application of core principles of chemistry edexcel new specification halogens
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Application ofCore principlesOf chemistry
Edexcel newSpecification
Halogens
Group 7 – the halogens
The elements in group 7 of the periodic table, on the right, are called the halogens.
fluorine
chlorine
bromine
iodine
astatine
I
Br
Cl
F
At
Inorganic chemistry of group 7 (limited to chlorine, bromine and iodine)
aRecall 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 solutionii Disproportionation reactions with cold and hot alkali, eg hot potassium hydroxide with iodine to produce potassium iodate(V)
cMake predictions about fluorine and astatine and their compounds based on the trends in the physical and chemical properties of the halogens.
Content of the specification
K
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 solutionii Silver halides with sunlight and their solubility in aqueous ammonia solutioniii Hydrogen halides with ammonia and with water (to produce acids)
Content of the specification
Halogen Electronic configuration
F 1s22s22p5
Cl 1s22s22p63s23p5
Br 1s22s22p63s23p63d104s24p5
I 1s22s22p63s23p63d104s24p64d105s25p5
At 1s22s22p63s23p63d104s24p64d104f145s25p65d106s66p5
Electronic configuration
a) recall the characteristic physical properties of the elements
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
Electronegativity is the relative tendency of an atom to attract bonding electrons towards itself in a covalent bond.All halogenshigh electronegativity values high tendency to attract an additional electron to achieve the stable octet electronic configuration highest among the elements in the same period
High ElectronegativityHigh Electronegativity
HalogenElectronegativity
value
F
Cl
Br
I
At
4.0
3.0
2.8
2.5
2.2
First ionisation energyFirst 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
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
Electron Electron
AffinityAffinity 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
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.
High Electron AffinityHigh 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)-
Electron Affinity down the group
Halogen F Cl Br I At
E.A. kJ/mol1
-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
F(g) + e- F(g)
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
poisonous and smelly.
brittle and crumbly when solid
What are the general properties of the halogens?
All the halogens are:
They become darker in colour down the group:
non-metals and so do not conduct electricity
is pale yellow
is yellow-green
is grey
is red-brown
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
-101
-7
114
-34
59
184
gas
gas
liquid
solid
All halogens have seven electrons in their outer shell. This means that:
What is the electron structure of the halogens?
fluorine
2,7
chlorine
2,8,7
bromine
2,8,8,7
They can easily obtain a full outer shell by gaining one electron.
They have similar chemical properties.
They all gain an electron in reactions to form negative ions with a -1 charge.
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. This means that the outer shell gets
further away from the nucleus and is shielded by more electron shells.
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.
This is why the reactivity of the halogens decreases going down group 7.
dec
reas
e in
rea
ctiv
ity
F
Cl
Br
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.
This means that all halogens exist as diatomic molecules:F2, Cl2, Br2 and I2.
+ F F F F
Oxidising ability of the halogens
• Fluorine (F2)
• Chlorine (Cl2)
• Bromine (Br2)
• Iodine (I2)
Decreasing oxidising ability
Decreasing reactivity
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
Physical properties
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
Reducing ability of the halides
Fluoride (F-) Chloride (Cl-) Bromide (Br-) Iodide (I-)
Increasing reducing ability
Variation in Chemical Variation in Chemical PropertiesProperties
Reactivity : F2 > Cl2 > Br2 > I2
React by gaining electrons
Oxidizing power : F2 > Cl2 > Br2 > I2
Oxidizing power decreases down the group
Properties of the Halogens
F
Cl
Br
I
At
Colour
Green
Orange
Grey/black
State
Gas
Liquid
Solid
Yellow
Black Solid
Gas
Decreasin
g reactivity
Increasin
g m
olecu
lar size
Increasin
g d
ensity
Appearances of halogens
at room temperature and
pressure: chlorine ,bromine
iodine
chlorine
iodine
bromine
a )
Appearance of solutions of the elements in water and hydrocarbon solvents
Reaction of halogens with water
The soluability of halogens in water decreases down the groupa) 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
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.
HalogenColour
in pure form in waterin 1,1,1-trichloroethane/
cyclohexane
F2 Pale yellow Pale yellow Pale yellow
Cl2 Greenish yellow Pale yellow Yellow
Br2 Reddish brown Yellow Orange
I2 Violet black
Yellow (only
slightly soluble)
Brown in KI(aq)
Violet
Colours of halogens in pure form and in solutions
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)
Halogens are non-polar so they are more soluble in hydrocarbon
solvents than in water (see chapter 2.3), as shown in fig. 2.5.13. (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.
Water layer
Non-polar/hydrocarbon solvent example cyclohexane layer
Chlorine Bromine Iodine
Colours of halogens in 1,1,1-trichloroethane:(a) chlorine; (b) bromine; (c) iodine
(a) (b) (c)
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
Aqueous chlorine
oxidizes green iron(II) ions to yellowish brown iron(III) ions
Reactions with Iron(II) IonsReactions with Iron(II) Ions
2Fe2+(aq) + Cl2(aq) 2Fe3+(aq) + 2Cl–(aq)
Aqueous bromine oxidizes green iron(II) ions to yellowish brown iron(III)
Reactions with Iron(II) IonsReactions with Iron(II) Ions
Iodine
a mild oxidizing agent
not strong enough to oxidize iron(II) ions.
2Fe2+(aq) + Br2(aq) 2Fe3+(aq) + 2Br–(aq)
All halogens(except I2) oxidize Fe2+ to Fe3+
Half reactionStandard electrode
potential (V)
Cl2(aq) + 2e– 2Cl–(aq)
Br2(aq) + 2e – 2Br–(aq)
Fe3+(aq) + e– Fe2+(aq)
I2(aq) + 2e– 2I–(aq)
+1.36
+1.07
+0.77
+0.54
X2(aq) + 2Fe2+(aq) 2X(aq) + 2Fe3+(aq)
( X = F, Cl, Br)
I2(aq) + 2Fe2+(aq) No reaction
Reactions with reducing agents example Fe2+
Reactions of halogens with iron(II) ionsHalogens
Reactant
Chlorine Bromine Iodine
Iron(II) ions The green iron(II)
ions are oxidized
to yellowish
brown iron(III)
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
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.
F2 reacts explosively even in the dark at 200C
Cl2 reacts explosively in sunlight
Br2 reacts moderately on heating with a catalyst
I2 reacts slowly and reversibly even on heating
Reactions with hydrogen
X2 + H2(g) 2HX(g)
Explain the extreme reactivity of fluorine in terms of the bond enthalpies of F–F and H–F bonds.
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.
F2 + H2(g) 2HF(g)
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
-1
Physical properties of halogens and hydrogen halides
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.
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.
All halogens(except I2) oxidize S2O32 to SO4
2
4X2(aq)+S2O32(aq)+5H2O(l) 8X(aq)+10H+(aq) +2SO4
2(aq)
(X = F, Cl, Br)
I2(aq) + 2S2O32(aq) 2I(aq) + S4O6
2(aq)
S
O
OO-
S
SS
O
O
O-
S
O
OO-
O-+5
+5 +60
0
S
S
OO-
O-+4
0Used in iodometric titration
ii) disproportionation reactions with cold and hot alkali, eg hot potassium hydroxide with iodine to produce potassium iodate(V)
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
Reactions with AlkalisReactions with Alkalis
Fluorine is passed through a cold and very dilute (2%) sodium hydroxide solution
oxygen difluoride (OF2) is formed
2F2(g) + 2NaOH(aq) 2NaF(aq) + OF2(g) + H2O(l) 0 cold, very dilute –1 –1
When fluorine is passed through a hot and concentrated sodium hydroxide solution
oxygen is formed instead
2F2(g) + 4NaOH(aq) 4NaF(aq) + O2(g) + 2H2O(l)
0 –2 –1 0 hot, concentrated
(Fluorine)
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) 0 cold, dilute –1 +1
reacts with hot and concentrated sodium hydroxide solution to form sodium chloride and sodium chlorate(V)
3Cl2(aq) + 6NaOH(aq) 5NaCl(aq) + NaClO3(aq) + 3H2O(l)
0 hot, concentrated –1 +5
Chlorine
Disproportionation
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) 2NaBr(aq) + NaBrO3(aq)
The chemical equation for the overall reaction:
3Br2(aq) + 6NaOH(aq) 5NaBr(aq) + NaBrO3(aq) + 3H2O(l)
0 cold, dilute –1 +5
Bromine
behaves similarly as bromine
Except that the reaction with a cold and dilute alkali
reversible3I2(aq) + 6KOH(aq) 5KI(aq) + KIO3(aq) + 3H2O(l) 0 cold, dilute –1 +5
Iodine
What are the products, other than water, when chlorine is passed through cold, dilute aqueous sodium hydroxide solution?A NaCl and NaClOB NaClO and NaClO3
C NaCl and NaClO3
D NaClO and NaClO4
c) 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
Iodine/thiosulfate titrations
One common oxidising agent is iodine:
I2 (aq) + 2e- 2I- (aq)
The reducing agents you met included thiosulfate:
2S2O3 2- (aq) S4O6
2-(aq) + 2e-
Combining these two, the overall ionic equation for the reaction of iodine with thiosulfate is:
I2 (aq) + 2S2O3 2- S4O6
2-(aq) + 2I- (aq)
This reaction is the basis of the iodine/thiosulfate titration.
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.
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 20.00 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
= 0.002 mol of thiosulfate
From the equation for the reaction of iodine with thiosulfate above, you can see that 0.002 mol of thiosulfate will react with 0.001 mol of I 2 . So 0.001 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 0.001 mol of iodine is given by:
number of moles IO 3- = 0.001 mol / 3 = 0.000 333 mol
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 = 0.0713 g x 10 = 0.713 g
Percentage purity of potassium iodate(V)
fig. 2.5.14 Titrating to find the percentage of copper in brass.
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)
HALIDES
Boiling temperature of 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
Hydrogen fluoride has abnormally high Hydrogen fluoride has abnormally high
boiling temperature and melting boiling temperature and melting
temperature among the hydrogen halides temperature among the hydrogen halides
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)
Reactions with ConcentratedReactions with ConcentratedSulphuric(VI) AcidSulphuric(VI) Acid
i) potassium halides with concentrated sulfuric acid
Bromides and iodides
do not give hydrogen bromide and hydrogen iodide respectively
sulphur dioxide or hydrogen sulphide is formed
Action of concentrated sulphuric(VI) acid on
sodium bromide sodium iodide
Chlorides NaCl(s) + H2SO4(l) NaHSO4(s) + HCl(g)
BromidesNaBr(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) 2NaHSO4(s) + SO2(g) + Br2(g) + 2H2O(l)
IodidesNaI(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) 8NaHSO4(s) + H2S(g)
+ 4I2(g) + 4H2O(l)
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.
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.
Action of concentrated sulphuric(VI) acid on halides
Halide ion
Action of concentrated sulphuric(VI) acid
ProductConfirmatory 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)
Action of concentrated sulphuric(VI) acid on halides
Halide ion
Action of concentrated sulphuric(VI) acid
ProductConfirmatory 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)
Action of concentrated sulphuric(VI) acid on halidesHalide
ionAction of concentrated
sulphuric(VI) acidProduct
Confirmatory test ofthe 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) 8NaHSO4(s)+H2S(g)+4I2(g) +4H2O(l)
HF + H2SO4
HCl + H2SO4
2HBr + H2SO4 Br2 + SO2 + 2H2O
2HI + H2SO4 I2 + SO2 + 2H2O
6HI + H2SO4 3I2 + S + 4H2O
8HI + H2SO4 3I2 + H2S + 4H2O
HAt + H2SO4
HF + H2SO4
HF + H2SO4
?
-1
-1
-1
-1 -2
+4
+4
0
+6
+6
+6
+6 0
0
0
0
Power of reduction
The reaction of chlorides with concentrated sulphuric(VI) acidused for the preparation of hydrogen chloride in the laboratory
Hydrogen bromide and hydrogen iodide cannot be prepared in this way
Reactions with ConcentratedReactions with ConcentratedSulphuric(VI) AciSulphuric(VI) Acidd
Reactions with Phosphoric(V) AcidReactions 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
Redox or Displacement reactions
Cl2(aq) + 2Br(aq) 2Cl(aq) +Br2(aq)
Cl2(aq) + 2I(aq) 2Cl(aq) + I2(aq)
Br2(aq) + 2I(aq) 2Br(aq) + I2(aq)
I2(aq) + I(aq) I3(aq)
(yellow) (brown)
More reactive Less reactive
0 0-1
0
0
0
0
-1
-1
i) halides with halogen
Aqueoussolution
Halogen added
F2 Cl2 Br2 I2
F– No reaction No reactionNo
reaction
No
reaction
Cl–
A pale
yellow
solution is
formed (Cl2
is formed)
No reaction
No
reaction
No
reaction
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 reactionNo
reaction
I–
A yellowish
brown
solution is
formed
(I3 is formed)
A yellowish
brown
solution is
formed
(I3 is formed)
A yellowish
brown
solution is
formed (I3
is formed)
No
reaction
Reactions of halide ions with halogens
Reactions with Silver Ions Reactions with Silver Ions
• Aqueous solutions of chlorides, bromides and iodides
give precipitates when reactingwith acidified silver nitrate solution
i) Halides in silver nitrate solution
Reactions with Silver Ions 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
AgI(s)AgBr(s)AgCl(s)
Two confirmatory tests for halides
1. Adding NH3(aq) to the AgX ppt
2. Exposing AgX ppt to sunlight
ii) silver halides with sunlight and their solubilitiesin aqueous ammonia solution
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
• When exposed to sunlight
light2AgCl(s) 2Ag(s) + Cl2(g)
2AgBr(s) 2Ag(s) + Br2(l)light
2AgI(s) No reactionlight
silver bromide turns yellowish grey
silver iodide remains yellow
silver chloride turns grey
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
Action of acidified silver nitrate solution on halides
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
iii) hydrogen halides with ammonia
Reaction of hydrogen halides with waterAll 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 2HCl(aq)
Hydrogen chloride hydrochloric acid ( strong acid)
2HBr(g) + H2O 2HBr(aq)
Hydrogen bromide hydrobromic acid ( stronger than HBr)
2HI(g) + H2O 2HI(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.
iii) hydrogen halides with water to form acid
Uses of Chlorine
Chlorine is used mostly to kill bacteria or as a bleach. Chlorine bleaches a piece of universal indicator paper white.
Summary
Water layer
Non-polar/hydrocarbon solvent example cyclohexane layer
Chlorine Bromine Iodine
X2(aq) + 2Fe2+(aq) 2X(aq) + 2Fe3+(aq)
X = F, Cl, Br
X2(aq) + 2S2O32-(aq) 2X(aq) + S4O6
2(aq)
Oxidation of Fe2+ and S2O32-
F2 + P PF5
Cl2 + P PCl3 + PCl5
Br2 / I2 + P PBr3 / PI3
Oxidation of Posperous
Reaction with alkali (disproportination reaction)
Dilute/cold alkali Conc. Hot alkali
NaOH + Cl2 NaCl + NaOCl + H2O NaCl + NaClO3 + H2O
Oxidation number of halogen
0 -1 +1 -1 +5
All halogens reacts the way shown below
Decreasing oxidising ability down the group
Decreasing reactivity down the group
Halide reactions
NaCl + H2SO4 NaHSO 4 + HCl
NaBr + H2SO4 NaHSO 4 + HBr
HBr + H2SO4 Br2 + SO2 + H2O
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
2HI + H2SO4 I2 + SO2 + 2H2O
6HI + H2SO4 3I2 + S + 4H2O
8HI + H2SO4 3I2 + H2S + 4H2O
Ion
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)
Aqueoussolution
Halogen added
F2 Cl2 Br2 I2
F– No reaction No reactionNo
reaction
No
reaction
Cl–
A pale
yellow
solution is
formed (Cl2
is formed)
No reaction
No
reaction
No
reaction
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 reactionNo
reaction
I–
A yellowish
brown
solution is
formed
(I3 is formed)
A yellowish
brown
solution is
formed
(I3 is formed)
A yellowish
brown
solution is
formed (I3
is formed)
No
reaction
Reactions of halide ions/Potassium halides with halogens (Disproportination reaction)
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
iii) hydrogen halides with ammonia
Reaction of hydrogen halides with waterAll 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 2HCl(aq)
Hydrogen chloride hydrochloric acid ( strong acid)
2HBr(g) + H2O 2HBr(aq)
Hydrogen bromide hydrobromic acid ( stronger than HBr)
2HI(g) + H2O 2HI(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.
iii) hydrogen halides with water to form acid
The bonding in gaseous hydrogen halides is best described asA 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.
Fountain experiment for dissolution of hydrogen halides
Copper(ii) thiosulphate titrationsCopper(ii) thiosulphate titrationsGeneral theoryGeneral 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) 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) S4O6
2-(aq) + 2I¯(aq)
therefore moles of S2O32- = moles of Cu2+(aq)
Cu + e- Cu / 2I I2 + 2e-2+ + -
Practical details
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.
A
A
Practical details
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) + 4I¯(aq) 2CuI(s) + I2(aq) off white solid
A B
B
+2 -1 +1 -1 0
Practical details
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) S4O6
2-(aq) + 2I¯(aq)
A B C
C
0 -1
Practical details
Starch solution is added near the end point.
Starch gives a dark blue colouration in the presence of iodine.
A B C D
D
Practical details
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.
A B C D E
E
Titration number 1 (trial) 2 3 4
Burette reading(final) / cm
21.45 41.35 21.95
Burette reading(initial) / cm
1.20 21.45 21.95
Volume ofCu (aq) used / cm
20.00 19.80
Titre used tocalculate mean ( )
Mean titre- Average titre
19.9 cm
3
3
32+
To find the approximate end
point
Complete the table and indicate with a tick ( ) those titres most suitable for calculating a mean titre.
3
Titration number 1 (trial) 2 3 4
Burette reading(final) / cm
21.45 41.35 21.95 41.75
Burette reading(initial) / cm
1.20 21.45 1.95 21.95
Volume ofCu (aq) used / cm
20.25 19.90 20.00 19.80
Titre used tocalculate mean ( )
Mean titre- Average titre
19.9 cm
3
3
32+
To find the approximate end
point
Complete the table and indicate with a tick ( ) those titres most suitable for calculating a mean titre.
3
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
Typical calculationTypical calculation
Number of water molecules of crystallisation1 titrate a known mass of copper compound or a known volume of a solution2 calculate the moles of S2O3
2- needed3 according to the equation… moles of Cu2+ = moles of S2O3
2-
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 water9compare the ration of moles of… water : moles of CuSO4 to find x
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 0.100 mol dm sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
Calculation – Example 1Calculation – 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 0.100 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) S4O6
2-(aq) + 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
Calculation – Example 1Calculation – 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 0.100 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) S4O6
2-(aq) + 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3
Calculation – Example 1Calculation – 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 0.100 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) S4O6
2-(aq)+ 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = 25.00 x 0.100 / 1000 = 2.50 x 10-3
moles of Cu2+ in 25cm3 = moles of S2O32- = 2.50 x 10-3
Calculation – Example 1Calculation – 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 0.100 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) S4O6
2-(aq)+ 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-
moles of Cu2+ in 25cm3 = moles of S2O32- = 2.50 x 10-3
moles of Cu2+ in 250cm3 = 2.50 x 10-3 x 10= 0.025
mass of Cu2+ in 250cm3 = 0.025 x 63.5 = 1.588g
Calculation – Example 1Calculation – Example 1
-3
Calculation – Example 1Calculation – Example 13.00g 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 0.100 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) S4O6
2-(aq)+ 2I¯(aq)
you get moles of S2O32- = moles of Cu2+(aq)
moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3
moles of Cu2+ in 25cm3 = moles of S2O32- = 2.50 x 10-3
moles of Cu2+ in 250cm3 = 2.50 x 10-3 x 10= 0.025
mass of Cu2+ in 250cm3 = 0.025 x 63.5 = 1.588g
% of Cu in the alloy = 1.588 / 3.00 x 100 = 52.91%
-3
Iodine can react with sodium hydroxide solution to form NaIO3(aq), according to the equation below.
3I2(aq) + 6NaOH(aq) 5NaI(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.
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
+
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
2+- 3+
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.
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
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.
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