Halogens

Fluorine (F2): very pale yellow gas. It is highly reactive

Chlorine : (Cl2) greenish, reactive gas, poisonous in high concentrations
Bromine (Br2) : red liquid, that gives off dense brown/orange poisonous fumes
Iodine (I2) : shiny grey solid sublimes to purple gas.

Element Melting Point Boiling Point Physical State

Trends in melting point and boiling point (oC) (oC)
Increase down the group
Fluorine -220 -188 Gas
As the molecules become larger they have more
electrons and so have larger London forces Chlorine -101 -35 Gas
between the molecules. As the intermolecular
forces get larger, more energy has to be put into Bromine -7 59 Liquid
overcoming the forces. This increases the melting
and boiling points. Iodine +114 184 Solid

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

Reactions of halogens
The oxidation reactions of halide ions by halogens

A halogen that is a strong oxidising agent will The oxidising strength decreases down the group.
displace a halogen that has a lower oxidising Oxidising agents are electron acceptors.
power from one of its compounds.

Chlorine will displace both bromide and iodide ions. Bromine will displace iodide ions.

Chlorine (aq) Bromine (aq) Iodine (aq)

Potassium Very pale green Yellow solution, no Brown solution,

chloride (aq) solution, no reaction no reaction
reaction

Potassium Yellow solution, Cl Yellow solution, no Brown solution,

bromide (aq) has displaced Br reaction no reaction

Potassium Brown solution, Cl Brown Solution, Br Brown Solution,

iodide (aq) has displaced I has displaced I no reaction

The colour of the solution in the test tube shows which free halogen
is present in solution.
Chlorine =very pale green solution (often colourless),
Bromine = yellow solution
Iodine = brown solution (sometimes black solid present)

All the halide salt solutions (KI, NaBr, KCl etc.) are colourless I2 (aq)
Br2 (aq)

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 Observations if an organic solvent is added
Chlorine (aq) Bromine (aq) Iodine (aq)
The colour of the organic
Potassium colourless, no yellow, no purple, no solvent layer in the test tube
chloride (aq) reaction reaction reaction shows which free halogen is
present in solution.
Potassium yellow, Cl has yellow, no purple, no Chlorine = colourless
bromide (aq) displaced Br reaction reaction Bromine = yellow
Iodine = purple
Potassium purple, Cl has purple, Br has purple, no
iodide (aq) displaced I displaced I reaction

Cl2 (aq)+ 2KBr (aq) Br2 (aq)+ 2KCl (aq)

Cl2 (aq)+ 2KBr (aq) Br2 (aq)+ 2KCl (aq)
Br2 (aq)+ 2KI (aq) I2 (aq)+ 2KBr (aq)

Cl2(aq) + 2Br – (aq)  2Cl – (aq) + Br2(aq)

Cl2(aq) + 2I – (aq)  2Cl – (aq) + I2(aq)
Br2(aq) + 2I – (aq)  2Br – (aq) + I2(aq)

Reduction half equation (gaining electrons) Oxidation half equation (losing electrons)
Cl2 (aq) + 2e-  2Cl- (aq) 2Br- (aq) Br2 (aq)+ 2e-

The oxidation reactions of metals and metal ion by halogens.

In all reactions where halogens are
2Na  2Na+ + 2e- reacting with metals, the metals are
Br2(l) + 2Na (s)  2NaBr (s) Br2 + 2e-  2Br - being oxidised

3Cl2(g) + 2 Fe (s)  2 FeCl3 (s) Br2(l) + Mg (s)  MgBr2 (s)

Cl2(g) + 2Fe2+ (aq)  2 Cl- (aq) + 2Fe3+ (aq) Chlorine and bromine can oxidise Fe2+ to Fe3+.
Iodine is not strong enough an oxidising agent to do this
2I- (aq) + 2Fe3+ (aq)  I2 (aq) + 2Fe2+ (aq) reaction. The reaction is reversed for iodine

The disproportionation reactions of chlorine.

Disproportionation is the name for a reaction where an element simultaneously oxidises and reduces.

Chlorine with water: Chlorine is both simultaneously reducing and oxidising

Cl2(aq) + H2O(l) ⇌ HClO(aq) + HCl (aq) changing its oxidation number from 0 to -1 and 0 to +1

If some universal indicator is added to the solution it will first The pale greenish colour of these solutions is
turn red due to the acidity of both reaction products. It will then due to the Cl2
turn colourless as the HClO bleaches the colour.

Chlorine is used in water treatment to kill bacteria. Chlorine can be used to treat drinking water and the water in
swimming pools. The benefits to health of water treatment by chlorine outweigh its toxic effects.

Reaction with water in sunlight

If the chlorine is bubbled through water in the presence of bright sunlight a different reaction occurs.
2Cl2 + 2H2O  4H+ + 4Cl- + O2
The same reaction occurs to the equilibrium mixture of chlorine water. The greenish colour of chlorine water fades
as the Cl2 reacts and a colourless gas (O2) is produced.
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 Reaction of halogens with cold dilute NaOH solution:
Cl2, Br2, and I2 in aqueous solutions will react with cold sodium hydroxide. The colour of the halogen solution
will fade to colourless

Cl2(aq) + 2NaOH(aq)  NaCl (aq) + NaClO (aq) + H2O(l)

The mixture of NaCl and NaClO is used as bleach and to disinfect and kill bacteria.

Reaction of halogens with hot dilute NaOH solution:

With hot alkali disproportionation also occurs but the halogen that is oxidised goes to a higher
oxidation state.

3Cl2 (aq) + 6 NaOH(aq)  5 NaCl (aq) + NaClO3 (aq) + 3H2O (l)

3I2 (aq) + 6KOH (aq)  5 KI (aq) + KIO3 (aq) + 3H2O (l)

3I2 (aq) + 6OH- (aq)  5 I- (aq) + IO3- (aq) + 3H2O (l)

In IUPAC convention the various forms of sulfur and chlorine compounds where oxygen is
combined are all called sulfates and chlorates with relevant oxidation number given in roman
numerals. If asked to name these compounds remember to add the oxidation number.

NaClO: sodium chlorate(I)

NaClO3: sodium chlorate(V)
K2SO4 potassium sulfate(VI)
K2SO3 potassium sulfate(IV)

Thiosulfate redox titration A starch indicator is added near the end

The redox titration between I2 and thiosulfate S2O32- is a point when the iodine fades a pale yellow
common exercise. to emphasise it.
With starch added the colour change is
2S2O32-(aq) + I2 (aq)  2I- (aq) + S4O62-(aq) from blue/black to colourless
yellow/brown sol colourless sol

The starch should not be added until nearly all the iodine has reacted because the blue
complex formed with high concentrations of iodine is insoluble and does not re-dissolve as
more thiosulfate is added.

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The reaction of halide salts with concentrated sulfuric acid.

The halides show increasing power as Explanation of differing reducing power of halides
reducing agents as one goes down the A reducing agent donates electrons.
group. This can be clearly demonstrated in The reducing power of the halides increases down group 7
the various reactions of the solid halides with They have a greater tendency to donate electrons.
concentrated sulfuric acid. This is because as the ions get bigger it is easier for the
outer electrons to be given away as the pull from the nucleus
on them becomes smaller.

Fluoride and Chloride

The H2SO4 is not strong enough an oxidising reagent to oxidise the chloride and
fluoride ions. No redox reactions occur. Only acid-base reactions occur.

NaF(s) + H2SO4(l) NaHSO4(s) + HF(g)

Observations: White steamy fumes of HF are evolved.
NaCl(s) + H2SO4(l)  NaHSO4(s) + HCl(g)
Observations: White steamy fumes of HCl are evolved.

These are acid–base reactions and not redox

reactions.
H2SO4 plays the role of an acid (proton donor).

Bromide
Bromide ions are stronger reducing agents than chloride and fluoride ions. After the initial acid-
base reaction the bromide ions reduce the sulfur in H2SO4 from +6 to + 4 in SO2

Acid- base step: NaBr(s) + H2SO4(l)  NaHSO4(s) + HBr(g)

Redox step: 2HBr + H2SO4  Br2(g) + SO2(g) + 2H2O(l)

Overall equation: combining two steps above:

2NaBr + 3H2SO4  2NaHSO4 + SO2 + Br2 + 2H2O Reduction product = sulfur dioxide

Ox ½ equation 2Br -  Br2 + 2e-

Re ½ equation H2SO4 + 2 H+ + 2 e-  SO2 + 2 H2O

Note the sulfuric acid acts as an acid in the first step producing HBr and
then acts as an oxidising agent in the second redox step.

Observations: White steamy fumes of HBr are evolved.

Orange fumes of bromine are also evolved and a
colourless, acidic gas SO2

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Iodide
Iodide ions are the strongest halide reducing agents. They can reduce the sulfur
from +6 in H2SO4 to + 4 in SO2, to 0 in S and -2 in H2S.

NaI(s) + H2SO4(l)  NaHSO4(s) + HI(g)

2HI + H2SO4  I2(s) + SO2(g) + 2H2O(l)
6HI + H2SO4  3 I2 + S (s) + 4 H2O (l)
8HI + H2SO4 4I2(s) + H2S(g) + 4H2O(l)

Ox ½ equation 2I -  I2 + 2e-
Re ½ equation H2SO4 + 2 H+ + 2 e-  SO2 + 2 H2O
Re ½ equation H2SO4 + 6 H+ + 6 e-  S + 4 H2O
Reduction products = sulfur dioxide, sulfur
Re ½ equation H2SO4 + 8 H+ + 8 e-  H2S + 4 H2O and hydrogen sulfide

Note the H2SO4 acts as an acid in the first step producing HI and then
acts as an oxidising agent in the three redox steps.

Observations:
White steamy fumes of HI are evolved.
Black solid and purple fumes of Iodine are
also evolved
A colourless, acidic gas SO2
A yellow solid of sulfur
H2S (hydrogen sulfide), a gas with a bad egg
smell,

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The reactions of halide ions with silver nitrate.
This reaction is used as a test to identify which halide ion The role of nitric acid is to react with any carbonate
is present. The test solution is made acidic with nitric impurities present to prevent formation of the
acid, and then silver nitrate solution is added dropwise. precipitate Ag2CO3. This would mask the desired
observations.
2 HNO3 + Na2CO3  2 NaNO3 + H2O + CO2
Fluorides produce no precipitate

Chlorides produce a white precipitate

Ag+(aq) + Cl- (aq)  AgCl(s)

Bromides produce a cream precipitate

Ag+(aq) + Br- (aq)  AgBr(s)

Iodides produce a pale yellow precipitate

Ag+(aq) + I- (aq)  AgI(s)

Effect of light on silver halides

The precipitates ( except AgI) darken in sunlight forming silver. This reaction is used in photography to form
the dark bits on photographic film.

Effect of ammonia on silver halides

The silver halide precipitates can be treated with ammonia solution to help differentiate
between them if the colours look similar:

AgCl AgBr AgI

Addition of Dissolves Does not Does not
aqueous dissolve dissolve
ammonia

Addition of Dissolves Dissolves Does not

concentrated dissolve
ammonia

The solubility of the silver halides in ammonia decreases in the order: AgF > AgCl > AgBr > AgI

Silver chloride dissolves in dilute ammonia to form a complex ion

AgCl(s) + 2NH3(aq) [Ag(NH3)2]+ (aq) + Cl- (aq)
Colourless solution

Silver bromide dissolves in concentrated ammonia to form a complex ion

AgBr(s) + 2NH3(aq) [Ag(NH3)2]+ (aq) + Br - (aq)
Colourless solution

Silver iodide does not react with ammonia – it is too insoluble.

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Hydrogen Halides

The reactions of the elements with hydrogen

F2(g) + H2 (g)  2 HF (g) All the halogens react with hydrogen gas to produce hydrogen halides.
Br2(l) + H2 (g)  2 HBr (g) The reactions decrease in speed and vigour going down the group.
I2(s) + H2 (g) ⇌ 2 HI (g)

Thermal stability of halides: Hydrogen Iodide will decompose if a hot nichrome wire is plunged into it. Purple
vapour of iodine will be seen.
2 HI (g)  H2 (g) + I2 (g)
A very hot wire would also decompose Hydrogen bromide
The general trend is the hydrides become less stable going down the group. This can be explained by the
decreasing size of the H-Hal bond energy going down the group. This is because as the halogen atoms become
bigger their bond length is longer. The bonding pair of electrons gets further from the halogen nuclei.

Producing hydrogen halides

Hydrogen halides are made in the laboratory by the
reaction of solid sodium halide salts with phosphoric acid
This is the apparatus used
to make the hydrogen
NaCl(s) + H3PO4(l)  NaH2PO4(s) + HCl(g) halide using phosphoric
acid.
Observations: White steamy fumes of the hydrogen halides
Notice the downward
are evolved.
delivery which is used
The steamy fumes of HCl are produced when the HCl meets because the hydrogen
the air because it dissolves in the moisture in the air halides are more dense
than air

Phosphoric acid is not an oxidising agent and so does not oxidise HBr and HI. Phosphoric acid is more suitable for
producing hydrogen halides than the ones with concentrated sulfuric acid to make HCl, HBr, and HI because there are
no extra redox reactions taking place and no other products.

Hydrogen halide

Solubility in water : The water quickly

The hydrogen halides are all rises up the tube
soluble in water. They dissolve
to form acidic solutions.

All the hydrogen halides react readily with

ammonia to give the white smoke of the
ammonium halide. This can be used as a test for the
presence of hydrogen halides.
HCl(g) + NH3 (g)  NH4Cl (s)
HBr(g) + NH3 (g)  NH4Br (s)
HI(g) + NH3 (g)  NH4I (s)

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