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The halide which does not give a precipitate with ${\text{AgN}}{{\text{O}}_3}$ is:
A. ${{\text{F}}^ - }$
B. ${\text{C}}{{\text{l}}^ - }$
C. ${\text{B}}{{\text{r}}^ - }$
D. ${{\text{I}}^ - }$

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Last updated date: 22nd Jun 2024
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Answer
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Hint: We will write the products of the reactions of halides with silver nitrate. Then check the solubility of the products. The product which will be soluble, the halide ion of that product will not give a precipitate with silver nitrate.

Complete Step by step answer: Silver nitrate reacts with halides and forms silver halide and nitrate ion.
The products of the reaction of each halide ion with silver nitrate are as follows
${{\text{F}}^ - }\, + {\text{AgN}}{{\text{O}}_3}\, \to \,\,\,{\text{AgF}}\,{\text{ + }}\,{\text{NO}}_3^ - $
Fluoride forms silver fluoride on reacting with silver nitrate.
${\text{C}}{{\text{l}}^ - }\, + {\text{AgN}}{{\text{O}}_3}\, \to \,\,\,{\text{AgCl}}\,{\text{ + }}\,{\text{NO}}_3^ - $
Chloride forms silver chloride on reacting with silver nitrate.
${\text{B}}{{\text{r}}^ - }\, + {\text{AgN}}{{\text{O}}_3}\, \to \,\,\,{\text{AgBr}}\,{\text{ + }}\,{\text{NO}}_3^ - $
Bromide forms silver bromide on reacting with silver nitrate.
${{\text{I}}^ - }\, + {\text{AgN}}{{\text{O}}_3}\, \to \,\,\,{\text{AgI}}\,{\text{ + }}\,{\text{NO}}_3^ - $
Iodide forms silver iodide on reacting with silver nitrate.
The solubility of all the products are explained as follows:
All the silver halides are insoluble in water except silver fluoride.
The reason for solubility of fluoride is,
The size of the fluoride ion is very small and the electronegativity is very high. Due to high electronegativity and small size, the fluoride is soluble in water. The fluoride also forms hydrogen bonding with water which makes it soluble in water.
The other halides are large and do not form hydrogen bonding, so they are insoluble.
Silver fluoride is soluble so, fluoride will not give a precipitate with ${\text{AgN}}{{\text{O}}_3}$.

Therefore, option (A) ${{\text{F}}^ - }\,$ is correct.

Note: The reaction of halides with silver nitrate is used for the quantitative analysis of halides. Silver fluoride is soluble in water. The remaining three halides give different colours. Silver chloride forms a white precipitate. Silver bromide forms a cream colour precipitate. Silver iodide forms a yellow precipitate. Silver forms three types of the compound with elemental silver, silver monofluoride, silver difluoride and silver monofluoride.