEquilibrium internuclear distance; for a covalent $\mathrm{AB}$ bond, $R=r_A+r_B$Higher bond order generally means shorter bond
Covalent radiusHalf the bond length between identical bonded atomsBonded situation
van der Waals radiusHalf the separation of identical nonbonded atoms in a solidNonbonded situation; larger than covalent radius
Bond angleAngle between bonding orbitals around a central atomHelps identify molecular shape
Bond enthalpyEnergy to break one mole of a specified bond type in gaseous moleculesLarger value → stronger bond
Mean bond enthalpyTotal dissociation enthalpy divided by number of bonds brokenUsed for polyatomic molecules; successive bonds may differ
Bond order (Lewis)Number of bonds between two atomsSingle = 1, double = 2, triple = 3

Trend: bond order ↑ → bond enthalpy ↑ and bond length ↓. In polyatomic molecules, the same bond can have different dissociation enthalpies in different chemical environments.

3. Resonance, polarity and dipole moment

Resonance — several valid Lewis canonical forms with the same nuclear arrangement contribute to one real resonance hybrid.

  • Use when one Lewis structure cannot match the observed bonding (e.g. equal bonds in $\mathrm{O_3}$ or $\mathrm{CO_3^{2-}}$).
  • Hybrid energy is lower than that of any individual canonical form; bond character is averaged.
  • Canonical forms are not separate real structures and are not in equilibrium with one another.

Bond polarity — equal sharing between identical atoms gives a nonpolar covalent bond; unequal sharing toward the more electronegative atom gives a polar covalent bond.

$$\mu=Qr$$

$Q$ = charge magnitude; $r$ = separation of charge centres. Dipole moment is a vector; $1\ \mathrm{D}=3.33564\times10^{-30}\ \mathrm{C,m}$. Molecular dipole = vector sum of bond dipoles (and the lone-pair contribution).

Molecule patternNet dipole
Symmetric bonds cancel (e.g. $\mathrm{BeF_2, BF_3, CCl_4}$)$0$
Bent or pyramidal arrangement (e.g. $\mathrm{H_2O, NH_3}$)Usually nonzero
$\mathrm{NH_3}$ vs $\mathrm{NF_3}$$\mathrm{NH_3}$ larger: lone-pair contribution reinforces N–H bond resultants; in $\mathrm{NF_3}$ it opposes N–F resultants

Fajans’ rules: greater covalent character when the cation is smaller, the cation charge is higher, or the anion is larger. For same-size, same-charge cations, a transition-metal-type $(n-1)d^nns^0$ configuration polarises more than a noble-gas configuration.

4. VSEPR shapes

VSEPR rule — valence-shell electron pairs spread out to minimise repulsion; multiple bonds count as one electron domain (a “super pair”).

$$\mathrm{lp-lp>lp-bp>bp-bp}$$

Electron domains around central atomElectron-pair arrangementExample with no lone pair
2Linear, $180^\circ$$\mathrm{BeCl_2}$
3Trigonal planar, $120^\circ$$\mathrm{BF_3}$
4Tetrahedral, $109.5^\circ$$\mathrm{CH_4}$
5Trigonal bipyramidal$\mathrm{PCl_5}$
6Octahedral$\mathrm{SF_6}$
Central-atom domainsMolecular shapeExample / angle
3 bonding + 1 lone pairTrigonal pyramidal$\mathrm{NH_3}$, $107^\circ$
2 bonding + 2 lone pairsBent$\mathrm{H_2O}$, $104.5^\circ$
2 bonding + 1 lone pairBent$\mathrm{SO_2}$, about $119.5^\circ$
4 bonding + 1 lone pairSee-sawLone pair prefers equatorial position
3 bonding + 2 lone pairsT-shapedLone pairs prefer equatorial positions

Lone pairs occupy more space than bonding pairs → bond angles shrink from ideal values. In trigonal bipyramidal arrangements, equatorial lone-pair placement reduces repulsions; axial bonds experience more repulsion and are longer/weaker than equatorial bonds.

5. Valence-bond theory, overlap and hybridisation

Valence-bond picture — covalent bonding forms when half-filled atomic orbitals overlap and opposite-spin electrons pair. As atoms approach, attraction lowers potential energy until the equilibrium distance; closer approach then raises energy sharply. Greater effective overlap generally gives a stronger bond.

Bond typeOrbital overlapKey feature
$\sigma$Head-on along internuclear axis; $s-s$, $s-p$ or $p-p$Greater overlap; stronger than $\pi$
$\pi$Sideways overlap of parallel $p$ orbitalsElectron clouds above and below the internuclear axis

A double bond contains $1\sigma+1\pi$; a triple bond contains $1\sigma+2\pi$. In phase/sign diagrams, $+$ and $-$ mark wave-function phase, not charge; effective overlap requires suitable phase and orientation.

Hybridisation — mixing valence orbitals of similar energy to make equivalent, directionally arranged hybrid orbitals. Number of hybrids = number of orbitals mixed; promotion is not always required, and filled orbitals may participate.

HybridisationOrbitals mixed → hybridsArrangement / angleChapter examples
$sp$$s+p\rightarrow2$Linear, $180^\circ$$\mathrm{BeCl_2}$; carbon in $\mathrm{C_2H_2}$
$sp^2$$s+2p\rightarrow3$Trigonal planar, $120^\circ$$\mathrm{BCl_3}$; carbon in $\mathrm{C_2H_4}$
$sp^3$$s+3p\rightarrow4$Tetrahedral, $109.5^\circ$$\mathrm{CH_4}$; also $\mathrm{NH_3,H_2O}$
$sp^3d$$s+3p+d\rightarrow5$Trigonal bipyramidal$\mathrm{PCl_5}$
$sp^3d^2$$s+3p+2d\rightarrow6$Octahedral$\mathrm{SF_6}$
$dsp^2$$d+s+2p\rightarrow4$Square planar$[\mathrm{Ni(CN)_4}]^{2-}$

$sp^3$ examples: $\mathrm{CH_4}$ has four bonding hybrids; $\mathrm{NH_3}$ has three bonds + one lone pair; $\mathrm{H_2O}$ has two bonds + two lone pairs. In $\mathrm{C_2H_4}$, each C–C double bond is $1\sigma+1\pi$; in $\mathrm{C_2H_2}$, the C–C triple bond is $1\sigma+2\pi$.

6. Molecular-orbital theory

MO formation (LCAO): atomic orbitals combine when their energies are similar, symmetry matches about the molecular axis, and overlap is effective. Combining $n$ atomic orbitals gives $n$ molecular orbitals.

$$\psi_{\mathrm{bonding}}=\psi_A+\psi_B \qquad \psi_{\mathrm{antibonding}}=\psi_A-\psi_B$$

Molecular orbitalElectron density / energyEffect
BondingDensity between nuclei; lower energyStabilises molecule
Antibonding ($*$)Node between nuclei; higher energyDestabilises molecule
$\sigma$ MOSymmetric around bond axisIncludes end-on combinations
$\pi$ MONot symmetric around bond axisSide-on $p$ combinations; density above/below axis

Fill MOs by increasing energy, following Aufbau, Pauli and Hund rules. For second-period homonuclear molecules:

  • $\mathrm{B_2,C_2,N_2}$: $\sigma_{1s}<\sigma^{1s}<\sigma{2s}<\sigma^{2s}<\pi{2p_x}=\pi_{2p_y}<\sigma_{2p_z}<\pi^_{2p_x}=\pi^{2p_y}<\sigma^*{2p_z}$
  • $\mathrm{O_2,F_2}$: same order through $\sigma^{2s}$, then $\sigma{2p_z}<\pi_{2p_x}=\pi_{2p_y}<\pi^{2p_x}=\pi^*{2p_y}<\sigma^*_{2p_z}$

$$\mathrm{MO\ bond\ order}=\tfrac12(N_b-N_a)$$

$N_b$ = bonding-MO electrons; $N_a$ = antibonding-MO electrons. Positive bond order → stable; zero or negative → unstable. Larger bond order generally means shorter, stronger bonding. Unpaired MO electrons → paramagnetic; all paired → diamagnetic.

SpeciesBond orderStability / magnetism
$\mathrm{H_2}$1Stable; diamagnetic
$\mathrm{He_2}$0Unstable
$\mathrm{Li_2}$1Stable; diamagnetic
$\mathrm{Be_2}$0Unstable
$\mathrm{C_2}$2Stable; diamagnetic; two $\pi$ bonds in the MO account
$\mathrm{O_2}$2Stable; paramagnetic, two unpaired electrons

7. Hydrogen bonding

Hydrogen bond — attraction between H attached to a strongly electronegative atom (F, O or N) and an electronegative atom on another site. The bonded H becomes partially positive; the electronegative atom is partially negative. Weaker than a covalent bond.

TypeWhere it formsExamples
IntermolecularBetween moleculesHF, alcohols, water
IntramolecularWithin one molecule$o$-nitrophenol

Hydrogen bonding is strongest in solids and weakest in gases; it substantially affects structure and properties.

8. Traps

  • Octet rule applies to every atom → Check incomplete, odd-electron and expanded-octet cases.
  • Formal charge is the atom’s actual charge → It is Lewis-structure bookkeeping; the ion’s net charge belongs to the whole ion.
  • Canonical resonance forms are structures the molecule switches between → The actual molecule is one resonance hybrid.
  • Polar bonds guarantee a polar molecule → Molecular geometry can make bond-dipole vectors cancel.
  • All electron-pair repulsions are equal → $\mathrm{lp-lp>lp-bp>bp-bp}$; account for lone pairs before assigning shape.
  • A double bond is two $\sigma$ bonds → It is $1\sigma+1\pi$; a triple bond is $1\sigma+2\pi$.
  • More antibonding electrons increase stability → Use $\tfrac12(N_b-N_a)$; zero bond order means unstable in this model.
  • $\mathrm{O_2}$ is diamagnetic because it has a double bond → MO theory leaves two unpaired electrons, so it is paramagnetic.

Flashcards

10 cards on Chemical Bonding and Molecular Structure. Flip them here, or read them all below.

Chemical Bonding and Molecular Structure flashcards1 of 10

Say the answer in your head, then flip.

All 10 cards

  1. 1

    In the octet-rule model, atoms form bonds by.

    Transfer or sharing of valence electrons so atoms attain stable outer-shell arrangements

  2. 2

    What are the three main kinds of exceptions to the octet rule?

    Incomplete octet, odd-electron species, and expanded octet

  3. 3

    How do bond order, bond strength, and bond length generally change together?

    Bond order rises, bond enthalpy generally rises, and bond length falls

  4. 4

    Write the relation between dipole moment $\mu$, charge magnitude $Q$, and charge separation $r$.

    $\mu=Qr$

  5. 5

    In VSEPR, what are the ideal electron-pair arrangements for 2, 3, 4, 5, and 6 electron domains around a central atom, in that order?

    Linear; trigonal planar; tetrahedral; trigonal bipyramidal; octahedral

  6. 6

    What is the VSEPR order of electron-pair repulsions, from strongest to weakest?

    Lone pair–lone pair $>$ lone pair–bond pair $>$ bond pair–bond pair

  7. 7

    How does the overlap that forms a sigma bond differ from the overlap that forms a pi bond?

    A sigma bond comes from head-on overlap along the internuclear axis; a pi bond comes from sideways overlap of parallel orbitals.

  8. 8

    Match $sp$, $sp^2$, and $sp^3$ hybridisation with their ideal geometry and bond angle.

    $sp$: linear, $180^\circ$; $sp^2$: trigonal planar, $120^\circ$; $sp^3$: tetrahedral, $109.5^\circ$

  9. 9

    Write the molecular-orbital bond-order formula using bonding electrons $N_b$ and antibonding electrons $N_a$.

    $\frac{1}{2}(N_b-N_a)$

  10. 10

    What is a hydrogen bond, and which atoms are involved?

    An attraction between a hydrogen atom bonded to F, O or N and an electronegative atom elsewhere; it is weaker than a covalent bond and can be intermolecular (HF, water) or intramolecular (o-nitrophenol)

Quiz

5 questions, each with the reason behind its answer.

1 of 5

A central atom has three bonding pairs and one lone pair in its valence shell. Which molecular shape is expected?

Answer key
  1. 1

    A central atom has three bonding pairs and one lone pair in its valence shell. Which molecular shape is expected?

    Answer: A. Trigonal pyramidal

    Four electron pairs point towards the corners of a tetrahedron, but only the three bonded atoms define the shape. With one lone pair, the molecule is trigonal pyramidal, as in $\mathrm{NH_3}$.

  2. 2

    Which molecule has polar bonds but no overall dipole moment because its bond dipoles cancel?

    Answer: A. $\mathrm{CO_2}$

    In linear $\mathrm{CO_2}$, the two equal C–O bond dipoles point in opposite directions, so their vector sum is zero.

  3. 3

    In ethene, what is the composition of the carbon–carbon double bond?

    Answer: B. One sigma bond and one pi bond

    A double bond contains one bond formed by head-on overlap along the internuclear axis (a sigma bond) and one formed by sideways overlap (a pi bond).

  4. 4

    Molecular orbital theory predicts that $\mathrm{O_2}$ is attracted to a magnetic field. Which feature of its electron arrangement accounts for this?

    Answer: C. It has two unpaired electrons

    The molecular orbital arrangement of $\mathrm{O_2}$ leaves two electrons unpaired in antibonding pi orbitals. Unpaired electrons make the molecule paramagnetic, so it is attracted to a magnetic field.

  5. 5

    True or false: a molecule with resonance forms keeps switching between them, spending part of its time in each.

    Answer: B. False

    False. The canonical forms are contributing representations, not separate structures that the molecule alternates between; the molecule is represented by one resonance hybrid.

Make your own for this chapter

Open this chapter of Class 11 Chemistry Part I in Classly and get your own cheat sheet, flashcards or quiz in about a minute. Free to start.

Try Cheat Sheet freeTry Flashcards freeTry Quiz Me free

Questions about Chemical Bonding and Molecular Structure

What are the most important topics in Class 11 Chemical Bonding?

Focus on Lewis structures with formal charge, exceptions to the octet rule, VSEPR shapes, dipole moment, hybridisation, sigma and pi bonds, and molecular orbital bond order and magnetism. Hydrogen bonding and resonance also appear often in school exams and in JEE and NEET questions.

How do I find the shape of a molecule using VSEPR?

Count the electron domains around the central atom, treating a multiple bond as one domain. That gives the electron-pair arrangement, such as tetrahedral for four domains. Then ignore the lone pairs to name the molecular shape, remembering that lone pairs repel more and squeeze bond angles, as in $\mathrm{NH_3}$ at $107^\circ$ and $\mathrm{H_2O}$ at $104.5^\circ$.

How do I find the bond order and magnetism of a molecule using molecular orbital theory?

Fill the molecular orbitals in order of increasing energy, then use bond order $=\tfrac12(N_b-N_a)$. A positive value means the molecule is stable. If any electron is unpaired the molecule is paramagnetic, and if all are paired it is diamagnetic, which is why $\mathrm{O_2}$ is paramagnetic. You can also make your own flashcards or quiz for any chapter in Classly.

Why can a molecule with polar bonds still have zero dipole moment?

Dipole moment is a vector, so the bond dipoles add as vectors. In a symmetric molecule such as $\mathrm{CO_2}$, $\mathrm{BF_3}$ or $\mathrm{CCl_4}$ they point in opposite directions and cancel. In a bent or pyramidal molecule like $\mathrm{H_2O}$ or $\mathrm{NH_3}$ they do not cancel, so the molecule is polar.