Chemical Tests: Halide Ions
F-, Cl-, Br- , I- (oxidation state= -1, due to their charge)
Reagent: Acidified Silver Nitrate solution
-WHY?
· to remove any impurities, E.g carbonate ions (CO32-) as CO2
· nitric acid (not HCl-contains CL- ions)
Colour of precipitate Observation with dilute NH3 Observation with concentrated NH3
K chloride White Precipitate dissolves (clear) -
K bromide Cream Precipitate remains (no change) Precipitate dissolves (clear)
K iodide Yellow Precipitate remains (no change) Precipitate remains (no change)
F ions form a SOLUBLE salt-no precipitate
AgNO3 (aq)+ KCl (aq)-> AgCl(s) + KNO3 (aq)
Ionic eq:
Ag+(aq) + Cl-(aq) -> AgCl (s) (precipitate)
From F2 -> I2, oxidising P decreases as:
· Atoms get bigger so e- is further away from + nucleus
· Shielding increases. So extra e- shells
Halide ions are – charged so will be oxidised (lose e-) OILRIG
Reducing Agents (have e- to give) -lose e-, become oxidised
From F2 à I2, reducing P increases as:
· Ions get bigger, so e- to be lost is further away from the + nucleus
· The e- to be lost is more shielded from nucleus
Redox explains displacements:
e.g Cl2 + 2Br- -> 2Cl- + Br2
ox.state: 0 -1 -> -1 0
0 -> -1 reduction (Cl2 is ox. Agent)
-1 -> oxidation (Br- id Red. Agent)
F-, Cl-, Br- , I- (oxidation state= -1, due to their charge)
Reagent: Acidified Silver Nitrate solution
-WHY?
· to remove any impurities, E.g carbonate ions (CO32-) as CO2
· nitric acid (not HCl-contains CL- ions)
Colour of precipitate Observation with dilute NH3 Observation with concentrated NH3
K chloride White Precipitate dissolves (clear) -
K bromide Cream Precipitate remains (no change) Precipitate dissolves (clear)
K iodide Yellow Precipitate remains (no change) Precipitate remains (no change)
F ions form a SOLUBLE salt-no precipitate
AgNO3 (aq)+ KCl (aq)-> AgCl(s) + KNO3 (aq)
Ionic eq:
Ag+(aq) + Cl-(aq) -> AgCl (s) (precipitate)
From F2 -> I2, oxidising P decreases as:
· Atoms get bigger so e- is further away from + nucleus
· Shielding increases. So extra e- shells
Halide ions are – charged so will be oxidised (lose e-) OILRIG
Reducing Agents (have e- to give) -lose e-, become oxidised
From F2 à I2, reducing P increases as:
· Ions get bigger, so e- to be lost is further away from the + nucleus
· The e- to be lost is more shielded from nucleus
Redox explains displacements:
e.g Cl2 + 2Br- -> 2Cl- + Br2
ox.state: 0 -1 -> -1 0
0 -> -1 reduction (Cl2 is ox. Agent)
-1 -> oxidation (Br- id Red. Agent)