Chapter 4: Chemical Bonding and Structure

Question 1

State the 3 types of bonding

Answer 1

1) Ionic: between metals and non-metals
2) Covalent: between non-metals
3) Metallic: metals

Rule: elements far apart on the periodic table form ionic compounds (vastly different electronegativities), elements close together form covalent compounds (similar electronegativities)

Question 2

Define cations, anions and transition metals

Answer 2

CATIONS: metals that lose electrons
ANIONS: metals that gain electrons
- Atoms gain/lose electrons to become isoelectronic with the nearest noble gas

*Exception: Transition metals
- Can form more than 1 ion, and may not become isoelectronic with a noble gas

Question 3

Explain how ionic compounds are formed

Answer 3

IONIC COMPOUNDS: formed when the electrons lost by a cation are donated and gained by an anion (electron transfer)
- No. of electrons lost = no. of electrons gained

Question 4

State the formulas of common polyatomic ions and work out the formulas of ionic compounds from their ionic charges

Answer 4

Common polyatomic ions:
1) Hydroxide: OH⁻
2) Hydrogencarbonate: HCO₃⁻
3) Nitrate: NO₃⁻ (vs. nitride N³⁻ and nitrite NO₂⁻ )
4) Carbonate: CO₃²⁻
5) Sulfate: SO₄²⁻ (vs. sulfide S²⁻ and sulfite SO²⁻)
6) Phosphate (PO₄³⁻)

Formulas of ionic compounds from ionic charges:
Criss-cross (charge of other ion becomes number of atoms)

Question 5

State the structure and bonding of an ionic compound

Answer 5

An ionic compound has a giant ionic lattice structure, where strong electrostatic forces of attraction exist between oppositely-charged cations and anions

Question 6

Explain the 4 physical properties of ionic compounds in terms of structure and bonding

Answer 6

1) High melting and boiling points
- Large amount of energy needed to overcome strong EFA between oppositely-charged ions in an ionic lattice structure
- Eg. M.P/B.P of MgO vs NaCl:
M.P of MgO will be higher than NaCl, as the EFA between the Mg²⁺ and O²⁻ ions are stronger than the EFA between Na⁺ and Cl⁻ ions, requiring more energy to overcome

2) Low volatility
- Large amount of energy needed to overcome strong EFA between ions

3) Non-conductor of electricity in the solid state, conductor in the molten/aqueous state
- Solid state: Strong EFA between ions hold them in their fixed positions in the ionic lattice structure –> lack of mobile charge carriers to conduct electricity
- Molten/aqueous state: Ionic lattice structure is broken –> strong EFA holding ions in their fixed places are overcome –> free-moving ions are able to act as mobile charge carriers

4) Soluble in water, not soluble in organic solvents
- Water: organic solvent –> releases energy as it hydrates the ions –> pays back the energy needed to break apart the ionic lattice
- Non-polar solvents: do not pay back the large amounts of energy needed to break the ionic lattice

Question 7

State the structure and bonding in metals

Answer 7

A metal has strong electrostatic forces of attraction between the positive ions in the metallic lattice and the sea of delocalised electrons

Question 8

Explain the 4 physical properties of metals in terms of structure and bonding

Answer 8

1) High M.P/B.P
Eg. M.P of Na vs Mg: Mg has a higher M.P:
- There are 2 delocalised electrons per cation in Mg, as compared to 1 delocalised electron per cation in Na, resulting in stronger EFA between the metallic lattice and the delocalised electrons that require more energy to overcome
- The Mg²⁺ ion has a smaller IR than Na⁺ –> delocalised electrons are closer to and are more strongly attracted to the nucleus of the cation in Mg

2) Ductile/malleable
- Non-directional metallic bonding: Metal ions in the lattice attract the delocalised electrons in all directions, and this attractive force is undisrupted by layers of the metal lattice sliding over each other
- Vs. brittle ionic solids: displacement of one layer of the ionic lattice will cause like charges to repel and break apart the lattice

3) Good conductor of heat

4) Good conductor of electricity
- Delocalised electrons are able to act as mobile charge carriers

Question 9

Define alloys and explain their difference in properties from pure metals

Answer 9

ALLOYS: homogenous mixture of 2 or more metals
- Usually stronger and stiffer than pure metals: a differently sized cation in the metallic lattice will prevent layers of cations from sliding over each other

Question 10

Define a covalent bond, state the different types of covalent bonds and define a lone pair

Answer 10

COVALENT BOND: electrostatic attraction between the negatively-charged shared pair of electrons and the positively-charged nuclei of the atoms that are bonded

• Single bond: 1 shared pair of electrons
• Multiple bonds: 2 or 3 shared pairs of electrons
• Lone pair: pair of electrons uninvolved in the covalent bonds

Question 11

Explain the inverse relationship between bond strength and bond length

Answer 11

Eg. A triple bond is the strongest but shortest bond
- As the AR increases, the distance between the electron pair and the nuclei of the atoms increase, weakening the attraction between them, creating a longer but weaker bond

Question 12

Define electronegativity and use electronegativity to predict polarity

Answer 12

ELECTRONEGATIVITY: a measure of attraction of an atom in a molecule for the electron pair in the covalent bond of which it is a part
- However, electronegativity of the 2 atoms sharing the electron pair may not be equal (the nuclei may not attract the shared electron pair equally)

NON-POLAR BOND:
- Atoms have equal electronegativity (attract the electron pair equally)

POLAR BOND:
- Atoms have different electronegativities (one atom attracts the electron pair more strongly than the other, causing the electron pair to lie more closely to that atom)
- Results in the formation of a dipole moment (δ⁻ for the more EN atom, δ⁺ for the less EN atom)

Question 13

Explain the octet rule and exceptions to the octet rule

Answer 13

OCTET RULE: most covalent compounds have a full valence shell with 8 electrons

*Exceptions:
1) BF₃ (B only has 3 electrons, and can only form 3 covalent bonds with F to have 6 electrons)
2) BeCl₂ (Be only has 2 electrons and can only form 2 covalent bonds with Cl to have 4 electrons)

3) SF₆ (S has 6 electrons and can expand its octet by forming 6 covalent bonds with F to have 12 electrons)

Question 14

Define a coordinate/dative covalent bond

Answer 14

Coordinate/dative covalent bond: when both electrons in the shared electron pair come from the same atom
- Must have a donor atom with a lone pair of electrons
- Must have a receiver atom with an empty orbital

Eg. NH₃ (N has 5 electrons: 3 existing covalent bonds + 1 lone pair) + H⁺ –> NH₄⁺

Eg. H₂O (O has 6 electrons: 2 existing covalent bonds + 2 lone pairs) + H⁺ –> H₃O⁺

Eg. NH₃ (N has 5 electrons: 3 existing covalent bonds + 1 lone pair) + BF₃ (has 6 electrons)

Eg. CO (O donates 2 electrons to C to form a triple bond)

Question 15

Draw Lewis structures of covalent compounds
(Eg. CO₃²⁻, PCl₄⁺)

Answer 15

CO₃²⁻: C forms a double bond with 1 O and single bonds with the other 2 Os, which gain 1 electron each resulting in a 2- charge

PCl₄⁺: P has 5 electrons, but loses 1 after forming 4 single bonds with Cl, resulting in a + charge

Question 16

Define resonance structures

Answer 16

RESONANCE STRUCTURES: when more than 1 Lewis structure can be drawn for the same covalent compound by alternating the position of the double bonds and lone pairs

*2 possible structures for SO₂:
1) S does not expand its octet by forming a double bond with 1 O and a coordinate bond with the other O
2) S expands its octet by forming double bonds with both Os

Question 17

Identify the shapes and bond angles of covalent molecules

Answer 17

1) Draw out the Lewis structure of the compound
2) Count the number of electron domains on the central atom (single bonds, multiple bonds and lone pairs = 1 electron domain)
3) Identify the basic shape
- Linear: 180°
- Trigonal planar: 120°
- Tetrahedral: 109.5°
4) Determine actual shape based on number of lone pairs
- 3 b.p + 0 l.p: Trigonal Planar (120°)
- 2 b.p + 1 l.p: Bent (117°)
- 4 b.p + 0 l.p: Tetrahedral (109.5°)
- 3 b.p + 1 l.p: Trigonal pyramidal (107.3°)
- 2 b.p + 2 l.p: Bent (104.5°)

Question 18

Predict whether a molecule will be polar or non-polar using its molecular shape

Answer 18

• POLAR BOND: when the bonded atoms have different electronegativities, creating a dipole moment
• POLAR MOLECULE: must have an overall dipole moment
  (polar bonds can cancel out if the molecule has a symmetrical shape)

Non-polar molecules:
1) Linear
2) Trigonal planar
3) Tetrahedral

Polar molecules:
1) Bent (117°)
2) Trigonal pyramidal (107.3°)
3) Bent (104.5°)

Question 19

State the 4 allotropes of carbon

Answer 19

1) Diamond
2) Graphite
3) Graphene
4) Fullerene

Structures: Macromolecular structures (carbon atoms are joined together differently)

Question 20

Use the structures and bonding of the macromolecular structures to explain their physical properties

Answer 20

DIAMOND:
- Each C atom is bonded tetrahedrally to 4 other C atoms
1) Hard
2) High M.P/B.P
3) Insoluble in water and non-polar solvents
(Strong covalent bonds between C atoms that require large amounts of energy to overcome)
4) Non-conductor of electricity
(Lack of mobile charge carriers)

GRAPHITE:
- Each C atom is joined to 3 others in a trigonal planar array
- Layer structure (strong covalent bonds between C atoms, weak IMFA between layers)
1) High M.P/B.P
2) Insoluble in water and non-polar structures
(Strong covalent bonds that require large amounts of energy to overcome)
3) Good lubricants (weak IMFA)
4) Conducts electricity (electron lone pairs not involved in the 3 covalent bonds can act as mobile charge carriers)

GRAPHENE:
similar to graphite

FULLERENE:
1) Insoluble in water but soluble in some organic solvents
2) Non-conductor of electricity (electrons cannot move from one molecule to the next)

SILICON DIOXIDE (SiO₂):
1) High M.P/B.P

Question 21

Distinguish between intramolecular and intermolecular forces

Answer 21

Intramolecular forces: covalent bonds joining atoms to form molecules

Intermolecular forces: weaker London dispersion forces between molecules (overcome: results in change of state)

Question 22

Explain how London dispersion forces arise in non-polar molecules

Answer 22

LONDON FORCES: temporary instantaneous dipole-induced dipole interactions (id-id)
- Electrons of an atom are in constant motion and are never symmetrically distributed about the nucleus, creating an instantaneous dipole in an atom which will induce an opposite dipole in a neighbouring atom, creating attractive forces between atoms

London forces grow stronger as:
1) Number of electrons increase
- Larger instantaneous dipole –> larger induced dipole –> stronger attractive force between atoms
- Increased AR –> weaker attractive force between nucleus and valence electrons –> larger induced dipole
2) As Mᵣ increases

Question 23

Explain the differing intermolecular forces in polar molecules

Answer 23

POLAR MOLECULES: both London forces and permanent dipole-permanent dipole (pd-pd) interactions exist between molecules

• Polar and non-polar molecule of similar Mᵣ:
  Polar molecule will have higher M.P/B.P, as pd-pd interactions are stronger than London forces

Question 24

Explain hydrogen bonding

Answer 24

HYDROGEN BONDING: intramolecular or intermolecular bonding between H atoms bonded to F/O/N atoms and a lone pair on a F/O/N atom
- Molecules that are only able to participate in intramolecular hydrogen bonding will have lower M.P/B.P than molecules able to participate in intermolecular hydrogen bonding

Question 25

Predict the relative M.P/B.P of substances

Answer 25

In substances of similar Mᵣ:
Hydrogen bonding > pd-pd > London forces

Question 26

Predict solubility of substances

Answer 26

General: a solute will dissolve in a solvent if the types of intermolecular forces are similar

Whether a solute will dissolve in a solvent depends on:
1) Energy needed to overcome intermolecular forces in solute and solvent
2) Energy released when intermolecular forces are formed between solute and solvent molecules
3) If energy released can pay back the energy needed

Eg.
- Pentane is soluble in hexane, as both contain London forces, where the energy released when London forces are formed between pentane and hexane pay back the energy needed to overcome the London forces in pentane and hexane

• Pentane is not soluble in water, as the energy needed to overcome the London forces in pentane and the hydrogen bonds in water are not paid back when London forces are formed between pentane and water (since hydrogen bonds are stronger than London forces)
• Usually, only substances that can participate in hydrogen bonding can dissolve in water
  *Exception: Long-chained alcohols
• Hydrogen bonding present in the O-H group, but length of hydrocarbon chain prevents both sides of the chain from hydrogen bonding to each other
• Energy needed to break the hydrogen bonds between water molecules are not paid back by the London forces formed between water molecules and the hydrocarbon portion of the alcohol