
Which of the following has the greatest reducing power?
A. $HBr$
B. $HI$
C. $HCl$
D. $HF$
Answer
164.4k+ views
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.
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.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE
