What is the oxidation state of bromine in \[HBr{{O}_{3}}\]?
(A) –3
(B) –1
(C) 1
(D) 3
(E) 5
Answer
265.2k+ 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 Main Mock Test 2025-26: Principles Related To Practical

JEE Main 2025-26 Experimental Skills Mock Test – Free Practice

JEE Main 2025-26 Electronic Devices Mock Test: Free Practice Online

JEE Main 2025-26 Mock Tests: Free Practice Papers & Solutions

JEE Main 2025-26: Magnetic Effects of Current & Magnetism Mock Test

JEE Main Statistics and Probability Mock Test 2025-26

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

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Marks vs Rank 2025 - Predict Your IIT Rank Based on Score

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

NCERT Solutions For Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts Of Chemistry - 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

