4.2 [Group 7] Flashcards
Why are fluorine and astatine ignored when considering the properties of Group 7?
Fluorine is ignored because it behaves differently from the other Group 7 elements and Astatine is ignored because it only exists in radioactive isotopes.
What is the important information about the Group 7 elements?
Melting and boiling temperatures increase down the group. Reactivity decreases down the group. Electronegativity decreases down the group.
What are the trends in melting and boiling temperatures in Group 7?
All the halogens exist as diatomic molecules, Melting and Boiling temperatures depend on the strength of the intermolecular forces of attraction between the molecules, also known as London forces.
As the 2 atoms in the diatomic molecule are identical the pair of electrons forming the covalent bond between them are shared equally between the 2 atoms. Means the halogen molecules are non-polar. However as the positive charges of the protons in the 2 nuclei are in fixed positions, but the electron density in a halogen molecule continuously fluctuates. Sometimes the centre of positive and negative charge don’t coincide the situation results in a temporary dipole or instantaneous.
Whyt does melting and boiling points increase down Group 7?
If 2 halogen molecules interact if the molecule on the left becomes an instantaneous dipole then it will cause an induced dipole in the molecule on the right. The results in forces of attraction between the 2 molecules.
Forces of attraction describe as instantaneous - induced dipole attraction. These are the intermolecular forces of attraction that exist between the molecules. These weak forces increase as the number of electrons and the size of the electron cloud increase. So the forces increase in strength down Group 7 as the number of electrons in the molecule increases. Explaining the increase in melting and boiling temperatures down Group 7.
How are equations for changes in state wrote?
Using equations for physical changes are used less often than chemical. Have to use the correct state symbols and the formulae.
For example:
When Bromine is left at room temperature it gives off a brown vapour boiling temperatures is not much higher than room temp.
Br2(l) > Br2(g)
When iodide is warmed most of it changes into a vapour without melting. This is called sublimation.
I2(s) > I2(g)
What are the trends in electronegativity?
Ability of an atom to attract the pair of electrons in a covalent bond.
Electronegativity depends on:
Nuclear charge - bigger the nuclear charge the higher the electronegativity
Distance between the nucleus and the bonding pairs of electrons - shorter the distance the higher the electronegativity
the shielding effect of electrons in higher energy levels - the fewer the energy levels, the higher the electronegativity
Electronegativity of the Group 7 elements is the highest in any group in PT. Electronegativity decreases fluorine is the highest.
What are the trends in reactivity in Group 7?
Fluorine is an extremely reactive element, and reactivity decreases down Group 7. Because of their high electronegativity most reactions of the halogens involve them acting as oxidising agents and gaining electrons (reduced) to form negative ions or become the slightly negative part of a polar molecule. The decreasing reactivity down the group can therefore be explained by the same factors that affect decreasing electronegativity down the group.
How do Group 7 elements react with metals in Group and 2?
Reactions are most vigerous between elements at the bottom of Group 1 and 2, and elements at the top of Group 7. So the most vigerous reactions are between caesium and fluorine, and the last vigerous between beryllium and iodine.
The products of these reactions are salts - ionic solids that are usually white.
All of these reactions involve electron transfer to the halogen, so they are redox reactions in which the halogen acts as the oxidising agent.
The oxidation number of the halogen decreases from 0 to -1 and the oxidation number of the metal increases from 0 to +1 or +2 (G1 and 2)
Lithium + Chlorine > Lithium Chloride
2Li + Cl2 > 2LiCL
How do Halogen /halide displacement reactions occur?
A more reactive halogen can displace a less reactive halogen from one of its compound.
Chlorine displaces bromine and iodine
Bromine displaces iodine but not chlorine
Iodine does not displace either chlorine or bromine.
These reactions occur in aqueous solution, so any reaction that occurs is indicated by a colour change. One problem in interpretating colour changes in these reactions is the similarly of some colour and the variation in colour with concentration.
Bromine in liquid state is red-brown but dissolved in water it might be orange or yellow depending on concentration.
Iodine dissolved in water may also appear brown at some concentrations.
When doing these reactions a good idea to add an organic solvent (cyclohexane) after the reaction, and then shake the tube. Halogens are more soluble in cyclohexane than in water. Halogen dissolves in the organic upper layer, where its colour change can more easily be seen.
Explain why chlorine displaces bromine?
Cl2 + 2NaBr > 2NaCl + Br2
Example of a redox reaction. The reacting halogen decreases its oxidation number from 0 to -1 and the reacting halide increases its oxidation from -1 to 0
The decreasing reactivity of chlorine, bromine and iodine in the reactions above can be explained using the same factors, Chlorine is the most reactive because:
It is the smallest atom, so the incoming electron gets closer to and is more attracted by the protons in the nucleus
It has the smallest number of complete inner energy levels of electrons, so the incoming electron experiences the levels of repulsion.
What is a disproportionation reaction?
Disproportionation is a type of reaction in which one element undergoes both oxidation and reduction at the same time.
How does chlorine react with water?
When chlorine is added to water it dissolves to form a solution that is called chlorine water. Some of the dissolved chorine also reacts to form a mixture of 2 acids.
2 acids that are formed are HCl and Chloric(I) acid. Its formula is HClO, but HOCl is also commonly used. Both acids are colourless solutions, so there is no visible change in this reaction.
CL2 + H20 > HCl + HClO
Chlorine is reduced from 0 to -1
Chlorine is oxidised from 0 to +1
How does chlorine react with a cold alkali?
When chlorine is added to cold dilute aqueous sodium hydroxide, it reacts to form the salts of the acid in the equation before. these salts are sodium chloride and sodium chlorate(I) also known as sodium hypochlorite.
Cl2 + 2NaOH > NaCl + NaClO + H20
Chlorine is reduced from 0 to -1
Chlorine is oxidised from 0 to +1
What are the trends for oxidising power and reducing power?
Oxidising power increases up Group 7 highest at Fluorine lowest at Astatine.
Reducing power Increases down Group 7, Lowest at fluoride highest at Asatide
How does Chlorine react with hot alkali?
When chlorine is added to hot concentrated sodium hydroxide solution is reacts to form sodium chloride and a different product, Sodium chlorate (V)
3Cl2 + 6NaOH > 5NaCl + NaClO3 + 3H20
Chlorine is reduced from 0 to -1
Chlorine is oxidised from 0 to +5
How can the reduction of a halide be represented by a half equation?
2X- > X2 + 2E-
What is the equation for partial ionisation of sulfuric acids?
Sulfuric acid is an acid but when it is concentrated it contains very few ions
HSO4- >< H+ + HSO4-
the reversible arrow the position of equilibrium lies well to the right. In the concentrated acid this first ionisation is far from complete.
HSO4- >< H+ + SO42-
Sulfuric acid especially when concentrated can act as an oxidising agent as well as an acid. When it acts as an oxidising agent it is reduced but the extent of its reduction and the products formed depend on the species being oxidised
3 possible reduction products are
Sulfur, sulfur dioxide , hydrogen sulfide.
How can the 3 different half equations be represented the oxidising action?
1)H2SO4 + 2H+ + 2e- > 2H20 + SO2
2)H2SO4 + 6H+ + 6e- > 4H2O + S
3) H2SO4 + 8H+ + 8e- > 4H20 + H2S
In reaction 1, the decrease in oxidation number (+6 to +4) is 2 which is the same as the number of H+ Ions and electrons in the equation
The pattern is similar to the 2 other equations the decrease in oxidation number of the is the same as the number of the H+ ions and electrons in the half equations
What are the observation and products when concentrate sulfuric acid is added to sodium chloride?
Observations: Misty fumes
Products: Hydrogen Chloride
What are the observation and products when concentrate sulfuric acid is added to sodium bromide?
Observations: Misty fumes, brown fumes, colourless gas with choking smell
Products: Hydrogen bromide, bromine and sulfur dioxide
What are the observation and products when concentrate sulfuric acid is added to sodium iodide?
Observations: Misty fumes, purple fumes or black solid, colourless gas with choking smell, yellow solid and colourless gas with rotten egg smell.
Products : hydrogen iodide, iodine, sulfur dioxide, sulfur and hydrogen sulfide.
What will the 3 different tube show that contain the sodium halides?
One tube will contain no colour. These are the misty fumes of hydrogen chloride formed from sodium chloride
One tube will contain brown fumes. This is bromine formed from sodium bromide.
The final tube will contain a black solid this is iodine formed from sodium iodide.
What oxidation and reduction will occur in these reactions?
With sodium chloride, the sulfuric acid behaves as only an acid, not as an oxidising agent. This is because chloride ions have low reducing powers
With sodium bromide, the greater reducing power of bromide ions causes the sulfuric acid to be reduced like the half equation 1
With sodium iodide, the much the greater the reducing power of the iodide ions causes the sulfuric acid to be reduced as in half equation 1,2 and 3
What is the half equation for sodium chloride and concentrated sulfuric acid?
Reactions between sodium chloride and sulfuric acid can be represented by one equation because no redox reactions are occurring.
NaCl + H2SO4 > NaHSO4 + HCl
What is the half equation for sodium bromide and concentrated sulfuric acid?
The formation of misty fumes of hydrogen bromide can be represented by
NaBr + H2SO4 > NaHSO4 + HBr
One redox reaction occurs formation of sulfur dioxide 2 relevant half equations.
2Br- > Br2 + 2e-
H2SO4 + 2H+ + 2e- > 2H20 + SO2 2e- cancels out
2Br- +H2s04 +2H+ > 2H20 + SO2 + Br2
What is the half equation for sodium iodide and concentrate sulfuric acid?
Formation of misty fumes of hydrogen iodide can be represented by analogous for sodium chloride.
NAI + H2SO4 > NaHSO4 + HI
table of observations show 3 redox reactions occur so the situation is complicated. Construct an equation showing the formation of sulfur dioxide same way as sodium bromide.
2I- > I2 + 2e- and H2SO4 + 6H+ + 6e- > 4H20 + S
Add them together by multiplying by 3 so 6e- cancel out giving
6I- + H2S04 + 6H+ > 4H20 + S + 3I2
So the equation represented the oxidation of misty fumes of hydrogen iodide
Use the same method to construct an equation to represent the oxidation of the misty fumes of hydrogen iodide to form hydrogen sulfide.
How can halides be tested for in solution?
These tests depends on the very low solubility of silver halides in water and their different solubility’s in aqueous ammonia.
The reagent is silver nitrate solution, but dilute nitric acid is added beforehand to make sure that any anions (especially carbonate ions) are removed as they would also form precipitate.
if a precipitate is obtained, it is then usual to add some ammonia solution this can be dilute or concentrated.
What are the results for adding silver nitrate solution to halide ions in solution?
Chloride ions: White precipitate
Bromide ions: Cream precipitate
Iodide ions: Yellow precipitate
What are the results for adding dilute aqueous ammonia to halide ions in solution?
Chloride ions are soluble
Bromide ions are insoluble
Iodide ions are insoluble
What are the results for adding concentrated aqueous ammonia to halide ions in solution?
Chloride ions are soluble
Bromide ions are soluble
Iodide ions are insoluble
Why can’t the silver nitrate test be done for fluoride ions in aqueous solution?
Silver fluoride is soluble in silver nitrate
What is the general ionic formula for the formation of the silver nitrate precipitate?
Ag+ + x- > AgX
AgNO3 + NaCl > AgCl + NaNO3
What is the advantages with with the dilute aqueous ammonia test?
All 3 colours of the precipitate in silver nitrates are similar for the halide ions its not easy to tell the difference so the better idea is to use the aqueous ammonia test because the precipitate all have different solubilities.
Silver chloride dissolves readily in both dilute and concentrate aqueous ammonia
Silver bromide dissolves readily in concentrate aqueous ammonia but not in dilute ammonia.
Silver iodide dissolves in neither
Dissolving of the precipitate occurs because of the formation of a complex ion
How do hydrogen halides act as acids?
All of the hydrogen halides are colourless gases and exist as polar diatomic molecule
They readily react with water to form acidic solution all of which are colourless.
What happens when Hydrogen fluoride chloride bromide and iodide react with acids?
Hydrogen fluoride > hydrochloric acid
Hydrogen chloride > hydrochloric acid
Hydrogen bromide > hydrobromic acid
Hydrogen iodide > hydiodic acid
