Question

The correct order of reducing power of halide ions is:
Options-
(A) $C{{l}^{-}}>B{{r}^{-}}>{{I}^{-}}>{{F}^{-}}$
(B) $C{{l}^{-}}>{{I}^{-}}>B{{r}^{-}}>{{F}^{-}}$
(C) $B{{r}^{-}}>C{{l}^{-}}>{{I}^{-}}>{{F}^{-}}$
(D) ${{I}^{-}}>B{{r}^{-}}>C{{l}^{-}}>{{F}^{-}}$

Answer 1

Hint: Write the electronic configuration of the elements to understand how the size of the atom reduces down the group. Halogens exist as a diatomic molecule. So, the element having high electronic repulsion will easily accept electrons thereby gaining stable octet configuration and reducing electronic repulsion and having high reducing power.

Complete answer:
The elements of group 17 are commonly referred to as halogen and group 17 is called halogen group. It consists of fluorine, chlorine, bromine, iodine and a radioactive element astatine.
In this question we will not consider astatine due to its radioactive properties.
We know that as we move down the group, the atomic size increases due to increase in the number of shells. Due to this the electrons are spread out and the electron density goes on decreasing.
As the electron density decreases in volume, the inter electronic repulsions also reduce in the molecule.
Based on the above we get to know that fluorine molecules will be the strongest oxidising agent due to small size and significant electronic repulsions. On the other hand, iodine will be the weakest oxidising agent.
Keeping this in mind, we can make the following conclusions:
- fluorine ion will be the weakest reducing agent
- Iodine ion will be the strongest reducing agent
Thus, the correct order of reducing power of halide ions is ${{I}^{-}}>B{{r}^{-}}>C{{l}^{-}}>{{F}^{-}}$.

Therefore, the correct answer is option (D).

Note:
In order to check our explanation and understanding about the reducing power of halide ions, we can recheck it with the electrode potential values for the oxidation of halide ions to their respective ions. The potential values are mentioned below for your reference.
$\begin{align}
  & {{\text{E}}^{0}}_{{{I}^{-}}/{{I}_{2}}}\text{ = -0}\text{.54 V} \\
 & {{\text{E}}^{0}}_{B{{r}^{-}}/B{{r}_{2}}}\text{ = -1}\text{.07 V} \\
 & {{\text{E}}^{0}}_{C{{l}^{-}}/C{{l}_{2}}}\text{ = -1}\text{.36 V} \\
 & {{\text{E}}^{0}}_{{{F}^{-}}/{{F}_{2}}}\text{ = -2}\text{.87 V} \\
\end{align}$