Question

Among $C{{l}^{-}},B{{r}^{-}},{{I}^{-}}$ , the correct order for being oxidise to dihalogen is
A. ${{I}^{-}}>C{{l}^{-}}>B{{r}^{-}}$
B. $C{{l}^{-}}>B{{r}^{-}}>{{I}^{-}}$
C. ${{I}^{-}}>B{{r}^{-}}>C{{l}^{-}}$
D. $B{{r}^{-}}>{{I}^{-}}>C{{l}^{-}}$

Answer 1

Hint: In this question, we have to answer the correct order of the ions being oxidised easily. Oxidation state is the charge of an atom if all of its bonds to other atoms were fully charged.

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.
We know all the elements of the halogen family show -1 oxidation state and fluorine is the most electronegative element.

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.

The tendency of the halogens to form ionic compounds increases in the following order :-
Astatine < iodine < bromine < chlorine < fluorine
So fluorides are more stable than chlorides, bromides or iodides.
Hence, fluorine ions will have the lowest oxidation state.
Iodine will exist with the strongest oxidation state.
Thus the correct order for oxidation to dihalogen is
${{I}^{-}}>C{{l}^{-}}>B{{r}^{-}}$
Thus, Option (A) is correct.

Note: Students become confused in the oxidising agent and in oxidation state. Remember that oxidation is the process of losing or gaining electrons or losing hydrogens and oxidising agent is the substance which donates electrons or oxygen or gains hydrogen.