Question

The correct order of reducing power of halide ions is:
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

We must know that the reducing power is the strength of a species to reduce the other element. It can also be stated as the ability to gain electrons. As we move up the periodic table in the group the reducing power of the respective ions decreases.

Complete step by step answer:
Let’s start with discussing the reducing power for better understanding of the question. Reducing power is the strength of a species to reduce the other element. It can also be stated as the ability to gain electrons. The species with high reducing power easily gain the electron as compared with the other species. Reducing power can also be described as the ability to donate a proton i.e. H+ ion. Species which easily donate protons or accept electrons have a higher reducing power.
Coming to the question we are given the halide ions and we need to arrange them in reducing power order. So we need to check that out of these ions which oxidises the fastest. It is clear that Iodine has the lowest reducing potential out of all these ions and hence can be easily oxidised. So, iodine has the highest reducing power of all the ions present. As we move up the periodic table in the group the reducing power of the respective ions decreases. Hence the correct order will be
${I^ - } < B{r^ - } < C{l^ - } < {F^ - }$
Hence, the answer to this question is option D.

Note:
We must know that the reducing power helps us understand which substance is the best for using in a reaction where we want to reduce the other substance for reaction to go forwards or for some other uses. Similar to reducing power there is a concept of oxidizing power which defines the ability of a substance to gain protons or to lose electrons.