Why does silver iodide have a higher melting point than vanillin?
The attractive forces holding the molecules of silver iodide together (intermolecular forces) are stronger than those in vanillin, therefore they require more energy to break them. The attractive forces between two molecules of silver iodide are much stronger than the attractive forces between two molecules of vanilin. This is due to the different types of bonds found in each molecule - silver iodide molecules contain ionic bonds, which are very strong, while vanilin molecules contain covalent bonds which are a lot weaker. Since the attractive forces are higher in silver iodide, it requires a lot more energy (i.e. heat) to break these attractive forces in order to melt silver iodide, therefore it has a much higher melting point than vanilin. The bonding of atoms.
Why silve iodideAgI a compound used in photography has much higher melting point than vanillin c8h8O3 a sweet smelling compound used in flavorings?
How do you make silver iodide with the help of silver nitrate and potassium iodide with suitable condition?
No. While vanillin is an aldehyde, which should react with Tollens' reagent to precipitate silver metal, vanillin does not "pass" Tollens' test. Tollens' reagent is very basic (sodium or potassium hydroxide). Vanillin has a phenolic hydrogen (OH bonded to a phenyl ring) which is slightly acidic. Vanillin will react first with the excess hydroxide ions in solution to form a phenoxide salt, which will not participate in the silver-precipitating reaction.