as topics

Question 1

state meaning of term periodicity

Answer 1

repeating patterns of physical/chemical properties going across periods

Question 2

why does K have largest atomic radius in period 4 (2)

Answer 2

• smallest no of protons so smallest nuclear charge
• but same number of shells so similar shielding

Question 3

general trend for first IE across a period and why

Answer 3

increase
increasing no protons as e- added to same shell, same shielding

Question 4

what r the anomalies to the increase in 1st IE

Answer 4

• small drop bw Mg+Al bc Mg has outer e- in in 3s but Al has outer e- in 3p which is higher in energy so easier to remove
• bw P+S, S’s e- is being paired w another e- in same 3p orbital - spin pair repulsion - so easier to remove 2nd e- in orbital

Question 5

how does MP/BP change across period 3

Answer 5

Na,Mg,Al = metallic bonding = strong bonding gets stronger the more e-s there are in outer shell that get released to sea of e-s, smaller sized ion w greater +ve charge also makes bonding strong
Si = macromolecular - many strong CBs bw atoms - high energy needed to break
Cl2, S8, P4 = simple molecular - weak VdWs bw molecules - little energy needed to break them - low mp+bp} more e- = stronger VdWs bw molecules - higher MP/BP
Ar = monoatomic - weak VdWs

Question 6

trend in AR down grp 2

Answer 6

Atomic radius increases down the group.
As one goes down the group, the atoms have more shells of electrons making the atom bigger.

Question 7

trend in MP down grp 2

Answer 7

Melting points decrease down the group. The metallic bonding weakens as the atomic size increases. The distance between the positive ions and delocalized electrons increases. Therefore the electrostatic attractive forces between the positive ions and the delocalized electrons weaken.

Question 8

trend in 1st IE down grp 2

Answer 8

The outermost electrons are held more weakly because they are successively further from the nucleus in additional shells.
In addition, the outer shell electrons become more shielded from the attraction of the nucleus by the repulsive force of inner shell electrons.

Question 9

how does Mg react w O

Answer 9

Mg burns with a bright white flame. The MgO appears as a white powder.
2 Mg + O2  2 MgO

Question 10

what does Mg ribbon have on it and why does it have to be removed

Answer 10

Magnesium ribbon will often have a thin layer of magnesium oxide on it formed by the reaction with oxygen in the air.
2 Mg + O2  2 MgO
The magnesium oxide needs to be removed by emery paper before doing reactions with magnesium ribbon.
If testing for reaction rates with Mg and acid, an un-cleaned Mg ribbon would give a false result because both the Mg and MgO would react but at different rates.
Mg + 2 HClMgCl2 + H2
MgO + 2 HClMgCl2 + H2O

Question 11

how does Mg react in steam

Answer 11

Magnesium reacts in steam to produce magnesium oxide and hydrogen. The Mg would burn with a bright white flame. The MgO appears as a white powder.
Mg(s) + H2O(g)  MgO(s) + H2(g)

Question 12

how does Mg react in warm water

Answer 12

Mg will also react with warm water, giving a different magnesium hydroxide product.
Mg + 2 H2OMg(OH)2 + H2
This is a much slower reaction than the reaction with steam and there is no flame.

Question 13

how do other grp 2 metals react w cold water

Answer 13

fizzing, (more vigorous down group)
the metal dissolving, (faster down group) the solution heating up (more down group)
with calcium a white precipitate appearing (less precipitate forms down group with other metals)

Question 14

why is Ti useful and whats it used for

Answer 14

because it is abundant, has a low density and is corrosion resistant – it is used for making strong, light alloys for use in aircraft.

Question 15

why can Ti not be extracted w carbon or by electrolysis

Answer 15

Titanium cannot be extracted with carbon because titanium carbide (TiC) it is formed rather than titanium.
Titanium cannot be extracted by electrolysis because it has to be very pure

Question 16

Steps in extracting titanium

Answer 16

1. TiO2 (solid) is converted to TiCl4 (liquid) at 900C:
   TiO2 +2Cl2 +2CTiCl4 +2CO
2. The TiCl4 is purified by fractional distillation in an argon atmosphere.
3. The Ti is extracted by Mg in an argon atmosphere at 500C:
   TiCl4 + 2Mg  Ti + 2 MgCl2

Question 17

Titanium is expensive because: 4

Answer 17

The expensive cost of the magnesium
2. This is a batch process which makes it expensive because the process is slower (having to fill up and empty reactors takes time) and requires more labour and the energy is lost when the reactor is cooled down after stopping
3. The process is also expensive due to the argon, and the need to remove moisture (because TiCl4 is susceptible to hydrolysis).
4. High temperatures required in both steps

Question 18

why tio2 converted ticl4

Answer 18

TiO2 is converted to TiCl4 as it can be purified by fractional distillation, TiCl4 being molecular (liquid at room temperature) rather than ionic like TiO2 (solid at room temperature).

Question 19

what is CaO used for

Answer 19

Calcium oxide can be used to remove SO2 from the waste gases from furnaces (e.g. coal fired power stations) by flue gas desulfurisation. The gases pass through a scrubber containing basic calcium oxide which reacts with the acidic sulfur dioxide in a neutralisation reaction.
SO2 + CaOCaSO3 calcium sulfite
The calcium sulfite which is formed can be used to make calcium sulfate for plasterboard.

Question 20

what do grp 2 hydroxides appear as

Answer 20

white precipitates

Question 21

Magnesium hydroxide is used .. and why

Answer 21

in medicine (in suspension as milk of magnesia) to neutralise excess acid in the stomach and to treat constipation.
Mg(OH)2 + 2HClMgCl2 + 2H2O
It is safe to use because it is so weakly alkaline. It is preferable to using calcium carbonate as it will not produce carbon dioxide gas.

Question 22

Ca(OH)2 use

Answer 22

to neutralise acidic soils. test for carbon dioxide. The limewater turns cloudy as white calcium carbonate is produced.
Ca(OH)2 (aq) + CO2 (g)  CaCO3 (s) + H2O(l)

Question 23

BaSO4 is used .. why

Answer 23

in medicine as a ‘Barium meal’ given to patients who need x-rays of their intestines. The barium absorbs the x-rays and so the gut shows up on the x-ray image. Even though barium compounds are toxic, it is safe to use here because barium sulfate’s low solubility means it is not absorbed into the blood.

Question 24

how to test for presence of sulphate ion

Answer 24

BaCl2 solution acidified with hydrochloric acid is used as a reagent to
test for sulfate ions.
If acidified barium chloride is added to a solution that contains sulfate ions a
white precipitate of barium sulfate forms.

Question 25

why is Hal needed in sulphate ion test

Answer 25

The hydrochloric acid is needed to react with carbonate impurities that are often found in salts which would form a white barium carbonate precipitate and so give a false result. You could not use sulfuric acid because it contains sulfate ions and so would give a false positive result. 2HCl + Na2CO3  2NaCl + H2O + CO2 Fizzing due to CO2 would be observed if a carbonate was present.

Question 26

when to use vacuum filtration? what is it

Answer 26

Use if larger amounts of solid are formed. The apparatus is connected to a water pump which will produce a vacuum. Buchner funnel Buchner flask (has thicker glass walls than a normal flask to cope with the vacuum )Filter paper, Air outlet to water pump

Question 27

how does fluorine appear

Answer 27

v pale yellow gas

Question 28

cl appearance

Answer 28

greenish gas

Question 29

br appearance

Answer 29

red liquid, gives off dense brown/orange fumes

Question 30

iodine appearance

Answer 30

shiny grey solid - sublimes to purple gas

Question 31

trend in MP/BP down grp 7

Answer 31

Increase down the group As the molecules become larger they have more electrons and so have larger van der waals forces between the molecules. As the intermolecular forces get larger more energy has to be put into break the forces. This increases the melting and boiling points.

Question 32

trend n electronegativity in grp 7

Answer 32

Electronegativity is the relative tendency of an atom in a molecule to attract electrons in a covalent bond to itself. As one goes down the group the electronegativity of the elements decreases. As one goes down the group the atomic radii increases due to the increasing number of shells. The nucleus is therefore less able to attract the bonding pair of electrons.

Question 33

does oxidising strength decrease/increase down grp 7

Answer 33

DECREASE

Question 34

cl/br/I colour in soln

Answer 34

cl = v pale green soln br = yellow soln I = brown soln

Question 35

why is nitric acid added when identifying halides

Answer 35

to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations 2 HNO3 + Na2CO3 2 NaNO3 + H2O + CO2

Question 36

does reducing strength decrease/increase down grp 7, why

Answer 36

INCREASE A reducing agent donates electrons. The reducing power of the halides increases down group 7 They have a greater tendency to donate electrons. This is because as the ions get bigger it is easier for the outer electrons to be given away as the pull from the nucleus on them becomes smaller.

Question 37

how do F- + Cl- react w H2So4

Answer 37

F- and Cl- ions are not strong enough reducing agents to reduce the S in H2SO4. No redox reactions occur. Only acid-base reactions occur. NaF(s) + H2SO4(l)NaHSO4(s) + HF(g) Observations: White steamy fumes of HF are evolved. NaCl(s) + H2SO4(l)NaHSO4(s) + HCl(g) Observations: White steamy fumes of HCl are evolved. These are acid – base reactions and notredoxreactions.H2SO4 playsthe role of an acid (proton donor).

Question 38

br- + H2SO4, reaction equations, REDOX equations, observations, sulfuric acid role

Answer 38

Br- ions are stronger reducing agents than Cl- and F- and after the initial acid-base reaction, the bromide ions reduce the sulfur in H2SO4 from +6 to + 4 in SO2 Overall equation: combining two steps above: 2NaBr + 3H2SO4  2NaHSO4 + SO2 + Br2 + 2H2O Acid- base step: NaBr(s) + H2SO4(l)  NaHSO4(s) + HBr(g) Redox step: 2 H+ + 2 Br - + H2SO4 Br2(g) + SO2(g) + 2 H2O(l) Observations: White steamy fumes of HBr are evolved. orange fumes of bromine are also evolved and a colourless, acidic gas SO2 Ox 1⁄2 equation 2Br -Br2 + 2e- Re1⁄2equationH2SO4 +2H+ +2e-  SO2 +2H2O Note the H2SO4 plays the role of acid in the first step producing HBr and then acts as an oxidising agent in the second redox step.

Question 39

I- + h2so4

Answer 39

I- ions are the strongest halide reducing agents. They can reduce the sulfur from +6inH2SO4 to+4inSO2,to 0 in S and -2 in H2S. Observations: White steamy fumes of HI are evolved. Black solid and purple fumes of Iodine are also evolved A colourless, acidic gas SO2 A yellow solid of sulfur H2S (Hydrogen sulfide), a gas with a bad egg smell, NaI(s) + H2SO4(l)NaHSO4(s) + HI(g) 2 H+ + 2 I- + H2SO4I2(s) + SO2(g) + 2 H2O(l) 6H+ +6I- +H2SO4 3I2 +S(s)+4H2O(l) 8H+ +8I- +H2SO44I2(s) + H2S(g) + 4H2O(l) Ox 1⁄2 equation 2I -  I2 + 2e- Re1⁄2equationH2SO4 +2H+ +2e-  SO2 +2H2O Re1⁄2equationH2SO4 +6H+ +6e-  S+4H2O Re1⁄2equationH2SO4 +8H+ +8e-  H2S+4H2O Reduction products = sulfur dioxide, sulfur and hydrogen sulfide Note the H2SO4 plays the role of acid in the first step producing HI and then acts as an oxidising agent in the three redox steps.

Question 40

chlorine w water reaction

Answer 40

Cl2(g) + H2O(l)⇌ HClO(aq) + HCl(aq)

Question 41

what happens if cl bubbled thru water in sunlight, observation

Answer 41

2Cl2 + 2H2O4H+ + 4Cl- + O2 The greenish colour of chlorine water fades as the Cl2 reacts and a colourless gas (O2) is produced.

Question 42

what happens if UI added to reaction mixture of Cl2 + h20

Answer 42

If some universal indicator is added to the solution it will first turn red due to the acidity of both reaction products. It will then turn colourless as the HClO bleaches the colour.

Question 43

what is chlorine used for

Answer 43

water treatment to kill bacteria. It has been used to treat drinking water and the water in swimming pools. The benefits to health of water treatment by chlorine outweigh its toxic effects.

Question 44

Reaction of chlorine with cold dilute NaOH solution, observation

Answer 44

Cl2,(and Br2, I2) in aqueous solutions will react with cold sodium hydroxide. The colour of the halogen solution will fade to colourless. Cl2 (aq) + 2 NaOH (aq)  NaCl (aq) + NaClO (aq) + H2O (l) The mixture of NaCl and NaClO is used as bleach and to disinfect/ kill bacteria.