If I had dilute Sodium hydroxide solution dropwise until in excess to a solution containing X ions, and I observed a green precipitate soluble in excess to form a green solution. I added few drops of Hydrogen peroxide solution, and I observed a yellow precipitate of Y ions. I added Butan-1-ol followed by dilute Sulphuric acid and formed a blue solution of M ions in organic layer.
State the ions;
(a). X
(b). Y
(c). M
[✓Hint: Think of confirmatory test of Chromium ions]

The answer is nitrogen-containing bases

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Answer ozone (o3) is made of three atoms of oxygen, it is a molecule;