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Summary Notes 3.2.3 - Group 7(17), the halogens $6.46
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Summary Notes 3.2.3 - Group 7(17), the halogens

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Summary Notes 3.2.3 - Group 7(17), the halogens

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  • August 10, 2022
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Taylor’s Notes Group 7 Notes

Physical Properties
Halogen Appearance of Element at RTP Element in Aqueous Solution Melting Point Boiling Point
Fluorine Yellow Gas Pale Yellow Solution -220 -118
Chlorine Green Gas Green Solution -101 -35
Bromine Red-brown Liquid Orange Solution -7.2 59
Iodine Dark Grey Solid Brown Solution 114 184

Melting Point and Boiling Points
They all exist as X2 and as you go down the group the atomic radius increases as the molecular size increases and so there
is more surface area overlap so the dispersion forces are stronger

Chemical Properties of Halogens
Halogens tend to act as oxidising agents by gaining electrons

X2 + 2e-  2X-

Halogens tend to be oxidising agents as halogens will undergo reduction to obtain a full outer shell

The Oxidising Ability of halogens decreases down Group 7

Iodine is a more powerful oxidising agent than chlorine as iodine has a larger atomic radius and it has more shielding of
nuclear charge meaning that the electron added is less attracted to the nucleus

Ionic Equations
Equation F2 + 2Br-  2F- + Br2 Equation Cl2 + 2I-  2Cl- + I2
Oxidised 2Br- Oxidised 2I-
Reduced F2 Reduced Cl2
Oxidising Agent F2 Oxidising Agent Cl2
Reducing Agent 2Br- Reducing Agent 2I-


There is no reaction when bromine is added to chloride ions as bromine is a relatively poor oxidising agent and so cannot
oxidise the chloride ions

Chemical Properties of Halides
Halides tend to act as reducing agents by donating electrons

2X-  X2 + 2e-

Halides tend to be reducing agents as halides will undergo oxidation to obtain a full outer shell

The Reducing Ability of halides increases down Group 7

Chloride is a weaker reducing agent than iodide as chloride has a smaller ionic radius and there is less shielding of nuclear
charge meaning there is a stronger attraction, so chloride less readily loses electrons

Reaction of Sulphuric Acid with Chloride
When Sulphuric Acid is added to a Potassium Chloride, Hydrogen Chloride and Sulphuric Acid are formed

KCl + H2SO4  KHSO4 + HCl

The Observations for this Reaction are:
 White steamy fumes of Hydrogen Chloride

This reaction is not a redox reaction as both sulphur and chlorine are remaining in the +6 and -1 Oxidation States

The Chloride ion is not a strong enough reducing agent to have a further reaction

, Taylor’s Notes Group 7 Notes


Reaction of Sulphuric Acid with Bromide
When Sulphuric Acid is added to Potassium Bromide, Hydrogen Bromide and Sulphur Dioxide are formed

KBr + H2SO4  KHSO4 + HBr

Bromide ions are strong enough reducing agents to reduce the Sulphuric Acid to Sulphur Dioxide

The Half Equation for Bromide to Bromine = 2Br-  Br2 + 2e-

The Half Equation for Sulphuric Acid to Sulphur Dioxide = 2H+ + H2SO4 + 2e-  SO2 + 2H20

The Overall Equation for the Reaction of Bromide Ions with Sulphuric Acid = 2Br- + 2H+ + H2SO4 Br2 + SO2 + 2H2O

The Observation for this Reaction are:
 Red Brown Liquid
 Orange Fumes

Reaction of Sulphuric Acid with Iodide
When Sulphuric Acid is added to Potassium Iodide which forms Hydrogen Iodide and Sulphuric Acid

KI + H2SO4  KHSO4 + HI then 2HI + H2SO4  I2 + 2H2O + SO2

Iodide ions can also produce Sulphur and Hydrogen Sulphide

The Half Equation for Iodide to Iodine = 2I-  I2 + 2e-

The Half Equation for Sulphuric Acid to Sulphur = 6H+ + H2SO4 + 6e-  S + 4H2O

The Overall Equation for the Reaction of Iodine Ions with Sulphuric Acid which forms Sulphur:
6I- + 6H+ + H2SO4  3I2 + S + 4H2O

The Half Equation for the Sulphuric Acid to Hydrogen Sulphide = 8H+ + H2SO4 + 8e-  H2S + 4H2O

The Overall Equation for the Reaction of Iodide Ions with Sulphuric Acid which forms Hydrogen Sulphide (H2S):
8I- + 8H+ + H2SO4  4I2 + H2S + 4H2O

The Observations for these Reactions are:
 Yellow Solid (Sulphur)
 Rotten Egg smell from Colourless Gas (Hydrogen Sulphide)
 Dark Grey Solid and Purple Fumes (Iodine)

Identification of Halide Ions
Add Nitric Acid to remove Carbonate Impurities and then add Silver Nitrate which will cause a precipitate to form

The Effect of Ammonia Solution to the Precipitate can then be used to confirm the result

Halide HNO3 + AgNO3 Dilute NH3 Concentrated NH3
Cl- White Precipitate Precipitate Dissolves Precipitate Dissolves
Br- Cream Precipitate No Change Precipitate Dissolves
I- Yellow Precipitate No Change No Change

The Ionic Equation for the Test for Chloride Ions = Cl- + Ag+  AgCl

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