Question

Which of the following has the greatest reducing power?
A. $HBr$
B. $HI$
C. $HCl$
D. $HF$

Answer 1

Hint: Reducing power is defined as the power of a species to reduce another substance and itself gets oxidised. As the size of the halogen atom increases, reducing power also increases.

Complete Step by Step Answer:
For halogen acids as the bond dissociation energy decreases the reducing power of that acid increases because the species which can easily dissociated can be easily oxidised and helps to reduce another species.
Among halogen acids as the size of the halogen atoms increases from fluorine to iodine the bond with the small hydrogen atom becomes weak due weak orbital overlap. Thus the halogen species that have weak bonds possess a very low bond dissociation energy.

So, among the halogen acids hydrogen iodide has the weakest bond that is formed between the small hydrogen and big iodine atom. It has the lowest bond dissociation energy among the halogen acids. Thus it has the greatest reducing power. Thus $HI$ has the greatest reducing power.
Thus the correct option is B.

Note: The hydrogen atom combining with different halogen or group $17$ elements forms halogen acids. Among the halogen acids hydrogen iodide has the lowest bond dissociation energy so, it can easily donate its hydrogen atom and acts as the strongest acid among the halogen acids. Hydrogen fluoride is least acidic among halogen acids.