Question

Iodoform gives a precipitate with $AgN{O_3}$ on heating but chloroform does not because.
A. $C - I$ bond in iodoform is weak and $C - Cl$ bond in chloroform is strong
B. Chloroform is covalent
C. Iodoform is ionic
D. None of the above

Answer 1

The atomic size of the iodine atom is greater than the atomic size of the chlorine atom. If the atomic size is more, that means the bond length of the larger atom is also more. Thus, the greater the bond length, the more the reactivity of the atom is and thus, the compound shall yield a product after reacting with the other reactant.

Complete step by step answer:
The reaction of iodoform with silver nitrate is as follows:
$CH{I_3} + AgN{O_3}\xrightarrow{\Delta }AgI \downarrow$
The silver iodide formed after the iodoform reacted with silver nitrate, is a yellow precipitate and marks the reaction of the two reactants. This reaction takes place because of the weaker bond strength of the $C - I$ bond. The size of iodine is the largest in its group and thus, the reactivity of the $C - I$ bond will also be the highest.
In the case of the reaction of chloroform and silver nitrate in the presence of high temperature, no reaction will take place and thus, no precipitate is formed.
$CHC{l_3} + AgN{O_3}\xrightarrow{\Delta } \times$
This reaction does not yield any product because of the strong bond strength of the $C - Cl$ bond. The energy is not sufficient to break the $C - Cl$ bond from the chloroform. Thus, due to weaker reactivity, it does not yield any product.

Thus, the correct option is A. $C - I$ bond in iodoform is weak and $C - Cl$ bond in chloroform is strong.

Note:
The size of the halogens decreases down the group and with that, the strength of the carbon-halogen bond also decreases. Thus, the reactivity of these bonds increases down the group. Down the group, the effective nuclear charge decreases as there is an increase in the number of shells of the atoms in the halogens.