Application of Core principles Of chemistry Edexcel new Specification Halogens.

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


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)


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)


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


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



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)




From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+I2(aq)

2S2O32-(aq) + I2(aq) S4O6

2-(aq) + 2I¯(aq)


moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3




From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+ I2(aq)

2S2O32-(aq) + I2(aq) S4O6

2-(aq)+ 2I¯(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



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)


moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-


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


-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)


moles of S2O32- = 0.100 x 25.00 / 1000 = 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.

Application of Core principles Of chemistry Edexcel new Specification Halogens.

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