C2 Bonding

Question 1

Electronegativity across the period

Answer 1

Nuclear charge increases while shielding effect remains relatively constant. Effective nuclear charge increases, hence electronegativity increases across the period.

Question 2

Electronegativity down the group

Answer 2

Elements in the same group experience roughly the same effective nuclear charge as both nuclear charge and shielding effect increase down the group. However, as the number of quantum shells increases, atomic radius increases. Hence, electronegativity decreases down the group.

Question 3

Factors affecting degree of covalent character in ionic bonding

Answer 3

polarising power of the cation
- Cations that are small and highly charged have high charge densities and high polarising power.

polarisability of the anion
- Anions that are relatively large have high polarisability. Their valence electrons are further from and less strongly attracted by the nucleus so the electron cloud is easily distorted by a cation, resulting in greater covalent character in the ionic bonding.

• Metal + non-metal atom -> metal cation has high charge and small radius -> high charge density + high polarising power
• Anion has large radius -> high polarisability
• Cation attracts outermost electrons or electron cloud of anion + distorts or polarises electron cloud of anion -> partial sharing of electrons -> ionic bonding with covalent character or even a covalent molecule

Question 4

Factors affecting dispersion forces (instantaneous dipole-instantaneous dipole attraction)

Answer 4

the number of electrons in the molecule
-> The larger the number of electrons, the more polarisable the electron cloud, the stronger the dispersion
forces.

the surface area of contact between adjacent molecules
-> The larger the surface area of contact, the more easily induced dipoles are formed, the stronger the dispersion forces.

Question 5

Conditions for hydrogen bonding to occur (sometimes might test)

Answer 5

the molecule must contain a hydrogen atom bonded to a highly electronegative atom (F, O, N),

must have an atom (F, O, N) with a lone pair of electrons.

Question 6

factors affecting strength and extent of hydrogen bonding

Answer 6

Strength:
-> The electronegativity of the atom that H atom is bonded to
More electronegative, the stronger the hydrogen bonding

Extent:
-> the number of H atoms bonded to F, O or N and the number
of lone pairs available.
More lone pairs, the more extensive the hydrogen bonding

Question 7

Using hydrogen bonding, explain why Boiling point: HF > NH3: (Strength)

Answer 7

F is more electronegative than N, the electronegativity difference between F and H is larger than that between N and H, hence the +ve charge on H in HF is larger than that in NH3. More energy is needed to overcome the stronger hydrogen bonding between HF molecules than that between NH3 molecules hence HF has a higher boiling point.

Question 8

Using hydrogen bonding, explain why Boiling point: H2O > NH3 and H2O > HF (Extensivity)

Answer 8

H2O has two lone pairs of electrons and two electron deficient hydrogen atoms hence H2O can form on average two hydrogen bonds per molecule. On the other hand, NH3 has only one lone pair while HF has only one electron deficient hydrogen atom, hence NH3 and HF can only form one hydrogen bond per molecule. More energy is needed to overcome the more extensive hydrogen bonding in H2O than that in NH3 and HF hence H2O has a higher boiling point.

Question 9

Explain why isomer 2-nitrophenol has a lower boiling point (with hydrogen bonding)

Answer 9

2-nitrophenol is able to form intramolecular hydrogen bonding so the intermolecular hydrogen bonding is less extensive. Its isomer 4-nitrophenol is only able to form intermolecular hydrogen bonding. Less energy is needed to break the less extensive intermolecular hydrogen bonding in 2-nitrophenol hence it has a lower boiling point.

*Intermolecular hydrogen bonding to be broken for boiling

Question 10

Dimerisation of carboxylic acids in non-polar solvent

Answer 10

In water, ethanoic acid (CH3CO2H) forms hydrogen bonds with H2O molecules.
In non-polar solvent such as benzene (unable to form hydrogen bonding), the carboxylic acid molecules form dimers via hydrogen bonding. Hence, the observed Mr is 120 instead of 60 (expected value for CH3CO2H).

Question 11

Why ice is less dense than liquid water

Answer 11

In ice, each H2O molecule is hydrogen bonded to four other molecules in a roughly tetrahedral arrangement, producing an open lattice, with empty spaces between the H2O molecules. The more random arrangement of hydrogen bonding in liquid water takes up less space as the H2O molecules are closer together.
For the same mass, the lattice structure of ice occupies a larger volume, hence the lower density.

Question 12

Why ethanol can dissolve in water

Answer 12

Energy released when forming hydrogen bonds between ethanol and water molecules is comparable to the energy needed to overcome the existing hydrogen bonding between ethanol molecules and the existing hydrogen bonding between water molecules.

Question 13

Why hexane CH3(CH2)4CH3, is insoluble in water

Answer 13

The energy released when forming weak permanent dipole-induced dipole interactions between polar water and non-polar hexane molecules is insufficient to overcome the dispersion forces between hexane molecules and the stronger hydrogen bonding between water molecules.

Question 14

Why Sodium chloride, NaCl, dissolves in water

Answer 14

• Both Na+ and Cl– ions are able to form ion-dipole attractions with water molecules. The ion-dipole attractions are relatively strong and very extensive. The energy released when forming ion-dipole interactions between Na+ and H2O and between Cl–
• and H2O is comparable to the energy required to overcome the hydrogen bonding between water molecules and ionic bonding between Na+ and Cl– ions in NaCl.

Question 15

Why Some ionic compounds are insoluble in water (eg AgCl)

Answer 15

The energy released from ion-dipole interactions depends on the charge densities of the ions. Ions with lower charge densities (e.g. Ag+, Cl−) form weaker ion-dipole interactions so less energy is released

Question 16

Determine whether
I. CH3CH2CO2− Na+
II. CH3CH(OH)CH3
III. CH3COCH3
IV. CH3CH2CH2CH3
V. CH3CH(CH3)2
is soluble in water and the electrical conductivity of the aqueous solution formed (if any).

Answer 16

CH3CH2CO2− Na+ is soluble in water because its ions can form ion-dipole interactions with H2O molecules. Since the aqueous solution contains ions which act as mobile charge carriers, the aqueous solution conducts electricity.

CH3CH(OH)CH3 and CH3COCH3 are soluble in water because the OH group in CH3CH(OH)CH3 and O atom in CH3COCH3 forms hydrogen bonding with H2O molecules. The aqueous solution does not conduct electricity because there are no mobile charge carriers such as electrons or ions. Valence electrons in CH3CH(OH)CH3 and CH3COCH3 are not mobile as they are involved in covalent bonding.

CH3CH2CH2CH3 and CH3CH(CH3)2 (hydrocarbons) are non-polar molecules which do not form favourable interactions with water molecules. Energy released from permanent dipole-induced dipole interactions between polar water molecules and non-polar hydrocarbon molecules is insufficient to overcome instantaneous dipole-induced dipole interactions (or dispersion forces) between hydrocarbon molecules and stronger hydrogen bonding between water molecules. CH3CH2CH2CH3 and CH3CH(CH3)2 do not dissolve in water and hence do not form aqueous solutions.

Question 17

Propanone (CH3COCH3) and water are miscible. Explain this with a suitable diagram to show the interactions formed between propanone and water.

Answer 17

CH3COCH3 forms hydrogen bonding with water. The energy released is able to compensate the energy needed to overcome the weaker permanent dipole-permanent dipole interactions between CH3COCH3 molecules, and the hydrogen boding between water molecules.
+ diagram

Question 18

Explain why solid iodine is soluble in organic solvent such as CCl4 but insoluble in water

Answer 18

Both I2 and CCl4 are non-polar simple molecules. The energy released when forming dispersion forces between I2 and CCl4 molecules is comparable to the energy needed to overcome existing disperson forces between I2 molecules and between CCl4 molecules. Hence, I2 is soluble in CCl4.

Water exists as polar simple molecules. The enery released when forming permanent dipole-nduced dipole interactions between polar water and non-polar I2,molecules is insufficient to overcome the dispersion forces between I2 molecules and the stronger hydrogen bonding betwen water molecules. Hence, I2 is insoluble in water.

Question 19

Simple molecular lattices: physical properties, forces of attraction

Answer 19

FOA: Held by comparatively weak intermolecular forces

Physical properties:
Low melting points and low boiling points
-> Element Argon has a simple atomic structure while hydrogen chloride has a simple molecular structure.
Stronger permanent dipole-permanent dipole attractions (and dispersion forces) exist between polar HCl molecules. Less energy is needed to overcome the weaker dispersion forces between argon atoms so it has a lower boiling point.

Soluble in organic solvents
-> organic solvents eg propanone (CH3COCH3), tetrachloromethane (CCl4) and benzene (C6H6)

May be soluble in water (only some)
-> some can form hydrogen bonds with water

Unable to conduct electricity
-> no presence of mobile charge carriers
-> some simple covalent substances are able to react with water to give aqueous ions. This allows the resulting solution to conduct electricity, for example: HCl(g) + H2O(l) → Cl–(aq) + H3O+(aq).

Question 20

Giant Molecular (or Giant Covalent) Lattices FOA and physical properties

Answer 20

FOA: held together by strong covalent
bonds

Diamond: each (sp3 hybridised) C atom is covalently bonded to four other C atoms, in a tetrahedral
arrangement.

Silicon(IV) oxide (also known as quartz), each Si atom is covalently bonded to four O atoms, and each O atom is bonded to two Si atoms. The formula SiO2 is an empirical formula.

Graphite, each (sp2 hybridised) C atom is covalently bonded to three other C atoms, forming a two-dimensional layer of hexagonal carbon rings. Each C atom in graphite has a 2p orbital containing one electron

Question 21

Explain in terms of structure and bonding the following:
(a) Argon has a lower boiling point than hydrogen chloride.
(b) Hydrogen iodide has a higher boiling point than hydrogen bromide.

Answer 21

• Argon has a simple atomic structure while hydrogen chloride has a simple molecular structure.
• Stronger permanent dipole-permanent dipole attractions (and dispersion forces) exist between polar HCl molecules*.
• Less energy is needed to overcome the weaker dispersion forces between argon atoms* so it has a lower boiling point.
• Both hydrogen iodide and hydrogen bromide are polar simple covalent molecules.
• Permanent dipole-permanent dipole attractions and dispersion forces exist between the molecules.
• HI molecules have more electrons so more energy is needed to overcome the stronger dispersion forces between the molecules. Hence, HI has a higher boiling point than HBr.

Strategy:
• State the structures
• Identify the type of attractions broken during boiling; keywords “between HCl molecules”, “between Ar atoms”
• Compare the strength of relevant attractions (pd-pd vs dispersion forces; HCl & Ar have comparable dispersion forces due to the same no. of electrons) & energy needed

Strategy:
• State the structures
• Identify the type of attractions broken during boiling; keywords “between molecules” or “intermolecular”
• Compare the strength of relevant attractions (in this case dispersion forces; pd-pd attraction is “irrelevant” as H─Br bond is more polar than H─I bond so pd-pd between HBr molecules is stronger which would result in a higher, rather than a lower, boiling point for HBr) & energy needed.

Question 22

Suggest the electronegativity for Se by comparison with that given for oxygen. EYA. [2]

Answer 22

• Se has an electronegativity value of 2.6. (accept any value between 0.7 – 3.5)
• Both the nuclear charge and shielding effect are greater for Se than O, causing the effective nuclear charge to be similar for Se and O.
• However, Se has more quantum shells than O, thus the nucleus of Se attracts electrons in a covalent bond more poorly towards itself, so Se’s electronegativity value is less than 3.5 / smaller than that of O’s.
• Se has fewer quantum shells than Fr, so Se attracts electrons in a covalent bond towards itself more strongly as compared to Fr, resulting in a larger electronegativity value / more than 0.7 for Se.

State a value, specific
Nuclear charge, shielding effect, effective nuclear charge
Number of quantum shells
Link to physical property: ability of the atom to attract electrons in a covalent bond to its nucleus

Question 23

Explain, in terms of structure and bonding, the differences in melting points of NaCl and HCl.

Answer 23

A has <describe> while B has <describe>. [giant/simple ionic / covalent / metallic / atomic structure / lattice]
<Larger/Smaller> amount of energy is required to overcome the <strength> <describe> between (particles of A) than the <strength> <describe> between (particles of B).
Hence, A has a higher/lower mp / bp than B.</describe></strength></describe></strength></describe></describe>

Answering techniques:
- Specific to Question
- Compare and explain differences in melting points of NaCl vs HCl
- Describe structure and bonding of NaCl
- Describe structure and bonding of HCl
- Compare strengths of bonding that is broken
[Link to physical property] Compare energy needed to overcome the type of bonding
- Correctly states the structure of both compounds
- Correctly identifies the bonding that is overcome when each compound melts
- Correctly compares the strengths of bonding and hence energy needed to overcome the bonding

NaCl has giant ionic lattice structure while HCl has simple molecular / simple covalent structure.
Larger amounts of energy is required to overcome the stronger ionic bonding between Na+ and Cl− ions for NaCl than the weaker permanent dipole-permanent dipole interactions and dispersion forces between molecules of HCl.
Hence, NaCl has a higher melting point than HCl.

Question 24

Diamond’s structure and physical properties

Answer 24

Structure: giant molecular; each (sp3 hybridised) C atom is covalently bonded to four other C atoms, in a tetrahedral arrangement.

Electrical conductivity:
-> unable to conduct electricity
-> there are no mobile charge carriers. All the valence electrons are
held between the atoms and are not free to move. Graphite is a good conductor of electricity (and heat) as the delocalised electrons are free to move along the layers

Hardness: very hard, strong and non-malleable
-> rigid structure held tgt by strong covalent bonds

Solubility in solvents: insoluble in water and organic solvents
-> Attractions between the solvent
molecules and carbon atoms are not strong enough to overcome the strong covalent bonds between the carbon atoms.

Usage: cutting tools and oil rig drills

Question 25

Graphite’s structure and physical properties

Answer 25

Giant molecular structure. Within each layer: - each C atom is covalently bonded to three other C atoms, forming a layer of hexagonal carbon rings. - Overlap of 2p orbitals (of each C atom) give rise to delocalised bonding system. Between layers: - dispersion forces hold the layers together good conductor of electricity (and heat) -> the delocalised electrons are free to move along the layers -> The delocalised electrons are unable to jump from one layer to the next so graphite does not conduct electricity in the direction at right angles to the layers. soft and slippery -> layers can slide over each other due to the dispersion forces between them insoluble in water and organic solvents -> Attractions between the solvent molecules and carbon atoms are not strong enough to overcome the strong covalent bonds between the carbon atoms. Examples: batteries and electrolysis lubricant and pencils

Question 26

Across period 2, describe the structures, bonding, MPs and electrical conductivity.

Answer 26

Na -> Al: -> Giant metallic structures -> Strong metallic bonding between cations and the sea of delocalised valence electrons - MP: relatively high. From Na to Al, melting point increases as more energy is needed to overcome the stronger metallic bonding due to -> the increase in number of valence electrons contributed per atom -> the increase in charge density of the cation (due to increasing charge and decreasing ionic radius/cationic size). - Good electrical conductors -> delocalised electrons act as mobile charge carriers -> Number of valence electrons contributed per atom increases from Na to Al, thus electrically conductivity increases. —— Si: Giant molecular structure Many strong covalent bonds between Si atoms MP: very high -> Si has a giant molecular structure and a very high melting point as a lot of energy is needed to overcome the many strong covalent bonds between the Si atoms during melting. -> Semiconductor of electricity. Si is a metal load (element with both properties of metals and non-metals) P4 -> Cl2: Simple molecular structures - Weak dispersion forces between the molecules (explain mp and bp) - Covalent bonds holding the valence electrons. (Explains electrical conductivity) - Low MP -> Non-polar P4, S8, and Cl2 have simple molecular structures. Ar has a simple atomic structure. Little energy is needed to overcome the weak dispersion forces between the molecules or between argon atoms hence the low melting points. -> The number of electrons and hence strength of dispersion forces decreases from S8 to P4 to Cl2 to Ar. -> Hence, energy needed to overcome the dispersion forces during melting and the resulting melting point decreases in the same order. - Do not conduct electricity -> no presence of mobile charge carriers in the molecules or argon atoms —— Ar: Simple atomic structures Weak dispersion forces between argon atoms. Covalent bonds holding the valence electrons. Low MP -> Non-polar P4, S8, and Cl2 have simple molecular structures. Ar has a simple atomic structure. Little energy is needed to overcome the weak dispersion forces between the molecules or between argon atoms hence the low melting points. -> The number of electrons and hence strength of dispersion forces decreases from S8 to P4 to Cl2 to Ar. -> Hence, energy needed to overcome the dispersion forces during melting and the resulting melting point decreases in the same order. - Do not conduct electricity -> no presence of mobile charge carriers in the molecules or argon atoms

Question 27

Explain why magnesium oxide has a higher melting point than sodium chloride, quoting relevant data from the data booklet.

Answer 27

Quote ionic radius data. Mg2+ has a smaller radius than Na+ while O2- has a smaller radius than Cl-. Mg2+ and O2- have higher charges than Na+ and Cl- respectively. Lattice energy = k(q+xq-)/r+ + r- Lattice energy of MgO is larger in magnitude More energy is needed to overcome the stronger ionic bonding in MgO.

Question 28

Ionic bond

Answer 28

electrostatic attraction between oppositely charged ions.

Question 29

Covalent bond

Answer 29

electrostatic attraction between the positively charged nuclei (of both bonded atoms) and their shared electrons

Question 30

Metallic bond

Answer 30

electrostatic attraction between a lattice of positive ions and the sea of delocalised electrons

Question 31

Electronegativity

Answer 31

of an atom is a measure of its ability to attract the electrons in a covalent bond to itself. The numerical values shown in the table show the relative ability of the atom to attract electrons in a covalent bond to itself. Therefore, oxygen has the greatest ability to do so, followed by chlorine, and carbon.

Question 32

Lattice energy

Answer 32

defined as the heat evolved when 1 mole of a pure ionic solid is formed from its constituent gaseous ions.

Question 33

Bond energy

Answer 33

average amount of energy required to break 1 mole of a covalent bond in the gaseous state to form gaseous atoms.

Question 34

Bond pair

Answer 34

An electron pair that is shared between bonded atoms is called a bond pair.

Question 35

Lone pair

Answer 35

An electron pair that is in the valence shell but not shared with another atom is called a lone pair.

Question 36

The VSEPR theory

Answer 36

states that the shape of a molecule or ion is determined by repulsions between electron pairs (or electron groups). 1. The electron pairs around the central atom of a molecule arrange themselves as far apart as possible so as to minimise their mutual repulsion. 2. The repulsion between electron pairs decreases in strength in the following order: lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion.

Question 37

Deduce the hybridisation state of all the carbon and nitrogen atoms in glycine.

Answer 37

C in -CHz- is bonded to 4 other atoms, there are 4 bond pairs and 0 lone pair, thus there are 4 electron groups around the central C atom in -CHz-, C in -CH2- is sp' hybridised, it is tetrahedral about central C in -CH2-, 109.5° C in -COOH (carboxylic acid group) is bonded to 3 other atoms, there are 3 bond pairs and 0 lone pair, thus there are 3 electron groups around the central C atom in -COOH, C in -COOH is sp2 hybridised, it is trigonal planar about central C in -COOH, 120°. N in -NH (primary amine group) is bonded to 3 other atoms, but it also has a lone pair of electrons (N is in group 15), there are 3 bond pairs and 1 lone pair, thus there are 4 electron groups around the central N atom, N is sp' hybridised, it is trigonal pyramidal about central N, 107°

Question 38

Antimony, Sb, is in Group 15 of the Periodic Table. It forms a series of salts which contains the SbF5n– anion, the structure of which is a square-based pyramid. Deduce the total number of valence electrons around the antimony atom, the value of n and the oxidation number of Sb in this ion.

Answer 38

Square-based pyramidal means there are 5 bond pairs and 1 lone pair around central Sb atom hence the total number of valence electrons around Sb is 12. Since Sb is Group 15, Sb has 5 valence electrons. Therefore, 2 Sb–F bonds are dative bonds which means n = 2 Let x be the oxidation number of Sb. Since the oxidation number of each F is −1, x + 5(−1) = –2, x = +3

Question 39

Using the Valence Shell Electron Pair Repulsion (VSEPR) theory, explain the bond angles in the molecules of the compounds NF3, NH3 and BF3. NF3: 102dg, NH3:: 107dg, BF3: 120dg

Answer 39

BF3 have 3 bond pairs and no lone pairs, it is trigonal planar about central B atom. NF3 and NH3 each have 3 bond pairs and 1 lone pair, both are trigonal pyramidal about central N atom. With an additional electron pair, bond angles for NF3 and NH3 are smaller than BF3 to minimise repulsion by arranging the electron pairs around the central atom as far as possible. As F is more electronegative than H, the electron density of the bond pairs in NF3 are lower around the central N atom, compared to NH3. Therefore, the bond pair – bond pair repulsion is weaker around the central N atom in NF3 than in NH3. Hence, the bond angle in NF3 is smaller than NH3.

Question 40

Suggest why the melting point of magnesium nitride is higher compared to that of magnesium oxide.

Answer 40

From lattice energy (𝑞+ × 𝑞−)/(𝑟+ + 𝑟 −), the effect of charges is more significant than the effect of ionic size. Since N3– has a larger charge than O2–, the lattice energy of Mg3N2 would have a larger magnitude than that of MgO. Hence, more energy is needed to overcome the stronger ionic bonding in Mg3N2, resulting in a higher melting point. Note: The “charge density” of an ion per se is not relevant to the discussion of lattice energy. The anionic size (N3− > O2−) would suggest that the oxide would have the larger lattice energy, whereas the anionic charge (N3− > O2−) would suggest the opposite. The anionic charge is the more important factor, making the lattice energy of the nitride larger than that of the oxide.

Question 41

State the bond energy values of the C − H bond and C − Cl bond using the Data Booklet. Explain why they are different.

Answer 41

C − H bond: 410 kJ mol–1 C − Cl bond: 340 kJ mol–1 H atom is smaller than Cl atom so the C − H bond is shorter than C − Cl bond. The overlap between the orbitals is more effective for the C − H bond so the bond is stronger and the bond energy is larger.

Question 42

Explain why both PVC and magnesium oxide can act as insulators.

Answer 42

In the covalent compound PVC, the valence electrons are held between the atoms and are not free to move. In the ionic compound MgO, the ions (Mg2+ and O2– ) are held in fixed positions in an ionic lattice. Hence, both are not able to conduct electricity as there are no mobile charge carriers.

Question 43

Explain why copper can conduct electricity

Answer 43

Copper is a good electrical conductors because it contains delocalised electrons which act as mobile charge carriers.

Question 44

Explain why copper can be drawn into wires

Answer 44

When a force is applied to a piece of copper metal, the 'sea' of delocalised electrons prevents repulsion between the cations as they move past one another and metallic bonding remains intact. Hence, copper is ductile and can be drawn into wires.

Question 45

Suggest why MgO needs to be encased in copper tubing.

Answer 45

Magnesium oxide is flaky and powdery so it needs to be encased in copper tubing to prevent it from flaking. (Concept: ionic compounds are brittle)

Question 46

How would you expect the magnitudes of the lattice energies of the oxides of the Group 2 elements to vary down the group?

Answer 46

Lattice energy = constant x (Q+ x Q-)/(r+ + r-) Down the group, the radii of the Group 2 cations increase, while the cationic charge remains constant. Since the anionic radius and charge remain constant as well, the magnitude of the lattice energies of the Group 2 oxides decreases down the group.

Question 47

Account for the difference in boiling points of SO3 and SO2, which are 45dgC and -10dgC respectively. [2]

Answer 47

Both compounds have simple covalent/molecular structures however. SO3 has a higher boiling point than SO2 because SO3 has stronger (intermolecular) instantaneous dipole-induced dipole attractions/dispersion forces due to its larger electron cloud size/greater polarisability (SO2 has weaker intermolecular permanent dipole-dipole attractions) Hence, more energy is required to overcome the bonding in SO3.

Question 48

Explain why CO2 has no overall dipole moment [1], COSe has a greater dipole moment than COS [1].

Answer 48

CO2 is linear and hence the dipoles cancel out. C=S bond is more polar than C=Se (Since S is more electronegative than Se). There is a smaller difference between the dipole moment of C=O and C=S than that between C=O and C=Se.

Question 49

MgO is preferred to PVC as an insulator in fire alarm cabling due to its high melting point. Suggest another reason why MgO is preferred to PVC.

Answer 49

MgO is not flammable, unlike PVC. or MgO is safe when heated, unlike PVC which may emit toxic (hydrogen chloride) fumes.

Question 50

Explain what is meant by sigma and pi bonds. Illustrate your answers with suitable diagrams.

Answer 50

A sigma bond is formed by head-on overlap of 2 orbitals. A pi bond is formed by side-on overlap of 2 orbitals.

Question 51

Sketch qualitatively, the variation in net dipole moment for CCl4, CH3I and CH3Cl. Explain your answer briefly.

Answer 51

CCl4 is a non-polar molecule. Though C–Cl bond is polar, CCl4 has a tetrahedral shape and thus all dipole moments cancel each other and there is overall zero net dipole moment. CH3I and CH3Cl are polar molecules with net dipole moments. However, chlorine is more electronegative than iodine and C–Cl bond is more polar than C–I bond. Hence, the net dipole moment is greater in CH3Cl than in CH3I.

Question 52

The boiling point decrease in the order, CCl4 > CH3I > CH3Cl. Explain why is this so.

Answer 52

The number of electrons (or the size of the electron cloud) of CCl4 > CH3I > CH3Cl and thus the polarisability of the electron cloud of CCl4 > CH3I > CH3Cl. Hence the strength of intermolecular dispersion forces in CCl4 > CH3I > CH3Cl. Amount of energy required to overcome stronger dispersion forces is greater, thus boiling point is in the order CCl4 > CH3I > CH3Cl. A complete answer should compare the following points: • No. of electrons in the molecules • Size of electron cloud, hence polarisability of electron cloud (larger e cloud more polarisable) • Strength of dispersion forces • Amount of energy required to overcome the intermolecular forces during boiling

Question 53

Account for the differences in melting point in terms of structure and bonding. In order: Highest mp: CH3CH2CO2^- Na+ CH3CH(OH)CH3 CH3COCH3 CH3CH2CH2CH3 Lowest mp: CH3CH(CH3)2

Answer 53

Melting point of CH3CH2CO2− Na+ > CH3CH(OH)CH3 > CH3COCH3 > CH3CH2CH2CH3 > CH3CH(CH3)2. CH3CH2CO2− Na+ has a giant ionic lattice structure with strong electrostatic forces between CH3CH2CO2− and Na+ ions (ionic bonding) which requires the largest amount of energy to overcome, thus it has the highest melting point. CH3CH(OH)CH3 is a simple covalent polar molecule with hydrogen bonding between its molecules* while CH3COCH3 is a simple covalent polar molecule with intermolecular* permanent dipole-permanent dipole interactions. Hydrogen bonding is stronger than permanent dipole-permanent dipole interactions and requires more energy to overcome, thus melting point of CH3CH(OH)CH3 is higher than that of CH3COCH3. CH3CH2CH2CH3 and CH3CH(CH3)2 exist as simple covalent non-polar molecules with intermolecular* dispersion forces which are weaker than permanent dipole-permanent dipole interactions, thus their melting points are the lower than that of CH3COCH3. Both CH3CH2CH2CH3 and CH3CH(CH3)2 have the same number of electrons but CH3CH2CH2CH3 is a straight chain molecule while CH3CH(CH3)2 is a branched molecule. Thus CH3CH(CH3)2 has a smaller surface area of contact between adjacent molecules, and its intermolecular dispersion forces is weaker and requires less energy to overcome. Note: • Comparison of strength of attraction (and specify type) broken during melting are necessary. • Covalent bonds in simple covalent molecules are not broken during melting. • *It is necessary to mention “between the molecules” or “intermolecular” for simple covalent molecules. • Energy comparisons are necessary when comparing melting points. • You may choose to ignore the relative molecular mass, Mr given as strength of dispersion forces depends on no. of electrons which affects the size and polarisability of the electron cloud.

Question 54

Even though ICl is more polar than IBr, IBr has a higher boiling point than ICl. Explain this dilemma as clearly as you can. [2]

Answer 54

Based on the relative strengths of pdpd alone, IBr should have the lower boiling point. However, the electron cloud size of the IBr is big enough such that the idid is stronger than the IMF present between ICl molecules.

Question 55

Like CaCO3, Al2O3 is a major component of naturally occurring rocks. Explain clearly, in terms of bonding and structure, why Al2O3 is insoluble in water. [3]

Answer 55

Al2O3 has a giant ionic lattice structure with its Al3+ cations and O2- anions held by strong ionic bonds. Energy released during the formation of the ion-dipole interactions between water and the ions in Al2O3 is insufficient to overcome the strong ionic bonds between the Al3+ cations and O2- anions hence Al2O3 is insoluble in water.

Question 56

CaCl2 is a solid with a high melting point (775dgC) whereas AlCl3 sublimes at 178dgC. Explain the difference in melting points between these 2 chlorides in terms of their structure and bonding. [3]

Answer 56

CaCl2 has giant ionic structure. Large amount of energy required to overcome strong ionic bonds between Ca2+ and Cl- ions. AlCl3 has simple molecular structure. Smaller amount of energy required to overcome weaker intermolecular dispersion forces. Hence melting points between of CaCl2 is higher than AlCl3.

Question 57

Explain, in terms of structure and bonding, the difference in the boiling points for both of CS2 and COS. [2]

Answer 57

Both CS2 and COS have simple molecular structures. CS2 has a larger number of electrons than COS. More energy is required to overcome the stronger instantaneous dipole-induced dipole interactions between CS2 molecules than the permanent dipole-induced dipole interactions between COS molecules. Hence,CS2 has a higher boiling point.

Question 58

Explain, in terms of structure and bonding, the difference in boiling points between: I. CH3OCH3 and CH3CH2OH, II. CH3OCH3 and CH3SCH3.

Answer 58

Note: 1. Specific to the question: compare data differences in bp 2. Describe the structure for each compound 3. Describe the bonding, IMF specifically for each compound 4. Compared the strengths of the bonding that is broken 5. Compared energy needed to overcome types of bonding. Both CH3OCH3 and CH3CH2OH are simple covalent, polar molecules. More energy is needed to overcome the stronger hydrogen bonding (and dispersion forces) between molecules of CH3CH2OH, as compared to the weaker permanent dipole-permanent dipole interactions (and dispersion forces) between molecules of CH3OCH3, hence CH3CH2OH has a higer boiling point. II. Both CH3OCH3 and CH3SCH3 are simple covalent, polar molecules with permanent dipole- permanent dipole (pd-pd) interactions and dispersion forces between molecules. Despite CH3OCH3 having stronger pd-pd interactions than CH3SCH3, the larger electron cloud size of CH3SCH3 leads to much stronger dispersion forces for CH3SCH3 than CH3OCH3. More energy is needed to overcome the stronger intermolecular forces for CH3SCH3 overall, resulting in a higher boiling point for CH3SCH3.

Question 59

(b) With the aid of a suitable diagram, explain why CH3OCH3 is miscible with CH3CH2OH. Refer to forces of attraction between particles in your explanation.

Answer 59

Note: 1. Diagram drawn 2. Refer to IMF between relevant particles 3. Describe solute-solvent interaction 4. Describe solute-solute and solvent-solvent interactions 5. Compared energy releasedwhe forming solute-solvent interaction against energy needed to overcome solute-solute and solvent-solvent interactions CH3OCH3 is miscible with CH3CH2OH as favorable hydrogen bonding can be formed between them [1]. The energy released when forming hydrogen bonds between CH3OCH3 and CH3CH2OH is greater than or comparable to the energy needed to overcome permanent dipole-permanent dipole interactions between CH3OCH3 and hydrogen bonding between CH3CH2OH [1].

Question 60

By reference to the type and extent of relevant intermolecular forces, explain as fully as you can the differences in boiling points between water and ammonia, between ammonia and hydrogen chloride, and between hydrogen chloride and methane.

Answer 60

Water forms an average of 2 hydrogen bonds per molecule, while ammonia forms only 1 hydrogen bond per molecule on average. More energy is needed to overcome the more extensive hydrogen bonding in water, resulting in a higher boiling point for water. In addition, as O is more electronegative than N, the delta+ charge on H is larger in water than that in ammonia. Note: To determine the average no. of hydrogen bonds per molecule, count the no. of H covalently bonded to N/O/F, and the no. of lone pairs on N/O/F. Go by the limiting factor. More energy is needed to overcome the stronger hydrogen bonding between ammonia, as compared to weaker permanent dipole-permanent dipole interactions between hydrogen chloride, resulting in higher boiling point for ammonia. More energy is needed to overcome the permanent dipole-permanent dipole interactions and dispersion forces between hydrogen chloride, as compared to just dispersion forces between methane, resulting in higher boiling point for hydrogen chloride

Question 61

Suggest reasons for the higher solubility of hydrogen chloride as compared to that of ammonia.

Answer 61

HCl is a strong acid and completely dissociates in water to form H3O+ and Cl− ions. Favorable ion-dipole interactions can be formed between each of these ions and water molecules. (This would suffice for a 1m explanation for HCl.) The energy released when forming these ion-dipole interactions is greater than the energy needed to overcome hydrogen bonding between water molecules, and the permanent dipole-permanent dipole interactions and dispersion forces between HCl molecules (as well as the H−Cl covalent bonds since the molecules dissociate in this context). Ammonia is a weak base and only partially ionises in water to form some NH4+ and OH− ions. The largely unionised NH3 molecules can only form weaker hydrogen bonding with water molecules, as compared to the stronger ion-dipole interactions which are formed between dissociated ions and water molecules. (An explanation needs to be given for NH3 too, not just HCl.)

Question 62

Suggest reasons for the much higher solubilities of ammonia and hydrogen chloride, as compared to that of methane.

Answer 62

Favorable hydrogen bonding or ion-dipole interactions can be formed between ammonia and water molecules, or between H3O+ or Cl− ions and water molecules, resulting in high solubilities for ammonia and hydrogen chloride respectively. (An explanation needs to be given for NH3 and HCl, not just methane.) No/very little (to address 0.0014 mol dm-3 solubility) favorable interactions can be formed between non-polar methane and polar water molecules. (This would suffice for a 1m explanation for methane.) The energy released when forming dispersion forces (or permanent dipole-induced dipole interactions) between water and methane is insufficient to overcome stronger hydrogen bonds between water and dispersion forces between methane.