Use Silver nitrate.
A yellow precipitate of silver iodide will form /
AgNO3 + I^- = AgI(s) + NO3^-
This is one of the classic tests for halogens.
lenpollock
Due to presence of chloride ion
as ions they are named Iodide.... charge is -1
Iodine gains one electron to form iodide ion. The size of iodide ion is larger than the size of iodine atom.
no its not, the iodide ion is
BaI2 is formed from Ba2+ and 2 I-
It is not the anions (e.g. iodide) that are responsible for the flame test color, rather the cations such as sodium ion, potassium ion and calcium ion give you different colors.
Due to presence of chloride ion
Iodide ion : I-
It's a qualaitative test to test for the presence of the borate anion in the solution
Roughly, the iodide ion appears as shown below: .. - :I: ..
Testing for Iodide, I-Sodium iodide, NaI, is the source of iodide anion for this experiment.Reaction with bleach involves three steps. The brown color shows the presence of I3- ions.Hypochlorite ion yields chlorine:OCl- (aq) + Cl- (aq) + H2OCl2 (aq) + 2 OH-Chlorine reacts with iodide anion:Cl2 (aq) + 2 I- (aq)I2 (aq) + 2 Cl- (aq)Triiodide ion is formed:I2 (aq) + I- (aq) I3- (aq)Starch reacts with iodine and iodide to form a characteristic blue/black complex. A corn starch packing peanut is shown here.Silver ion reacts with iodide to form silver iodide, AgI.Ag+ (aq) + I- (aq) AgI (s)Reaction of iodide with sulfuric acid produces hydrogen sulfide gas and brown triiodide solution in a series of reactions:I- (aq) + H2SO4 (aq) HI (aq) + HSO4- (aq)8 HI + H2SO4 (aq) H2S (g) + 4 I2 (aq) + 4 H2OI2 (aq) + I- (aq) I3- (aq)Again, starch is used to confirm the presence of iodine and iodide.