Question

Find the number of halides which are insoluble in water
\[AgI\] , \[AgBr\], \[AgF\] , \[AgCl\], $NaCl$

Answer 1

Hint: We have to know that the halogen family belongs to group 17 in periodic table and they are responsible for the formation of halides. Halides are formed by the halogen family elements. A salt is said to be soluble in water if it dissociates itself into ions or may lead to hydrogen bond formation.

Complete answer:
The given salts are halides of different ions. In the question it is asked to find the number of halides that are insoluble in water. Out of all the halides given, only sodium chloride is the salt that is completely soluble in water. It is the most common reaction of dissolving sodium chloride in water which forces itself to dissociate in ions. The all other halides given are of the same element that is silver. All of these halides are insoluble in water due to its covalent character since these halides are easily polarized thus it can be insoluble in water. Out of the halogen family fluorine is more electronegative atom than chlorine than bromine and the one which is least electronegative is iodine atom. Thus silver iodide has more covalent bond and is very less polar thus they do not form hydrogen bonds with water due to which they tend to be insoluble in water. So due to this electronegativity nature, halides cannot form hydrogen bonds with water thus all these salts are insoluble in water.
\[AgI\] , \[AgBr\], \[AgF\] , \[AgCl\]all these halides are insoluble in water and \[\left( {NaCl} \right)\] sodium chloride is soluble in water.

Note:
We must have to know that electronegativity is the property in which an atom can be attracted to share electrons. Also the solubility depends on the electronegativity nature of elements, since electronegativity decreases down the group. Out of the halogen family fluorine is more electronegative atom than chlorine than bromine and the one which is least electronegative is iodine atom.