What is the oxidation state of bromine in \[HBr{{O}_{3}}\]?
(A) –3
(B) –1
(C) 1
(D) 3
(E) 5
Answer
249.9k+ views
Hint: The oxidation states of all of the atoms in a neutral molecule must add up to zero. With this in mind, try and calculate the oxidation state of Br in the given compound.
Step-by-Step Solution:
Let us first understand what the oxidation state of an element in a compound really is before moving onto the calculation of Bromine in this particular compound.
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component.
The terms “oxidation state” and “oxidation number” are often used interchangeably. The transfer of electrons is described by the oxidation state of the molecule. One might mistake formal charge for oxidation state but they are different. Oxidation state is commonly used to determine the changes in redox reactions and is numerically similar to valence electrons, but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.
Now, we know that the oxidation state of Hydrogen is +1 and that of Oxygen is -2 in a compound.
Applying this to \[HBr{{O}_{3}}\],
$\begin{align}
& 0=(1\times 1)+(OS\text{ of Bromine)}\times \text{1+(-2 }\times \text{ 3)} \\
& \Rightarrow \text{0 = 1 + (OS of Bromine) - 6} \\
& \Rightarrow \text{0 = -5 + (OS of Bromine)} \\
& \Rightarrow 5\text{ = OS of Bromine} \\
\end{align}$
Therefore, the oxidation state of Bromine in \[HBr{{O}_{3}}\] is +5, making the answer to this question e) 5.
Note: Chemists use oxidation numbers (or oxidation states) to keep track of how many electrons an atom has. Oxidation numbers don’t always correspond to real charges on molecules, and we can calculate oxidation numbers for atoms that are involved in covalent (as well as ionic) bonding.
Step-by-Step Solution:
Let us first understand what the oxidation state of an element in a compound really is before moving onto the calculation of Bromine in this particular compound.
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component.
The terms “oxidation state” and “oxidation number” are often used interchangeably. The transfer of electrons is described by the oxidation state of the molecule. One might mistake formal charge for oxidation state but they are different. Oxidation state is commonly used to determine the changes in redox reactions and is numerically similar to valence electrons, but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.
Now, we know that the oxidation state of Hydrogen is +1 and that of Oxygen is -2 in a compound.
Applying this to \[HBr{{O}_{3}}\],
$\begin{align}
& 0=(1\times 1)+(OS\text{ of Bromine)}\times \text{1+(-2 }\times \text{ 3)} \\
& \Rightarrow \text{0 = 1 + (OS of Bromine) - 6} \\
& \Rightarrow \text{0 = -5 + (OS of Bromine)} \\
& \Rightarrow 5\text{ = OS of Bromine} \\
\end{align}$
Therefore, the oxidation state of Bromine in \[HBr{{O}_{3}}\] is +5, making the answer to this question e) 5.
Note: Chemists use oxidation numbers (or oxidation states) to keep track of how many electrons an atom has. Oxidation numbers don’t always correspond to real charges on molecules, and we can calculate oxidation numbers for atoms that are involved in covalent (as well as ionic) bonding.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Chemistry Chapter 9 - Hydrocarbons - 2025-26

CBSE Notes Class 11 Chemistry Chapter 5 - Thermodynamics - 2025-26

CBSE Notes Class 11 Chemistry Chapter 8 - Organic Chemistry Some Basic Principles And Techniques - 2025-26

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

