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Of the halides ions , \[{I^ - }\] is the most powerful reducing agent.
(A) True.
(B) False.

Last updated date: 20th Jun 2024
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Hint: Reducing agent after carrying out reduction itself gets oxidised. On moving down the group ,size increases then the tendency to lose electrons also increases and hence reducing character increases.

Complete answer:Halides are the negative ions of halogens.Reducing agents are the substances which reduce other elements and get oxidised themselves.The reducing character is the ease with which an element loses electrons.
Halides have negatively charged.They want to lose their charge and get oxidised. Due to the presence of one extra electron they act as good reducing agents.
On moving down the group,the size of the halide ions increases. Each of them have eight electrons in their valence shell. The electrons in fluoride are held tightly by the nucleus as compared to others. The electrons in iodide are most loosely held because of its large size. Hence, it is most easy for the iodide ion to lose electrons.
Hence,we can observe a trend here that on moving down the group in case of halogens, the reducing character of elements increases. Fluoride is the least strong reducing while iodide is the most strong reducing agent.
The reducing order of the halides is - $F < Cl < Br < I$

Hence , the iodide ion is the most powerful reducing agent among other halides.

Note:Reducing power depends upon size. The other factors on which reducing power can depend are:
1. Ionisation enthalpy
2. Electro- positivity
3. Electronegativity of elements.