number of electrons in molecules increases down group = stronger Van der Waals forces
this increases chances of dipoles arising within a molecule which would induce dipoles in neighbouring molecules = strengthens intermolecular bond
Chlorine has a low boiling point becuase the forces between the molecules are weak.
Explain how these forces arise between molecules of chlorine. (3)
random movement of electrons in one molecule creates a dipole
induces a dipole in a neighbouring molecule
these temporary dipoles attract
Trend in bond enthalpies in group 7
decrease going down the group
due to the increasingatomic radius, increased shielding and reduced attraction between the nucleus and bonding pair of electrons
exception to this is fluorine.
it has such a smallatomic radius that bonding and non-bonding electrons repel each other, reducing the bondenthalpy.
Trend in oxidising ability of halogens
oxidising ability decreases down group 7
atomic radius increases down group = weaker force of attraction between nucleus and outer shell of electrons
increasedshielding effect down group
means it’s harder to gain an electron as oxidising agents are electron acceptors
Fluorine is the strongestoxidising agent of group 7 because it has least number of electrons, so very small atomicradius so the attraction between nucleus and outer electrons is strong. This and the least shielding effect means it’s easier to accept an electron into its outermost shell.
Trend in electronegativity in group 7
decreases going down the group
as the atomic radius increases, the shared pair of electrons in the covalent bond are further from the nucleus, reducing the strength of the attraction.
increasing atomic radius also means increasing shielding effect, again reducing the electronegativity.
Trend in reducing ability of halide ions
Going down group 7, the halide ions become strongerreducing agents.
This is because the ionic radiusincreases so the outer electrons are lessstrongly attracted.
Further down the group electrons are more easilylost, making them better reducing agents.
Chloride ions with sulphuric acid
Solid sodium chloride reacts with concentrated sulphuric acid. This reaction is performed in a fume cupboard due to the HCl fumes produced. Sodium hydrogensulfate is also formed.
NaCl (s) + H2SO4 (l)→ NaHSO4 (s) + HCl (g)
This is not a redox reaction as no oxidation states change in the process.
The chloride ion is too weak a reducing agent to change the oxidation state of sulphur.
Give an equation for the reaction of solid sodium bromide with concentrated sulfuric acid to form bromine.
State one observation made during this reaction.
2H2SO4 + 2NaBr → Na2SO4 + SO2 + Br2 + 2H2O
orange/brown fumes/solution
Solid sodium iodide reacts with concentrated sulfuric acid to form iodide and sulfur in a redox reaction
Give a half equation to show the conversion of iodide ions to iodine.
Give a half-equation to show the conversion of sulfuric acid to sulfur.
Give an overall equation for this redox reaction.
Identify one other sulfur-containing reduction product formed when solid sodium iodide reacts with concentrated sulfuric acid.
2I- → I2 + 2e-
H2SO4 + 6H+ + 6e- → S + 4H2O
6H+ + 6I- + H2SO4 → 3I2 + S + 4H2O
SO2 or H2S
State one observation when solid sodium chloride reacts with concentrated sulfuric acid.
Give an equation for the reaction.
State the role of chloride ions in the reaction.
Misty or steamy or white fumes/gas
NaCl + H2SO4 → NaHSO4 + HCI
Base OR proton acceptor
Give an equation for the redox reaction between the solid sodium bromide and concentrated sulfuric acid.
Explain, using oxidation states, why this is a redox reaction.
2NaBr + 2H2SO4 → Na2SO4 + Br2 + SO2 + 2H2O
Br changes oxidation state from -1 to 0 and is oxidised
S changes oxidation state from +6 to +4 and is reduced
State what is observed when aqueous chlorine is added to sodium bromide solution.
Give an ionic equation for the reaction.
Yellow or orange solution
Cl2 + 2Br- → 2Cl- + Br2
Solid sodium chloride reacts with concentrated sulfuric acid.
Give an equation for this role.
State the role of sulfuric acid in this reaction.
NaCl + H2SO4 → NaHSO4 + HCl
Protondonor
Fumes of sulfur dioxide are formed when sodium bromide reacts with concentrated sulfuric acid.
For this reaction
give an equation
give one other observation
state the role of the sulfuric acid.
2NaBr + 2H2SO4 → Na2SO4 + SO2 + Br2 + 2H2O
brown gas/fumes or orange gas/fumes
Oxidising agent
Reaction of iodide ions with sulphuric acid
NaI (s) + H2SO4 (l)→ NaHSO4 (s) + HI(g)
2H+ + 2I- + H2SO4 (l)→ I2(g) + SO2(g) + 2H2O (l)
6H+ + 6I- + H2SO4 (l)→ 3I2(g) + S(s) + 4H2O (l)
Overall equation for steps 2 and 3:
8H+ + 8I- + H2SO4 (l)→ 4I2(g) + H2S(s) + 4H2O (l)
Fumes of HI are white in colour.
Iodine fumes are purple in colour.
Hydrogen sulphide has a rotten egg smell.
Chlorine undergoes a redox reaction called as disproportionation, where chlorine is both oxidised and reduced.
Reaction of chlorine with water for watertreatment
Chlorine is used in water treatment to kill bacteria and make water safe for drinking.
Chlorine reacts with water to form hydrochloric acid and chloric (I) acid.
Cl2 (aq) + H2O (l) → HCl (aq) + HClO (aq)
Chloric (I) acid decomposes slowly in water to form reactiveoxygen molecule that kills the bacteria in water.
HClO → HCl +[O]
Chlorine may react with organic compounds in water to form products that could be hazardous to health. It is important to use sodiumhydroxide to remove chlorine in water before using it in aquariums as chlorine could be harmful for aquatic organisms.
An alternative to direct chlorination is to add solid sodium or calcium chlorate (I) which will dissolve in water to form the chlorate ions.
In order to shift the equilibrium to the right and maintain the concentration of HClO, the water must be slightly acidic, but not so much as to endanger people.
Under alkaline conditions, the equilibrium will shift to the left.
Reaction of chlorine with sodium hydroxide
Chlorine reacts with cold sodium hydroxide to form sodium chloride and sodium chlorate (I) in a redox reaction