4. Inorganic Chemistry (GROUP 7 HALOGENS)
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Give the ionic equation between silver nitrate and halide ions during the precipitate reaction which tests for halides
Ag+ (aq) + X- (aq) either Cl,Br,I —> AgX (s) either AgCl,AgBr,AgI
Describe how to test for halide ions and the observations you would see
Add dilute nitric acid to solution - prevents formation of silver precipitate by removing any interfering anions that might be present eg: carbonate ions, CO3 2- ions Add silver nitrate Would form a precipitate: Chlorine ions - white precipitate Bromide ions - cream precipitate Iodide ions - yellow precipitate –> Afterwards, can add ammonia which will dissolve precipitate to better understand which halogen is present
Describe what you would see when you add silver halide precipitates with ammonia
Chlorine - dissolves in concentrated AND dilute ammonia Bromine - ONLY dissolves in CONCENTRATED ammonia Iodine - doesn’t dissolve in NEITHER concentrated or dilute ammonia
Give the trend for electronegativity down group 7
Electronegativity down group 7 decreases as the atomic radius increases due to higher number of shells and more shielding so the bonding electrons in the covalent bond are further from the nucleus so there is less attraction so the bonding electrons are pulled less to the nucleus
Give the trend for reactivity down group 7
Reactivity down group 7 decreases as the atom gets larger with more shells so the outer shell electron is more shielded from the nucleus resulting in decreased electrostatic forces of attraction so it is harder to attract an electron
Describe the trend for melting and boiling point for group 7
Increase down the group As the molecules become larger they have more electrons and so have larger London forces so more energy required to break the forces. This increases the melting and boiling points
Given that melting point increases down group 7, give the appearance of fluorine, chlorine, bromine and iodine at room temperature
Fluorine - pale yellow gas Chlorine - pale green gas Bromine - orange liquid Iodine - gray solid (easily warmed to give purple vapours)
Give the colour of the following halogens in water (after they have been displaced by a more reactive halogen): Chlorine Bromine Iodine
Chlorine - pale green often colourless solution Bromine - yellow solution Iodine - brown solution
Give the colour changes when an organic solvent like cyclohexane has been added to aqueous halogens (this is done after displacement reactions)
chlorine: pale green to pale yellow bromine: yellow to orange iodine: brown to purple
Give the colour of bromide ions
Colourless
Give the equation for the reaction between chlorine and cold and hot sodium hydroxide -NaOH
The colour of solution fades to colourless forming e.g. NaClO and NaCl and H20: Cl2(aq) + 2NaOH(aq) –> NaCl (aq) + NaClO (aq) + H2O(l) • hot is similar however halogen reaches a higher oxidation state: 3Cl2 +6NaOH→5NaCl+NaClO3 +3H2O
Give the disproportionation reaction between chlorine and water
Cl2 + H2O –> HClO + HCl
Give the uses of HClO, NaClO and HClO3
HClO - kills bacteria in water to sterilise water NaClO - used in bleach NaClO3 - used in herbicides
Describe the trend of the reducing power of halides
Increases down the group They have a greater tendency to donate electrons. 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 is smaller. This means that the elements at the top of group 7 are strong oxidising agents, not reducing agents
What would be produced and what would you see with the reaction of: Fluorine and sulfuric acid Chlorine and sulfuric acid
Steamy white fumes of hydrogen halide F- and Cl- ions are not strong enough reducing agents to reduce the S in H2SO4 . No redox reactions occur. Only acid-base reactions occur. Equation: NaCl + H2SO4 → NaHSO4 + HCl
What would be produced and what would you see with the reaction of: Bromine and sulfuric acid
Br- ions are stronger reducing agents than Cl- and F- and reduce the S in H2SO4 from +6 to +4 in SO2 Equations: NaBr + H2SO4 → NaHSO4 + HBr 2HBr + H2SO4 → Br2 + SO2 + 2H2O Produces white steamy fumes of HBr Reddish brown fumes of bromine are also evolved and SO2 (sulphur dioxide) a colourless, acidic gas
What would be produced and what would you see with the reaction of: Iodine and sulfuric acid
I- 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. Equations: NaI + H2SO4 → NaHSO4 + HI 2HI + H2SO4 → I2 + SO2 + 2H2O 6HI + H2SO4 —> 3I2 + S + 4H2O 8HI + H2SO4 —> 4I2+ H2S + 4H2O White steamy fumes of HI are evolved. Purple vapours 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
A portion of the organic liquid P is added to the red‑brown aqueous solution formed after bromine displaces iodine from potassium iodide. The mixture is shaken and then allowed to stand for several minutes. State what observations would be expected, including any changes in colour.
When the organic liquid P is added and the mixture is shaken, the iodine distributes itself between the two layers. Iodine is a non-polar molecule, so it is more soluble in the non-polar organic solvent than in polar water. Therefore: Iodine moves from the aqueous layer into the organic layer The organic layer changes from colourless to purple/violet because dissolved iodine is purple in organic solvents The aqueous layer becomes colourless because it now contains less iodine After standing, two separate layers are seen because the organic liquid and water are immiscible.