3.1

Question 1

3. 1.1) a) How is the periodic table arranged in terms of:
   i) 1. Proton number
   ii) 2. Periods
   iii) 3. Groups

Answer 1

1. The periodic table is the arrangement of elements by increasing proton number.
2. There are trends in chemical and physical properties across a period - these trends are repeated across each successive period (periodicity), a result of the repeating pattern of electron configuration.
3. All the elements in a vertical group have similar chemical properties (due to the the same number of electrons in the outer shell and the same type of orbitals).

Question 2

3.1.1 b) i) The periodic trend in electron configuration across period 2 and 3 can be compared. Explain this in terms of the the first element, the second element and the fifth element of each period.

Answer 2

The difference between the two successive periods is the increase of one electron shell. Consider the electron configuration of:
- the first element in each period (Li = [He]2s1 and Na = [Ne]3s1)
- the second element in each period (Be = [He]2s2 and Mg = [Ne]3s2)
- the fifth element in each period (N = [He]2s2, 2p3 and P = [Ne]3s2, 3p3).
Elements in each group have the same number of electrons in their outer shell. Therefore the outer shell electron configuration across a period is repeated with each successive period - this is why the trend in chemical and physical properties across a period is also repeated with each successive period.

Question 3

3.1.1 b) ii) How can elements be classified in terms of orbitals?

Answer 3

Elements can be classified into the s-, p- or d- block. This allows you to determine which orbital the outer shell electron is in (e.g. the s-block elements have an outer shell electron configuration of s1 or s2, and the p-block elements have an outer shell electron configuration of s2p1 to s2p6.

Question 4

3.1.1 c) What is the first ionisation energy?

Answer 4

The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms (to form one mole of +1 ions).

Question 5

3.11 c) Which factors affect ionisation energy and how?

Answer 5

1) Nuclear charge: as the number of protons in the nucleus increases, the positive nuclear charge also increases - the electrostatic attraction between the positive nucleus and the negative outer shell electrons is therefore stronger, and so more energy is required to remove the electron. The ionisation energy increases.
2) Electron shielding: as the number of inner shells increases, more negative electrons repel each other and this electron shielding results in weaker electrostatic attraction between the positive nucleus and the negative outer electrons. Less energy is required to remove the outer shell electron. The ionisation energy decreases.
3) Atomic radius: a larger atomic radius results in the outer electron being further away - so the electrostatic attraction between the negative outer electron and the positive nucleus is weaker, and less energy is required to remove the electron. The ionisation energy decreases.
[Note: while the atomic radius increases with the addition of shells, an increase in protons will draw the electrons closer to the nucleus, decreasing the atomic radius]

Question 6

3.11 c) i) What is the trend in ionisation energy across a period?

Answer 6

The first ionisation energies across a periodic table show a general increase - this is a result of:
1) The number of protons in the nucleus, and therefore the nuclear charge, increases - this results in stronger electrostatic attraction between the negative outer electron and the positive nucleus.
2) This increase in protons, nuclear charge, and electrostatic attraction causes the electrons to be pulled in closer to the nucleus, decreasing the atomic radius. This further strengthens the electrostatic attraction between the nucleus and the outer electron.
3) The number of shells doesn’t change as you go across a period - electrons are added to the same shell - so the electron shielding effect is relatively similar for each element in a period.
This allows the attraction between the outer electron and the positive nucleus to strengthen as you go across a period, making it harder to remove an electron, and therefore increasing the ionisation energy.

Question 7

3.1.1 c) i) What is the first exception to the trend in first ionisation energy across period 2 and 3?

Answer 7

There is a small decrease in first ionisation energy between group 2 and 3. In period 2, the decrease is between Be and B, and in period 3, the decrease is between Mg and Al. In group 2 (Be and Mg), the outermost electron is in an s-orbital, but in group 3 (B and Al), the outermost electron is in a p-orbital:
- p-orbitals have higher energy, and so are found further away from the nucleus, than s-orbitals - less energy is required to remove the outer electron from a p-orbital (overriding the increased nuclear charge) and so the ionisation energy drops.

Question 8

3.1.1 c) i) What is the second exception to the trend in first ionisation energy across period 2 and 3?

Answer 8

There is a small decrease in first ionisation energy between group 5 and 6. In period 2, the decrease is between N and O, and in period 3, the decrease is between P and S. In group 5 (N and P), the electron is removed from a singly occupied orbital, whereas in group 6 (O and S), the electron is removed from an orbital containing two electrons: the repulsion between these two electrons makes it easier remove an electron from the shared orbital, thereby decreasing the ionisation energy.

Question 9

3.11 c) i) What is the trend in ionisation energy down a group?

Answer 9

The first ionisation energy decreases down a group - this is the result of:
Atomic radius: the number of shells increases as you go down a group, resulting in the atomic radius therefore increasing - the outer electron is further away from the nucleus, and so the electrostatic attraction is weaker between the two.
Electron shielding: as the number of inner shells increases going down a group, the shielding effect on the outermost electron increases (as the increasing number of electrons repel each other), resulting in weaker electrostatic attraction between the outer electron and the nucleus.
While the nuclear charge does increase going down a group, due to more protons, this effect is overridden and less energy is required to remove the electrons going down a group.

Question 10

3.11 c) ii) How can successive ionisation energies show shell structure?

Answer 10

Successive ionisation energies are a measure of the amount of energy required to remove each electron, in turn, from an element.
Successive ionisation energies tend to increase: this is because electrons are being removed from an increasingly positive ion, and there is less repulsion amongst the remaining electrons, resulting in a stronger electrostatic attraction between the outer electron and the nucleus.
Successive ionisation energy can indicate how many electrons their are in each shell. This is because there is a large jump in ionisation energy when electrons are being removed from different shells - the electron is in a shell closer to the nucleus, with less inner shells (and therefore less electron shielding) resulting in a stronger electrostatic attraction between the outer electron and the nucleus.

Question 11

3.1.1 d) i) What is metallic bonding? ii) What structures do metals form?

Answer 11

Metallic bonding is the strong electrostatic attraction between metal cations (positive ions) and delocalised electrons.
All metals form giant metallic lattices, where the delocalised electrons - from the outermost shell - move through the closely packed cations.

Question 12

3.1.1 f) Explain the melting and boiling points of giant metallic lattices in terms of structure and bonding.

Answer 12

Metals have high melting and boiling points: a large amount of energy is required to overcome the strong electrostatic attraction between the metal cations and the delocalised electrons.
As the charge on the cation increases, the number of delocalised electrons per ion also increases. For example, Mg2+ has 2 delocalised electron per ion, and experiences more electrostatic attraction than Na+, which only has one delocalised electron per ion.
The stronger the electrostatic attraction, the higher the melting point.
The size of the metal ion is also a factor: a smaller ionic radius will hold the delocalised electrons closer to the nuclei.

Question 13

3.1.1 f) Explain the physical properties of giant metallic lattices in terms of structure and bonding.

Answer 13

Metals tend to have a high melting and boiling point - this is because a large amount of energy is required to overcome the strong electrostatic attraction between the positive cations and the negative electrons.
Metals are good electrical conductors - this is because the delocalised electrons act as mobile charge carriers.
Metals do not dissolve in solution - this is because the electrostatic attraction between the positive cations and the negative electrons is too strong to break.
Metals are both ductile and malleable, a result of the delocalised electrons allowing metal ions to slide past each other.

Question 14

3.1.1 e) Explain what solid giant covalent lattices (of carbon and silicon) are?

Answer 14

Solid giant covalent lattices (or macromolecular structures) of carbon (e.g. diamond, graphite and graphene) and silicon (e.g. silicon dioxide) are networks of atoms bonded by strong covalent bonds.

Question 15

3.1.1 f) Explain the physical properties of diamond in terms of structure and bonding.

Answer 15

In diamond, each carbon atom is covalently bonded to four other carbon atoms - the atoms arrange themselves in a tetrahedral shape, forming a giant covalent lattice.
Diamond has a very high melting point and is insoluble in water - this is because a large amount of energy is required to break these strong covalent bonds.
However, diamond is unable to conduct electricity: there are no mobile charge carriers.

Silicon also forms a giant covalent lattice, with similar properties to carbon i.e. each silicon atom is able to form four strong, covalent bonds.

Question 16

3.1.1 f) Explain the physical properties of graphite in terms of structure and bonding.

Answer 16

In graphite, each carbon atom is bonded to three other carbon atoms, arranged in sheets (or layers) of interlocking hexagonal rings. The fourth outer electron of each carbon is delocalised.
The delocalised electrons act as mobile charge carries, allowing graphite to conduct electricity.
The layers are bonded together by weak induced dipole-dipole forces (which are easily broken) - this allows the layers to slide over each other, resulting in graphite’s malleable properties.
The layers are relatively far apart, compared to the length of the covalent bonds at leasy, so graphite is less dense than diamond.
Due to the strong covalent bonds between the carbons in a layer, graphite has a very high melting point.
Graphite is insoluble - the covalent bonds are too strong to be broken by weak interactions with water.

Question 17

3.1.1 f) Explain the physical properties of graphene in terms of structure and bonding.

Answer 17

Graphene is just one sheet, or layer, of carbons atoms (with each carbon atom bonded to three others, forming interlocking hexagonal rings).
The graphene layer, one atom thick, means it is transparent, flexible and low in mass (i.e. light).
The delocalised electrons can move quickly above and below the sheet, making it the best known electrical conductor.
The delocalised electrons also strengthen the covalent bonds between the carbons atoms, making it extremely strong.

Question 18

3.1.1 f) Explain the physical properties of giant covalent lattices (in general) in terms of structure and bonding.

Answer 18

A large amount of energy is required to break the strong covalent bonds, so naturally they have very high melting and boiling points.
With the exception of graphite and graphene, they can not conduct electricity - there are no freely moving particles to carry the charge.
They are insoluble in both polar and non-polar solvents (due to the strong covalent bonds).

Question 19

3.1.1 g) Explain the variation in melting points across period 2 and 3 in terms of structure and bonding.

Answer 19

Between groups 1 and 4, the melting points are the highest, steadily increasing as you go across the group - these elements have giant lattice structures:
- Li, Be, Na, Mg and Al all form giant metallic structures. In fact, the metallic bonds increase as you go across a period (the ionic charge increases, the ionic radius decreases, and the number of delocalised electrons per ion increases).
- B, C and Si form giant covalent structures, where each successive group has more electrons with which to form covalent bonds.
Between groups 4 and 5, there is a sharp decrease in melting points. This marks the change from giant lattice structures to simple molecular structures.
- Because the elements between groups 5 and 8 form simple molecular structures, the weak intermolecular forces between the molecules require little energy to be broken. They all have consistently low melting and boiling points.

Question 20

3.1.2 a) Explain the outer shell electron configuration of group 2 elements, and how they can achieve noble gas configuration.

Answer 20

Each group 2 element has two outer shell electrons in the s2 sub-shell - in order to achieve a noble gas configuration, each atom loses the 2 electrons in a redox reaction. As strong reducing agents, group 2 elements can be oxidised to form 2+ ions::
M → M+ + e-
M+ → M2+ + e-
Overall: M → M2+ + 2e-

Question 21

3.1.2 b) i) Give the redox reaction for group 2 elements reacting with oxygen and describe the product formed.

Answer 21

The group 2 elements react vigorously when burnt in oxygen, producing solid white oxides (with the general formula: MO - containing M2+ and O2- ions):
2M(s) + O2(g) → 2MO(s)
- The metal is oxidised and loses 2 electrons.
- The oxygen is reduced and gains 2 electrons.

Question 22

3.1.2 b) ii) Explain the relative reactivities of the group 2 elements by their redox reaction with water.

Answer 22

The group 2 elements react with water to form a metal hydroxide (general formula: M(OH)2 ). Hydrogen gas is produced as a byproduct:
M(s) + 2H2O → M(OH)2(aq) + H2(g)
- The metal is oxidised, losing 2 electrons.
- 2 of the 4 hydrogen atoms are reduced, with each gaining 1 electron (to form H2).
Moving further down the group, the reactivity increases and each metal reacts more vigorously - Beryllium doesn’t even react, in fact, and Magnesium reacts very slowly; Calcium reacts steadily, Strontium reacts fairly fast and Barium reacts rapidly.

Question 23

3.1.2 b) iii) Explain the relative reactivities of the group 2 elements by their redox reaction with dilute acids.

Answer 23

The group 2 elements react with dilute acids to form a salt and water:
M(s) + 2HCl(aq) → MCl2(aq) + H2(g)
Moving further down the group, each metal reacts more vigorously - Beryllium doesn’t even react, in fact, and Magnesium reacts very slowly; Calcium reacts steadily, Strontium reacts fairly steadily and Barium reacts rapidly.

Question 24

3.1.2 c) Explain the trend in reactivity of group 2 elements, in terms of the first and second ionisation energies.

Answer 24

As you go down group 2, the first and second ionisation energies decrease (this is a result of increasing atomic radius and increased shielding). This makes it easier for elements to lose 2 electrons and form cations - so reactivity increases going down the group.

Question 25

3.1.2 d) Explain the action of water on group 2 oxides, including the trend in reactivity.

Answer 25

Group 2 oxides can react with water to form metal hydroxides:
MO(s) + H2O(l) → M(OH)2(aq) [or Ca2+(aq) + 2OH-(aq)]
The metal hydroxides are slightly soluble in water, and form alkaline solutions (releasing OH- ions).
The solubility of hydroxides in water increases as you go down the group (when a hydroxide is more soluble than another, it will release more OH- ions, creating an alkaline solution with a higher pH).
In fact, Be is insoluble in water (it doesn’t even form a metal oxide), and magnesium oxide reacts slowly, producing a hydroxide that isn’t very soluble. The other solutions are strongly alkaline however (with pH 10 - 12).

Question 26

3.1.2 e) Give examples of the following group 2 compounds neutralising acids:
i) Ca(OH)2
II) Mg(OH)2 and CaCO3

Answer 26

Group 2 compounds (e.g. oxides, carbonates and hydroxides) can act as bases.
1. Calcium hydroxide (or ‘lime’) is used in agriculture to neutralise acidic soil:
Ca(OH)2(s) + 2HCl(aq) → CaCl(aq) + 2H2O(l)
2. Antacids, such as magnesium hydroxide (Mg(OH)2) and calcium carbonate (CaCO3) can treat indigestion, which is the build up of acid:
Mg(OH)2(s) + 2HCl(aq) → 2H20(l) + MgCl2(aq)
CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)

Question 27

3.1.3 a) The halogens exist as diatomic molecules (X2), having relatively low melting and boiling points. Going down the group, explain the trend in boiling point.

Answer 27

As you move down the group, their boiling points increase (and their physical states change from gas to liquid, to solid). Each successive element has an extra shell of electrons, so the size and relative mass of the atoms increase as you go down the group. This increase in electrons results in stronger and more frequent induced dipole-dipole forces, which requires more energy to break (thereby increasing boiling point).

Question 28

3.1.3 b) Employing you knowledge of their outer shell electron configuration, how do halogens achieve noble gas configuration?

Answer 28

Halogens have an outer shell s2p5 electron configuration. To achieve a noble gas configuration, they must gain one electron (i.e. in a redox reaction), forming 1- ions.

Question 29

3.1.3 c) Explain the trend in reactivity for group 7 by describing a method using displacement reactions.

Answer 29

Halogens are highly electronegative - this makes them strong oxidising agents. However, their reactivity, and ability to oxidise, decreases moving down the group. This is made apparent through redox reactions between aqueous solutions of halide ions (e.g. KCl(aq), KBr(aq), or KI(aq)) and aqueous solutions of halogens (e.g, Cl2(aq), Br2(aq) and I2(aq) ): a more reactive halogen will oxidise and displace the halide ions of a less reactive halogen, leaving the least reactive halogen in an aqueous solution. Because different halogens form different coloured solutions - in both water and cyclohexane (an organic solvent) - the least reactive halogen will colour the solvent at the end of the reaction:
Cl2 is pale green in both water and cyclohexane.
Br2 is orange in both water and cyclohexane.
I2 is brown in water and violet in cyclohexane.

Question 30

3.1.3 c) Illustrate displacement reactions with halogens to determine reactivity.

Answer 30

If you mix chlorine water with either potassium iodide solution or potassium bromide solution, the chlorine will oxidise and displace both ions, leaving the other to dissolve in the solvent:
Cl2(aq) + 2KI/2Br(aq) → 2KCl(aq) + I2/Br2(aq)
or the ionic equation:
Cl2(aq) + 2I-/2Br-(aq) → 2Cl(aq) + I2/Br2(aq)
Bromine water can only oxidise and displace potassium iodide solution (it doesn’t react with potassium chloride):
Br2(aq) + 2KI(aq) → 2KBr(aq) + I2(aq)
or the ionic equation:
Br2(aq) + 2I-(aq) → 2Br(aq) + I2(aq)
Iodine water can’t oxidise and displace either.

Question 31

3.1.3 d) Explain the trend in reactivity for group 1.

Answer 31

Halogens become less reactive, or less oxidising, as you go down the group. It becomes harder to form 1- ions, a result of
- the increasing atomic radius (due to the successive addition of shells)
- and the increased electron shielding (again, a result of more inner shells).
This weakens the electrostatic attraction between the outer shell and the positive nucleus, making it harder for larger atoms to attract an electron and form the 1- ion (despite increased charge on the nucleus).

Question 32

3.1.3 e) Explain the term disproportionation.

Answer 32

Disproportionation is the oxidation and reduction of the same element in a redox reaction.

Question 33

3.1.3 e) i) Give a use for the reaction between chlorine and water.

Answer 33

Chlorine reacts with water disproportionately to form a mixture of hydrochloric acid and chloric (I) acid:
CL2(g) + H2O(l) → HCl(aq) + HClO(aq)
For each chlorine molecule, one chlorine atom is reduced from 0 to -1 (in HCl) and the other is oxidised from 0 to +1 (in HClO).
Aqueous chloric (I) acid dissociates in solution to make chlorate (I) ions (also called hypchlorite ions):
HClO(aq) + H2O(l) → ClO-(aq) + H3O+(aq)
or HClO(aq) → ClO-(aq) + H+(aq)
Chlorate (I) ions kill bacteria - so chlorine can be used in water purification.

Question 34

3.1.3 e) ii) Which disproportionate reaction forms bleach?

Answer 34

Chlorine reacts with cold, dilute aqueous sodium hydroxide, forming sodium chlorate (I) solution, NaClO - otherwise known as bleach:
2NaOH(aq) + Cl2(g) → NaClO(aq) + NaCl(aq) + H2O(l)
For each chlorine molecule, one chlorine atom is reduced from 0 to -1 (in NaCl) and the other is oxidised from 0 to +1 (in NaClO).

Question 35

3.1.3 f) Compare the benefits and risks of chlorine, as well as considering any ethical implications.

Answer 35

+ It kills disease-causing microorganisms, such as bacteria.
+ It prevents the growth of algae, eliminates bad tastes and smells, and removes discolouration caused by organic compounds.
- Hazards of toxic chlorine gas - it irritates the respiratory system if breathed in.
- Chlorine can react with other compounds in water to form chlorinated hydrocarbons - many of which are carcinogens.
Ethical considerations: people’s right to make their own choices is taken away (i.e. we don’t get a say in chlorinated water).

Question 36

3.1.3 g) How can you test for halide ions?

Answer 36

To test for halide ions, first add dilute nitric acid (to remove any other ions in the mixture). Then add silver nitrate solution (AgNO3(aq)), forming a silver halide precipitate (AgX(s)). Note the colour of any precipitate formed.
Ag+(aq) + X-(aq) → AgX(s)
If the colour is hard to distinguish, add aqueous ammonia (first dilute, then concentrated). Note the solubility of the precipitate.
AgCl = white, soluble in both NH3 solutions
AgBr = cream, soluble in concentrated NH3 only
AgI = yellow, insoluble in both NH3 solutions

Question 37

3.1.4 a) i) How can you test for carbonate ions {anions}?

Answer 37

To test for carbonates (CO3 2-), add a diluted strong acid (e.g. nitric acid). Carbon dioxide gas should be produced, and effervescence observed:
e.g. CaCO3(s) + HCl(aq) → CO2(g) + H2O(l) + CaCl2(aq)
ionic equation:
CO3 2-(s) + 2H+(aq) → CO2(g) + H2O(l)
To test for CO2, bubble the gas through limewater (Ca(OH)2) -if carbon dioxide is present, it will react to form a fine white precipitate, turning the limewater:
CO2 (g) + Ca(OH)2 (aq) → CaCO3 (s) + H2 (l)

Question 38

3.1.4 i) How can you test for sulfate ions {anions}?

Answer 38

To test for sulfates (SO4 2-), add dilute HCl, followed by barium chloride solution (BaCl2) - a white precipitate of barium sulfate should be produced BaSO4:
e.g. Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
ionic equation:
Ba2+(aq) + SO4 2-(aq) → BaSO4(s)

Question 39

3.1.4 ii) How can you test for ammonium ions {cations}?

Answer 39

To test for ammonium ions (NH4+), add sodium hydroxide solution and heat gently:
e.g. NH4Cl(aq) + NaOH(aq) → NH3(aq) + NaCl(aq) + H2O(l)
ionic equation:
NH4+(aq) + OH-(aq) → NH3(g) + H2O(l)
Ammonia gas (NH3), is be produced. Test the gas with damp red litmus paper - in the presence of an alkaline (such as ammonia) it should turn blue.

Question 40

3.1.4 When testing for anions, what is the sequence you should carry out your tests in? Why is this needed?

Answer 40

Test in the following order: carbonates, then sulfates, then halides.

1) Test for carbonates first - add dilute nitric acid (and not dilute sulfuric acid or dilute hydrochloric acid). If carbonate ions are present, and effervescence is observed, continue adding the dilute nitric acid until the bubbling has stopped and all carbonate ions have been removed.
2) Test for sulfate ions only after you’ve tested for carbonate ions: this is because barium ions also react with carbonate ions to form a white precipitate of barium carbonate (BaCO3). If you intend to carry out a halide test afterwards, use barium nitrate (Ba(NO3)2) - this is to prevent introducing chloride ions to your solution. Remove any and all sulfate ions by filtering out the barium sulfate precipitate if it has formed.
3) Test for halide ions only after you’ve tested for sulfate/carbonate ions: this is because silver ions react with sulfate/carbonate ions to form a precipitate of silver sulfate (Ag2SO4) or a precipitate of silver carbonate (Ag2CO3).