1. Halogens Edexcel new Specification Application of Core principles
Of chemistry
Edexcel new
Specification

2. Group 7 – the halogens
The elements in group 7 of the periodic table, on the right, are called the halogens. I Br Cl F At
fluorine
chlorine
bromine
iodine
astatine

3. Content of the specification
Inorganic chemistry of group 7 (limited to chlorine, bromine and iodine) a
Recall the characteristic physical properties of the elements limited to the appearance of solutions of the elements in water and hydrocarbon solvents b
Describe and carry out the following chemical reactions of halogens:
i Oxidation reactions with metal and non-metallic elements and ions such as iron(II) and iron(III) ions in solution
ii Disproportionation reactions with cold and hot alkali, eg hot potassium hydroxide with iodine to produce potassium iodate(V) c
Make predictions about fluorine and astatine and their compounds based on the trends in the physical and chemical properties of the halogens.

4. Content of the specificationK
Carry out an iodine/thiosulfate titration, including calculation of the results and evaluation of the procedures involved, eg determination of the purity of potassium iodate(V) by liberation of iodine and titration with standard sodium thiosulfate solution describe and carry out the following reactions:
i Potassium halides with concentrated sulfuric acid, halogens and silver nitrate solution
ii Silver halides with sunlight and their solubility in aqueous ammonia solution
iii Hydrogen halides with ammonia and with water (to produce acids)

5. Electronic configuration
Halogen
Electronic configuration F
1s22s22p5 Cl
1s22s22p63s23p5 Br
1s22s22p63s23p63d104s24p5 I
1s22s22p63s23p63d104s24p64d105s25p5 At
1s22s22p63s23p63d104s24p64d104f145s25p65d106s66p5

6. a) recall the characteristic physical properties of the elements

7. All halogens
outermost shell electronic configuration of ns2np5
one electron short of the octet electronic configuration
In the free elemental state
form diatomic molecules
complete their octets by sharing their single unpaired p electrons

8. Electronegativity value
High Electronegativity
Electronegativity is the relative tendency of an atom to attract bonding electrons towards itself in a covalent bond.
All halogens
high electronegativity values high tendency to attract an additional electron to achieve the stable octet electronic configuration highest among the elements in the same period
Halogen
Electronegativity value F Cl Br I At
4.0
3.0
2.8
2.5
2.2

9. First ionisation energy
First ionization energy of the halogens decreases down the group as the atomic radius increases down the group so shielding effect by the inner electrons increases down the group

10. Melting and Boiling Points: increases down the group
 The melting and boiling points increase down the group because of the van der Waals forces. The size of the molecules increases down the group. This increase in size and increase in number of electrons means an increase in the strength of the van der Waals forces. 
F < Cl < Br < I < At

11. Electron Affinity
The general decrease in electron affinity
the atomic size increases
the number of electrons shells down the group increases
the effective nuclear charge decreases
tendency of the nuclei of halogen atoms to attract additional electrons decreases

12. Electron Affinity Its value indicates the ease of formation of anions
Fluorine atom
very small atomic size
energy is required to overcome the repulsion between the additional electron and the electrons present in the electron shell -
F(g) + e F(g)
High Electron Affinity
Electron affinity is the enthalpy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state.

13. Electron Affinity Halogen F Cl Br I At E.A. kJ/mol1 -322 -349 -335
 down the group
F(g) + e F(g)
Halogen F Cl Br I At
E.A. kJ/mol1
-322
-349
-335
-295
-270
The number of electron shells and size of atoms  down the group
The nuclear attraction for the additional electron  down the group
Electron affinity  from Cl to I

14. Why fluorine break the trend ??
The reason that the electron affinity is not as high as might otherwise be predicted for fluorine is that it is an extremely small atom, and so it's electron density is very high. So the repulsion between the incoming electron and seven electrons in the second shell reduces energy liberated by the attraction between the incoming electron and the nucleus

15. What are the general properties of the halogens?
All the halogens are:
non-metals and so do not conduct electricity
brittle and crumbly when solid
poisonous and smelly.
They become darker in colour down the group:
is pale yellow
is yellow-green
is red-brown
is grey

16. What is the physical state of the halogens?
The melting and boiling temperatures of the halogens increase down the group, as the molecules become larger.
Halogen
Relative size
Tm(°C)
Tb(°C)
State
-220
-118
gas
-101
-34
gas -7 59
liquid
114
184
solid

17. What is the electron structure of the halogens?
All halogens have seven electrons in their outer shell. This means that:
fluorine
2,7
They can easily obtain a full outer shell by gaining one electron.
chlorine
2,8,7
They all gain an electron in reactions to form negative ions with a -1 charge.
bromine
2,8,8,7
They have similar chemical properties.

18. How does electron structure affect reactivity?
The reactivity of alkali metals decreases going down the group. What is the reason for this?
The atoms of each element get larger going down the group. F
decrease in reactivity
This means that the outer shell gets further away from the nucleus and is shielded by more electron shells. Cl
The further the outer shell is from the positive attraction of the nucleus, the harder it is to attract another electron to complete the outer shell. Br
This is why the reactivity of the halogens decreases going down group 7.

19. How do halogen molecules exist?
All halogen atoms require one more electron to obtain a full outer shell and become stable.
Each atom can achieve this by sharing one electron with another atom to form a single covalent bond. +  F
This means that all halogens exist as diatomic molecules: F2, Cl2, Br2 and I2.

20. Oxidising ability of the halogens
Fluorine (F2)
Chlorine (Cl2)
Bromine (Br2)
Iodine (I2)
Decreasing oxidising ability
Decreasing reactivity

21. Physical properties Going down the group Atomic size increases
the electron clouds of the molecules become larger
more polarizable
Melting abd boiling tenperature increases
Electron affinity decreases from clorine to iodine

22. the greater the number of electrons, the stronger the dispersion forces and the higher the boiling points/melting points of the substance
 larger group 7 molecules have larger, more dispersed electron clouds that lead to increased polarizability and therefore, stronger attraction between atoms

23. Fluoride (F-) Chloride (Cl-) Bromide (Br-) Iodide (I-)
Reducing ability of the halides
Fluoride (F-)
Chloride (Cl-)
Bromide (Br-)
Iodide (I-)
Increasing reducing ability

24. Variation in Chemical Properties
Reactivity : F2 > Cl2 > Br2 > I2
React by gaining electrons
Oxidizing power : F2 > Cl2 > Br2 > I2
Oxidizing power decreases down the group

25. Properties of the Halogens
Colour
State
Decreasing reactivity
Increasing molecular size
Increasing density F Cl Br I At
Gas
Yellow
Gas
Liquid
Solid
Green
Orange
Grey/black
Black
Solid

26. chlorine
bromine
Appearances of halogens
at room temperature and
pressure: chlorine ,bromine
iodine
iodine

27. a )
Appearance of solutions of the elements in water and hydrocarbon solvents

28. Reaction of halogens with water
The soluability of halogens in water decreases down the group
a) F2 oxidises H2O to O2 gas in a very exothermic reaction.
2F2(g) + 2H2O(l) → O2(g) + 4HF(g)
b) Cl2 dissolves in H2O and some hydrolysis occurs. A pale yellow solution of 'chlorine water' is formed which is a mixture of two acids. No O2 is evolved.
Cl2(g) + H2O(l) → HCl(aq) + HOCl(aq)
Cl2(g) + H2O(l) → 2H+(aq) + Cl- (aq) + CLO-
Chlonic(I) acid is the substance which gives a solution of chlorine its bleaching properties

29. c) Br is only slightly soluble in H2O and there is less hydrolysis.
Br2(l) + H2O(l) → HBr(aq) + HOBr(aq)
d) I2 is virtually insoluble in H2O. It is however soluble in KI solution due to the formation of the triiodide anion.
I2(s) + I-(aq) → I3-(aq)
Note: All halogens are more soluble in non-polar solvents such as CCl4.

30. Colours of halogens in pure form and in solutions
in water
in 1,1,1-trichloroethane/ cyclohexane F2
Pale yellow
Cl2
Greenish yellow
Yellow
Br2
Reddish brown
Orange I2
Violet black
Yellow (only slightly soluble)
Brown in KI(aq)
Violet

31. Halogens
non-polar molecules
not very soluble in polar solvents (such as water)
but very soluble in organic solvents (such as 1,1,1-trichloroethane, cyclohexane)

32. Halogens are non-polar so they are more soluble in hydrocarbon
solvents than in water (see chapter 2.3), as shown in fig (see next slide)
In each tube there is water (lower layer) and cyclohexane
(upper layer) . Chlorine is added to the first tube, bromine to the
second and iodine to the third.
You will notice, especially with bromine and iodine, that most of the halogen has dissolved in the cyclohexane layer.

33. Water layer Chlorine Bromine Iodine
Non-polar/hydrocarbon solvent example cyclohexane layer
Water layer
Chlorine
Bromine
Iodine

34. (a)
(b)
(c)
Colours of halogens in 1,1,1-trichloroethane: (a) chlorine; (b) bromine; (c) iodine

35. b ) describe and carry out the following chemical reactions of
halogens:
i ) oxidation reactions with metal and non-metallic elements
and ions such as iron(II) and iron(III) ions in solution

36. Reactions with Iron(II) Ions
Aqueous chlorine
oxidizes green iron(II) ions to yellowish brown iron(III) ions
2Fe2+(aq) + Cl2(aq) Fe3+(aq) + 2Cl–(aq)

37. Reactions with Iron(II) Ions
Aqueous bromine oxidizes green iron(II) ions to yellowish brown iron(III)
2Fe2+(aq) + Br2(aq) Fe3+(aq) + 2Br–(aq)
Iodine
a mild oxidizing agent
not strong enough to oxidize iron(II) ions.

38. Standard electrode potential (V)
Reactions with reducing agents example Fe2+
All halogens(except I2) oxidize Fe2+ to Fe3+
Half reaction
Standard electrode potential (V)
Cl2(aq) + 2e– Cl–(aq)
Br2(aq) + 2e – Br–(aq)
Fe3+(aq) + e– Fe2+(aq)
I2(aq) + 2e– I–(aq)
+1.36
+1.07
+0.77
+0.54
X2(aq) + 2Fe2+(aq) X(aq) + 2Fe3+(aq)
( X = F, Cl, Br)
I2(aq) + 2Fe2+(aq)  No reaction

39. Reactions of halogens with iron(II) ions
Reactant
Chlorine
Bromine
Iodine
Iron(II) ions
The green iron(II) ions are oxidized to yellowish brown iron(III) ions
The solution remains green since iron(II) ions are not oxidized by iodine

40. Reactions with non-metals
Chlorine reacts with most non-metals to form molecular chlorides.
Hot silicon, for example, reacts to form silicon tetrachloride, SiCl4(l), and phosphorus produces phosphon1s trichloride, PCl3).
However, chlorine does not react directly with carbon, oxygen or nitrogen.

41. Reactions with hydrogen
X2 + H2(g) HX(g)
F2 reacts explosively even in the dark at 200C
Cl2 reacts explosively in sunlight
Br2 reacts moderately on heating with a catalyst
I2 reacts slowly and reversibly even on heating

42. Explain the extreme reactivity of fluorine in terms of the bond enthalpies of F–F and H–F bonds.
F2 + H2(g) HF(g)
Fluorine has an exceptionally small F-F bond enthalpy.
Thus, the activation energy of its reaction with hydrogen is also exceptionally small.
Hydrogen fluoride has the highest bond enthalpy among the hydrogen halides.
Thus, the formation of HF from H2 and F2 is the most exothermic.
The energy released from the reaction further speeds up the reaction.

43. Physical properties of halogens and hydrogen halides-1
Bond
Bond length nm
Bond energy k J mol
F-F
0.142
158
Cl-Cl
0.199
243.4
Br-Br
0.228
192.9
I-I
0.267
151.2
At-At -1
Bond
Bond length nm
Bond energy k J mol
H-F
0.092
568
H-Cl
0.127
432.0
H-Br
0.141
366.3
H-I
0.161
298.3
H-At

44. Reactions with phosphorus
F2 + P PF5
Cl2 + P PCl3 + PCl5
Br2 + P PBr3
I2 + P PI3
F2 is the strongest oxidizing agent, it always oxidizes other elements to their highest possible oxidation states.

45. Reactions with phosphorus
F2 + P PF5
Cl2 + P PCl3 + PCl5
Br2 + P PBr3
I2 + P PI3
Br2 and I2 are NOT strong enough to oxidize P to its highest possible oxidation state.

46. All halogens(except I2) oxidize S2O32 to SO42
4X2(aq)+S2O32(aq)+5H2O(l)  8X(aq)+10H+(aq) +2SO42(aq)
(X = F, Cl, Br)
I2(aq) + 2S2O32(aq) I(aq) + S4O62(aq) +5 +6 +4
Used in iodometric titration

47. ii) disproportionation reactions with cold and hot alkali, eg hot potassium hydroxide with iodine to produce potassium iodate(V)

48. Reactions with Alkalis
The reactions between halogens and aqueous alkalis
disproportionation (except fluorine)
react differently under cold / hot and dilute / concentrated conditions
their reactivities decrease down the group

49. (Fluorine)
Fluorine is passed through a cold and very dilute (2%) sodium hydroxide solution
oxygen difluoride (OF2) is formed
2F2(g) + 2NaOH(aq) NaF(aq) + OF2(g) + H2O(l) cold, very dilute – –1
When fluorine is passed through a hot and concentrated sodium hydroxide solution
oxygen is formed instead
2F2(g) + 4NaOH(aq) NaF(aq) + O2(g) + 2H2O(l) – – hot, concentrated

50. Chlorine
reacts with cold and dilute sodium hydroxide solution to form sodium chloride and sodium chlorate(I) (also called sodium hypochlorite)
Cl2(aq) + 2NaOH(aq) NaCl(aq) + NaOCl(aq) + H2O(l) cold, dilute –
reacts with hot and concentrated sodium hydroxide solution to form sodium chloride and sodium chlorate(V)
3Cl2(aq) + 6NaOH(aq) NaCl(aq) + NaClO3(aq) + 3H2O(l)
hot, concentrated –
Disproportionation

51. Bromine undergoes similar reactions with alkalis as chlorine
sodium bromate(I) formed is unstable
disproportionates to form sodium bromide and sodium bromate(V) readily at room temperature and pressure
reversible
Br2(aq) + 2NaOH(aq) NaBr(aq) + NaOBr(aq) + H2O(l)
cold, dilute
3NaOBr(aq) NaBr(aq) + NaBrO3(aq)
The chemical equation for the overall reaction:
3Br2(aq) + 6NaOH(aq) NaBr(aq) + NaBrO3(aq) + 3H2O(l)
cold, dilute –

52. Iodine behaves similarly as bromine
Except that the reaction with a cold and dilute alkali
reversible
3I2(aq) + 6KOH(aq) KI(aq) + KIO3(aq) + 3H2O(l)
cold, dilute –
What are the products, other than water, when chlorine is passed through cold, dilute aqueous sodium hydroxide solution?
A NaCl and NaClO
B NaClO and NaClO3
C NaCl and NaClO3
D NaClO and NaClO4

53. carry out an iodine/thiosulfate titration, including calculation of the results and evaluation of the procedures involved, eg
determination of the purity of potassium iodate(V) by liberation of iodine and titration with standard sodium thiosulfate solution

54. Iodine/thiosulfate titrations
One common oxidising agent is iodine:
I2 (aq) + 2e-  2I- (aq)
The reducing agents you met included thiosulfate:
2S2O3 2- (aq)  S4O62-(aq) + 2e-
Combining these two, the overall ionic equation for the reaction of iodine with thiosulfate is:
I2 (aq) + 2S2O3 2-  S4O62-(aq) + 2I- (aq)
This reaction is the basis of the iodine/thiosulfate titration.

55. Finding the purity of potassium iodate(V)
Iodate(V), IO 3- , reacts as an oxidising agent:
IO 3-(aq) + 5I- (aq) + 6H+ (aq)  3I2 + 3H 2O(l)
If iodate(V) ions are added to excess potassium iodide solution, iodine is liberated. The mixture can be titrated with standard sodium thiosulfate solution using starch as indicator.

56. A weighed sample of impure potassiun1 iodate(V) solid (0
A weighed sample of impure potassiun1 iodate(V) solid (0.80 g) is dissolved in water and made up to 250 cm3 of solution in a volumetric flask.
25.0 cm3 of this solution are added to excess potassium iodide solution, to liberate iodine.
In the titration cm3 of 0.1 mol dm-3 sodium thiosulfate were needed to react with the liberated iodine.
20.00 x 0.1 mol sodium thiosulfate
1000
= mol of thiosulfate

57. From the equation for the reaction of iodine with thiosulfate above, you can see that mol of thiosulfate will react with mol of I 2 . So mol is the quantity of iodine that was liberated by 25.0 cm3 of impure potassium iodate(V) solution.
From the equation for the reaction of iodate(V) with iodide above, you can see that the number of moles of iodate(V) required to liberate mol of iodine is given by:
number of moles IO 3- = mol / 3 = mol

58. The mass of potassium iodate(V) in the sample weighed out is 10 times this, since 25 cm3 of the 250 cm3 solution was used in the reaction:
mass KIO3 = g x 10 = g
Percentage purity of potassium iodate(V)

60. fig. 2.5.14 Titrating to find the percentage of copper in brass.

61. HALIDES d) describe and carry out the following reactions:
i) potassium halides with concentrated sulfuric acid, halogens and silver nitrate solution
ii) silver halides with sunlight and their solubilities in aqueous ammonia solution
iii )hydrogen halides with ammonia and
with water (to produce acids)

62. Boiling temperature of Halides

63. Hydrogen fluoride has abnormally high boiling temperature and melting temperature among the hydrogen halides
Because of hydrogen bionds between hydrogen and flurine
from clorine to idoine it increases as number of electrons increase so van der Waals forces increase

64. Reactions with Concentrated Sulphuric(VI) Acid
i) potassium halides with concentrated sulfuric acid
Reactions with Concentrated Sulphuric(VI) Acid
Concentrated sulphuric(VI) acid an oxidizing acid
exhibits both oxidizing and acidic properties
On treatment with concentrated sulphuric(VI) acid
fluorides and chlorides give hydrogen fluoride and hydrogen chloride respectively
NaF(s) + H2SO4(l) NaHSO4(s) + HF(g)
NaCl(s) + H2SO4(l) NaHSO4(s) + HCl(g)

65. Action of concentrated sulphuric(VI) acid on
Bromides and iodides
do not give hydrogen bromide and hydrogen iodide respectively
sulphur dioxide or hydrogen sulphide is formed
sodium bromide
sodium iodide
Action of concentrated sulphuric(VI) acid on

66. Chlorides NaCl(s) + H2SO4(l) NaHSO4(s) + HCl(g) Bromides
NaBr(s) + H2SO4(l) NaHSO4(s) + HBr(g)
2HBr(g) + H2SO4(l) SO2(g) + Br2(g) + 2H2O(l)
The chemical equation for the overall reaction is
2NaBr(s) + 3H2SO4(l) NaHSO4(s) + SO2(g)
+ Br2(g) + 2H2O(l)
Iodides
NaI(s) + H2SO4(l) NaHSO4(s) + HI(g)
8HI(g) + H2SO4(l) H2S(g) + 4I2(g) + 4H2O(l)
8NaI(s) + 9H2SO4(l) NaHSO4(s) + H2S(g)
+ 4I2(g) + 4H2O(l)

67. Which of the following statements about the elements in Group 7 is incorrect?
A They all show variable oxidation states in their compounds.
B They all form acidic hydrides.
C Electronegativity decreases as the group is descended.
D They all exist as diatomic molecules.

68. Bromides and iodides
do not react in the same way as fluorides and chlorides
 the hydrogen bromide and hydrogen iodide produced are oxidized by concentrated sulphuric(VI) acid to bromine and iodine respectively
Hydrogen chloride is not oxidized by concentrated sulphuric(VI) acid
In solution the halide ions act as reducing agents, the strongest ability increases down the group. HI is the strongest reducing agent.
HI in will reduce H2SO4 to H2S.
HBr will only reduce H2SO4 to SO2.

69. Action of concentrated sulphuric(VI) acid on halides
Halide ion
Action of concentrated sulphuric(VI) acid
Product
Confirmatory test of the product
Cl–
Steamy fumes are formed
No green gas is evolved even on heating
HCl
Dense white fumes are formed with aqueous ammonia
It turns blue litmus paper red but not bleached
NaCl(s) + H2SO4(l) NaHSO4(s) + HCl(g)

70. Action of concentrated sulphuric(VI) acid on halides
Halide ion
Action of concentrated sulphuric(VI) acid
Product
Confirmatory test of the product
Br–
Steamy fumes are formed
A pungent smell is detected
A brown gas is evolved on warming
HBr
White fumes are formed with aqueous ammonia
SO2
It turns orange dichromate(VI) solution green
Br2
A red colour is observed when adding hexane
NaBr(s) + H2SO4(l) NaHSO4(s) + HBr(g)
2HBr(g) + H2SO4(l) SO2(g) + Br2(g) + 2H2O(l)

71. Action of concentrated sulphuric(VI) acid on halides
Halide ion
Action of concentrated sulphuric(VI) acid
Product
Confirmatory test of the product I–
Steamy violet fumes are formed
A bad egg smell is detected HI
White fumes are formed with aqueous ammonia
H2S
It turns lead(II) ethanoate paper black I2
A violet colour is observed when adding hexane
NaI(s) + H2SO4(l) NaHSO4(s) + HI(g)
8HI(g) + H2SO4(l) H2S(g) + 4I2(g) + 4H2O(l)
The chemical equation for the overall reaction is
8NaI(s) +9H2SO4(l) NaHSO4(s)+H2S(g)+4I2(g) +4H2O(l)

72. ? Power of reduction HF + H2SO4 HCl + H2SO4
2HBr + H2SO Br2 + SO2 + 2H2O
2HI + H2SO I2 + SO2 + 2H2O
6HI + H2SO I2 + S + 4H2O
8HI + H2SO I2 + H2S + 4H2O
HAt + H2SO4
reducing power increase -1 +6 +4 -1 +6 +4 -1 +6 -1 +6 -2 ?

73. Reactions with Concentrated Sulphuric(VI) Acid
The reaction of chlorides with concentrated sulphuric(VI) acid used for the preparation of hydrogen chloride in the laboratory
Hydrogen bromide and hydrogen iodide cannot be prepared in this way
Reactions with Phosphoric(V) Acid
Phosphoric(V) acid- H3PO4
not an oxidizing agent reacts with halides to form the corresponding hydrogen halides
a general method to prepare hydrogen halides in the laboratory

74. Redox or Displacement reactions
i) halides with halogen
Redox or Displacement reactions -1
More reactive
Less reactive
Cl2(aq) + 2Br(aq) Cl(aq) +Br2(aq)
Cl2(aq) + 2I(aq) Cl(aq) + I2(aq)
Br2(aq) + 2I(aq) Br(aq) + I2(aq) -1 -1
I2(aq) I(aq) I3(aq)
(yellow) (brown)

75. A pale yellow solution is formed (Cl2 is formed)
Reactions of halide ions with halogens
Aqueous solution
Halogen added F2
Cl2
Br2 I2 F–
No reaction
Cl–
A pale yellow solution is formed (Cl2 is formed)

76. Reactions of halide ions with halogens
Aqueous solution
Halogen added F2
Cl2
Br2 I2
Br–
A yellow solution is formed (Br2 is formed)
A yellow solution is formed (Br2 is formed)
No reaction I–
A yellowish brown solution is formed (I3 is formed)
A yellowish brown solution is formed (I3 is formed)

77. Reactions with Silver Ions
i) Halides in silver nitrate solution
Reactions with Silver Ions
Aqueous solutions of chlorides, bromides and iodides
 give precipitates when reacting with acidified silver nitrate solution

78. Reactions with Silver Ions
Ag+(aq) + Cl–(aq)  AgCl(s) white ppt
Ag+(aq) + Br–(aq)  AgBr(s) pale yellow ppt
Ag+(aq) + I–(aq)  AgI(s) yellow ppt

79. AgCl(s)
AgBr(s)
AgI(s)

80. Two confirmatory tests for halides
ii) silver halides with sunlight and their solubilitiesin aqueous ammonia solution
Two confirmatory tests for halides
Adding NH3(aq) to the AgX ppt
Exposing AgX ppt to sunlight

81. AgX(s) dissolve in NH3(aq) due to the formation of soluble complex ions.
AgCl(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + Cl(aq)
AgBr(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + Br(aq)
AgI(s) + 2NH3(aq)  No reaction
Solubility in NH3(aq)  down the group

82. When exposed to sunlight
 silver chloride turns grey
light
2AgCl(s)  2Ag(s) + Cl2(g)
 silver bromide turns yellowish grey
2AgBr(s)  2Ag(s) + Br2(l)
light
 silver iodide remains yellow
2AgI(s)  No reaction
light

83. Action of acidified silver nitrate solution on halides Ion
Action of acidified AgNO3 solution on halides
Confirmatory test of the product
Effect of adding aqueous ammonia
Effect of exposure to sunlight
Cl–
A white ppt is formed
The white ppt dissolves
The solution turns grey
Br–
A cream ppt is formed
The pale yellow ppt slightly dissolves
The solution turns yellowish grey I–
A yellow ppt is formed
The yellow ppt does not dissolve
The solution remains yellow

84. iii) hydrogen halides with ammonia
All hydrogen halides are white steamy fumes when expose to the air
HF(g) HCl(g) HBr(g) HI(g)
HF(g) + NH3(g) NH4F(s) white smoky fume
ammonium fluoride
HCl(g) + NH3(g) NH4Cl(s) white smoky fume
ammonium chloride
HBr(g) + NH3(g) NH4Br(s) white smoky fume
ammonium bromide
HI(g) + NH3(g) NH4I(s) white smoky fume
ammonium iodide

85. Reaction of hydrogen halides with water
iii) hydrogen halides with water to form acid
Reaction of hydrogen halides with water
All the hydrogen halides dissolves in water to give strong acids except hydrogen fluoride.
2HF(g) + H2O H2F2(aq)
Hydrogen fluoride hydrofluoric acid ( weak acid)
2HCl(g) + H2O HCl(aq)
Hydrogen chloride hydrochloric acid ( strong acid)
2HBr(g) + H2O HBr(aq)
Hydrogen bromide hydrobromic acid ( stronger than HBr)
2HI(g) + H2O HI(aq)
Hydrogen iodide hydroiodic acid ( Strongest acid)
All the hydrogen halides are very soluble in water. Their solutions are acidic due to the formation of H+ ions.

86. Uses of Chlorine
Chlorine is used mostly to kill bacteria or as a bleach. Chlorine bleaches a piece of universal indicator paper white.

87. Summary

88. Water layer Chlorine Bromine Iodine
Non-polar/hydrocarbon solvent example cyclohexane layer
Water layer
Chlorine
Bromine
Iodine

89. Oxidation of Fe2+ and S2O32-
X2(aq) + 2Fe2+(aq)  2X(aq) + 2Fe3+(aq)
X2(aq) + 2S2O32-(aq)  2X(aq) + S4O62(aq)
Decreasing oxidising ability down the group
X = F, Cl, Br
Oxidation of Posperous
Decreasing reactivity down the group
F2 + P  PF5
Cl2 + P  PCl3 + PCl5
Br2 / I2 + P  PBr3 / PI3
Reaction with alkali (disproportination reaction)
All halogens reacts the way shown below
Dilute/cold alkali
Conc. Hot alkali
NaOH + Cl2 
NaCl + NaOCl + H2O
NaCl + NaClO3 + H2O
Oxidation number of halogen

90. reducing power increase
Halide reactions
NaCl + H2SO4  NaHSO 4 + HCl
reducing power increase
NaBr + H2SO4  NaHSO 4 + HBr
HBr + H2SO4  Br2 + SO2 + H2O
2HI + H2SO4  I2 + SO2 + 2H2O
6HI + H2SO4  3I2 + S + 4H2O
8HI + H2SO4  3I2 + H2S + 4H2O
Compound
Color if produced
HCl/ HBr/ HI
steamy fumes.. White smoke with ammonia gas
SO2
Not visible but pungent smell
Br2
Brown gas I2
Violet gas
H2S
Smell of bad egges S
Yellow solid

91. Action of acidified AgNO3 solution on halides
Confirmatory test of the product
Effect of adding conc. and dil. ammonia
Effect of exposure to sunlight
Cl–
A white ppt is formed
Soluble in conc. and dil. ammonia
The solution turns grey
Br–
A cream ppt is formed
Soluble in conc. ammonia Insoluble in dil. ammonia
The solution turns yellowish grey I–
A yellow ppt is formed
Insoluble in conc. and dil. ammonia
The solution remains yellow (No reaction)

92. A pale yellow solution is formed (Cl2 is formed)
Reactions of halide ions/Potassium halides with halogens (Disproportination reaction)
Aqueous solution
Halogen added F2
Cl2
Br2 I2 F–
No reaction
Cl–
A pale yellow solution is formed (Cl2 is formed)

93. Aqueous solution Halogen added F2 Cl2 Br2 I2
Reactions of halide ions/Potassium halides with halogens (Disproportination reaction)
Aqueous solution
Halogen added F2
Cl2
Br2 I2
Br–
A yellow solution is formed (Br2 is formed)
A yellow solution is formed (Br2 is formed)
No reaction I–
A yellowish brown solution is formed (I3 is formed)
A yellowish brown solution is formed (I3 is formed)

94. iii) hydrogen halides with ammonia
All hydrogen halides are white steamy fumes when expose to the air
HF(g) HCl(g) HBr(g) HI(g)
HF(g) + NH3(g) NH4F(s) white smoky fume
ammonium fluoride
HCl(g) + NH3(g) NH4Cl(s) white smoky fume
ammonium chloride
HBr(g) + NH3(g) NH4Br(s) white smoky fume
ammonium bromide
HI(g) + NH3(g) NH4I(s) white smoky fume
ammonium iodide

95. Reaction of hydrogen halides with water
iii) hydrogen halides with water to form acid
Reaction of hydrogen halides with water
All the hydrogen halides dissolves in water to give strong acids except hydrogen fluoride.
2HF(g) + H2O H2F2(aq)
Hydrogen fluoride hydrofluoric acid ( weak acid)
2HCl(g) + H2O HCl(aq)
Hydrogen chloride hydrochloric acid ( strong acid)
2HBr(g) + H2O HBr(aq)
Hydrogen bromide hydrobromic acid ( stronger than HBr)
2HI(g) + H2O HI(aq)
Hydrogen iodide hydroiodic acid ( Strongest acid)
All the hydrogen halides are very soluble in water. Their solutions are acidic due to the formation of H+ ions.

96. The bonding in gaseous hydrogen halides is best described as
A mainly covalent with an increasing tendency towards ionic as you go down the group.
B mainly covalent with an increasing tendency towards ionic as you go up the group.
C mainly ionic with an increasing tendency towards covalent as you go down the group.
D mainly ionic with an increasing tendency towards covalent as you go up the group.

97. Fountain experiment for dissolution of hydrogen halides

98. Copper(ii) thiosulphate titrations
General theory
Copper(II) compounds can be analysed by a redox titration.
The general procedure is that excess potassium iodide solution is added to a neutral solution of copper(II). This liberates iodine according to the equation below and the amount of iodine is found by titration with sodium thiosulphate solution.
Just before the end-point, several drops of starch solution are added and one continues the titration until the blue colour just disappears and an off-white precipitate remains.
2Cu2+(aq) + 4I¯(aq) CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I¯(aq)
therefore moles of S2O32- = moles of Cu2+(aq) - 2+
Cu + e Cu / 2I I2 + 2e- +

99. (alternatively dissolve a known mass of solid in water)
Practical details A A
Pipette a known volume of a solution of Cu2+ ions into a conical flask.
(alternatively dissolve a known mass of solid in water)
Neutralise the solution by adding sodium carbonate solution dropwise until a feint precipitate starts to form.

100. Add excess potassium iodide solution to liberate iodine.
Practical details
A B
Add excess potassium iodide solution to liberate iodine.
The copper(II) is reduced to copper(I) and half the iodide ions are oxidised to iodine.
2Cu2+(aq) I¯(aq) CuI(s) + I2(aq)
off white solid B

101. The iodine is reduced back to iodide ions.
Practical details
A B C C
Titrate with a standard solution of sodium thiosulphate until the solution lightens. DO NOT ADD TOO MUCH.
The iodine is reduced back to iodide ions.
2S2O32-(aq) I2(aq) S4O62-(aq) I¯(aq)

102. Starch solution is added near the end point.
Practical details
A B C D
Starch solution is added near the end point.
Starch gives a dark blue colouration in the presence of iodine. D

103. Record the volume added and repeat to obtain concordant results.
Practical details
A B C D E E
Continue with the titration, adding the sodium thiosulphate dropwise until the blue colour disappears at the end point. This indicates that all the iodine has reacted.
Record the volume added and repeat to obtain concordant results.

104. To find the approximate end point
Mean titre- Average titre
19.9 cm 3
Titration number
1 (trial) 2 3 4
Burette reading
(final) / cm
21.45
41.35
21.95
(initial) / cm
1.20
Volume of
Cu (aq) used / cm
20.00
19.80
Titre used to
calculate mean ( ) 3 3 2+ 3
Complete the table and indicate with a tick ( ) those titres most suitable for calculating a mean titre.

105. To find the approximate end point
Mean titre- Average titre
19.9 cm 3
Titration number
1 (trial) 2 3 4
Burette reading
(final) / cm
21.45
41.35
21.95
41.75
(initial) / cm
1.20
1.95
Volume of
Cu (aq) used / cm
20.25
19.90
20.00
19.80
Titre used to
calculate mean ( ) 3 3 2+ 3
Complete the table and indicate with a tick ( ) those titres most suitable for calculating a mean titre.

106. The difference between the student’s mean titre and the accurate value was not due to the limitations in the accuracy of the measuring instruments.
Suggest one possible reason for this difference.
Judgement (of colour change) at end point / adding starch too early in the titration / jet of burette not filled

107. Typical calculation Number of water molecules of crystallisation
1 titrate a known mass of copper compound or a known volume of a solution
2 calculate the moles of S2O32- needed
3 according to the equation… moles of Cu2+ = moles of S2O32-
4 calculate the number of moles of Cu2+
5 calculate the number of moles of CuSO4
moles of CuSO4 = moles of Cu2+ (there is one Cu2+ in every CuSO4)
6 calculate the mass of copper sulphate by multiplying the moles of copper sulphate by the molar mass of copper sulphate (CuSO4)
7 calculate mass of water(= mass of CuSO4.xH2O - mass of CuSO4)
8 divide the mass of water by 18 to find the number of moles of water
compare the ration of moles of… water : moles of CuSO4 to find x

108. Calculation – Example 1
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a mol dm sodium thiosulphate solution
Calculate the percentage of copper in the alloy.

109. Calculation – Example 1
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a mol dm sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations 2Cu2+(aq) + 4I¯(aq) 2CuI(s)+ I2(aq)
2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)

110. Calculation – Example 1
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a mol dm sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+I2(aq)
2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = x / 1000 = x 10-3

111. Calculation – Example 1
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a mol dm sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+ I2(aq)
2S2O32-(aq) + I2(aq) S4O62-(aq)+ 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = x / 1000 = x 10-3
moles of Cu2+ in 25cm3 = moles of S2O32- = x 10-3

112. Calculation – Example 1
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a mol dm sodium thiosulphate solution
Calculate the percentage of copper in the alloy. -3
From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) S4O62-(aq)+ 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = x / = x 10-
moles of Cu2+ in 25cm3 = moles of S2O32- = x 10-3
moles of Cu2+ in 250cm3 = x x 10 =
mass of Cu2+ in 250cm3 = x = g

113. Calculation – Example 1
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a mol dm sodium thiosulphate solution
Calculate the percentage of copper in the alloy. -3
From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+ I2(aq)
2S2O32-(aq) + I2(aq) S4O62-(aq)+ 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = x / = x 10-3
moles of Cu2+ in 25cm3 = moles of S2O = x 10-3
moles of Cu2+ in 250cm3 = x x 10 =
mass of Cu2+ in 250cm3 = x 63.5 = g
% of Cu in the alloy = / x 100 = %

114. 3I2(aq) + 6NaOH(aq) 5NaI(aq) + NaIO3(aq) + 3H2O(l)
Iodine can react with sodium hydroxide solution to form NaIO3(aq), according to the equation below.
3I2(aq) + 6NaOH(aq) NaI(aq) + NaIO3(aq) + 3H2O(l)
Which of the statements about the reaction is false?
A The oxidation number of some iodine atoms goes up.
B At high temperatures NaIO(aq) also forms.
C Sodium ions are spectator ions.
D The oxidation number of some iodine atoms goes down.
Which of these statements about fluorine is not correct?
A It is a gaseous element at room temperature and pressure.
B It can react with chloride ions to form chlorine.
C It forms salts with Group 1 metals.
D It is less electronegative than chlorine.

115. 1 Which is a true statement about hydrogen fluoride?
A It is more acidic than hydrogen chloride.
B It is oxidised by concentrated sulfuric acid.
C It disproportionates in water.
D It forms white fumes with gaseous ammonia.
2 The reaction OCl− + Cl− + 2H Cl2 + H2O is
A disproportionation
B acid–base
C redox
D synthesis +

116. - Which statement is not true?
A Chlorine oxidises bromide ions to bromine.
B Hydrogen iodide is more polar than hydrogen bromide.
C Astatide ions, At , reduce Fe ions to Fe ions.
D Silver chloride is soluble in concentrated aqueous ammonia.
Using starch as an indicator, the colour change at the end point in an iodine/thiosulfate titration is from
A red-brown to colourless
B blue to colourless
C colourless to blue
D yellow to colourless - 3+ 2+

117. Which mixture gives steamy fumes?
A hydrogen chloride and ammonia
B chlorine and potassium bromide
C concentrated sulfuric acid and silver bromide
D silver chloride and aqueous ammonia
Which statement about hydrogen bromide is not true?
A It gives a brown solution with chlorine water.
B It gives a cream precipitate with aqueous silver nitrate.
C It has a higher boiling temperature than hydrogen chloride.
D It reacts in a disproportionation reaction with cold aqueous sodium hydroxide.

118. In the estimation of the concentration of an oxidising agent, excess potassium iodide is added before titrating with sodium thiosulfate.
The reason for this is
A all the oxidising agent must react
B the reaction between the oxidising agent and iodide ions is reversible
C to ensure that the reaction is complete
D to ensure that all the iodine atoms in the compounds are converted to iodine molecules

119. Which of the following is not a true statement about hydrogen iodide?
A It forms steamy fumes in moist air.
B It dissolves in water to form an acidic solution.
C It forms a cream precipitate with silver nitrate solution.
D It forms dense white smoke with ammonia.