
Among the halogens, the one which is oxidised by nitric acid is :
A. Fluorine
B. Iodine
C. Chlorine
D. Bromine
Answer
225k+ views
Hint: Nitric acid is a very powerful oxidising agent. In the halogens family, only iodine can be oxidised by the nitric acid due to its existence as the strongest reducing agent. Now we discuss why nitric acid can only be oxidised with iodine.
Complete Step by Step Solution:
We know group 17 are referred as halogens and are known as halogen groups. This group consists of fluorine, chlorine, bromine, iodine and a radioactive element astatine. Let us discuss the reducing power of the halogens as:
We know as we move down the group, the atomic size goes on increasing because of the increase in the number of shells. Due to this, electron density goes on decreasing. So the fluorine molecule will be the strongest oxidising agent due to its small size. On the other hand, iodine exists as the weakest oxidising agent.
So fluorides are more stable than chlorides, bromides or iodides.
Hence, fluorine ion will be the lowest reducing agent.
Iodine will exist with the strongest reducing agent.
So as the reducing properties increase as we move down the group, so iodine is oxidised by nitric acid. Nitric acid oxidises iodine to iodic acid according to the following reaction :
${{I}_{2}}+10HN{{O}_{3}}\to 2HI{{O}_{3}}+10N{{O}_{2}}+4{{H}_{2}}O$
Thus, Option (B) is correct.
Note: Remember that $HI{{O}_{4}}$ cannot be formed when the iodine reacts with nitric acid because of higher oxidation state and the nitric acid cannot oxidise iodine to $HI{{O}_{4}}$.
Complete Step by Step Solution:
We know group 17 are referred as halogens and are known as halogen groups. This group consists of fluorine, chlorine, bromine, iodine and a radioactive element astatine. Let us discuss the reducing power of the halogens as:
We know as we move down the group, the atomic size goes on increasing because of the increase in the number of shells. Due to this, electron density goes on decreasing. So the fluorine molecule will be the strongest oxidising agent due to its small size. On the other hand, iodine exists as the weakest oxidising agent.
So fluorides are more stable than chlorides, bromides or iodides.
Hence, fluorine ion will be the lowest reducing agent.
Iodine will exist with the strongest reducing agent.
So as the reducing properties increase as we move down the group, so iodine is oxidised by nitric acid. Nitric acid oxidises iodine to iodic acid according to the following reaction :
${{I}_{2}}+10HN{{O}_{3}}\to 2HI{{O}_{3}}+10N{{O}_{2}}+4{{H}_{2}}O$
Thus, Option (B) is correct.
Note: Remember that $HI{{O}_{4}}$ cannot be formed when the iodine reacts with nitric acid because of higher oxidation state and the nitric acid cannot oxidise iodine to $HI{{O}_{4}}$.
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