Chemical Bonds and Molecular Geometry
Covalent, ionic, and metallic bonds; polarity and electronegativity; molecular geometry (hybridization and VSEPR); resonance and intermolecular forces.
Complete Theory
4A covalent bond is the sharing of one or more pairs of electrons between two atoms (valence bond theory). Bonds can be single (one pair, e.g. ), double (two pairs, e.g. ) or triple (three pairs, e.g. ). The Morse curve describes potential energy as a function of internuclear distance: the minimum corresponds to the bond length (equilibrium distance) and the well depth to the bond energy (how much energy is needed to break the bond).
Electronegativity () is the tendency of an atom to attract shared electrons in a bond, measured on the Pauling scale (0.7 to 4.0). The electronegativity difference between two atoms determines the bond type:
Real-world applications: molecular polarity determines solubility ("like dissolves like": polar substances in polar solvents, nonpolar in nonpolar). Surfactants (soaps) have a polar hydrophilic head and a nonpolar hydrophobic tail, enabling grease emulsification in water.
Electronegativity () is the tendency of an atom to attract shared electrons in a bond, measured on the Pauling scale (0.7 to 4.0). The electronegativity difference between two atoms determines the bond type:
- : pure covalent (nonpolar) — electrons are shared equally (e.g. )
- : polar covalent — electrons are shifted toward the more electronegative atom, creating a dipole (e.g. )
- : ionic — the more electronegative atom completely strips away one or more electrons (e.g. )
Real-world applications: molecular polarity determines solubility ("like dissolves like": polar substances in polar solvents, nonpolar in nonpolar). Surfactants (soaps) have a polar hydrophilic head and a nonpolar hydrophobic tail, enabling grease emulsification in water.
The ionic bond forms by complete electron transfer from one atom to another: the atom that loses electrons becomes a cation (positive charge), the one that gains them becomes an anion (negative charge). The two ions are held together by electrostatic attraction (Coulomb's law). Ionic compounds are crystalline solids at room temperature with high melting points (e.g. NaCl melts at 801°C).
The lattice energy is the energy released when gaseous ions assemble to form an ionic crystal. It is very large and negative (e.g. -788 kJ/mol for NaCl), and depends on:
Real-world applications: ionic compounds like NaCl (table salt), CaCO₃ (calcium carbonate, marble) and NaOH (sodium hydroxide) are ubiquitous. Electrical conductivity in the molten state or in solution arises from mobile ions (electrolytes), the principle behind batteries and electrolysis.
The lattice energy is the energy released when gaseous ions assemble to form an ionic crystal. It is very large and negative (e.g. -788 kJ/mol for NaCl), and depends on:
- Ionic charges (, ): higher charges → greater lattice energy (e.g. with and has roughly 4× the lattice energy of NaCl)
- Interionic distance (): smaller ions → shorter distance → stronger attraction
- Madelung constant (): depends on the crystal lattice geometry (e.g. NaCl has a face-centered cubic structure with )
Real-world applications: ionic compounds like NaCl (table salt), CaCO₃ (calcium carbonate, marble) and NaOH (sodium hydroxide) are ubiquitous. Electrical conductivity in the molten state or in solution arises from mobile ions (electrolytes), the principle behind batteries and electrolysis.
VSEPR theory (Valence Shell Electron Pair Repulsion) is based on a simple principle: electron pairs (both bonding and lone) in the valence shell of a central atom arrange themselves as far apart as possible to minimize electrostatic repulsion. The total number of electron domains (bonding pairs + lone pairs) determines the base geometry.
Hybridization explains how atomic orbitals combine to form equivalent bonding orbitals:
Resonance: when a single Lewis structure is insufficient to describe a molecule, multiple resonance limit structures are used. The real structure is a resonance hybrid (average of the limit structures), with delocalized electrons. Classic examples:
Hybridization explains how atomic orbitals combine to form equivalent bonding orbitals:
- (2 domains) → linear geometry (180°), e.g. , , (acetylene)
- (3 domains) → trigonal planar geometry (120°), e.g. , , (ethylene)
- (4 domains) → tetrahedral geometry (109.5°), e.g. , ; with lone pairs the geometry distorts (e.g. pyramidal 107°, bent 104.5°)
Resonance: when a single Lewis structure is insufficient to describe a molecule, multiple resonance limit structures are used. The real structure is a resonance hybrid (average of the limit structures), with delocalized electrons. Classic examples:
- Benzene (): 2 Kekulé structures, all C-C bonds are equivalent (order 1.5)
- Carbonate ion (): 3 resonance structures, each C-O bond has order 4/3
The metallic bond is described by the "sea of electrons" model: metal atoms lose their valence electrons, which become delocalized and move freely among the metal cations arranged in a regular lattice. This delocalization explains the characteristic properties of metals:
Intermolecular forces (increasing strength):
- Electrical and thermal conductivity — free electrons carry charge and heat
- Ductility and malleability — ions can slide without breaking the metallic bond
- Luster — free electrons interact with light (reflection)
Intermolecular forces (increasing strength):
- London (dispersion) forces: instantaneous dipoles from temporary fluctuations in the electron cloud. Universal (present in all molecules), increase with molecular mass and surface area. Example: noble gases liquefy only through these forces.
- Dipole-dipole (Van der Waals): attraction between the positive pole of one polar molecule and the negative pole of another. Stronger than London forces (for molecules of comparable mass).
- Hydrogen bond: the strongest dipole-dipole interaction, formed when hydrogen is bonded to a very electronegative atom (F, O, N). Explains water's anomalous properties (high boiling point, ice density < liquid water), the double helix structure of DNA (H-bonds between complementary bases), and protein structure.
Worked Examples
2Example 1Geometry and polarity of and
Given
: central C, 2 double bonds, no lone pairs
: central O, 2 single bonds, 2 lone pairs
Find
Geometry and polarity of each molecule
Step-by-step solution
1In , carbon has hybridization with 2 electron domains, yielding linear geometry (180°). The two C=O bonds generate dipole vectors of equal magnitude but opposite direction that cancel, making the molecule nonpolar (). CO₂ is a colorless, odorless asphyxiant gas whose atmospheric increase causes the greenhouse effect.
2In , oxygen has hybridization with 4 electron domains (2 O—H bonds and 2 lone pairs). The base geometry is tetrahedral, but due to the lone pairs (which occupy more space than bonds), the effective geometry is bent with an angle of 104.5° (instead of 109.5°). The O—H electronegativity difference and the asymmetric geometry produce a net dipole moment → polar molecule ( D).
✓ Final result: linear, nonpolar (); bent 104.5°, polar ( D)
Example 2Bond type from
Given
, ,
Find
Bond type in NaCl and HCl
Step-by-step solution
1Calculate the electronegativity difference for NaCl: . Since , electron transfer is complete: sodium gives up an electron becoming Na⁺ and chlorine accepts it becoming Cl⁻ → ionic bond. NaCl is common table salt.
2For HCl: . Since , the bond is polar covalent: electrons are shared but shifted toward chlorine (more electronegative), creating partial dipoles . HCl dissociates completely in water (strong acid).
✓ Final result: NaCl ionic (); HCl polar covalent ()
Exercises with Solutions
3Exercise 1Hybridization and geometryMedium
Problem to solve
Determine the hybridization and geometry of methane , ammonia , and water .
Given data
CH₄: central C with 4 C—H bondsNH₃: central N with 3 N—H bonds + 1 lone pairH₂O: central O with 2 O—H bonds + 2 lone pairs
Step-by-step solution
1: 4 electron domains (4 bonds, 0 lone pairs) → hybridization , perfect tetrahedral geometry with 109.5° angles. All four positions are equivalent.
2: 4 electron domains (3 bonds, 1 lone pair) → hybridization , base geometry tetrahedral but the lone pair occupies more space → effective geometry trigonal pyramidal with 107° angles (slightly less than 109.5°).
3: 4 electron domains (2 bonds, 2 lone pairs) → hybridization , effective geometry bent with 104.5° angle. The two lone pairs further compress the angle.
✓ Final answer: CH₄ tetrahedral (109.5°); NH₃ pyramidal (107°); H₂O bent (104.5°) — all
Exercise 2Intermolecular forcesMedium
Problem to solve
Why does water ( g/mol) boil at 100°C while methane ( g/mol) boils at −161°C, despite having nearly the same molar mass?
Given data
H₂O: M=18 g/molCH₄: M=16 g/mol
Step-by-step solution
1Water forms hydrogen bonds: each H₂O molecule can form up to 4 hydrogen bonds with neighboring molecules (two through its H atoms, two through O's lone pairs). These bonds are strong (about 20 kJ/mol each) and require significant energy to break, explaining the high boiling point.
2Methane has only London forces (very weak, a few kJ/mol) because it is a nonpolar molecule with essentially nonpolar C—H bonds. London forces increase with mass and surface area, but CH₄ has small mass and is compact.
3Over 260°C difference at comparable mass: the most striking proof of the importance of hydrogen bonding. Without H-bonds, water would boil at about −80°C and life as we know it would not exist.
✓ Final answer: Hydrogen bonding in H₂O (up to 4 per molecule) requires far more energy to separate molecules than the weak London forces of CH₄
Exercise 3ResonanceHard
Problem to solve
Explain why the three C–O bonds in the carbonate ion all have the same length, about 129 pm (intermediate between a single C–O bond, 143 pm, and a double bond, 122 pm).
Given data
CO₃²⁻: central C, 3 O atoms around, charge -2
Step-by-step solution
1A single Lewis structure would show one C=O double bond and two C—O single bonds (with negative charge on the two single-bonded O), implying bonds of different lengths, contradicting experimental data.
2There are 3 equivalent resonance limit structures, where the double bond rotates among the three oxygens. The real structure is a resonance hybrid with electrons delocalized over the entire ion.
3In the resonance hybrid, each C—O bond has bond order 4/3 (1.33), the average of single (1) and double (2): (1+1+2)/3 = 4/3. This intermediate bond order explains the intermediate length (129 pm) and the negative charge equally distributed over the three oxygens.
✓ Final answer: By resonance: 3 equivalent structures → hybrid with bond order 4/3, intermediate length, delocalized charge
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