Question

Strongest reducing agent among the following is:
A.$F^{-}$
B.$C{l}^{-}$
C.$B{r}^{-}$
D.$I^{-}$

Answer 1

Hint:To answer this question, we need to recall the electronic configurations of the given elements. Halogens exist naturally as diatomic molecules. A reducing agent itself gets oxidized. Oxidation is the removal of electrons from the atom. So, the element that has the highest electronic repulsion in the atom would accept electrons more easily and gain a stable noble gas-like configuration and reduce the electronic repulsions. Thus, that element would have the lowest reducing power.

Complete step by step solution:
We know that in a period, as we move downwards from the top to bottom, the atomic size increases because of an increase in the number of shells of the atom. As a result of the increasing size, the electrons are spread out and the electron density decreases.
Fluorine atoms are smallest in size and thus would be the strongest oxidizing agent because of high electron repulsions. Whereas, iodine ions are large in size and have the lowest electron repulsions out of all the given ions. Thus, it has the least tendency to lose an electron.
Thus, we can say that iodide ions are the strongest reducing agents among the giving compounds.

So, the correct answer is D.

Note:
In order to find out the reducing power of the given elements, we can also compare the values of their standard electrode potentials. Iodine has the highest electrode potential value and thus is the weakest oxidising agent or the strongest reducing agent. Whereas, fluorine has the most negative electrode potential and is thus the strongest oxidising agent or the weakest reducing agent.