Question

Which of the following compounds will precipitate most readily with \[AgN{O_3}\]​?
A. \[CC{l_3}CHO\]
B. \[CHC{l_3}\]
C.\[{C_6}{H_5}C{H_2}Cl\]
D. \[CH{I_3}\]

Answer 1

Hint: Among the compounds given, the compound that will readily precipitate out with silver nitrate (\[AgN{O_3}\]) will be the one that can let go of its halogen atom readily. Higher the ability of the halogen to leave more readily will be the precipitation with silver nitrate.

Complete Step by Step Solution:
Silver nitrate (\[AgN{O_3}\]) usually reacts with halides (\[{X^ - }\]) to form silver halide precipitates. This reaction is even used to detect the presence of halides in mixtures of salts.
\[AgN{O_3}(aq) + {X^ - }(aq) \to AgX(s) + N{O_3}^ - (aq)\]

All the compounds given as options possess a halogen substituent bonded to a carbon atom. The compound in which the halogen atom is more likely to leave as a halide will be the one that will react most readily with silver nitrate to form the corresponding silver halide precipitate.

Looking at the options, we notice that three of the given compounds have a carbon-chlorine (\[C - Cl\]) sigma bond while the last one has a carbon-iodine (\[C - I\]) sigma bond. We know that the strength of a sigma bond depends on the extent of overlap of the valence orbitals. We also know that orbitals of similar symmetry (shape and size) will have a greater overlap than orbitals of dissimilar symmetry.
The valence orbital of carbon is 2p, that of chlorine is 3p and that of iodine is 5p. Therefore, in a \[C - Cl\]sigma bond, the orbital overlap is 2p-3p while in a\[C - I\] bond it is 2p-5p. The size difference between 2p and 3p orbitals is much less than the difference between 2p and 5p orbitals. Thus, the 2p-3p overlap is much stronger than the 2p-5p overlap which makes the\[C - Cl\]sigma bond much stronger than the\[C - I\] sigma bond. This means that iodine is much more likely to leave as an iodide ion (\[{I^ - }\]) than chlorine. Thus, iodoform (\[CH{I_3}\]) can precipitate much more readily \[AgN{O_3}\]than the other given compounds.
\[CH{I_3}(aq) + AgN{O_3}(aq) \to AgI(s) + CH{I_2}N{O_3}(aq)\]

Thus, option D is correct.

Note: Another reason for iodine acting as a much better leaving-group than chlorine is that the iodide ion is much larger than the chloride ion due to which it can disperse its excess negative charge over a larger surface area and become much more stable.