Types of Chemical Bonds

Understand ionic, covalent, and metallic bonds, including how electronegativity differences determine bond type and properties.

Types of Chemical Bonds

Why Do Atoms Bond?

Atoms form chemical bonds to achieve lower potential energy and greater stability.

Key principle: Atoms bond to achieve a stable electron configuration, typically a filled valence shell (octet rule for most elements, duet for hydrogen).

Three Main Types of Bonds

1. Ionic Bonds

Definition: Electrostatic attraction between oppositely charged ions

Formation:

  • Complete transfer of electrons from metal to nonmetal
  • Metal loses electrons → becomes cation (+)
  • Nonmetal gains electrons → becomes anion (-)
  • Opposite charges attract

Characteristics:

  • Form between metals and nonmetals
  • Large difference in electronegativity (ΔEN > 1.7)
  • Form crystalline lattice structures
  • High melting and boiling points
  • Conduct electricity when molten or dissolved (ions are mobile)
  • Brittle (layers can shift and repel)

Example: NaCl (sodium chloride)

\ceNa−>Na++e−\ce{Na -> Na+ + e-} (sodium loses 1 electron) \ceCl+e−−>Cl−\ce{Cl + e- -> Cl-} (chlorine gains 1 electron) \ceNa++Cl−−>NaCl\ce{Na+ + Cl- -> NaCl} (ionic bond forms)

Electronegativity:

  • Na: 0.9
  • Cl: 3.0
  • ΔEN = 2.1 (large difference → ionic)

2. Covalent Bonds

Definition: Sharing of electron pairs between atoms

Formation:

  • Electrons are shared between atoms (not transferred)
  • Both atoms contribute to shared electron pair
  • Occurs between nonmetals
  • Can be single, double, or triple bonds

Types of Covalent Bonds:

a) Nonpolar Covalent:

  • Equal sharing of electrons
  • ΔEN < 0.4
  • Symmetrical electron distribution
  • Example: H₂, O₂, N₂, Cl₂

b) Polar Covalent:

  • Unequal sharing of electrons
  • 0.4 < ΔEN < 1.7
  • More electronegative atom attracts electrons more
  • Creates partial charges (δ+ and δ-)
  • Example: H₂O, HCl, NH₃

Characteristics of Covalent Compounds:

  • Lower melting/boiling points than ionic (generally)
  • Do not conduct electricity (no charged particles)
  • Can be gases, liquids, or solids at room temperature
  • Soft and flexible (molecular compounds)

Example: H₂O (water)

Each H shares electrons with O:

  • O has 6 valence electrons
  • Each H has 1 valence electron
  • 2 covalent bonds form (O-H bonds)
  • O achieves octet, H achieves duet

Electronegativity:

  • H: 2.1
  • O: 3.5
  • ΔEN = 1.4 (moderate difference → polar covalent)

3. Metallic Bonds

Definition: Attraction between metal cations and delocalized electrons

Formation:

  • Metal atoms release valence electrons
  • Electrons form "sea" of mobile electrons
  • Metal cations are embedded in electron sea
  • Electrons are delocalized (shared among all atoms)

Characteristics:

  • Occur in pure metals and alloys
  • Conduct electricity in solid state (mobile electrons)
  • Conduct heat well
  • Malleable and ductile (layers can slide)
  • Lustrous (shiny)
  • Variable melting points (depends on metal)

Example: Sodium metal (Na)

  • Each Na atom contributes 1 valence electron to "sea"
  • Na⁺ ions arrange in crystal lattice
  • Electrons move freely throughout structure

Visualization:

\ce[Na+][Na+][Na+]\ce{[Na+][Na+][Na+]} Electron sea: e−e−e−e−e−\text{Electron sea: } e^- e^- e^- e^- e^-

Electronegativity and Bond Type

Electronegativity (EN): Ability to attract bonding electrons

Pauling Scale: 0.7 (Cs) to 4.0 (F)

Predicting Bond Type

Use difference in electronegativity (ΔEN):

| ΔEN Range | Bond Type | Example | |-----------|-----------|---------| | 0 - 0.4 | Nonpolar covalent | H₂, Cl₂ | | 0.4 - 1.7 | Polar covalent | H₂O, NH₃ | | > 1.7 | Ionic | NaCl, MgO |

Note: These are guidelines; bonding is a spectrum, not discrete categories.

Example Calculations:

1. H-Cl bond:

  • H: 2.1, Cl: 3.0
  • ΔEN = |3.0 - 2.1| = 0.9
  • Polar covalent

2. Na-Cl bond:

  • Na: 0.9, Cl: 3.0
  • ΔEN = |3.0 - 0.9| = 2.1
  • Ionic

3. C-H bond:

  • C: 2.5, H: 2.1
  • ΔEN = |2.5 - 2.1| = 0.4
  • Nonpolar covalent (borderline)

Bond Polarity

Polar bonds have unequal electron distribution:

  • More electronegative atom has partial negative charge (δ-)
  • Less electronegative atom has partial positive charge (δ+)
  • Creates dipole moment

Representation:

\ceH−Cl\ce{H-Cl}

With electronegativity values:

  • H (2.1) → δ+
  • Cl (3.0) → δ-

Dipole arrow: Points from δ+ to δ-

\ceH−>[δ+][δ−]Cl\ce{H ->[ \delta^+][\delta^-] Cl}

or

\ceH−>[+][−]Cl\ce{H ->[+][-] Cl}

Bond Strength

Factors Affecting Bond Strength

  1. Bond order: Triple > Double > Single
  2. Atomic size: Smaller atoms form stronger bonds
  3. Electronegativity difference: Greater difference can mean stronger bonds (to a point)

Bond Length vs. Bond Energy

Inverse relationship:

  • Shorter bond → Stronger bond → Higher bond energy
  • Longer bond → Weaker bond → Lower bond energy

Example: Carbon bonds

| Bond | Length (pm) | Energy (kJ/mol) | |------|-------------|-----------------| | C-C | 154 | 348 | | C=C | 134 | 614 | | C≡C | 120 | 839 |

As bond order increases: length decreases, energy increases

Lewis Structures and Bonding

Lewis structures show:

  • Valence electrons as dots
  • Bonds as lines (2 electrons per line)
  • Lone pairs as pairs of dots

Steps to draw:

  1. Count total valence electrons
  2. Connect atoms with single bonds
  3. Complete octets (start with outer atoms)
  4. Place remaining electrons on central atom
  5. Form multiple bonds if needed

Example: CO₂

Total valence electrons: 4 (C) + 6×2 (O) = 16

\ceO=C=O\ce{O=C=O}

  • C has 4 bonds (8 electrons) ✓
  • Each O has 2 bonds + 2 lone pairs (8 electrons) ✓
  • Total: 16 electrons ✓

Comparing Properties

| Property | Ionic | Covalent | Metallic | |----------|-------|----------|----------| | Elements | Metal + Nonmetal | Nonmetal + Nonmetal | Metal + Metal | | Electron transfer/sharing | Transfer | Sharing | Delocalized | | Melting point | High | Low to moderate | Variable | | Electrical conductivity (solid) | No | No | Yes | | Electrical conductivity (liquid) | Yes | No | Yes | | Solubility in water | Often soluble | Varies | Insoluble | | Hardness | Hard but brittle | Soft | Malleable |

Exceptions and Special Cases

1. Network Covalent Solids

  • Giant covalent structures (diamond, SiO₂)
  • Very high melting points
  • Very hard
  • Do not conduct electricity

2. Polar vs. Nonpolar Molecules

Bond polarity ≠ Molecular polarity

  • Molecule can have polar bonds but be nonpolar overall (CO₂, CCl₄)
  • Symmetry cancels out individual bond dipoles
  • Will discuss more in VSEPR theory

3. Transition Between Bond Types

Bonding is a continuum, not discrete:

  • No bond is 100% ionic (even NaCl has some covalent character)
  • No bond is 100% covalent (even H₂ has tiny dipole from quantum effects)

Summary

  1. Ionic bonds: Transfer of electrons, metal + nonmetal, ΔEN > 1.7
  2. Covalent bonds: Sharing of electrons, nonmetal + nonmetal, ΔEN < 1.7
    • Nonpolar: ΔEN < 0.4
    • Polar: 0.4 < ΔEN < 1.7
  3. Metallic bonds: Delocalized electrons, metal + metal
  4. Electronegativity difference predicts bond type
  5. Bond strength: Depends on bond order, atomic size, and electronegativity

📚 Practice Problems

1Problem 1medium

❓ Question:

Classify each of the following compounds as ionic, polar covalent, or nonpolar covalent: (a) NaCl, (b) H₂O, (c) Cl₂, (d) NH₃, (e) MgO. Explain your reasoning.

💡 Show Solution

Solution:

(a) NaCl - Ionic

  • Na (metal, EN = 0.93) + Cl (nonmetal, EN = 3.16)
  • ΔEN = 2.23 (>1.7 typically indicates ionic)
  • Complete electron transfer from Na to Cl

(b) H₂O - Polar covalent

  • H (EN = 2.20) + O (EN = 3.44)
  • ΔEN = 1.24 (between 0.5 and 1.7)
  • Unequal sharing of electrons, bent shape creates dipole moment

(c) Cl₂ - Nonpolar covalent

  • Cl (EN = 3.16) + Cl (EN = 3.16)
  • ΔEN = 0 (identical atoms)
  • Equal sharing of electrons, symmetric molecule

(d) NH₃ - Polar covalent

  • N (EN = 3.04) + H (EN = 2.20)
  • ΔEN = 0.84
  • Unequal sharing, trigonal pyramidal shape creates dipole

(e) MgO - Ionic

  • Mg (metal, EN = 1.31) + O (nonmetal, EN = 3.44)
  • ΔEN = 2.13 (>1.7)
  • Complete electron transfer from Mg to O

2Problem 2easy

❓ Question:

Classify the following bonds as ionic, polar covalent, or nonpolar covalent: (a) K-Cl, (b) C-O, (c) Cl-Cl. Use electronegativity values: K = 0.8, Cl = 3.0, C = 2.5, O = 3.5

💡 Show Solution

Solution:

Given: Electronegativity values and three bonds to classify Find: Bond type for each

Step 1: Calculate ΔEN for each bond

(a) K-Cl: ΔEN=∣3.0−0.8∣=2.2\Delta EN = |3.0 - 0.8| = 2.2

(b) C-O: ΔEN=∣3.5−2.5∣=1.0\Delta EN = |3.5 - 2.5| = 1.0

(c) Cl-Cl: ΔEN=∣3.0−3.0∣=0.0\Delta EN = |3.0 - 3.0| = 0.0

Step 2: Apply classification rules

| ΔEN Range | Bond Type | |-----------|-----------| | 0 - 0.4 | Nonpolar covalent | | 0.4 - 1.7 | Polar covalent | | > 1.7 | Ionic |

Step 3: Classify each bond

(a) K-Cl: ΔEN = 2.2

  • ΔEN > 1.7 → Ionic bond
  • K is metal, Cl is nonmetal ✓
  • K loses electron → K⁺
  • Cl gains electron → Cl⁻

(b) C-O: ΔEN = 1.0

  • 0.4 < ΔEN < 1.7 → Polar covalent bond
  • Both C and O are nonmetals ✓
  • O is more electronegative → partial negative charge (δ-)
  • C has partial positive charge (δ+)

(c) Cl-Cl: ΔEN = 0.0

  • ΔEN = 0 → Nonpolar covalent bond
  • Identical atoms share electrons equally
  • No charge separation
  • Example: Cl₂ molecule

Answer:

  • (a) K-Cl: Ionic
  • (b) C-O: Polar covalent
  • (c) Cl-Cl: Nonpolar covalent

Summary: Large ΔEN (>1.7) indicates electron transfer (ionic), moderate ΔEN (0.4-1.7) indicates unequal sharing (polar covalent), and small ΔEN (<0.4) indicates equal sharing (nonpolar covalent).

3Problem 3medium

❓ Question:

Classify each of the following compounds as ionic, polar covalent, or nonpolar covalent: (a) NaCl, (b) H₂O, (c) Cl₂, (d) NH₃, (e) MgO. Explain your reasoning.

💡 Show Solution

Solution:

(a) NaCl - Ionic

  • Na (metal, EN = 0.93) + Cl (nonmetal, EN = 3.16)
  • ΔEN = 2.23 (>1.7 typically indicates ionic)
  • Complete electron transfer from Na to Cl

(b) H₂O - Polar covalent

  • H (EN = 2.20) + O (EN = 3.44)
  • ΔEN = 1.24 (between 0.5 and 1.7)
  • Unequal sharing of electrons, bent shape creates dipole moment

(c) Cl₂ - Nonpolar covalent

  • Cl (EN = 3.16) + Cl (EN = 3.16)
  • ΔEN = 0 (identical atoms)
  • Equal sharing of electrons, symmetric molecule

(d) NH₃ - Polar covalent

  • N (EN = 3.04) + H (EN = 2.20)
  • ΔEN = 0.84
  • Unequal sharing, trigonal pyramidal shape creates dipole

(e) MgO - Ionic

  • Mg (metal, EN = 1.31) + O (nonmetal, EN = 3.44)
  • ΔEN = 2.13 (>1.7)
  • Complete electron transfer from Mg to O

4Problem 4hard

❓ Question:

(a) Explain why ionic compounds have high melting points while covalent compounds generally have low melting points. (b) Why does diamond have an extremely high melting point (~3550°C) despite being a covalent compound?

💡 Show Solution

Solution:

(a) Melting point comparison:

Ionic compounds (high MP):

  • Strong electrostatic attractions between oppositely charged ions
  • These ionic bonds extend throughout the crystal lattice (lattice energy)
  • Requires large amounts of energy to overcome these attractions
  • Example: NaCl melts at 801°C

Molecular covalent compounds (low MP):

  • Strong covalent bonds WITHIN molecules
  • Weak intermolecular forces BETWEEN molecules (dipole-dipole, London dispersion)
  • Only need to overcome weak IMFs to melt
  • Example: H₂O melts at 0°C, CH₄ melts at -182°C

(b) Diamond's exception: Diamond is a network covalent (atomic) solid, not a molecular covalent compound.

Structure: Each carbon atom is sp³ hybridized and covalently bonded to 4 other carbons in a tetrahedral arrangement, creating a 3D network extending throughout the crystal.

Why high MP: To melt diamond, you must break strong C-C covalent bonds (not just weak IMFs). The entire crystal is essentially one giant molecule held together by covalent bonds.

Other examples of network covalent solids: SiO₂ (quartz), SiC (silicon carbide), graphite

5Problem 5hard

❓ Question:

(a) Explain why ionic compounds have high melting points while covalent compounds generally have low melting points. (b) Why does diamond have an extremely high melting point (~3550°C) despite being a covalent compound?

💡 Show Solution

Solution:

(a) Melting point comparison:

Ionic compounds (high MP):

  • Strong electrostatic attractions between oppositely charged ions
  • These ionic bonds extend throughout the crystal lattice (lattice energy)
  • Requires large amounts of energy to overcome these attractions
  • Example: NaCl melts at 801°C

Molecular covalent compounds (low MP):

  • Strong covalent bonds WITHIN molecules
  • Weak intermolecular forces BETWEEN molecules (dipole-dipole, London dispersion)
  • Only need to overcome weak IMFs to melt
  • Example: H₂O melts at 0°C, CH₄ melts at -182°C

(b) Diamond's exception: Diamond is a network covalent (atomic) solid, not a molecular covalent compound.

Structure: Each carbon atom is sp³ hybridized and covalently bonded to 4 other carbons in a tetrahedral arrangement, creating a 3D network extending throughout the crystal.

Why high MP: To melt diamond, you must break strong C-C covalent bonds (not just weak IMFs). The entire crystal is essentially one giant molecule held together by covalent bonds.

Other examples of network covalent solids: SiO₂ (quartz), SiC (silicon carbide), graphite

6Problem 6medium

❓ Question:

Explain why sodium chloride (NaCl) has a much higher melting point (801°C) than water (H₂O, melting point 0°C), even though both compounds involve bonding between atoms.

💡 Show Solution

Solution:

Given: NaCl melting point = 801°C, H₂O melting point = 0°C Find: Explain the large difference

Step 1: Identify bond types

NaCl:

  • Na: metal (electronegativity = 0.9)
  • Cl: nonmetal (electronegativity = 3.0)
  • ΔEN = 2.1 → Ionic bond

H₂O:

  • H: nonmetal (electronegativity = 2.1)
  • O: nonmetal (electronegativity = 3.5)
  • ΔEN = 1.4 → Polar covalent bond

Step 2: Describe structure differences

NaCl:

  • Forms ionic crystal lattice
  • Each Na⁺ surrounded by 6 Cl⁻ ions
  • Each Cl⁻ surrounded by 6 Na⁺ ions
  • 3D network of strong electrostatic attractions
  • Must break many strong ionic bonds to melt

H₂O:

  • Forms discrete molecules (H-O-H)
  • Covalent bonds within molecules (strong)
  • Hydrogen bonds between molecules (weaker)
  • Must break intermolecular forces to melt, NOT covalent bonds

Step 3: Compare forces that must be overcome

To melt NaCl:

  • Must overcome ionic bonds between Na⁺ and Cl⁻
  • Very strong electrostatic attractions (∝ 1/r²)
  • Lattice energy is very high (~787 kJ/mol)

To melt H₂O:

  • Must overcome hydrogen bonds between molecules
  • Much weaker than ionic or covalent bonds (~20 kJ/mol per H-bond)
  • Covalent O-H bonds remain intact when ice melts

Step 4: Energy comparison

Breaking ionic bonds in NaCl:

  • Requires ~787 kJ/mol (lattice energy)
  • Very high temperature needed (801°C)

Breaking H-bonds in ice:

  • Requires ~6 kJ/mol (enthalpy of fusion)
  • Low temperature needed (0°C)

Answer:

NaCl has a much higher melting point because melting requires breaking strong ionic bonds in the crystal lattice, while melting ice only requires breaking weak hydrogen bonds between water molecules. The covalent bonds within water molecules remain intact during melting.

Key distinction:

  • Intramolecular forces (ionic/covalent bonds): very strong
  • Intermolecular forces (H-bonds, dipole-dipole, London): weaker

Melting ionic compounds requires breaking intramolecular forces, while melting molecular compounds typically only requires breaking intermolecular forces.

Additional note: If you had to break the O-H covalent bonds in water (like in decomposition), that would require much more energy (~460 kJ/mol per bond), comparable to ionic bonds.

7Problem 7hard

❓ Question:

Diamond (C) and graphite (C) are both composed entirely of carbon atoms but have very different properties. Diamond is extremely hard with a very high melting point (~3550°C), while graphite is soft and has a lower melting point (~3600°C but sublimes). Both conduct heat well, but only graphite conducts electricity. Explain these differences in terms of bonding and structure.

💡 Show Solution

Solution:

Given: Two allotropes of carbon with different properties Find: Explain differences using bonding and structure

Step 1: Describe diamond structure

Diamond:

  • 3D network covalent structure
  • Each C atom bonded to 4 other C atoms
  • Tetrahedral geometry (sp³ hybridization)
  • All electrons in localized C-C single bonds
  • No free electrons

Bonding diagram: \ceC−C−C−C\ce{C - C - C - C} (in 3D tetrahedral network)

Each C: 4 covalent bonds, no lone pairs

Step 2: Describe graphite structure

Graphite:

  • Layered structure
  • Each C atom bonded to 3 other C atoms (in plane)
  • Trigonal planar geometry (sp² hybridization)
  • Delocalized π electrons above and below planes
  • Fourth electron per C is delocalized
  • Layers held by weak London dispersion forces

Bonding diagram: \ceC=C−C=C\ce{C=C-C=C} (resonance, delocalized π system)

Each C: 3 σ bonds, 1 delocalized π electron

Step 3: Explain hardness

Diamond - extremely hard:

  • Must break strong C-C covalent bonds (348 kJ/mol)
  • 3D network means cutting requires breaking many bonds
  • All atoms rigidly locked in position
  • No planes of weakness

Graphite - soft:

  • Layers can slide past each other
  • Only weak London forces between layers (~10 kJ/mol)
  • C-C bonds within layers are strong
  • But layers separate easily (like a deck of cards)
  • This is why graphite is used as lubricant

Step 4: Explain electrical conductivity

Diamond - does NOT conduct:

  • All 4 valence electrons per C in localized bonds
  • No free electrons to carry charge
  • All electrons held tightly in σ bonds
  • Insulator

Graphite - conducts electricity:

  • Delocalized π electrons above/below planes
  • One electron per C is mobile within layers
  • Electrons can move parallel to layers
  • Similar to metallic bonding (electron sea within layers)
  • Conductor

Step 5: Explain melting point

Both have very high melting points:

  • Diamond: ~3550°C (actually sublimes at very high T)
  • Graphite: ~3600°C (sublimes above ~3700°C at low pressure)

Why both are high:

  • Both must break strong C-C covalent bonds
  • Network covalent structures throughout
  • Diamond: 3D network
  • Graphite: 2D networks in each layer

Why similar values:

  • Both involve breaking C-C bonds (~348 kJ/mol)
  • Graphite layers are strongly bonded internally
  • Weak interlayer forces don't affect melting (that's not what melts)

Step 6: Explain thermal conductivity

Both conduct heat well:

  • Diamond: Atoms in rigid 3D network vibrate and transfer energy
  • Graphite: Delocalized electrons + rigid layers transfer energy

Answer Summary:

| Property | Diamond | Graphite | Explanation | |----------|---------|----------|-------------| | Structure | 3D network, sp³ | Layers, sp² | Hybridization difference | | Hardness | Extremely hard | Soft | 3D network vs. sliding layers | | Electrical conductivity | No | Yes (in plane) | No free e⁻ vs. delocalized π electrons | | Melting point | Very high (~3550°C) | Very high (~3600°C) | Both require breaking C-C bonds | | Thermal conductivity | High | High | Rigid structure + (for graphite) mobile electrons |

Key insight: Same element (carbon), but different bonding arrangements (sp³ vs. sp²) and structures (3D network vs. layered) lead to dramatically different macroscopic properties. This demonstrates that properties depend not just on atomic composition, but on how atoms are connected.

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